If you are a student of my webpage I welcome you, but please keep this notice
in mind, and plan on re-reading this unit when this notice goes away, as that
is your signal that I consider it to be in finished form.
Here I try to give meaning to Lesson 016 with respect to Diode approximations
What is an approximation? A definition often tells you very little about any
significantly involved subject. I can however get the point across by giving
you an example of something you know well. The numerical constant Pi is a
non-repearing, irrational, number, that on occasion we choose to approximate
with a short string of digits like
Pi is approximately equal to 3.1415926
But eight digits, can be too precise for some applications, say for instance
you want to compute some rough idea of how many feet of cable are wrapped into
a given coil, that you wrapped around your thumb/forefinger, and your elbow.
The technique is you measure the mean diameter of the coil, probably with
a simple ruler, and count the number of times the cable crosses the same
place. Next you multiply that number, times the measured mean diameter, and
then times Pi. Since the measurement is of less precision than two
significant digits of accuracy, and the count of wraps of cable is an integer
that probably fails to take into account the fact that one third of an
additional loop of cable was not counted, your result, regardless how
precise your representation of Pi is, will be no more than two places of
accuracy. If you are doing the calculations in your head, because the
calculator is not within easy reach, using an approximation of Pi to only
two places of accuracy is more than adequate for this application.
To actually grind the number out to eight significant figures is really
stupid. If you disagree, I need only to ask you if you are trying to be
so precise, if you thought about whither your calculation was considering
the length of the plugs on the end of the cable, or not. If you failed to
even consider the next level of the approximation of the cables length,
the length of the plugs themselves, how can you rightfully be concerned
about the precision of your approximation of Pi. Hopefully at this point you
begin to see how futile, meaningless, and unnecessary such a high precision
approximation of Pi in this example is, and if you were prepared to answer
that one, did you do anything to insure that when you measured the mean
diameter of the coil, that the cable you were measuring was formed into a
perfectly round circle, if the circle isn't perfectly round to start with,
high precision Pi won't accurately reflect the length of the cable anyway,
whither you are including the plugs on the ends or not. Even more to the
point the cable, bundled into a coil in this manner, is lumpy, making it
impossible to form a perfectly round circle, that would warrant the use of
more than three significant digits of Pi.
If I'm going to measure the height of a building, and I make my measurement
before they paint the trim of the top of the building, making it three
one thousandths of an inch taller, because of the thickness of the paint
after it dries, does this omission void the accuracy of my measurement? What
about the temperature of the structural members of the building, failure to
take this into account will certainly give an incorrect assessment of the
buildings height, unless the building's overall expansion coefficient is
zero.
These silly, and extreme examples demonstrate why we have levels, or orders
of approximation in engineering. Measuring the height of a dime, before
versus after painting might be significant, but the height of a building
with or without that extra coat of paint is not significant. A painted dime
for instance may not be accepted as the dime it is, in a vending machine,
if the dime has been painted, but I think the error in the height of a
building resulting from painting will make little difference in whither
the building is salable.
If you are building a simple rectifier, with a single peak charging capacitor
that is powered from AC line voltage, eg. 120 volts AC, the small forward
bias potential of a few tenths of a volt, eg. seven tenths for silicon, will
have negligible effect on the voltage the capacitor charges to. On the other
hand the same circuit powered by a two volt RMS AC sinewave power source will
see a significantly lower voltage than you would normally expect as a direct
result of this same seven tenths of a volt lost in forward biasing the diode.
Here's the example of the line voltage
example above.
120 VAC RMS x 1.414 = 169.7 Volts Peak
169.7 Volts Peak - 0.7 Diode drop = 169 VDC
Your error had you not accounted for diode
drop would be...
169.7
--------- -1 * 100 = 0.414 percent
169.0
Now for the 2.0 VAC case
2.0 VAC RMS x 1.414 = 2.828 Volts Peak
2.828 Volts Peak - 0.7 Diode drop = 2.128 VDC
Your error had you not accounted for diode
drop would be...
2.828
--------- -1 * 100 = 32.89 percent
2.128
As you can see the 120 volt example was not appreciably affected by Diode
Junction Drop, but in a similar circuit powered by only 2.0 volts Diode
Junction Drop is significant.
The Junction Drop is not any kind of a battery, well ok if you shine enough
light directly onto a diode crystal, it will produce a small amount
electricity. What happens is photons fall on the the surface of the diode,
and some of them are absorbed, raising electrons out of their normal orbital
position. This disturbance reaches the junction, and upsets the equilibrium
between the electrons and their holes at the site of the junction. A small
electric voltage is formed that is below the Junction Potential, and since it
is being continuously replenished by a flood of photons current flows if a
path exists across the terminals of the Diode. This is the principle
by which solar cells operate. However in absents of something to replenish
the charges, such as the aforementioned photons, a Diode's forward junction
potential will quickly discharge to zero volts and current will cease.
As the next level approximation when we factor in the Junction Drop we can
draw a reasonable approximation by using our earlier model of a diode, and
a battery that has the same voltage as our Diode's Junction Potential.
This new model, or Next Level Approximation, can be drawn as follows.
ideal diode + -
\ | |
------------->|--------| |---------
/ | | 0.7 volt
You may have noticed I keep using the phrase "sustained current" and
wondered why. While the Junction Drop is not a battery a diode charged
right up to forward bias potential will deliver a short pulse of current
as the charges leak out into a load subsequently connected to this diode.
The diode is also a tiny capacitor, and the non-sustained current are those
charges leaking off to the load. But after they have done so, all current
flow ceases, and for current to be useful, in the sense that it performs
work, like the kind of thing a battery does, eg. able to turn the motors of
machinery, I can with conviction state, no useful current flows from the
junction of a diode, because that current is not sustained.
What it does mean is that in order to forward bias a diode, it takes an
additional voltage applied externally of at least the junction drop voltage.
Further when the diode is in forward bias mode the current passing through
it, multiplied by that junction drop voltage gives a wattage figure that
manifests itself as heat.
You can take advantage of the phenomenon to do a myriad of practical things.
A simple two diode clamp is possible constructed as follows will reliably
convert any suitably high enough AC voltage sinewave into something nearly
approaching a square wave.
Next I show a more versatile version of the diode clamp, in that it isn't
stuck at only producing an output of plus, and minus one junction drop.
Ok these second level approximations are going to give you an appreciation
of junction potential, but they are only an approximation that is based on
the junction potential being some set value, eg. 0.3 volts for Germanium,
or 0.7 volts for Silicon. In the next level of approximation we take a good
hard look at this assumption. It's not strictly true. First of all the
0.3 volt, or 0.7 volt, junction potential we have been using changes
slightly as a result of the quantity current flowing through the diode
changes. The junction voltage follows a negative exponential
curve as current through the diode increases, assuming there are no heating
effects. Additionally temperature plays a part that causes a relatively
linear change in junction voltage as long as we operate the diode near
room temperature. Solidstate physics follow mathematical relationships on
the absolute temperature scale, I.E. Degrees Kelvin. This scale begins at
zero thermal energy. There is nothing colder in the universe than this kind
of zero temperature. In fact it is impossible to ever reach absolute zero,
because no matter how hard you try some small amount of external heat still
will penetrate the walls of your Ice-Box. So we simply take measurements,
and work the problem backward, attempting to project the numbers that might
occur at absolute zero temperature. Or we use a convenient approximation,
based on room temperature. Stated in english it reads for a constant
applied current a diode that drops a given forward biased junction
voltage will decrease that voltage by 2.5 millivolts for every centigrade
degree the temperature is increased. If you wire up a circuit that provides
a constant, but low current to a diode, and very carefully measure the
junction potential, for moderate temperature extremes this diode can be
used as a functional thermometer. Moderate here meaning minus forty degrees
to plus 212 Fahrenheit, or 100 Centigrade. What I mean by low current is
low enough that self heating, caused by current passing through the diode
under test does not appreciably heat the temperature measuring device,
namely the diode. In a practical version of this kind of circuit, you would
likely use several diodes in series, making the voltage change larger, and
therefore more easily detected. Also you might pulse the current "on" just
long enough to make your measurement, to keep from heating the diode, and if
the device is battery operated this will extend battery life as well.
Notice that I added another caveat, I included the condition that the
current flowing through the diode was controlled, that is held constant.
Part of the reason this is necessary is that the diode is composed of
three regions.
When a positive voltage is applied externally from the "P" side of the diode
with the negative side of that same voltage source connected to the "N" side,
of the diode, the diode is said to be forward biased. Under forward biased
conditions, the external voltage creates an electric field that concentrates
across the depletion region. This field is of such a polarity that it
diminishes the effect of the inherent field set up by the depletion zone.
