# Add negative feedback to this 2-stage amplifier

I am designing this 350mW 2-stage amplifier. The amplifier has only a voltage amplification stage Q2 biased with a current source Q1+Q3, a buffer Q6 and an output stage Q4+Q5. The input will be an iPhone (1V p-p).

It works OK (in simulation of course). The THD at 1kHz measured in LTSpice is 0.48% for an overall gain of 4.8 under load. At higher frequencies is not great. For example THD at 10kHz is 3.7%.

I am aware that this is not probably ideal, but this is mainly a learning exercise.

I would like now to add global negative feedback, but I am not sure how to proceed. This is what I have tried. I have added a feedback resistor RNFB, a capacitor CNFB and reduced RS1 to 5K. The RNFB was found just by trial and error and my understanding is that the capacitor is needed to maintain the DC bias.

This seems to be a small improvement. For the same gain and peak-peak values THD is now 0.12%, the same for 5kHz, but worse at 10kHz (4.14%).

My understanding is that a differential amplifier input stage is usually used to add the feedback, but that will be my next "lesson" :-).

A few questions:

1. Is this the correct way to add global feedback?
2. Is this what it is considered current feedback?
3. How to calculate RNFB?
4. Is it possible to improve the THD at higher frequencies?
5. Finally, in this answer it was suggested that a single transistor could be used as a "differential input device". Could this work for this circuit?

EDIT 1 - Based on @jonk answer

The reason for choosing the bias method in stage 1 is the active load. I was not able to bias with the method C (from the answer) so I used the recommendations from this question. Was this wrong?

I have tried to apply the feedback proposed to my circuit.

However, results are weird. I have tested different values of RNFB and I have the same outcome.

Finally, I will replace the output stage with a push-pull in the next iteration.

• How are you measuring / calculating THD ? What formula / tool /steps are you using ? What happened to the gain with NFB ? What is the bandwidth with and without NFB ?
– AJN
Jun 25, 2021 at 13:04
• @AJN everything is simulated and calculated in LTSPice
– Rojj
Jun 25, 2021 at 13:17
• @Rojj The standard CE BJT stages are almost never used, in practice. They were used a lot when BJTs cost US\$5 each back when that bought you a huge, pricey meal at a fancy restaurant. Now that BJTs are a dime a dozen, not so much. That said, using what you are interested in using, I wrote an answer entirely about NFB that directly applies to your situation. (Except you are still using that UGLY emitter follower output stage. You need to lose that. It is the worst possible thing you can do, literally, and still get some kind of output.)
– jonk
Jun 25, 2021 at 15:57
• After these modifications, you do not have negative feedback anymore in your circuit. Do you see it?
– G36
Jun 27, 2021 at 12:52
• Also, why did you decide to use two emitter followers? Why cannot you simply use Q5, Q4 connected directly to the Q1 collector and remove Q4 from a circuit?
– G36
Jun 27, 2021 at 13:01

There is quite a variety of CE audio amplifier topologies using only one BJT.

Let start just with three comment DC operating point biasing methods. (I'm not showing the emitter circuit because that's a separate discussion.)

