I have been dealing with these differential amplifiers for weeks now and still don't understand the basic principles of its operation and how should it be used inside an audio amplifier.

I understand everything to the point when input (say AC) signal is applied to circuit. I completely do not understand how should negative feedback from output of amplifier back to non-inverting input do anything at all. All that distortion canceling within diff-amplifier pair negative feedback is totally unclear to me (in practical way mostly).

enter image description here

  • How do the two transistors improve amplifier's THD (Total Harmonic Distortion)? How is a part of that distortion being canceled by diff-amp? Does there have to be 180 degree phase shift back to negative feedback for distortion to get canceled (If yes, why)?

Any answer to one of this question with practical example would certainly improve my understanding of this "wonder".

  • \$\begingroup\$ The linearizing effect is pretty simple. The standard equation showing gain is \$G=\frac{A}{1+A\cdot F}\$. (You can find out how this is derived here: Negative-feedback amplifier.) If the value of \$A\cdot F\$ is always large regardless of the input voltage (which allows that the gain itself depends on the input voltage and is perhaps highly non-linear), compared to 1, then \$G\approx\frac{1}{F}\$. Which is roughly linear and a function only of the feedback percentage and no longer a function of the input (if it ever was.) \$\endgroup\$
    – jonk
    Dec 4, 2017 at 19:40
  • \$\begingroup\$ Are you talking about the cancellation of odd harmonic distortion in differential amplifier? \$\endgroup\$
    – sarthak
    Dec 4, 2017 at 19:47
  • \$\begingroup\$ @sarthak Does it matter which? Yes, usually the second and third harmonic should be important to cancel them out as much percentage as possible. \$\endgroup\$
    – lucenzo97
    Dec 4, 2017 at 20:26
  • \$\begingroup\$ You're trying to "see" 2 things in one go. I would suggest to split the two. The negative feedback is one thing. A differential pair is anotehr thing. Study how opamps are used with negative feedback and how the feedback compensates for distortion in opamps. The diff. pair is just part of the way that an opamp can be implemented. The same can be achieved with MOSFETs or Vacuum tubes (valves) as well. The real "trick" is in the feedback. \$\endgroup\$ Dec 4, 2017 at 20:29
  • \$\begingroup\$ Odd or even harmonics, it doesn't matter. Also differential amplifiers cannot cancel distortion. They can suppress common mode but that is something different. Only negative feedback can suppress distortion. This works the same on odd and even order distortion. Keno: sarthak is confusing you and himself as well. \$\endgroup\$ Dec 4, 2017 at 20:31

2 Answers 2


You can view the distortion as unwanted signals present at the output which was not present at the input of the amplifier.

And in general case, the negative feedback always reduces distortions.

In your amplifier, the Q1, Q2 as its name suggests working as a differential amplifier. And the job for this Diff amp is to amplify (only) the difference between the two its inputs.

The Q1 transistor is "watching/monitors" the input signal and the Q2 transistor is "watching/monitors" the output signal feedback via the R5 resistor.

And if any "distortion" is seen at the output ( generated output is not equal to desired output). The error signal will be "produced" in the diff amp. And the 180-degrees out of phase signal will be "generated" to cancel the distortions at the output (eg. to cancel +1V the amp will try to produce -1V ).

If Vout is larger than expected (Vin) the Q2 will reduce his Ic current. But at the same time, the Q1 Ic current will goes up. And Q3 and Q4 current will also increase so the Vout voltage will drop.

And to show you how this "error signal" seen at the diff amp input looks like.

I made this simple circuit in LTspice.

enter image description here

And the output voltage for "pure" sinewave at the input looks like this (together with the signal error in red):

enter image description here

  • \$\begingroup\$ And with RF1 we determine the proportion of negative feedback being applied back to Q2? \$\endgroup\$
    – lucenzo97
    Dec 4, 2017 at 20:41
  • \$\begingroup\$ So, why does the distortion get suppressed? One reason for this must be 180 phase shift to input of Q2, probably. And since there is was no such distortion on Q1 such is fed to Q2, only sine wave gets amplified and distortion gets suppressed to some point. Can you confirm this statement? \$\endgroup\$
    – lucenzo97
    Dec 4, 2017 at 20:46
  • \$\begingroup\$ Yes, you can say that. As for the feedback portion. The feedback factor ( feedback gain ) is defined as Vb_Q2/ Vout =R3/(RF1 + R3). \$\endgroup\$
    – G36
    Dec 4, 2017 at 20:57
  • \$\begingroup\$ Just to be clear, is node N004 at the base of Q2 in your simulation? \$\endgroup\$ Dec 4, 2017 at 21:18
  • \$\begingroup\$ Yes N004 is a voltage at the Q2 base. \$\endgroup\$
    – G36
    Dec 5, 2017 at 4:35

It seems like what you're having trouble understanding is about differential amplifiers and negative feedback in general rather than this specific circuit, so let's generalize this.