As a result the net field within the depletion zone is narrowed, and some
carriers are no longer prevented from diffusing to opposite sides of the
junction. A few electrons spill over from the "N" side to the "P" side
and become carriers on the P side. Conversely a few holes spill over
from the "P" side to the "N" side of the diode and become carriers on
the "N" side. This spill over is called carrier injection.
The carrier injection process causes the current in the forward biased "PN"
junction to increase exponentially with the applied voltage according to
the following equation:
_ _
| vd / n * vT |
id = Is * ( | e | -1)
|_ _|
Where:
vd = voltage applied to the diode
id = resulting current that flows through the diode
Is = scale current, also called saturation current
e = the base of the natural logarithms
vT = the thermal voltage
T = the temperature in degrees Kelvin
n = the emission coefficient
Scale current, AKA Saturation current, AKA Reverse Saturation current, is
a function of the donor and acceptor impurity concentrations in the diode,
as well as the diode temperature, the area of the junction and other fixed
constants. The value of Is typically lies in the
range 10^-8 to 10^-14 amperes for discrete silicon
devices. The value of Is can be as small as 10^-16 amperes
in an integrated circuit diode.
The thermal voltage vT is equal to Boltzmann's entropy
constant k times the temperature in degrees
Kelvin T, divided by the charge q of
a single electron in coulombs, sometimes referred to in physics as a constant
named Elementary charge.
vT = (k * T) / q
Where:
k = 1.38054 x 10^-23 J K^-1
q = 1.6022 x 10^-19 Coulombs
At room temperature, vT is approximately 25 millivolts. The
physical origin of the exponential diode equation is derived from first
principles.
I have saved discussion of n for last.
The constant n is called the emission coefficient. Its
value lies in the range of one to two, depending on the semiconductor used
to make the diode, the magnitude of the forward diode current, and the value
of the scale current. Specifically germanium diodes exhibit an emission
constant of very nearly 1.0, while Silicon is always very close
to 2.0. Since most devices you'll be working with are almost
always based on Silicon the emission constant for diodes is for all
practical purposes 2.0 for diodes.
Transistors are deliberately slanted in the gradient of their impurity
concentration. This affects a transistors emission coefficient considerably.
Silicon transistors exhibit an emission coefficient very
near 1.0 The significance of this is that one cannot expect to
compensate for thermal variations of a Silicon power transistor, by using
the junction drop of a diode thermally coupled to the case of the transistor
to provide bias voltage for the transistor. The reason this won't work is
that the transistor's bias voltage requirement at the emitter, base junction
changes at twice the rate of a diode made of the same material. In essence
what I am saying is that the two curves diverge, thus doing this produces a
circuit that is likely to go into thermal run away, and melt down.
In this range the diode is neither completely on, nor completely
off. Its nonlinear characteristic is sometimes a problem for circuits we
build, because nonlinearity introduces distortion into a signal path.
On the other hand however that same nonlinearity, because it is exponential,
or logarithmic, depending on which characteristic you reference, I.E.
current, or voltage, with respect to the other, can be used to null out
some other unavoidable exponential, or logarithmic distortion already present
in your circuit. Not only that, because adding logarithms performs
arithmetic multiplication, critically placed diodes in an amplifier circuit
can produce an analog multiplier, or wave shaper. You can convert a triangle
waveform into something approaching a sinusoid, by using a couple of diodes
as a wave shaper. The following is such an experiment, I used the Velleman
Oscilloscope to make measurements in the waveforms that follow.
Bipolar transistors:
In the simplest view, a transistor is nothing more than a current amplifier.
If there is adequate voltage available, and of the right polarity, at the
terminals of the emitter, and collector, the current that flows through the
collector, will ignoring leakage, saturation, and a host of other details,
be a preset multiple of current flowing through the emitter base region.
That trait, the multiplication of current, is amplification, in this specific
case current amplification, and in the jargon of electronic data books, this
characteristic has a name, we call it Beta and its symbol is the greek
letter Beta it looks somewhat like an upper case letter "B" with a
descender. This characteristic, and its attendant symbol usually has a
numerical value that falls between 25, and 300. I say usually, although
single transistors have been made for very specialized applications, at a
noticeably higher price with guaranteed minimum specified Betas
in excess of 1000 they tend to be low power, and rather low
voltage devices as well. On the opposite end of the spectrum, some
specialized transistors optimized for very high emitter to collector
voltage, tend to have Beta specifications as low as 7.0 and
they are also more expensive. Another term you will encounter when talking
about amplification, is Gain. In the case of Beta the
Gain being referred to is specifically current gain. Two or more
transistors wired up in a configuration called Darlington results in
current gain that is the arithmetic product of both transistors Betas.
For example, two transistors, with modest beta gains of 70, and 75 wired
up into a darlington pair configuration results in what appears to be
a single transistor with a beta gain of 5250. Manufacturers sell devices
called Darlington transistors wired up this way, and sealed
inside a single three lead plastic case. Using one of these saves you the
effort of wiring two transistors up in such a configuration. It is not
unusual to see darlington pair devices with betas in the tens of thousands.
It's not without a downside, since there are two emitter base junctions
effectively wired in series, the forward bias voltage of a darlington
pair is twice the forward junction drop of a single transistor, and as
an engineer you must take this into consideration, when designing circuits
based on these devices. Plane old transistors, the kind you think of then
you think of a non exotic device, are called bipolar transistors.
They are a device with two junctions, placed very, very close to each other
in an otherwise unremarkable block of pure intrinsic Silicon. The doping is
set up to go
from "P" to "N" and back to "P" again
or they can be made the other polarity where the doping goes
from "N" to "P" and back to "N" again.
Some things that influence the design of the doping concentration are that
to have a useful transistor you want reasonably high beta and you
would also like to have as high an operating voltage as possible. As I have
pointed out to you these two characteristics are at odds with each other.
The most common types of transistors strike a balance between these, and
other tradeoffs. Across the spectrum of transistor types that you find in
the data books, the tradeoffs I will teach you will as you read the
specifications almost jump right off the page, and when you see a device
that seems to violate the tradeoff mentality, it seems to have the best of
all worlds, you had better be suspicious of the one characteristic that
seldom finds its way into a data book... The price!
Ok enough already, here's a drawing of a biased "NPN" transistor.
I show the electron paths in blue, by the way electrons are blue
regardless what you may have heard about them being some other
color :-) They leave the battery at its negative terminal, travel
through the circuit and ultimately arrive at the positive return
battery terminal. I simultaneously show two instances of
either the same transistor, or if you prefer, two separate, but
absolutely identical transistors. However you view it I am illustrating
only one transistor here, so that I can show the effect of two different
levels of base bias in a side by side comparison.
The Beta gain of this transistor is 20.0 the
junction drop is 0.7 volts exactly. The transistor on the left
has 3 milliamperes of current flowing through its base, and this causes
60 milliamperes to flow through its collector, the 3 ma is arithmetically
multiplied by the Beta factor, and this results in the 60 ma collector
current. If you were to open the 3.3 K Ohm resistor, or short the base
lead to the negative battery terminal, the base current would stop, and
since the collector current is dependent on base current, the collector
current would also cease.
What is happening here:
In the above drawing the electrons flowing into the Emitter terminal
have at first only one path of current available to them. They flow across
the Emitter Base junction into the base region by forward
biasing what first appears to be a diode. Since the
Base Collector junction from this vantage point represents a
reverse biased diode jumping that gap is very unlikely, instead what
happens is those electrons exit the Base lead wire, traverse the
resistor, and return to the battery's positive terminal. This activity
sets up a condition that is regularly replenishing charge carriers in the
Base region. Due to the Base region being made very thin,
and very wide, and deep comparatively speaking, having such a large number
of charge carriers bouncing around in such a thin region, it becomes far
more likely that they will diffuse into the Collector region, than
travel down the long thin plane edgewise to make it to the Base
lead wire. Therefore the thinner, and wider, and deeper you make the
Base region, the higher the Beta gain the device will have.
If you turn on, Base current, Beta times that
amount of current can potentially flow through the Collector.
If you wire up a circuit that limits the current that is allowed to
flow through the Collector to a point below the Base
current multiplied by the Beta gain, the transistor is said to
be saturated. When a transistor is in saturation it exhibits an unusually
low voltage between the Emitter and Collector
in fact a transistor so biased can exhibit Emitter-Collector voltage
considerably below the forward bias voltage. This may seem to some of you
to defy logic. How could you possibly have a Emitter-Collector voltage
lower than the forward bias junction voltage. It would seem that for
electrons to get through the first junction, on their way to the second
junction there would have to be at least 0.7 volts to account
for the Emitter-Base junction, since it is in series with the crystal
region that holds the collector. The way I explain this, is that we are
concerned only about charge carriers migrating through the crystal.