simulate this circuit – Schematic created using CircuitLab

• A has the advantage that the DC biasing current is really tiny. So the DC quiescent current for the amplifier stage is almost entirely determined by $$\I_{\text{C}_Q}\$$. However, the circuit is poorly managed. It's DC quiescent operating point is almost entirely determined by vagaries of the BJT, such as variations from one device to another and variations due to temperature. It's sensitivity to variations in $$\\beta\$$ are such that $$\\%\,I_{\text{C}_Q}\approx \frac{R_\text{B}+R_\text{E}}{R_\text{B}+\left(\beta+1\right)R_\text{E}}\cdot \%\,\beta\$$. For a typical circuit of this type, this usually means that a 50% change in $$\\beta\$$ (quite possible with BJTs from the same bag) leads to about a 40% change in the DC quiescent operating point. Here, $$\A_v\approx\frac{R_\text{C}}{R_\text{E}}\$$ when $$\R_\text{E}\gg r_e^{\:'}\$$.
• B remedies much of the problems with the DC quiescent operating point. It's sensitivity to variations in $$\\beta\$$ are such that $$\\%\,I_{\text{C}_Q}\approx \frac{R_\text{B}+R_\text{C}+R_\text{E}}{R_\text{B}+\left(\beta+1\right)\left(R_\text{C}+R_\text{E}\right)}\cdot \%\,\beta\$$. For a typical circuit of this type, this usually means that a 50% change in $$\\beta\$$ leads to about a 15% change in the DC quiescent operating point. It's now dependent on $$\R_\text{C}\$$. So that's another factor to consider. This circuit also provides some added local NFB ($$\R_\text{E}\$$ also provides local NFB) because it brings some of the collector voltage (inverted and amplified by the BJT) back to the base, out of phase with the base's input signal. Here, $$\A_v\approx \frac{R_\text{C}}{R_\text{E}}\cdot\frac{1}{\left[1+\frac{ 1}{\frac{R_\text{E}}{R_\text{C}}+\frac{R_\text{B}+R_\text{C}}{\left(\beta+1\right)R_\text{E}}}\right]}\$$ when $$\R_\text{E}\gg r_e^{\:'}\$$. Lots of gain is tossed away, but the added NFB does improve THD.
• C usually sets up a bias current that is about 10 times greater than the required base current for either A or B above, so its quiescent operating current is about 10% more. But for that, you get a much more stable DC operating point and the freedom to select base voltage determined by a Thevenin voltage you control. I won't bother writing up the sensitivity equation, due to its length, but the results are that a 50% change in $$\\beta\$$ leads to about a 1% change in the DC quiescent operating point. This is one of the reasons why the slightly increased DC quiescent current is worth the trouble. It greatly reduces variations in the circuit operation due to vagaries related to the BJT. As per A above, $$\A_v\approx\frac{R_\text{C}}{R_\text{E}}\$$ when $$\R_\text{E}\gg r_e^{\:'}\$$. But lacking the added NFB of B, the THD will be higher than B.

I haven't mentioned input resistance, yet. A and B will have higher input impedance than C and so, depending on the source input impedance, C is more likely to attenuate the input signal a bit. So you can expect to get somewhat less voltage gain from C than from A, despite the simplified equations suggesting similar voltage gain between the two.

Bootstrapping is a way to greatly increase (more than a factor of 10 is easily achieved) the input resistance:

simulate this circuit

In the above case, I'm showing a commonly used emitter-leg with a bypassed resistor to increase AC voltage gain. This gets to another point: there are myriad topologies for a CE amplifier stage. And varying goals and priorities will shape the decisions made in selecting one from another.

However, that said, almost none of these are used anymore. The long-tailed pair differential pair as the input stage, with a stiffening current mirror for the collector loads, is almost always used. Even in discrete circuits. Not only does it provide better performance but it also provides a very convenient place to add global NFB (one of the bases of the diff-pair.)

But let's get back to the global NFB using something a lot closer to your configuration. Here's something to consider:

simulate this circuit

With $$\V_\text{CC}=10\:\text{V}\$$, $$\I_\text{Q}\approx 1.1\:\text{mA}\$$ and $$\r_e^{\:'}\approx 24\:\Omega\$$ for stage 1. So, for stage 1, $$\A_{v_{_1}}=\frac{R_{\text{C}_1}}{R_\text{E}+r_e^{\:'}}\approx 17.5\$$. Similarly, you'll find that the 2nd stage (AC grounded-emitter which, without global NFB would exhibit substantial distortion) has $$\I_\text{Q}\approx 1.3\:\text{mA}\$$ and that $$\A_{v_{_2}}=\frac{R_{\text{C}_1}}{r_e^{\:'}}\approx 180\$$. Combined, this appears to imply $$\A_v= 3150\$$, without global NFB.