The first thing we need to do is identify the feedback going from the output to the input, and then consider the circuit without it. I've highlighted the feedback network here: enter image description here

I'm also going to ignore C7, the compensation capacitor, which is very important but can be thought about later/separately.

I'm not looking to do the full analysis here, but the small signal resistance looking into the inputs is about \$\beta R1\$, the output resistance is about 15k (and is proportional to the Early voltage of Q4), and the differential gain \$A_{diff} \approx 15\cdot\beta\$ (also going to ignore the common-mode gain for now).

So then we can replace this with an equivalent looking like this:


simulate this circuit – Schematic created using CircuitLab

So what's the problem with this amplifier? It has a very high gain of about 1500, it has a pretty high input resistance of 150k. The problem is that its properties are very dependent on the transistors. A transistor with a β=200 when cold might have a β=50 when hot. A batch of transistors will have large variations in β. You wouldn't want, for example, your speakers to be four times louder on a hot day than on a cold day. 1500 (or 3000, or 750, depending on β) is also a lot of gain. Also the output resistance is kind of high at 15k.

Now that we have a simpler model, let's add the feedback loop back in. I'm going to assume that at the frequency we're interested in, C5 is an open circuit and C6 is a short circuit.


simulate this circuit

All we added here is a voltage divider. Since Rin=150k is so much larger than R7 that it is in parallel with, we can ignore it. So,

$$V_-=V_o \frac{\mathit{R7}}{\mathit{R5}+\mathit{R7}}$$


$$A_{diff}(V_+-V_-) = A_{diff}(V_{in} - V_o\frac{\mathit{R7}}{\mathit{R5}+\mathit{R7}})$$

The output current \$I_o\$ is

$$I_o = \frac{V_o}{R5+R7}$$ $$V_o = A_{diff}(V_+-V_-)-R_oI_o$$

so finally we have

$$V_o = A_{diff}(V_{in} - V_o\frac{\mathit{R7}}{\mathit{R5}+\mathit{R7}})-V_o\frac{R_o}{R5+R7}$$

Then move Vo to the left side,

$$V_o + A_{diff}V_o\frac{\mathit{R7}}{\mathit{R5}+\mathit{R7}} + V_o\frac{R_o}{R5+R7} = A_{diff}V_{in}$$

and solve for Vo

$$V_o = \frac{A_{diff}V_{in}}{(1 + \frac{R_o}{R5+R7} + A_{diff}\frac{\mathit{R7}}{\mathit{R5}+\mathit{R7}})}$$

But Adiff (1500 or so) is really big compared to 1, and Ro/R5+R7 too. So let's ignore those terms.

$$V_o = \frac{A_{diff}V_{in}}{A_{diff}\frac{\mathit{R7}}{\mathit{R5}+\mathit{R7}}}=V_{in}\frac{\mathit{R5}+\mathit{R7}}{\mathit{R7}} = V_{in}(1+\frac{\mathit{R5}}{\mathit{R7}})$$

This doesn't contain Adiff at all! Remember Adiff was heavily dependent on beta, which would have caused the amplifier to have very inconsistent properties. So by introducing feedback we've made the gain dependent almost exclusively on the value of some resistors, which can be very consistent.

Now I only talked about gain here, but the same applies to distortion. Consider if the amplifier had an open-loop gain of 1000 for Vo=0V and 2000 for Vo=5V. A signal large enough to go through both points would be distorted as the parts above Vo=5V would be amplified only half as much. With the negative feedback, the closed-loop gain would change very little as Vo went above 5V, it would still be approximately 1+R5/R7.

The factor by which this is reduced is how much larger \$A_{diff}\frac{\mathit{R7}}{\mathit{R5}+\mathit{R7}}\$ was than 1 when we decided to ignore the 1 term (and the output resistance term though we can come back to that). This is the open loop gain divided by the closed loop gain.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.