While it is true that it requires 0.7 volts to free charge
carriers to move within the crystal, once they are in motion the current
they can carry, is carried from the Emitter to the Collector
if they don't exit the Base lead first. They can and do drive the
Collector below the voltage of the Base with respect to
the Emitter. This is a charge phenomenon, in fact the Base-Collector
junction is still reverse biased, even though current flows into the
Collector.
If you turn on, and off the Base current, in a transistor that has
a current limited Collector such that Saturation is certain
the transistor's output voltage, measured from circuit ground, where the
Emitter is wired as the ground, or reference terminal, with the
output voltage measured at the Collector, the output will go from
one extreme, the transistor full on, and virtually zero volts at the
Collector, to the condition called cutoff, eg. the transistor
turned as far off as possible, meaning no current flows, and the voltage
measured at the Collector being nearly the voltage of the power
source. This action has been seen by some, to warrant calling the transistor
a controlled switch, in that it can switch current on or off. My feeling is
that such an explanation is an over simplification.
In truth there is no such thing as "digital" in the real world. All sources
of naturally occurring information originate as an analog signal, we
merely choose to represent this information as parallel digital data
out of necessity, or convenience, in an effort to escape the noise of the
analog world. Indeed the very circuit blocks we build that we call digital
are composed of analog components, and if not operated within the tightly
specified constraints the manufacturers lay out in the data books, and
app notes; these so-called "digital" devices show there true analog nature.
I'll give you one example. Some types of CMOS logic specify that a logic one
must never be below a specified voltage, about 2.5 volts, and
they also specify that a logic zero must never be above 0.8 volts.
Obviously to transition from a one to a zero, or vice versa, you must
violate this rule. So they make an exception to this rule, by telling you
in the data books that you must accomplish this within a very short period
of time, usually in the nanosecond range. Guess what happens if you
deliberately violate this rule. The device goes analog, producing output
voltage that is an analog voltage between a logic one and a logic zero on
average, and it tends to oscillate at very high frequency, higher than the
device is designed to withstand, and left in this state for very long often
results in the device self destructing, from overheating a tiny portion of the
silicon chip. Further if you look closely at the circuitry inside the chip
you see a very serious problem with placing on an input any voltage between
the permitted ranges for a logic one, or zero. The totem pole output
transistors, MOS-Fets in this case, both turn on simultaneously, creating
an electrical short circuit directly across the powersupply! No wonder it
self destructs. To understand anything in electronics, even digital, you
need to know how devices behave in the analog realm. Please heed this
warning, you will be insufficiently prepared to deal with the digital
realm, if you forgo the effort to adequately understand the analog realm.
I have seen people attempt this, thinking that if they simply read the
guidelines in the data books, and app notes, that's all they will ever need.
Then when they try to design a digital circuit of a challenging complexity
things go horribly wrong. The circuit misbehaves in ways they have no hope
of understanding.
The transistor in question is an "NPN" that piece of information is encoded
into the symbol by the direction the Emitter arrow faces. If it had
faced pointing toward the body of the transistor, that is the vertical line
which represents the silicon wafer, it would instead denote a "PNP" device.
To help you keep straight in your mind which is which, imagine the arrow
having a perpendicular line attached to the point of the arrow. This would
form the symbol for a diode, and that will indicate to you what polarity
the Emitter Base junction is, for both forward and reverse
bias.
If you only connect two terminals of a transistor at any one time, not a
very useful configuration, it will behave as if it were two diodes. Since
your ohm meter has only two lead wires with which to connect a component
to, simply leaving the third transistor lead wire dangling in the open
air, and connecting the leads of the ohm meter to any two lead wires of
a transistor, allows you to observe the two diode junctions formed inside
the transistor. The transistor is of course a far more interesting device
than this simple exercise lets on, but it is useful for both determining
whither the device is an "NPN" or a "PNP", and it clearly identifies
which lead wire of the transistor is the base lead. This information, in
the absents of any real documentation on the part, or even worse, in the
case where the transistor in question is House marked, that is
stamped with a number that has meaning only within the confines of the
factory that produced it; may indeed be the only way you can figure out
anything at all about the device.
Using a simple electro mechanical ohm meter, that derives its power from
a single 1.5 volt flashlight battery, a silicon junction drop
appears to be approximately half scale, regardless what ohms range you
set the meter to. This is not strictly true, but it tends to be more true,
than behaving like a resistance. The reason this happens is because
the 0.7 volt junction drop, is almost half the voltage of the
flashlight battery inside the meter. The reason it is not strictly true
is that the diode junction drop is as I've pointed out before a natural
logarithmic curve of the current, and the ohm meters short circuit current of
each successive range multiplier is a factor of ten different from the adjacent
setting. The point of all of this, is that you can read the junction
drop of a diode or transistor, directly off the dial, by using a linear
range, such as a voltage scale, and multiplying by a fixed constant, if
your particular meter lacks a 1.5 volt range, and then
subtracting your adjusted reading from the fullscale value,
namely 1.5 volts. Because the meter selector dial for the
ohms ranges offers up currents a factor of ten apart, if you know the
fullscale current of any one of them, you simply multiply by factors
of ten to get the others. If you know the fullscale current, for a given
range, if you multiply the percentage of the present needles deflection,
times the known full scale current, you can get the current that is
flowing through the diode under test. Without even turning the meter on,
by simply knowing that it is powered by a single 1.5 volt
flashlight battery, and picking a spot about half way in the middle of
the dial, looking for a point where both a resistance, and a voltage
intersect you can compute by using simple ratios the fullscale current,
for any range available to the selector switch. This unusual application
of a simple VOM, Volt Ohm Milliammeter thus makes possible a poor mans
Diode curve tracer, and two such meters, applied in this fashion, allow
you to accomplish what a Transistor curve tracer does. This may seem
like a lot of mental gyrations to save a mere $3000.00 but consider
that Transistor curve tracers are heavy pieces of electronic gear, and
working in the field, you are often making repairs while standing on
rickety scaffolding, that getting a heavy fancy curve tracer up there
to make one measurement might make you envious of the guy who really
knows how to make good use of the simpler, and lighter, piece of gear.
In the above exercise I show using a simple ohm meter, to measure diode
voltage to current relationships. I show you how to take the short circuit
current directly off the dial mathematically. You might ask why not
simply use another meter to measure it. If you derive this value
mathematically, and measure it, the two will not agree. My method is not
at fault here, your use of the second meter is. Why? Because the second
meter, set to measure current, drops a quarter of a volt at full scale.
Since we only have 1.5 volts to work with, that quarter of a volt
is a significant error.
In my case I have a Micronta Range Doubler 22-204C multimeter that uses
a 1.5 volt flashlight battery for all but the highest resistance
range that being the R x 10,000 range, which uses
a 9 volt transistor radio battery. So for this experiment I will
use the R x 1 -to- 1000 ranges inclusively.
The meter dial itself sports a scale intended for use with both voltage,
and current that spans zero to ten. This is convenient, as it permits me
to use this linear scale, as a decimal fraction, or percent scale.
Reading a 1N4001 rectifier diode, these are rated
at 1.0 amp RMS, so even the fullscale current of
the R x 1 scale is well within the safe operating range
of the diode. Upon measuring the diode on the ranges that are powered by
the 1.5 volt battery. The following decimal fraction values come
to light.
0.50 read on R x 1
0.59 read on R x 10
0.64 read on R x 100
0.72 read on R x 1000
To make them useful we must arithmetically convert the reading to voltage
across the diode, and current passing through it, using the following
formula.
E = (1 - Scl) * 1.5
I = (1 - Scl) * RngCur
Where:
E = Voltage across the diode
I = Current through the diode
Scl = the decimal fraction read off the dial
1.5 = full scale meter voltage, eg. the battery
RngCur = full scale meter current, for a given range.
Applying this bit of arithmetic treachery, yields...
Rx1 diode Volts = 0.75 = (1 - 0.50) * 1.5
Rx1 diode Amps = 0.075 = (1 - 0.50) * 0.150
Rx10 diode Volts = 0.615 = (1 - 0.59) * 1.5
Rx10 diode Amps = 0.00615 = (1 - 0.59) * 0.015
Rx100 diode Volts = 0.54 = (1 - 0.64) * 1.5
Rx100 diode Amps = 0.00054 = (1 - 0.64) * 0.0015
Rx1000 diode Volts = 0.42 = (1 - 0.72) * 1.5
Rx1000 diode Amps = 0.000042 = (1 - 0.72) * 0.00015
+----------------------------------------------
| *
0.07 | .
0.06 | .
0.05 | .
0.04 | .
0.03 | .
0.02 | .
0.01 | . *
0.00 +------------------------*------*--------------
0.0 0.1 0.2 0.3 0.4 0.5 0.6 0.7
The plot above is Current on the vertical axis, and
Voltage on the horizontal axis, of the above data.
