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 2.128As 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

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

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 coefficientScale 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

The thermal voltage

vT = (k * T) / q Where: k = 1.38054 x 10^-23 J K^-1 q = 1.6022 x 10^-19 CoulombsAt room temperature,

I have saved discussion of n for last.

The constant

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

Ok enough already, here's a drawing of a biased "NPN" transistor.

I show the electron paths in blue, by the way electrons

What is happening here:

In the above drawing the electrons flowing into the

If you turn on,

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

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

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 1000To 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 artNotice 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

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

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

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

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:

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

Another way to state this is to say that the transistor is loading down the current limiting

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,

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,

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 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 181It 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

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 emitterI covered thermal voltage

Two important things to note:

First understand that R'e is current dependent. This fact means two things,

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

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

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

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

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 - 70Look 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

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 / VzOk 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 = ------------ IzAs 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 = 17.11if 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 = 19.35and 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 = --------------- .050Hence 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 VoltsRed Line:

Where you place your

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 wattIn 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

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

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