I have been looking into audio amplifiers theory. So I understand that in the output you need a current gain in able to drive low impedance loads. But I am having difficulty in understanding why it is important to have a high voltage gain at low frequency for the IPS(input pair stage) and VAS (voltage amplifier stage).

Why is a high open loop gain so important for low distortion ?

What would happen if the open loop gain is lowered significantly ? by for example excessive emitter degeneration

Does anyone have an intuitive explanation?

  • 1
    \$\begingroup\$ IPS? VAS? What are they? \$\endgroup\$
    – Andy aka
    Aug 16, 2018 at 17:01

2 Answers 2


Anyone who claims that only the open loop gain matters for distortion doesn't understand how distortion in circuits with feedback works.

Let's say that you have an opamp with a open-loop gain (so at DC and very low frequencies) of 1000.

You want to use it to make an amplifier which amplifies an input signal 10 times.

To do this you would be using the opamp in a feedback configuration, don't worry about how the schematic would look like, that's not relevant yet.

enter image description here

To get the closed loop gain of 10 you would need to make \$\beta = 1/10\$

That then results in a closed loop gain of \$A_{ol} \beta = 1000*1/10 = 100\$

This is also called the "excess loopgain" as it is gain that is "too much", you need only 10 gain and you "waste" 100. But that does not mean that this excess loopgain is useless! In fact it is what decreases distortions created inside the loop.

Suppose that the opamp is slightly non-linear and instead of outputting 100 mV (when the opamp has a Vin = 0.1 mV, like so 0.1 mV * 1000 = 100 mV) it outputs only 90 mV. Oh dear, that's a 10% error !!

Luckily that's where the excess loopgain comes in. Since the opamp is monitoring it's own (distorted) output voltage with the (non distorted) input signal it can "detect" it's own distortion! This mechanism then reduces the distortion by the excess loopgain factor so in our case by a factor 100. So 10% distortion becomes 0.1%, that's much better.

Now what would happen if the opamp didn't distort 10% but we'd use a better opamp which also has an \$A_{ol}\$ of 100 but distorts 10x less so only 1%. Then the total distortion at the output will be 1% / 100 = 0.01%

So in conclusion: it is not only the loopgain that matters but also how much distortion the opamp adds. So a 10 higher \$A_{ol}\$ is pointless if the opamp also has a 10x higher distortion!

Both excess loopgain and intrinsic linearity of the opamp matter.

Disclaimer: I am aware that above numbers aren't 100% accurate, as commented below the actual excess loopgain is really 99 instead of 100. I prefer to explain with rounded off numbers to show the principle, not the exact values.

  • 1
    \$\begingroup\$ Just a note. In your first example, with \$A_{ol}=1000\$, \$\beta=0.099\$ and not \$0.1\$. The excess loopgain is \$99\$ and not \$100\$. In the second example, you appear to apply the entire \$A_{ol}\$, when dividing. But the excess loopgain in the second case with \$A_{ol}=100\$ has \$\beta=0.09\$ and an excess loopgain of \$A_{ol}\beta=9\$. \$\endgroup\$
    – jonk
    Aug 16, 2018 at 15:40
  • \$\begingroup\$ Thanks for your answer, I feel this answers my confusion. However, can you explain numerically/mathematically how you come to 0.1% error because of the excess loopgain @Bimpelrekkie \$\endgroup\$
    – Navaro
    Aug 16, 2018 at 16:00
  • \$\begingroup\$ Reading your answer again, I am actually kind of confused with the closed loop gain you mention ( should it not be Acl=Aol/(1+Aol). @bimpelrekkie \$\endgroup\$
    – Navaro
    Aug 16, 2018 at 16:24
  • \$\begingroup\$ Same as the 1% / 100 = 0.01%: 10% / 100 = 0.1 % \$\endgroup\$ Aug 16, 2018 at 16:24
  • 1
    \$\begingroup\$ the closed loop gain you mention Yes you are 100% correct and that is also Jonk's point: my numbers are not exact but rough estimates I will add that to the question. If you must, do the exact calculation. For me 99 and 100 are "the same". I mean, no one is going to notice the difference in a real circuit. \$\endgroup\$ Aug 16, 2018 at 16:27

It sounds like you are talking about open-loop gain. Semiconductors (and tubes/valves) are inherently non-linear devices. Non-linearity produces distortion.

Feed a pure sine wave into an amplifier with distortion and you will get harmonics that obviously were not there in the input signal.

One way to linearize an amplifier is to employ feedback. You compare the output of the amplifier (possibly scaled) to your input.

Feedback Diagram

Diagram from Wikipedia

If there is any difference between the two the amplifier's gain acts on the error (the difference between the input and output) to move the output to track the input. This forces the error closer to zero. Infinite gain at DC (e.g. an integrator) will force the DC error to zero.

So the more gain you have the tighter the tracking between the input and output, and non-linearities in the internal circuitry don't matter as much.

Emitter degeneration is actually a form of feedback, so that will linearize one particular gain stage, but if it's part of a larger amplifier with feedback it will reduce the open-loop gain and result in lower gain with more error and distortion over the frequency range where the degeneration is active.


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