# Simple power amplifier with feedback - designing procedure

I have a problem with understanding the whole feedback business in BJT power amplifier. So far, I only considered it in terms of thermal stability and $$\β\$$ uncertainty in simple 1-staged amps and was quite comfortable with it but now - in a full power amp - I fail.

Anyway, here's my problem. I do not know how to calculate the value of $$\R2\$$. The goal is to design a power amp that could produce $$\5W\$$ onto a $$\8 Ohm\$$ speaker at maximum input $$\1V_{pp}\$$. I did, however, established some value of $$\R2\$$ resistor but only using a real-time simulation (It was a blind adjusting, looking at the oscilloscope)

### My calculation method is this:

For a $$\5W\$$ at $$\8 Ohm\$$ I need about $$\6.5V_{RMS}\$$ so $$\18V_{pp}\$$ => power supply of $$\24V\$$ should do it.

At the maximum level, the A point voltage sits at around $$\22.4V\$$.

Assuming that T2s $$\β\$$ = $$\1000\$$, the $$\R3\$$ has to deliver $$\1.125mA\$$ so:

$$R3=\frac{(24V-22,4V)}{1.125mA}=1422 \longrightarrow 1,5k$$

Now, at quiescent state, A point voltage sits at $$\13.2V\$$ so current must be:

$$I_q=\frac{(24V-13.2V)}{R3}=7.2mA$$

Now, that gives me $$\I_b\$$ of $$\T1\$$ transistor (assuming $$\β=100\$$)

$$I_b=\frac{I_q}{β}=72µA$$

To achieve this current I calculate R1 ($$\V\$$ at B point = $$\12V\$$):

$$R1 = \frac{\frac{24V}{2}-0.65V}{I_b} \longrightarrow R1=160k$$

And that's it, I have no idea how to proceed on calculating (not blind shooting) the $$\R2\$$ value when input is +/-$$\0.5V\$$. Also, I don't seem to understand how to calculate the open-loop gain.

Yes, I have multiple books with formulas but.. well, it doesn't help.

Could somebody please help me sort that out? I am a pure hobbyist so there is no teacher to talk to...

P.S. Forgive the battery biasing Darlingtons - that's just for simplification

• +1 for including a schematic (not everyone does that!) and clearly stating the problem. I'll write an answer. Dec 17, 2020 at 12:51

Find the exact value of an $$\R_2\$$ resistor will not be an easy job.
The first approximation is to assume that we have an "ordinary op-amp based" amplifier. The inverting amplifier with a gain equal to: $$\frac{V_{OUT}}{V_{IN}} = - \frac{R_1}{R_2}$$

Thus, we can find $$\R_2\$$ value (first iteration):

$$R_2 = \frac{160k\Omega}{18} \approx 9k\Omega$$

But as it turns out the voltage gain is much less than the desired value.

Thus, to get the correct value let us try to find the amplifier input impedance.

But first, we need to find the open-loop gain. In this circuit, we have a common emitter amplifier plus the Darlington based emitter follower.

The first stage gain will be around

$$A_{V1} \approx \frac{R_3}{r_{e1}} = \frac{1.5k\Omega}{3.6\Omega} \approx 416V/V$$

Where :

$$\r_e = \frac{V_T}{I_E} \approx \frac{26mV}{I_E}\$$ (in room temeprature, due to $$\V_T\$$ temerature dependent)

The Darlington voltage gain will be current dependent, so let us assume (wild guess) that the second stage gain is $$\0.9\$$.

Therefore the open-loop gain is $$\A_{OL} = 416 \times 0.9 \approx 370 V/V\$$.

And this means that due to the Miller effect $$\R_1\$$ will be seen at the input as a much smaller resistor $$\R_M = \frac{R_1}{A_{OL} + 1} \approx 430\Omega\$$.

Miller's Theorem - Input Capacitance

Aditional the T1 input resistance is $$\R_{T1} = (\beta +1)r_e \approx 360\Omega\$$

Therefore $$\R_{IN} = R_M||R_{T1} \approx 200\Omega\$$.

So, to get a voltage gain around $$\18V/V\$$ we need to attenuate the input signal $$\\frac{370}{18} \approx 20\$$ times.

Knowing this and that the $$\R_2\$$ will form a voltage divider with amplifier input impedance we have a second guess about the $$\R_2\$$ resistor value.

$$R_2 \approx (20 - 1) \times 200\Omega \approx 3.8k\Omega$$ as this value is available in E24 series.