Asterisks "*" are actual measurements, dots "." are
my crude interpolation attempt using ASCII art
Notice that we are not reading the ohm scale of the meter to get this
information, in spite of the fact that we are using the ohms range selector
switch positions to actually subject the diode to mild voltage, and current
stresses that give us information about the diode's forward bias
characteristics. The thing that permits us to do this is that we are
applying an understanding of how a multimeter works, to a task it was
never intended for, but is nonetheless useful, and reliable. You can just
as easily apply this technique to a transistor by forward biasing its
Emitter Base junction with one ohm meter, while using another
ohm meter to observe the Voltage / Current relationship of the
Emitter Collector terminals. The beauty of this is that you
don't even require a bench supply to perform such a test.
There are of course fancy digital multimeters that will measure a transistors
beta gain. They have built right into the front panel, a cute eight pin
socket that you plug in the transistor to be tested into three of the holes
in the socket. The choice of which holes you use, hinges on whither the
transistor is of type "NPN" or "PNP", and what the order of the
Emitter / Base / Collector orientation is. I'm not
arguing against the purchase of such a device, just that you probably don't
want to rely on it. Using my simple ohm meter approach offers you one
really big advantage they don't provide. You can not only measure the beta
gain, but you can do so at, in this example, four different current settings.
If you are measuring beta, it's important to do that measurement at a current
that is near the actual current you will be subjecting the device to, when
you wire up your circuit. This is because the beta is somewhat current
dependent.
Since transistors amplify current, the voltage that is present between the
Emitter, and Collector has little to do with the current
flowing through the Collector this behavior is not at all like a
resistor, whose current is linearly voltage dependent. There is a thermal
effect though, but it can be nulled out with feedback. This is truly
something new, something that you haven't seen before, a device that within
a voltage range acts as a circuit element to limit current irrespective of
the voltage across it. It is within these limits a
constant current source a thing quite different from a
constant voltage source. The current it controls as I have
said is a function of Base Emitter current, and the
beta gain of the transistor. So what are these voltage limits
I am so carefully constraining this system to? The lower limit is that you
must be dropping enough Emitter Collector voltage that the
transistor is not in Saturation. Not even the onset of saturation.
Since saturation is lower than the junction voltage, by a few tenths of
a volt, a good rule of thumb is keep at least seven tenths of a volt across
the Emitter Collector terminals if you want the transistor to
act as a current limiter. At the high end the transistor will avalanche if
the Emitter Collector voltage is too great, the avalanche voltage
is considerably greater than the posted specification sheet for the
transistor, by its manufacturer, so simply staying below the absolute
maximum rating given you by the manufacturer will be one limit at the high
end. Another limit is thermal dissipation of the transistor's package.
If you draw 20 ma through the Base of a transistor, whose
Beta Gain is 20, the Collector current will
be 400 ma. If across that same transistor's
Emitter Collector terminals you are dropping 40 volts
the transistor is dissipating sixteen watts of heat! You can melt solder
on a package the size of a transistor dissipating that much heat. There are
several ways to deal with this, Heat sinks, these are metal finned heat
radiators that help spread the heat into the surrounding air, Fans, do
a much better job of this, and in extreme cases Water Cooling systems, not
very different from the ones in your Car's Radiator. This all goes back to
earlier lessons where I got rather "preachy" about watts. Basically you are
multiplying the Voltage, in this case across the Emitter Collector
connection, by the current flowing through the Collector it's as
simple as that. So now the mystery of the limits of voltage
min and max is solved. As long as you stay inside these extremes
the Collector will faithfully represent Beta times the
Base current, assuming of course you have the polarities right.
Because of a resistors linear voltage to current relationship the more
current you draw through a resistor, the more voltage it drops. If you place
a resistor in series with the collector of a transistor that is controlling
collector current via beta, times the base current, the voltage drop across
that resistor is now proportional to the collector current. The following is
an important heads up, don't forget this, or you will be very
confused. The resistor is connected between the powersupply and the
Collector since most measurements are relative to circuit ground
the affected point, the point where the collector, and resistor come together
exhibits a decreasing voltage with respect to an increasing
current. This does not mean the resistor is working backwards, if you were
to measure the voltage drop across the resistor itself, that voltage does
behave as you would expect. However in a practical observation, the meter
has the common terminal connected to the same circuit ground that all the
other equipment is ground referenced to. This is necessary equipment
technique, because as more and more pieces of equipment are connected to
a circuit, if the grounds are connected to arbitrary points in the circuit
electrical safety grounds tend to short things out, and those pieces of
non-grounded equipment can introduce confusing noise signals into sufficiently
high impedance circuits, that make reliable observations impossible.
In addition to these reasons, circuit blocks are typically connected to each
other sharing a common ground. So you as an engineer have to make these
transformations anyway. They take a little getting used to, but after a
while they come quite naturally. This characteristic, of exhibiting a
decreasing voltage at the collector, while the base current is
increasing or vice-versa, has a name, we call it inversion,
and it is crucial to many facets of electronics, things such as oscillators,
and the most fundamental gate in a computer, a thing called an inverter,
would not be possible if inversion were not possible.
Inversion:
In the left diagram in the above schematic the base current is limited by
a single 1000 ohm resistor that is fed by a powersupply that is
ten volts above the junction drop of approximately 0.7 volts.
I label this supply as 10.7 volts. This insures the base resistor
has approximately 10.0 volts across it. Thus to compute the base
current is simple ohms law, ten volts divided by the base resistors
resistance. Once the base current is known, with a known beta gain factor
we can compute the collector current by multiplying the two. Once the
collector current is known one simply multiplies that by the resistance of
the collectors series resistor, again using ohms law, to get the voltage
drop across the, in this example 25 ohm resistor. Then the final
step is to subtract the collector resistors voltage drop from the collectors
powersupply in this case conveniently set to exactly ten volts to
give 5.0 volts at the output terminal.
In the right hand diagram we reach into our box of resistors and grab
a 5000 ohm resistor, and bend its leads so that we can jumper
it across the existing 1000 ohm base resistor, to temporarily
increase the base current a little bit. This 5000 ohm resistor
being added is the Independent Variable of our experiment, and
it causes many effects. The final result, the Dependent Variable
ultimately exhibits a lower output voltage at the collector.
The gist of where this is going is that here we are observing a phenomenon
where an increase of input current, causes a reduction of output voltage.
We call this inversion. Also true is the opposite side of the same coin,
a reduction in input current, produces an increase in output voltage. This
second statement is really no different from the first statement, in fact you
can get the results from the first data set, to prove the second assertion
by simply reversing the order in which each scenario is presented.
It is important to the study of electronics to truly understand this.
The concepts developed here are used in circuits throughout the entire
discipline. One of the concepts I have slipped into the mix with out
mentioning it by name, is the idea of what exactly a
pull-up resistor is. If you really didn't get what inversion is
don't despair, my description of the concept of a pull-up resistor
may bring it all home to you.
Pull Up:
If you were to open the base lead, thus stopping all current flowing in the
base of the transistor, the collector current would also be similarly
reduced. Zero base current multiplied by any fixed factor for beta gain
produces zero collector current. Zero collector current means no voltage
drop across the 25 ohm resistor. When the resulting voltage drop
of zero is subtracted from the 10.0 volt powersupply, the output
voltage at the point where the collector, and the 25 ohm resistor
are joined, is also 10.0 volts. The resistor is said to have
pulled up the collector to the same voltage as the powersupply.
At the other extreme if you replace the 1000 ohm base resistor
with a 100.0 ohm resistor your base current shoots up
to 100 ma, and multiplying by a beta of 20.0 gives a collector
current of 2.0 whole Amperes. The 10.0 volt supply
limited by the 25 ohm resistor simply is not able to deliver
any more than 400 ma, so the transistor is saturated, with
about 0.2 volts between its collector and emitter. In this
example it is said that the transistor is pulling down as hard as it can,
against the pull-up resistor and has won the battle so
completely, that the pull-up resistor never had a chance.
Another way to state this is to say that the transistor is loading down
the current limiting collector resistor pulling its potential
all the way down to the ground potential, or at least nearly so.
In the previous case where we interrupted all current to the base,
the 25 ohm resistor wins the battle, because the transistor
is not loading down the 25 ohm current limited voltage at all
so it is pulling up the collector all the way to
the 10 volt supply. In the case where neither the
pull up nor the collector of the transistor is completely
winning or loosing, that is some state in between the two extremes, we
call this the analog or linear range of operation, and a
circuit that operates only at the two extremes is called digital or
switchmode operation. A circuit that operates in both digital
and analog at the same time, is called an analog circuit that is
overdriven and there are examples of such devices sold commercially,
one such device is the well known Guitar Fuzz box, and its variable
filtered cousin the Waa-Waa Pedal.
