Class A Amplifier Design - Part 2

USING EMITTER DEGENERATION

Now if you provide feedback in-phase you get an oscillator. What happens in this instance is some of the energy in the output is fed back to add to the input so we have sustained oscillations or a fly wheel effect.

This is most definitely NOT the goal here. We want to do the exact opposite.

We will take some of the output out-of-phase so that it subtracts from the input. This is called negative feedback. It is an enormous aid to stability. This is all toward our goal of a linear amplifier.

We will use that tactic in conjunction with emitter degeneration. Look back to the earlier discussion on the emitter by pass capacitor C3.

TIP: - if you trouble shoot individual stages in a defective amplifier (a.c., r.f. or i.f) and note that the signal coming out of a stage is significantly less than what it you would expect, given the magnitude of the input - suspect the emitter bypass capacitor (cathode by pass for a valve set). Sometimes they go open circuit. That means for all practical purposes there is no emitter by pass capacitor. This doesn't happen often and I've only encountered it twice in 40 years. But in fact it represents unwanted emitter degeneration.

So how do we put all these things together?.

Let us take a simple case of emitter degeneration (controlled). We will use our original amplifier Figure 3 which has had a minor change made to become Figure 4, here you will note that R3 has been split into two resistors, R3a and R3b.

R3a has no bypass capacitor while R3b is by-passed for a.c. or r.f.

Figure 4.

Now this is emitter degeneration. How do we use it to advantage?. Well let's assign a value of 500 ohms to R3a and a value of 4100 ohms to R3b, the combined value still equals the value of R3 in the original circuit. For d.c. conditions nothing has changed at all. Again noting I have not used standard values in my calculations but the nearest standard would be used in the real world. The differences resulting would not be significant and would be well within tolerances anyway. I just wanted to use the same numbers for illustrative purposes.

Originally we had an emitter voltage of 3.22V, a base voltage of 3.87V and a collector voltage of 7.24V, these d.c voltages still remain unchanged. But because of the voltage divider action of R3a and R3b (similar to the action of R1 and R2) we get a new voltage at the junction or point where R3a and R3b meet. What is this new voltage?.

Again using ohms law, if we have our standing collector current of 0.7 mA passing through R3b then there must be a voltage drop of 4100 * .0007 = 2.87 across it. That is the voltage at the point a and b resistors meet. But for a.c. purposes this point also represents ground because of the emitter by pass capacitor.

Well now, what happens to our 10 millivolt a.c. signal?.

READ THE NEXT TWO PARAGRAPHS AT LEAST 3 TIMES - you will need to.

Hitting the base with a 10 millivolt a.c. or r.f. signal (which we did before anyway - but I didn't bother to explain it to you then) means the voltage on the base, instantaneously rises from 3.87V to 3.88V. The emitter will try to follow to keep a constant difference of 0.65V and adjust to 3.23V, but the junction point is by-passed for a.c. or r.f., therefore the new emitter voltage (3.23V) minus the junction point voltage (2.87V) gives us (3.23 - 2.87) or a 0.36V voltage difference, which is divided by our un-by-passed emitter resistor R3a of 500 ohms. This gives us 0.36/500 or .00072A or 0.72 mA current flowing through it.

Now the collector current must have risen from 0.7 mA to 0.72 mA and this is an increase of 0.02 mA. This increase is across our load resistor of 6800 ohms, giving a voltage variation of (6800 * 0.00002) or 0.136V or 136 millivolts for an input of 10 millivolts!. Our gain has now become (0.136V/0.01V) or 136/10 which is 13.6 (22.67 dB) compared to a previously unstable 204 before.

Better still - compare the ratio of the load resistor R4 to the un-by-passed emitter resistor R3a - which is 6800/500 or 13.6

Well how about that - joy, oh joy. Irrespective of the transistor gain figures which vary wildly, even within the same batch, we can now program our gain to practical levels.

Also it can be proven that the input resistance or impedance of our amplifier now becomes Beta (a.c.) times Re which in this case was something like 90 * 500 ohms = 45000. This is a fairly high figure and it means the previous stage would not be loaded down.

Now of course comes the downside. All of this gain and power is being dissipated in R4 as heat. It isn't doing anything much. If the following stage had a matching impedance of 6800 ohms then the combined load would be 6800/2 or 3400 ohms. The gain would of course drop.

This brings us now to R.F. Amplifiers. If our R4 resistor was replaced with a choke or a coil, the gain would be restored. All our capacitors would need replacing with more suitable values for the frequency of interest. Typical values are 0.1 uF.

Why a choke or coil?. Because they are inductors and as such resist any change in current flowing flowing through them. All alternating signals would be dumped out through the coupling capacitor C3.