The situation will look more or less like this:

simulate this circuit – Schematic created using CircuitLab

$$\frac{V_{OUT}}{V_{IN}}=\frac{A_{OL}}{(1 + A_{OL})\times \frac{R_2}{R1} + (1 + \frac{R_2}{R_{T1}})}$$

Now, we can simulate the circuit and see the outcoming results.

And I hope that this is just a theoretical exercise and you don't want to build this circuit in real life? Mainly due to the lack of emitter resistors in the Darlington stage and the high possibility of a thermal runaway.

• Perfect! Thank You, $3.8k$ is exactly the value I came up with through simulation. I suspected it has something to do with input impedance and a voltage divider but I would never come up with Miller's theorem myself here and no, I am not going to build exactly THIS amplifier - I deliberately simplified it for analysis purposes and omitted some blocks like bias servo, source current, emitter resistors etc. One more thing - would it be any different if I used a voltage divider instead of single $R1$ resistor to bias $T1$? Would imput impedance be $Rm||R3||RT1$? Dec 17, 2020 at 19:44
• $R3$ would be of course a second resistor of a biasing voltage divider (not shown at current diagram) Dec 17, 2020 at 19:48
• You are right about the Ry (R3) resistor influence on Rin resistance. But typically R3 will have a value more than 10 times Rt1 so we can ignore the R3 influence.
– G36
Dec 17, 2020 at 20:05

From the base of T1 to the output, this is basically an inverting amplifier with a feedback resistor R1. So I can simplify this circuit to this:

simulate this circuit – Schematic created using CircuitLab

The amplifier "amp" contains all the transistors and what you need for biasing them. (Note that your amplifier doesn't have a real + input like my "amp" has. But that's OK, your amplifier responds to voltages relative to ground. My "amp" does that too as I connected its + input to ground).

Now it is easy, you want a voltage gain of

$$\\frac {V_{out}}{V_{in}}= \frac {18}{1} = 18\$$.

This is an inverting amplifier with feedback configuration (read more here) so the voltage gain is simply

$$\A = \frac {R_1}{R_2}\$$

So $$\ R_2 = \frac {R_1}{18}\$$ = 8.9 kohm

Note how the value of $$\R_2\$$ will not disturb the DC (biasing) that you so carefully calculated. There's a capacitor in series with $$\R_2\$$ so for the DC biasing nothing changes!

• Thank You for Your interest! However, I did some reading on the matter and I am familiar with the basic concepts (my amp is an inverting one and I am trying to apply a negative feedback to it) as well as the formula You posted ($R2/R1$) and it does not work here. After applying 8.9k resistor, the output voltage dropped significantly and the maximum swing is +/-$3.85V$ instead of the previous +-$9V$ . That's why I posted this question in a first place - I seemingly missing something important. The correct value of $R2$ is something like 3.5k-3.7k (picked experimentally) Dec 17, 2020 at 13:22
• tinyurl.com/y84tw7jl Give it a go... Dec 17, 2020 at 13:26
• If your amplifier can handle the 8 ohm load then this should not happen. I suggest that you "scale down" the load, make it 800 ohms. Do you now get the expected voltage gain? Dec 17, 2020 at 13:27
• same story... tinyurl.com/y8cwzer9 Dec 17, 2020 at 13:28
• I see you are using VERY large capacitor values. The 1 mF at the input with 8.9 kohms gives a time constant of 9 seconds. So more than 9 seconds are needed to properly charge that 1 mF capacitor and for the complete amplifier to bias properly! I changed it to 10 uF. I suggest that you set the input signal to a very small value ( 1 mV) and then check if all DC voltages and currents are as expected. Dec 17, 2020 at 13:34

You are missing a vitally needed resistor that fixes your circuit: -

Call the resistor I added RY (meaning R yellow). It is needed to resolve the problem of incorrect DC biasing (basically your current circuit has none so it won't work at all).

And, it might be a little fiddly to find the exact value of RY because your input stage isn't going to co-operate some of the time. A differential front end would be more preferable and then you wouldn't need RY. However, I think we can make an estimate of RY that proves to be satisfactory...

Thought experiment, if you applied a controllable DC voltage source to where RY is, what voltage do you need to set it at to get the quiescent DC voltage on the output to be roughly 12 volts (mid-rail)?