All that I have told you in the preceding section is factual, but in the
real world of discrete devices, not very practical. Integrated circuits
are another matter entirely, they get away with doing things
in IC design because the entire IC chip is so small
that the whole thing runs at the same temperature, and the concentration
of the doping compounds is uniform over the whole IC chip, again
in part because it is so small. The realities of working with discretes,
individual packaged transistors means several things. One transistor can
by dissipating more watts than another component on the board, be
running at a significantly different temperature than another. The
realities of mass producing transistors often results in wide variations
of beta gain from one lot to the next. If you read the databook, it is not
uncommon to see the spread of min/max beta gains for a given transistor
part number to vary by as much as 400 percent. Indeed often
they don't even spec the max beta for "economy" transistors at all.
What's an engineer to do. Well for one thing don't rely on beta, as a spec
that you can nail down. A statement such as this probably seems ludicrous
in light of how critical beta is to the operation of a circuit. The answer
to this seeming inconsistency is that to build a practical and reliable
circuit you need to introduce feedback to compensate for these wide
variations in beta gain. If you design your circuit conservatively, not
only will beta variations be brought under control, but some of your
thermal instabilities will also be covered as well. The more conservative
the design of a circuit, the more likely you are to be able to call out
dozens of different transistor numbers, as substitutes for a device that
may in the future be discontinued, no longer to be manufactured.
This section shows how to design simple conservative practical amplifiers
that require no critical, low tolerance, I.E. high precision, eg. expensive
components made of the element "unobtainium" :-)
This approach to design tends to make these circuits less vulnerable to wider
temperature extremes as well. Wann'a design a circuit that goes into your
automobile, you better think about the temperature extremes the cab of
your car can reach. If you drive to northern climates, in winter, at night
it is possible to experience temperatures below minus forty, and in the
summer a closed locked cab in some southern states, can melt a
vinyl LP record album so totally that is shrivels up into a ripely
Christmas Tree ornament that most resembles a Starfish in shape.
Seriously, do you really expect your circuit to work under those conditions?
Well GM, Ford, Chrysler, etc. do this sort of thing all the time. They test
their circuits in environmental testing chambers. You have such equipment
in your kitchen, namely the Refrigerator, and the Oven,
Not the Microwave, the conventional Oven, ok. :-)
Subjecting your circuit to the Microwave-Oven would be like testing to see
if it would withstand the spearhead of a solar flair! Yikes! On the bench
any source of heat or cooling can get you some preliminary results, a heat
lamp, an Ice Cube, the hot tip of a soldering iron, or
Freeze Spray sold in electronic parts shops for tracing thermally
intermittent components, can give you some idea how things will act, but
nothing beats the tried and true conventional Oven set to 150 degrees
Fahrenheit, and then to test the other extreme the Freezer section of your
Refrigerator. When doing the last test, beware of moisture condensation,
a sealed cabinet with a new or re-cycled Silica Gel packet generally
eliminates this problem. I have heard, but not yet myself confirmed the
efficacy of a technique of "baking" used Silica Gel Packets to
restore their Hygroscopic properties. Incidentally if you don't know what
Silica Gel Packets are, you've probably thrown away a dozen or
so of these things. They are the little packets that come shipped with
nearly every electronic device you buy, the only instructions on the
packet is a short message that says "Don't Eat" I never eaten one of
these so I'm not qualified to tell you what they taste like. They do
however aggressively absorb water vapor, and inside a sealed container they
dry out the air inside. Their mortal enemy is the humidifier, so they tend
to loose their effectiveness if left out in the normal room air. For
completeness sake I should mention simulators, such as Spice.
Simulators can if you don't make any mistakes, and if you don't accidentally
exceed the mathematical precision of a Spice library, accurately
predict what your circuit will do at various temperature extremes.
Relying on these software tools to tell you if you missed anything is one
thing, trying to use them to tweak your circuit will lead to irrational,
and inaccurate conclusions, because you are much more likely to exceed
known limits of precision in a given library without realizing it. They are
indispensable tools for IC design, but not that useful for
discreet component circuits. The problem is that you cannot rely on one
vendors Spice Library to be a perfect fit to another
manufacturers component, and since design using only one manufacturers
product line in any large complex circuit is all but impossible, eg.
they won't have a component you need, or they ask too much money for it,
your Spice simulation is not likely to accurately represent any
real circuit you build. On the bright side there is a GPL version
of Spice for what that's worth. In my presentation I endeavor to
show how to design a circuit conservatively enough that you can more easily
cope with these parameters, because you have a wide range of tolerance to
work within, without getting anywhere near the danger zone. Even if you
find yourself approaching the danger zone, because you are
"in the loop" you are keenly aware of the tradeoffs you were
considering at that time. The simulation approach has the problem that you
are not really designing the circuit, the computer is making countless
decisions for you, some of which if you knew what they were you'd be
horrified by the ghastly assumptions made on your behalf.
There are three basic configurations of the simple transistor amplifiers.
One each, in which for each case, a single different transistor wire is held
at some AC reference, eg. common, and the other two wires are used
to amplify signal. These configurations are given names based on which of
the three transistor leads are chosen as the AC ground reference,
or common reference. Their names are as follows, common Emitter,
common Collector, and perhaps somewhat less useful the
common Base configuration. You are expected to build these, and
given that I have designed them so that you are required to make as few
changes as possible to transform one of these amplifiers into another.
Since the circuit is conservatively designed none of the values are terribly
critical, meaning if the nearest value of say a resistor is within fifteen
percent your circuit will still work, just not as optimally as it would
have if the proper component had been used. My first example is the
common Emitter, and common Base, configurations
shown side by side, they are remarkably similar.
To discuss the AC paths of this circuit, for simplicity I will
assume that the output is driving a load such as an oscilloscope fitted with
a times ten probe, such that it's load is negligible, and that the input
is being driven with a signal generator whose output impedance is so low
as to be negligible, on the order of an ohm or maybe as much as two ohms.
I will also stipulate that the frequency of the signal is
between 400Hz and 10kHz to limit coupling losses in the
capacitors to one percent. For example the 200 uf capacitor in the
common base configuration at 400 Hz works out to less than
two ohms.
1
1.9894 = ---------------------------
2 * Pi * 400 Hz * .000200 f
Since it's in series with a 200 ohm resistor
the Xc introduces less than one percent of error. If you really
want to do this as a lab, and you have only an AC wall-wart or
Variac for a signal source, your signal source is by definition the
frequency your Electrical Power Utility's output frequency. If you are in
the US your line frequency is 60 Hz, elsewhere it varies
but most everywhere other than the US the line frequency
is 50 Hz. As a work around to allow you to use line frequency as
a signal source, I suggest that you multiply all of the capacitor values
by a factor of ten, so the 200 uf cap is replaced by
a 2000 uf cap. and so on.
Ok let's examine the common emitter configuration from
an AC perspective first. This is to be a second approximation
examination. The emitter circuit is from an AC point of view
a 2k and a 220 ohm in parallel from the emitter to ground
the parallel equivalent resistance is 198 ohms, I choose to call
it 200 ohm to make the figures easy. A 200 ohm emitter
impedance to ground looks from the base lead perspective, after you factor
in a beta gain of 20.0 like 4000 ohms. So to get the effective
base loading using a trick borrowed from thevenin, the AC base
circuit looks like a 13K, and a 9.1K, and the beta amplified emitter
impedance of 4k all wired in parallel. The AC base
is 2.29 K ohm. Now to examine the collector circuit. The collector
itself looks like a current source, and believe it or not it has an infinite
impedance. The only other component in the collector circuit that could
possibly act as a load is the 2K ohm resistor. Since
the AC component of input signal voltage is being duplicated at
the emitter, and the emitter to ground impedance this voltage is impressed
upon is 200 ohms, and since the impedance in the collector circuit
is 2K ohms, and nineteen twentieths, of the actual current to drive the emitter
impedance is in fact coming from the collector circuit, this circuit is a
voltage amplifier with a voltage gain of nearly ten to one. For all practical
purposes the gain of any conservatively designed common emitter amplifier
is the ratio of the total collector impedance divided by the total
emitter impedance.
Collector impedance
Av = -------------------
Emitter impedance
Where Av is an abbreviation for Amplification of Voltage. Without giving you
a bunch of formulas, I have using ohms law, and a few other techniques, shown
you how to derive all of this on your own. If you study this section
carefully, you can begin to design amplifiers of different gain factors,
and input/output impedances to suit your needs as they arise. Since this
amplifier has a voltage gain of ten, and can at most produce an output
signal of 3.0 volts peak, you require an input signal
of 0.3 volts peak applied to the input to fully drive the amplifier.
