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I am somewhat a noob with electronics. I do understand some stuff to an extent and can follow logic but this goes beyond my full understanding.

I have an audio system that I would like to modify in order to increase the output gain of it and increase the audio signal strength. I have a basic understanding of OpAmps after reading and watching some YouTube videos but I am not sure 100% about the circuit below.

My understanding is that depending on the configuration of the feedback loop, you make the OpAmp to behave as unity gain buffer or as an actual amplifier (nevermind the other tons of ways to use OpAmps). In the circuit below notice that it has a 680 pF cap (C72) in the feedback loop. What is the purpose of this cap? Is this in effect a unity gain configuration?

If I wanted to increase the gain, could I replace that cap with a resistor of "x" value?

I need to be 100% certain before attempting any modification, this are SMD components and quite small. If I go through this trouble I am hoping it will be a one way street (measure twice cut once).

enter image description here

enter image description here

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  • \$\begingroup\$ Two things. 1) a bipolar transistor is not a good device for muting audio, because of all of those negative half-cycles. Try a small-signal FET such as a 2N7000 / 7002. 2) when the mute is on, the opamp sees a 50 ohm load. Depending on the peak signal voltage, the output current could exceed the device rating. \$\endgroup\$
    – AnalogKid
    Sep 6, 2023 at 16:42

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That is not a simple gain stage; it is a very high performance, fully differential amplifier. As such, its gain equation is more complex than you might be familiar with. This configuration is used to cancel out common mode hum and noise. This is noise that is induced equally into both halves of the input signal pair, and is a common problem in audio systems.

https://en.wikipedia.org/wiki/Differential_amplifier#Operational_amplifier_as_differential_amplifier

https://en.wikipedia.org/wiki/Common-mode_signal

https://en.wikipedia.org/wiki/Common-mode_rejection_ratio

Based on applications information in the datasheet, my guess is that C72 provides feed-forward compensation to keep the circuit from breaking into high-frequency oscillation. Unfortunately, this solves one problem but introduces another. It changes the frequency response and phase characteristic of only the inverting half of the circuit. C73 is added to the non-inverting input circuit to create the same characteristic. This makes the input impedances of the two inputs equal as seen by the input signals, and is critical to achieve high common-mode rejection.

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  • \$\begingroup\$ Very useful answer. Thanks a lot. \$\endgroup\$ Sep 6, 2023 at 15:54
  • \$\begingroup\$ The input impedances of the two inputs aren't equal as seen by the input signals since AOUTL P sees the ground point over C74 while AOUTL N mainly sees the virtual ground on the negative opamp input. Only at frequencies significantly beyond the GBW of the opamp are the input impedances matched. See this treatise for several possible approaches to getting matched input impedances. \$\endgroup\$
    – user107063
    Sep 6, 2023 at 17:09
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In the circuit below notice that it has a 680pF Cap (C72) in the feedback loop. What is the purpose of this Cap?

These capacitors lower the bandpass of the opamp circuit.
With 680 pF, it should be: BP > 100 kHz.

If you want some gain, change the 3 kOhm to some higher ...
NB: stability unknown. (band will be reduced to 10 kHz until one reduces the 680 pF).

enter image description here

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  • \$\begingroup\$ Thanks for your response \$\endgroup\$ Sep 6, 2023 at 15:48
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This is a comparatively complex differential amplifier with frequency limitation. The input impedance is not independent from the gain, and for its functionality it is important that the symmetry of both inputs is maintained (even if this particular circuit does not offer identical input impedances on the incoming positive and negative signal lines: AOUTL P has a higher input impedance. Increasing the gain would require increasing both R75 and R76, however that would also increase the impedance imbalance between AOUTL P and AOUTL N, and there is also the R73/C72/R74/C73 compensation interspersed that may affect circuit stability.

I'd advise against touching this stage. If you want more gain, put a separate gain stage behind. If you are up for experiments, double both R75 and R76 but be prepared to put back the original values if things go south.

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  • \$\begingroup\$ Thank you very much. If it makes any difference AOUTL P/N are the direct outputs of a DAC (updated OP with that section of the diagram). I understand your comments and certainly a separate gain stage might be simpler but as you said, I feel I am up for experiments. This is how one learns, don't we? With that extra information of the DAC, is there any extra comments? \$\endgroup\$ Sep 6, 2023 at 15:45
  • \$\begingroup\$ @ChristianGraffe Frankly, for DAC output I'd expect a dedicated instrumentation amp rather than a normal opamp: the former includes on-die matched resistors that make for better common mode rejection because of the tight tolerance achievable in this manner. When experimenting with other resistor pairings, you need to use high-precision resistors preferably from the same batch. I see the learning value of broken clockwork; broken SMD circuits will do more to improve your soldering skills than your circuit design skills. \$\endgroup\$
    – user107063
    Sep 6, 2023 at 16:06
  • \$\begingroup\$ haha that is fair enough and thank you for the input. \$\endgroup\$ Sep 6, 2023 at 16:27
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Note that audio signal strength and gain are two different things.

The output of the amplifier is going to be limited by the supply voltage, in this case 5 V. That means that your output signal will be able to swing less than \$\pm\$2.5 V no matter what the gain is. The gain is going to determine what input level it will take to get the maximum output. Let's say for example an amplifier is capable of \$\pm\$2 V before it gets close enough to the supply rails to start clipping, and the gain is set at 10, that would mean you need a \$\pm\$200 mV input signal to get full output. Increasing the gain to 20 would not give you a \$\pm\$4 V output, it would just decrease the maximum input signal to \$\pm\$100 mV.

This will help if your input signal source is lower than the amplifier is designed for, you will get more output (within the limits) with less input signal. If you already are driving it close to it's maximum output, increasing the gain will just make it clip.

To find out how close to the supply rails you can get you should check the datasheet. Under Output Voltage Swing it gives you typical and max values for several different load impedances.

More gain may be what you want, but make sure before going to the trouble of modifying it.

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  • \$\begingroup\$ Yes sorry, what I meant is gain. The end result is that I want louder output so I can drive and even overdrive the input stage of a guitar amplifier. This is a guitar based audio circuit and the actual output signal range to be at instrument level is around the 200-300mV. So if I am understanding correctly your response, based on this it is likely I have enough headroom to increase the gain slightly. I just need an extra 50% or even less. \$\endgroup\$ Sep 6, 2023 at 15:51
  • \$\begingroup\$ @ChristianGraffe Yes,it looks like you should be okay with that. Check the update to my answer regarding output swing. \$\endgroup\$
    – GodJihyo
    Sep 6, 2023 at 17:04

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