Your diagram appears to be telling me this; if you have 646 mV in, you'll get 12.08 volts out so, choose RY so that it automatically produces this voltage ratio in conjunction with R1 (the feedback resistor). In other words: -

$$\dfrac{RY}{RY+R1} = \dfrac{0.646}{12.08}$$

If I've calculated it correctly, RY = 9.04 kΩ.

If your diagram was misleading me then you'll have to measure it - however, it will be around 0.7 volts hence why I used that value.

Solve that and you're done. R2 does not affect the DC bias point at all because it is in series with your input capacitor. R2 affects AC gain only.

Also, I don't seem to understand how to calculate the open-loop gain.

A reasonable approximation for a common emitter stage is Rc/Re i.e. the ratio of collector resistor to emitter resistor sets the gain. However, you don't have an Re so you have to estimate the BJT's internal value ($$\r_E\$$) and this is based on: -

$$r_E = \dfrac{0.026\text{ volts}}{Ie}\hspace{1cm}\text{At around 27° celcius}$$

So, if the current passing through T1 is 5 mA then $$\r_E\$$ is about 5 ohm. This makes the T1 stage have a voltage gain of 1500/5 = 300. However, you will have distortion when not using an external emitter resistor so I don't think this design is going to win the HiFi amp of the year award!

T2 and T3 are voltage followers and don't produce voltage gain. Their naturally high β (because they are Darlington BJTs) is not going to mean that they significantly load R3 (aka Rc).

To find R2 based on achieving 18 volts p-p on the output with 1 volt p-p on the input means you start with the voltage gain and that is clearly 18. Once you have got the biasing fixed up by using RY you then can recognize that point C (the base of T1) is now a viable virtual ground. Yeah, OK, you might say that the open-loop gain of maybe 300 isn't going to make it that good a virtual ground but, I don't think it will be bad at all: -

$$18 = \dfrac{160 \text{ kohm}}{R2}$$

If I've calculated it correctly, R2 = 8.89 kΩ.

Given that the base of T1 is a virtual earth, the presence of RY will be negligible to small. But, don't forget, T1 will have temperature dependencies that will make the bias point shift and, as previously mentioned, this can only be realistically overcome by using a pukka differential front-end. You'll also benefit by the open-loop gain having a couple of zeros added to it.

• If I get it correctly, You're suggesting using a voltage divider to bias T1. I know that's better but that's not the point. I need to know how to calculate $R2$ and it has nothing to do with biasing (but has a lot to suppressing the input signal as it serves as input impedance). As for open-loop gain - I don't think it's correct either. You're ignoring the push-pull input impedance which is in paralell with R3 and they together create a colector resistance. However, it's unclear to me how to calculate the push-pull input impedance. Dec 17, 2020 at 14:56
• @DawidW I'm still in the process of calculating R2 but it can't be done without explaining the fix (RY) to make it work. It won't work without RY. T2 and T3 are Darlington devices with an effective beta of maybe 5000. The load is 8 ohms hence the effective impedance they present back to R3 is 40,000 ohm and that modifies R3 from 1500 ohm to 1446 ohm. Is that a big difference really? Are you expecting BJTs to play-ball using a simple open-loop analysis? Not a chance, but, it's the best we can do and hope that closed-loop feedback makes things good. But you still need RY. Dec 17, 2020 at 15:03
• well, after applying $Ry$ = 9k and $R2$ = 8.9k the output voltage dropped horribly and now it's $7.6V_{pp}$. It's the same story as in the other answer (check the discussion there). See for Yourself: tinyurl.com/y9ntw5z4 The correct $R2$ in my circuit is ~ 3.7k and there's no need for another resistor for it to work then (at least in simulation). Basically, it's the same diagram I find in some electronics books. Dec 17, 2020 at 15:40
• @DawidW your previous question showed a similar amplifier configuration but they used a pot to set the bias point AND they also used a pull-down resistor (RY) like how I mentioned. You cannot expect perfection when trying to calculate BJT stuff. Not even simulators are that good are getting it right - that is why your previous circuit uses a pot and it's why most biasing stages for BJTs use a pot to set the quiescent current. Manufacturing variances per transistor are not that good. Dec 17, 2020 at 15:46
• ... Anyway, if you turn off the input signal, what DC voltage do you get at the mid-point of the amplifier? AND bear in mind that if you have used any other output transistor configuration than what is shown in your question, my calculations could be wrong. The devil is in the details @DawidW. If you really need this to be more perfect, you should add a differential amplifier at the front end. Dec 17, 2020 at 15:50