If for example you were considering using an AC Wall-Wart whose
output RMS voltage was 9.0 volts AC an attenuator would
need to be constructed to reduce the 9.0 volt RMS output
to 0.3 volts peak. May I suggest using a 1.5 ohm resistor
in series with a 62 ohm resistor wired across the Wall-Wart, and
using the voltage dropped across the 1.5 ohm as the signal source
to connect to the input of your amplifier. Note: These values do take into
account the RMS vs peak voltages involved. Better yet, use three, use
a 1.5 ohm, and a 39 ohm, and a 100 ohm rheostat,
eg. variable resistor, all wired in series, this will give you a knob that
you can turn to adjust the input signal over a fairly wide range.
Common Base:
As in the common emitter the AC impedance ratio of the collector
to the emitter circuit is for the most part what controls voltage gain
of the transistor amplifier itself, however inherent in the design of the
common base configuration is the fact that the emitter AC impedance
reducing circuit used to set the AC voltage gain now also forms
a voltage divider that attenuates the input signal before it reaches the
emitter. Both of these configurations have an AC voltage gain of
ten, but if you look closely at the AC paths from the emitter
to ground, the gain limiting resistor in the common emitter configuration
is 220 ohms and in the common base it's only 200 ohms.
This causes the common emitter amplifier to have a gain of ten as the
emitter impedance is 200 ohms when the 2K ohm resistor is
considered, on the other hand the emitter impedance through the generator,
remember I called out a very low impedance generator 1.5 ohms, so
that the generator's impedance could be ignored. Thus the combined emitter
impedance of the common base is only 181 ohms, and the works out to a voltage
gain of eleven to one, within the transistor section itself, however,
the 200 ohm resistor in series with the generator and emitter, through a
negligible 2000 uf cap. creates
a 10 to 11 voltage divider. When you multiply the ratios
of the front end attenuation, by the emitter/collector voltage gain, you
get exactly ten.
Terms
/ \
Attenuation Amplification
2000 2000
10 = ---- * ----
2200 181
It is important to mention that the base is an AC short to ground
in this common base configuration. The capacitor between the base and ground
essentially insures that the base is not free to move in voltage at the
whim of the emitter activity reduced by beta.
So if the ratio of Collector to Emitter impedance is what controls
the AC voltage gain, what if we create an AC short
from the emitter to ground, couldn't we expect a nearly infinite voltage
gain? If you did this the gain would in fact be very high, but it would
also be unpredictable. For instance in a common emitter circuit, with
a voltage gain of ten the effect of the transistor's beta gain on
the input impedance has little impact. For example an emitter
impedance of 200 ohm multiplied by an unusually low beta
of 20.0 appears to the base as 4000 ohms, significant,
but not serious. To ask for a voltage gain of 100 would require
an AC emitter impedance of 20 ohms, and once again
multiplying that by a beta of 20.0 again still our worst case
scenario works out to a 400 ohm AC base loading effect.
The total input impedance of our amplifier with a voltage gain
of 100 is now only 372 ohms. If it were being driven by
a previous stage with no load a voltage gain of 10 and and as in
our example a no load collector impedance of 2000 ohms, this 372 ohm
load when factored in reduces the gain of the previous stage to 1.57
hardly any gain at all. Remember Unity Gain a gain
of 1.0 is is the gain of a simple wire, no amplification, and any
body trying to sell you an amplifier with a gain of less than one, is in fact
selling you an attenuator, a device that reduces the signal, rather than
amplifying it.
All of engineering is mired in the art of managing tradeoffs. Aggressive
design presses you ever closer to the point where ever more factors must
be considered, to make a system work reliably over the varying conditions,
and manufactureable over time, and across vendors. Reducing emitter impedance
to set the gain higher you must begin to consider the internal spreading
resistance of the base region of the transistor itself, and the larger
factor, because it is not being amplified by beta gain, the internal emitter
base forward junction effective resistance, which is not really a resistance
at all, it is instead similar to the diode current, to voltage curve, and for
very small excursions of voltage, behaves somewhat like resistance.
These things have names R'b, and R'e while we strive to
make them as insignificant as possible using conservative design they do
affect the AC performance of our circuits, as you will see.
R'e is given by
vT
R'e = ------
Ie
Where:
R'e = effective internal
emitter resistance
vT = Thermal voltage
approx 25 mv at
room temperature
Ie = Current flowing
through the emitter
I covered thermal voltage vT earlier this section, it is equal to
Boltzmann's entropy constant k times the temperature in
degrees Kelvin T, divided by the charge q of
a single electron in coulombs, sometimes referred to in physics as a constant
named Elementary charge. At room
temperature, vT is approximately 25 millivolts.
Two important things to note:
First understand that R'e is current dependent. This fact means two
things, (1.) R'e changes depending on what circuit you place
the transistor into, and (2.) R'e changes a little bit as
the ebb, and flow of the AC signal goes through the amplifier. This
means that the amplifiers voltage gain is changing slightly with respect to
the instantaneous voltage seen at the base lead relative to ground. The higher
the base DC voltage the more AC voltage gain, the lower
the less. This is Logarithmic Distortion and it is a major
contributor to something called Total Harmonic Distortion, sometimes
abbreviated THD by the HiFi audio industry, and often cited by
"Golden Ear" Audiophiles as the top of their list of things to minimize,
and, or avoid. The way you keep this type of distortion to a minimum, is
to avoid allowing R'e to be a major component of your voltage gain.
In other words, keep the gain reasonable, and use conservative design
criteria. Add an extra stage if you need more gain.
The second thing to note, is that if a single stage common emitter, amp
results in inversion, what happens if you cascade two of them together, that
is, what do you get when you invert the previously inverted signal? You get a
normal, non-inverted signal. Three common emitter amplifiers chained together
result in an inverted signal again, and so on. Basically an odd number of
common emitter amplifiers, ends up inverted, and an even number makes a
non-inverted output. Note that I was careful to specify common emitter
amplifiers when talking about inversion. Common base, and common collector
amplifiers do not invert the signal!
Running the numbers of R'e on this simple two stage amplifier
from the quiescent point to near saturation, and then plugging
the R'e variance back into the gain equation you get
about 4.7 percent distortion, which is pretty bad, but when the
signal hits the second stage, it too distorts the signal, but because of
inversion the logarithmic nonuniformity is for the most part nulled out.
If you were to exaggerate this logarithmic distortion, of just the first
amplifier alone, and look at it on an oscilloscope, what you would see is
an output waveform whose top part of the wave was smashed down a little,
and the bottom, or negative tips of the wave were elongated somewhat. Now
flip that image upside down, and distort the result again exactly the same
way, and it's obvious why this is so, the elongated peaks are now on top,
so they get smashed, back toward their normal shape, and the smashed ones
now on the bottom, get stretched out a little, again back to their normal
shape as they go through the second stage of amplification. If you worked
very hard to keep the voltage gain in any pair of stages of amplification
as closely matched as possible, and the emitter currents are also matched
Logarithmic Distortion can be held to tiny fractions of a percent.
Golden Ears love this approach, they don't trust negative feedback, or
at least that is what they tell you, what's funny about this, is that the
external emitter impedance is in reality providing negative feedback.
They, or at least many of them refuse to see it that way. I know I've
probably offended an audiophile or two, please send your flames to
dev/null I am endeavoring to explain this subject, to people who
wouldn't understand the higher math anyway, if they are curious, and I'm
sure some of them will be by reading this tirade, when they reach the point
in their learning curve that they can comprehend the theory I'm quite
certain that they will be vital members of the audio news groups.
In the above two stage common emitter amplifier I did not attempt to match
these gain factors, because to match them you must take into account the
output loading of the final stage amplifier, and since I am specifying
it is not loaded balancing this, for minimum distortion is pointless.
The input stage on the left, when the input loading effect is factored in
from the stage on the right is the AC parallel combination of
the left transistors collector resistor, 2K ohm, the two base
resistors on the right hand transistor 13K ohm,
and 9.1k ohm, and the effective right hand transistor's emitter to
ground AC impedance, of 200 ohm multiplied by a beta gain
of 20.0 worst case, gives us 4K ohm. This comes
to 1.0674K ohm. Dividing that by the left transistor's emitter to
ground AC impedance of 200 ohms gives a total AC voltage
gain of 5.3372, and since the output circuit of the right hand
transistor is open circuit, its gain is given by 2K ohm divided
by 200 ohm, eg. 10.0 so total system voltage gain, of
both stages of amplification is the arithmetic product of each individual
stage. Thus 5.3372 times 10.0 is 53 or approximately
Av = 50 for nice round numbers, absolute worst case.
I show how I got the numbers I got, so that if you are fuzzy on any of the
details you can run the numbers yourself, and hopefully get the same numbers.
