A WIZARD'S ELECTRONICS COMPANION

AMPLIFIERS

Common Emitter Configuration for Transistors

This is used to achieve voltage gains, particularly. "Common emitter" refers to the fact that the emitter of the transistor is closest to being common to both the input and output signals. Normally the extra parts are dispensed with to show the emitter connected to ground directly, but we will put it into the Class A amplifier configuration as this is its usual application. The Class A output can be modeled as a resistor in series with a variable resistor which achieves the analog output voltage variations. The input and output capacitors Cb and Cc, chosen to have low impedences compared to the rest of your stage, are here to demonstrate DC isolation between stages of your circuit which has very specific voltage requirements for these nodes. The AC component (your signal) which rides upon the DC biasing as a wave on the water is the only part you're interested in, so this allows you to dispense with the DC part entirely without having to deal with it in the next stage which may not be feasible.

The DC base biasing from Ra and Rb, then, can be calculated independently of what came before. Rc will be limited on the low end by either the transistor's power dissipation or the impedence of the previous stage (the AC component will still have a certain impedence, a limit to the current you can use.) If the previous stage is also a transistor stage, look at the "signal" resistor on the output for the approximate impedence value, if it, too, is a common-emitter stage, it will be in the same place as Rc, respectively. Ra will form an AC voltage-divider with the Rc of the previous stage (and Cb, which we made large so we could ignore its impedence at the important frequencies). This will load the previous stage and will, of course, have to be taken into account when you designed the previous stage. You could also fudge things a bit by ignoring some of the available current. You would be losing a bit of power amplification, but you might just be able to mostly ignore the effect on the previous stage if it's critical by not loading it too much. For maximum use of power, try to match impedences.

Let us calculate a set of hypothetical part values as I describe the circuit. Initially, there is the source impedence of the previous stage, which we shall arbitrarily say was 1kohms. To match impedences would put Ra at about 1kohms also. Let us assume further that this is an audio amplifier capable of a frequency response of some 20-20kHz. Since the lowest frequencies will have the greatest difficulty passing the DC blocking capacitor Ra, 20Hz is value at which the capacitive reactance becomes critical. For this, match impedences again and say that the capacitive reactance must be 1kohms at 20Hz. Then, by the formula:

X[C] = 1 / (2 * pi * f * C) ==> C = 1 / (2 * pi * f * X[C]) = 1 / (2 * pi * 20Hz * 1kohms) = 8µF (10µF is a nice standard value)

Rb forms a voltage divider with Ra which gives your base bias voltage. The divider should have enough current in it to not get pulled too much by the loading effect of the base current. A good rule is to use ten times the base current to get the divider current. You can reduce the divider current, but be aware that the current demand will depend on the actual current gain of the transistor which may vary quite a bit--far more than the variation likely in your resistors, and you don't want to risk current-starving them so your circuit doesn't function at all. Matching impedences, then, would put Ra equal to the source impedence, or 1kohms.

The DC operating point of the base-emitter junction needs to be higher than the expected 0.7V (in a silicon bipolar transistor), but must not be so high it hogs the voltage swing range on the collector, for the base will be operating at 0.7V above that of the emitter and the collector cannot drop lower in voltage than about 0.3V above that of the emitter (unless forced by outside energy). Let's only steal about a third of the available voltage. The DC operating point would be:

Vq = supply / fraction = 10V / 3 = 3.3V

The current in the divider is then:

Ia = E / R = 3.3V / 1kohms = 3.3mA

The divider current is assumed to be able to support the load of the base easily, so we say ten times. That makes the base current:

Ib = 3.3mA / 10 = 330µA

Anywhere close to 0.7V in a silicon bipolar transistor (or 0.3V in a germanium one) will be seen during operation, so the voltage on Re will be that amount lower than the voltage on the base when the transistor is on. Re is calculated from an assumed minimum current gain times the allowed current in the base. Re stabilizes the current load as there is nothing really preventing this diode junction from hogging current if it is available. An emitter resistor's voltage will rise in proportion to the current being pushed through it, lowering the voltage drop base-to-emitter and gracefully shutting down the collector-to-emitter junction. Re, essentially, sets the current for the emitter and, thereby, in the collector. The voltage on the emitter is:

Ve = Vb - 0.7V = 3.3V - 0.7V = 2.6V (since we ignored the base-emitter voltage drop earlier, we're only stole about a quarter of the voltage)

Assume the current gain of a generic NPN bipolar transistor is 100 times. That puts the current being allowed for the emitter at:

Ie = Ib * (current gain) = 330µA * 100 = 30mA

The emitter resistor then becomes:

Re = Ee / Ie = 2.6V / 30mA = 87ohms

Select Rc so the middle of your useable collector voltage range (the quiescent voltage, or DC operating point when there is no signal) occurs when quiescent current flows in the collector. The available output voltage range is now:

Vcc - Ve = 10V - 2.6V = 7.4V

One-half would be:

7.4 / 2 = 3.7V (each way from quiescent)

The output quiescent voltage on the collector is then:

Vq = Ve + 1/2V(swing) = 2.6V + 3.7V = 6.3V

When the transistor is off, the collector voltage Vc goes all the way to 10V, and when fully on, down to:

2.6V + V(sat) = 2.6V + 0.3V = 2.9V (the observant will note we ignored that 0.3V when we estimated the quiescent collector voltage)

Since the output of the transistor stage can be modeled as resistors in series, we know the current in the branch is 30mA, and the voltage across Rc and the transistor are equal and both 3.7V. Rc thus is:

Rc = E / I = 3.7V / 30mA = 123ohms (at least a quarter-watt resistor is called for here. Those power ratings are based on free air circulation.)

The maximum current in the output, because we chose our quiescent operating point at the halfway mark, is going to be twice the quiescent current, or 60mA. In some other types of configurations, such as when there is a DC signal path (no DC blocking capacitors), the power has to be more carefully considered, but in our circuit, peak currents can only occur with frequency components down to 20Hz. This will create a certain amount of heating in the transistor die before the current must symmetrically swing back to zero. Given the efficiency of a Class A amplifier is low, it just adds to the case supporting using the average quiescent collector current to calculate our power dissipation. Yet another fact supporting this is our maximum power dissipation occurs when impedences are equal between source and load. If we imagine Rc is the source and the transistor is the load, maximum power is being dissipated in the transistor when its equivalent resistance is equal to Rc. The power dissipation in both the transistor and Rc is:

P = I * E = 30mA * 3.7V = 111mW (which is about as much as we dare use with complete comfort for typical small components of the 150-250mW range)

The power dissipation required of Re is:

P = I * E = 2.6V * 30mA = 78mW (an eighth-watt resistor might do it, but I would use a quarter watt here, too. A factor of at least two helps when it runs hot.)

Our output DC blocking capacitor Cc has to contend with a lower impedence than the input in order to be useful, 123ohms at our 20Hz frequency:

Cc = 1 / (2 * pi * f * X[C]) = 1 / (2 * pi * 20Hz * 123ohms) = 65µF (63µ is a common standard value)

A tenth of a watt is plenty to play with for cheap audio. A typical 2-inch diameter speaker, common in small audio consumer products is usually rated around a tenth- to a quarter-watt. An 8-ohm impedence is quite common. The figure shown gives three ways of utilizing our newly-designed Class A amplifier stage with a speaker.

The first replaces our collector resistor with a transformer primary of 100-ohms. These little audio transformers became common in the 1960s with the advent of low-voltage, battery-powered consumer items such as radios and tape-recorders, and are only about a half-inch in size. This is the most efficient approach (though the costliest) using 90%+ of the signal power. The one on the right saves on parts, but is rather inefficient. The DC current is not enough to harm the coil windings. Even though we are wasting quite a bit it, it might be enough for our purposes if our output impedence were low enough (we're wasting it all in the series resistor). The DC quiescent current is 30mA, and the DC operating point voltage is 3.7V (across the resistor). The DC power in the collector portion of the circuit is:

P(dc) = I * E = 30mA * 3.7V = 111mW

The RMS AC Power, voltage, and currents are rather dependent on amplitude (we assume maximum--it can't be higher than this) and waveshape (we'll assume a sinewave, which is good enough, all things considered.) The peak of our sinewave we already calculated at 3.7V. For a sinewave the RMS is:

Peak / SQRT(2) = 3.7V / 1.414 = 3.7 * .707 = 2.6V RMS AC.

But we don't see all this voltage because our load is only 8 ohms of it all! A better formula uses current and resistance. The RMS current is also based on the simplification of assuming a sinewave. 30mA is average, 60mA is maximum (or peak-to-peak), so RMS is:

Pk * 0.707 = 30mA * .707 = 21mA AC RMS

P(rms) = I(rms)^2 * R(spkr) = (21mA)^2 * 8ohms = 3.5mW RMS AC

With these gross inefficiencies, you won't hear much even from that two-inch speaker. The absolute efficiency, takes into account all the power pouring through that output stage. The power dissipated in the output stage is:

P = I * E = 30mA * 10V = 300mW

The true efficiency of the output stage is then:

Eff% = (Pwr out / Pwr in) * 100% = (3.5mW / 300mW) * 100% = 1.2%

Notice that the useable RMS power is only 111mW. Even with a 100% use of the signal we've created, the efficiency of the speaker (only at full power) for the useable portion of the signal is:

Eff%(max) = [P(ave) / (P(ave) + P(total))] * 100% = (111mW / 300mW) * 100% = 37%

The middle approach, like the transformer, blocks DC from getting to the speakers, but the current (and thus the power) is split in half to accomplish this, so it is, at best going to have half the efficiency of the "direct drive" approach--1.8%. Of course, if you lower your output impedence to around 8 ohms, this efficiency goes way, way up, but cannot be more than 37%.