But the input of the 2nd stage loads the output of the 1st stage. So, this attenuates by about $$\\frac{R_4\mid\mid R_5\mid\mid \left(\beta+1\right)r_e^{\:'}}{R_\text{C1}+R_4\mid\mid R_5\mid\mid \left(\beta+1\right)r_e^{\:'}}\approx 0.257\$$. So the total open-loop voltage gain is closer to $$\A_v\approx 800\$$.

I've added global NFB via $$\C_6\$$ and $$\R_f\$$. You can tell that this is NFB, because the output of the second stage is in-phase with the signal input, but amplified. When you apply an in-phase signal to the emitter of the BJT that has the signal directly applied to it, this counters the input. It does this because, when the signal input tries to raise the base voltage upwards, the 2nd stage output (which is in-phase with this) tries to also lift up the emitter of the input signal BJT. In short, it tries to make the emitter move up when the signal tries to move the base up. If they both move upwards together, then the BJT doesn't see a "signal" anymore. So this is NFB. We just need to make sure it doesn't completely oppose the input signal. That's all.

The closed loop voltage gain equation is:$$A_{\text{CL}}=\frac{A_{\text{OL}}}{1+A_{\text{OL}}\cdot B_\text{NFB}}$$Suppose you want $$\A_{\text{CL}}=50\$$. Then you need $$\B_\text{NFB}=1.875\%\$$.

Look closely and take note that $$\R_f\$$ drives $$\R_\text{E}\$$ and so this is just a voltage divider. The applied NFB here will be based upon $$\\frac{R_\text{E}}{R_\text{E}+R_f}\$$. We know that part of the first stage's emitter resistance is bypassed. But the remaining amount means that $$\R_\text{E}=180\:\Omega\$$ for these purposes. Solving this equation yields $$\R_f=9420\:\Omega\$$. I'd just use a common $$\10\:\text{k}\Omega\$$ resistor here. The resulting expected gain would then be $$\A_v\approx 53\$$. (About $$\34\:\text{dB}\$$.)

Close enough.

Why don't you pop that into a simulator? See what it does for you. Check out the THD, as well.

None of the above fixes your output drive stage, which is unconscionable. For an $$\8\:\Omega\$$ speaker, you will definitely want a 2-quadrant (push-pull) output stage. However, you seem stuck on the idea for now while you work out the rest. So I'll leave that problem for you to work on at a later time. The above discussion, I believe, already gets at the thrust of your main question.

As a final note: The above design is for educational purposes only. The voltage gain is likely too high both at very low and very high frequencies. It would be useful to design a bandpass filter to limit the frequencies that the amplifier may process.

Here's an example that significantly improves the input impedance and provides some modest control of gain near the audio band skirts.

simulate this circuit

(You may also notice the series capacitor + resistor across $$\R_f\$$.)

Best wishes.

Edit: (Fixed a schematic error that got by me when entering the last schematic.)

• Thanks for the answer and comments. I have added some details. I will replace the ugly second stage in the next iteration.
– Rojj
Jun 27, 2021 at 8:24
• @Rojj Without your design notes, it's less interesting to criticize your efforts. I didn't write a lot of notes above, but at least I wrote some and made some predictions on the basis of actual calculations. You might apply yourself more to helping us (me, or others) follow your design thinking process better than you have. I can read. But I want to see your thoughts as you lay out your design so that I can follow in your footsteps better. Just FYI. On the other hand, if you want to drop your design and understand mine better, I'd be happy to add more notes.
– jonk
Jun 27, 2021 at 9:59
• It is a heater, not a 0.35W amplifier. Even when it is not playing sounds it draws 1A and heats with 9W. Jun 27, 2021 at 15:03
• @Audioguru That output stage is an abomination. No question. But so long as it stays in Spice, the OP is safe. ;)
– jonk
Jun 27, 2021 at 17:16
• @jonk that makes sense. I will post separate questions for the various stages and they will not have that output stage configuration.
– Rojj
Jun 28, 2021 at 7:36