Attempting to do this, and failing to get the numbers to come out right,
probably means you, or I, goofed somewhere. If you think I goofed, you can
always E-mail me, but E-mail once sent, is exceedingly difficult to withdraw,
and if in fact you goofed, and then subsequently discover your mistake, you
are powerless to pull back E-mail that will reflect badly on you. What I am
trying to say here is that there is no such thing as a stupid question,
just some that are more difficult to live down. :-)
So far I have discussed the common emitter, in depth, and given some coverage
of the common base, I will cover more practical uses for the common base
later in this course, but the one thing of the basic three configurations
that I haven't discussed is the Common Collector circuit.
The Common Collector circuit, is better known throughout the
industry as the Emitter Follower, because it more aptly describes
what this circuit does. Remember my saying at the outset, that bipolar
junction transistors are in reality nothing more than current amplification
devices. An Emitter Follower takes advantage of this by
reproducing at the emitter a voltage slightly behind the input base voltage,
but at substantially lower input base current than the emitter is capably
driving the output load. The bulk of the energy to drive the load is emanating
from a power source that is connected to the collector, however the signal
characteristics of the base voltage, less the junction drop, are what is
manifest at the emitter, albeit at much greater current; again the current
is coming from the collector powersupply. What I mean by signal
characteristics of the base voltage, is things like regulation, line ripple,
or some intentionally injected AC signal voltage.
Still the nine volt battery is wasting away, at a 37 milliampere
rate. We can minimize this quite a bit by adding another transistor, although
we do suffer an additional junction drop, for the new, additional transistor.
The following circuit is your first serious electronic project. It is a
regulated variable DC powersupply. You will use this device
as long as you work with electronics. Even if you buy new, or design and
build better better ones, an old working variable powersupply never lies
idle, they do get used. You can never have enough of these gizmos, you will
find yourself series them, using one for biasing, while the other one is
providing main power, I even quick charge NiCad batteries on a bench supply
rather than wait for them to charge in the charger, which can take days.
Let me begin this section by saying the best way to learn engineering is by
doing it. The most important specification of any component is a
characteristic called price. While some characteristics fluctuate with
environmental conditions such a temperature, or vary widely from one lot
number to the next, those things can be scientifically addressed by
statistical analysis. The parameter Price however is a characteristic
largely dependent on a myriad of market conditions, such as Supply,
Demand, not to mention deliberate market manipulation, and since it
is not an electrical characteristic, that directly affects the operation
of the circuit, or system it is placed into, this critical parameter, is
often not available in data books. Be that as it may, you as an engineer
must consider this parameter when you design a circuit, for your design
to be complete. A good first level approximation that takes into account
the all important Price coefficient, involves looking at the system
as a whole, looking for a single component that is noticeably more expensive
than all the others combined. In this bench powersupply, that component is
probably the power transformer. If you happen to have a suitable transformer
on hand, or if you can find an inexpensive one at a surplus electronics
store, that will fill your needs, at one tenth the cost of a new one, unless
you are independently wealthy, the effort spent making that transformer
work, is less of an effort, than the effort you spend in your day job, to
earn enough money to buy the expensive new one. As I started this paragraph;
you learn by doing, if you go out and buy a transformer, that does not
require you to redesign the circuit given above, you may as well be building
a predesigned kit, from the point of view of how much you are likely to
learn. In lesson 014 about two thirds of the way down I give
some tips on using an existing transformer that you salvaged from some
other device, for use your project. I gave information on deliberately, and
safely setting the taps of a transformer Wrong to make it produce a
lower, or higher voltage than it was originally intended for. I then
introduced you to a myriad of diode and capacitor configurations, that
make possible the production of the desired DC voltage in spite of
the situation not being perfectly suited to your project. I do not
intend you to build this bench supply exactly as shown, you are expected
to save money by redesigning it, a little, here and there, using what you
already know. What you do need to know before you go off scrounging up a
transformer, is what the requirements of the raw DC powersupplies
are.
The Main supply:
This is the raw supply filtered by the 10,000 uf capacitor.
It powers both transistors, and the 1K ohm pull up resistor that
drives the two 1N4148 forward junction compensation diodes at the
right middle bottom of the diagram above. This main supply is responsible
for generating the output current of the entire bench supply. If the
transformer after rectification, and filtering is only capable of delivering
one ampere of current then that is the limit of your bench supply, and you
should fuse it no higher than one ampere. However the 2N3055 power
transistor is an awesome transistor, that can cost as little
as $1.50 US ordered form the likes of Mouser, Digikey, etal.
Here's what I mean by awesome, Power Dissipation 115 Watts derated
linearly to 200 deg C, at a rate of 0.66 Watts
per Centigrade degree. Maximum collector current of 15 Amperes Wow!
and collector emitter breakdown voltage of a whopping 70 volts, when
a 100 ohm resistor is placed across the emitter, and base, as I have
done in my design. Note in order to take advantage of this voltage you
cannot use the 2N2222A it breaks down at 40 volts, if you
want to build the high voltage version of this supply, I give you an
alternative transistor to use in place of the 2N2222A.
The 2N3036 is an 80 volt transistor, with only half the
beta of the 2N2222A, but the higher voltage bench supply is not
going to as as capable in its output current because of may factors, chief
among them power dissipation in the 2N3055 itself.
At 60 volts across the emitter, and collector of a power transistor
A mere 500 milliampere load will produce 30 watts of
waste heat in the power transistor, and a simple two cell flashlight bulb,
a number PR2 draws that much current. The compromise that one
usually makes when building the higher voltage version of this powersupply
is in the area of output current. With less output current, the lower beta
in that higher voltage alternative to 2N2222A is not really a
problem, however power dissipation in the 2N3036 used to substitute
the 2N2222A is likely to be a concern, you may even have to consider
devising some way to heatsink the 2N3036 driver transistor.
Part Absolute Maximum Minimum - Maximum
Number Breakdown voltage Beta
2N2222A Vce = 40 min Hfe @ 150ma = 100 - 300
2N3036 Vce = 80 min Hfe @ 150ma = 50 - 150
2N3055 Vce = 70 min Hfe @ 4.0 A = 20 - 70
Look over your stock of used transformers, and see if any you already have
on hand are usable for this project, in the voltage range you want to work
with. I also need to mention that the low current raw powersupply, the one
that powers the zener diode regulator circuit, need not be built as I have
shown here, and if you notice how low current that section of the raw
supply is you probably picked up on the notion that this raw supply does
very little work. In fact all it has to do is provide a few milliamperes
through the zener diode current limiting resistor R1 to
bias the zener diode into regulation, at a voltage high enough to allow you
to dial in the entire usable range available in the main supply. The
only critical thing about the zener supply is that it be a high enough
voltage to do this even during a brown-out. Translation... The zener supply
should be at least 33% higher than the zener's breakdown voltage.
And the zener's breakdown voltage should be at least as high as the unloaded
main supply. If you were to find a 110 Volt AC transformer
with two secondaries, a high current main winding, and a low current small
winding that was about one third the voltage of the main winding. If the
main winding serves your needs for the main raw DC supply the
smaller one could be used to build a small "pony" supply, that you wire in
series with the main DC supply. This adds the low voltage to
the high voltage, and insures the zener bias supply requirement is met.
Don't crowd too close to the breakdown voltage of the transistors, if you
suspect that your unloaded peak voltage will be 38 volts, don't
press your luck, use the 80 volt driver transistor.
Murphy's Law:
Anything that can go wrong, will go wrong.
If you fail to heed this warning, and you go ahead and build it using
a 2N2222A transistor, as soon as you plug it everything will
work fine for a few minutes, and the Air-Conditioner will click off, and
your house current wall socket voltage will increase a few percent, suddenly
your circuits main raw supply is now idling at 42 volts, and
the 2N2222A is breaking down, leaking an erratic two or three volts into
the output, even though the dial is turned all the way down. I would advise
staying under the breakdown voltage by at least ten percent, to allow for
small voltage surges, and other variations.
After you have decided what raw voltage you are going to shoot for, you must
decide what what the top end of your dial will set the powersupply output to.
If you set your Zener voltage, that is you order a zener diode part
number that has the desired reverse breakdown characteristic voltage, to a
voltage slightly above the raw supply, you will be able to dial in the full
range, the raw supply can deliver. Doing this has the drawback that the top
end of the available voltage, falls in and out of regulation at
the 120 Hz ripple frequency, or for those of you in Europe, and
elsewhere 100 Hz. If you set this Zener diode voltage a
little lower you will reduce the chance that it will allow ripple into the
output, but also limit the maximum usable output. Perhaps a nice compromise
would be to give it the full range, and then paint a red-line on the dial.
Now that you have selected your zener diode you need to compute the value
of R1, and R2, the knob. The most readily available
zener diodes are 1 watt, and 400 milliwatt, I advise you
to go for the 1 watt variety if at all possible. Knowing the
breakdown voltage of the zener, and its wattage you can compute what current
it will require to dissipate the full one watt. Once that is known, you can
compute the resistance of R1 by applying that current to what is
left of the raw zener supply after subtracting the zener voltage itself
from the loop. Be sure when doing this to consider the unloaded peak voltage
of the raw supply, because that is the condition the unit will idle at,
eg. spend most of its life doing.
Raw - Vz
R1 = ------------
Iz
Where:
Raw = Raw Zener supply
Vz = Zener diode rev-
erse breakdown
voltage
Iz = 1 Watt / Vz
Ok I've just instructed you to bias the Zener diode to dissipate at its
absolute maximum dissipation spec, this hardly sounds like conservative
design rules, now does it. Well I'm not finished yet. Next you are going
to compute the value of R2. I want you to draw away from the zener
diode one third of the current, it otherwise would be converting to heat.
Vz * 3
R2 = ------------
Iz
As a final step go back though all your calculations and use what you know
to calculate the wattages of resistors, and the highest possible voltages
that capacitors in your circuit are subjected to so that you may fully
specify all components of the system.
One of the nice things about designing your own circuit, is that you are
aware of all the tradeoffs you made in the design process. You go to the
electronic parts store, and the part that you want is brand spanking new,
and the proprietor of the electronic parts store is real proud of this one,
you can tell he really likes this one, because he has priced it so high
that only a five star General, somebody like General Dynamics :-)
is the only person on the planet who could ever afford it. However sitting
right beside it, is a prettier one, that would do the job, if you made
minimal changes to your circuit... Yep, been there done that.
Knowing how to engineer your own stuff has its advantages.
Here I give a practical version, using a real world transformer, and
other parts. Let me see now, I'll just scrounge around in my box of medium
sized transformers, here's one, it's a Radio Shack
model 273-1512 by the way don't go out and try to buy one of these
little gems, I seriously doubt they sell it any more, as it's pretty old.
Its Primary is 120 volts at 60 Hz, and it has a single
center tapped secondary winding rated 25.2 volts at 2 Amp
this info is printed right on the outer layer of protective insulation.
This means that if wired up as shown in the above circuit, the high current,
main raw supply will reach
((25.2V / 2) * (Sqrt 2)) - 0.7V = 17.11
if powered with 120 volts. However if we take into consideration
that it is possible for the line voltage to be as high as 135 volts
((135 / 120) * (25.2V / 2) * (Sqrt 2)) - 0.7V = 19.35
and the zener regulator circuit, the one that uses a 300 uf
capacitor for a filter, will attain twice that much
voltage 38.69 volts. Since this represents a fairly steep voltage
rise on the part of the 300 uf capacitor I would add an in-rush
limiting series resistor between the 1N4148 diodes and the
capacitor, to limit the in-rush current just a little, to protect
the 1N4148 diodes, since they can only withstand 500 ma
for a one second surge an 82 ohm resistor will do the trick
nicely. A 1N4747 zener diode breaks down at 20 volt,
and is rated at 1 watt, so its max current is 50 ma,
(38.69 raw peak) - (20.0 Vz)
R1 = ----------------------------
.050
(20.0 Vz) * 3
R2 = ---------------
.050
Hence R1 = 390, note the actual calculated value
is 373.8 ohm, but you don't spec a value like that, because you
can't afford to buy it, I try to stay with 5% values and if a
choice has to be made between 360, and 390 ohm, the
higher value, is a lower current, so that is the safe choice. Plugging
the 390 ohm resistor back into the voltage across it, gives
us 0.89568 watts, so we spec that R1 is a one watt
resistor. R2 works out to be 1200 ohm, but try to find
a 1.2K ohm pot in your local Radio Shack,
however 1K is very close, and its lower resistance pulls even
more heat off from the zener, a good thing to do. If R2 is set
to 1K ohm it will dissipate 400 milliwatts, again
the nearest you will find is a one half watt potentiometer, so one half
watt is all you need, that said you could use a full one watt pot, and
you might consider this if you find the one watt pot to be easier to
mount, and find a mating knob for it's shaft, or find the one watt version
to be of superior mechanical construction, such as a metal shaft, versus
a plastic shaft. This potentiometer gets a real workout mechanically,
so spending a little more money on the front panel control knob makes
sense, especially considering you will likely use this powersupply for
many years to come.
Xfmr = 25.2 volt, @ 2 amp
ZD = 1N4747
R1 = 390 ohm @ 1 Watt
R2 = 1K ohm pot @ 0.5 watt
Plus an 82 ohm 0.5 watt
in-rush diode protection res
Plus change the 1N4001 diodes
for a higher current diode
Forward current = 5.0 Amps
Reverse voltage = 50 Volts
Red Line:
Where you place your Red Line is dependent on line voltage, and
load current. An arbitrary choice of say 60% of the dial may not be
the best way to handle this. Using the Radio Shack
model 273-1512 for this example you should be able to achieve
output current of 4.0 Amperes, assuming you changed out
the 1N4001 diodes for a higher current diode, with useful output
voltage of zero to twelve volts. Since the point on the dial at which ripple
becomes a problem is mostly dependent on output load current, if you sample
the this point experimentally at several currents, you end up with several
places to put your Red Line mark on the dial. May I suggest
that you place several Red Line marks, one for each current.
This approach is useful, since you generally know in advance the approximate
current your load will draw, you simply look at the appropriate
Red Line mark for that current, and know that beyond that point
regulation is no longer assured. I show below a test setup that will allow
you to measure where the onset of ripple begins to interfere with regulation,
using either a headphone, or the speaker out of a transistor radio, held to
your ear so that you can detect the faint hint of ripple voltage riding on
top of the DC voltage, that this test setup blocks from reaching
the delicate headphone transducer element, by the judicious use of
an AC coupling capacitor. I make some assumptions about the
point in current which will elicit ripple from my experience in such matters,
in short I pull the load numbers out of the air for the value of load
resistors they are not all that critical, and you should measure the current
and voltage, where you first detect ripple, for each Red Line
you draw on the dial, number the Red Line mark on the dial itself
and then make a chart, that you paste to the top of the instrument, that
shows the red line, its current, and the still, but barely regulated voltage
you measured, at that current. When completed you will have a bench supply
that will allow you to push it right up to its very limit, and remain in
regulation, and if you need more juice, you can push beyond this limit,
albeit with loss of regulation, and clean DC filtration. Why might
you want to forgo regulation? I'll answer that with a question, do you need
pure regulated DC voltage to light up a light bulb, or run a motor,
probably not, and you might really appreciate a little more power.
Mark Load current Load
nmbr resistor
1 4.0 Amps 2.7 ohm 50 watt
2 2.0 Amps 6.2 ohm 25 watt
3 1.0 Amps 13 ohm 15 watt
4 0.5 Amps 27 ohm 7 watt
5 0.250 Amps 56 ohm 3 watt
In a pinch, you can use a resistor of inadequate wattage, for a short period
of time, if the spread between the required wattage, and the available
resistors wattage rating is not too great. The wider the spread the less
time you have to make your measurement, as every second that passes is
heating the resistor to higher, and higher temperatures. You can fan cool
a resistor, or simply blow on it. On occasion when testing computer
powersupplies that I repaired, the voltage was
low, 5, and 12 volts, I dug out an old coffee can, filled
it with water, hooked clip leads onto the resistors, and submerged
resistor and all in the water. This worked well enough that I was able to
run a lengthy burn in test, eg. many hours. However at the end of the test
the electro-chemical action of the water, made the resistor leads so brittle
that they broke off as soon as I handled them. So this is just a tip, not
a production technique. Also you might want to use a large high power
rheostat, (variable resistor) rather than guessing at where the two
intersecting lines cross, you can adjust the load as well to get currents
for your Red Line marks to represent exact currents. You will
see what I mean when you actually do this step. This is one those things
that is just too complicated to explain, but after you do it you will know
why I didn't attempt to explain it here.
The right side of the above drawing is a mockup of what the dial will look
like. If yours is not this fancy, after you have gotten the
Red Lines marked with something crude like a felt tip pen, fire
up Xpaint and make a nice front panel on your computer. If you are fortunate
enough to have a Color Printer, color is a nice touch. They make full sheet
sticky label stock, that will feed through your printer. Peel and stick,
the label down to the front panel, and then add a layer of transparent
laminating plastic over the top. Trim the excess, and presto, instant
professional front panel. About the actual
markings, 0, 5, and 10 volts are pretty
obvious, 20 is shown in inverse video to indicate that this
is only where 20 would be, but that the circuit cannot ever
achieve 20 volts.