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On the image below you can see an inverting amplifier with input capacity C1 and capacity C_F for general stability.

Inverting amplifier

Now my book shows me following frequency plot:

A_n ... Noise-Gain

A_D ... open-loop gain

A_U ... V_out / V_in --> inverting amplifier gain

enter image description here

You can clearly see that C1 is increasing the noise gain at some point. And C_f is stabilizing A_n and A_U.

But why exactly does C1 increase the noise gain exactly?

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2 Answers 2

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Consider the non-inverting op-amp circuit for a simpler entry point to understanding noise gain: -

enter image description here

I took this basic circuit off the internet and added two things: -

  1. En - a noise source in series with the non-inverting input
  2. Cin - the input capacitance of the inverting input to ground

Now consider what the gain of the circuit is. At DC it is clearly 1 + Rf/Rg (as per all the text books on the subject). However, if we made our input frequency high enough, the capacitive reactance of Cin would start to become more dominant than the impedance of resistor Rg and now the gain tends towards 1 + Rf/XCin. As frequency rises and rises the gain tends towards infinity (theoretically) and is limited only by the open-loop gain of the op-amp.

So if you used one of these circuits you would find that the internal op-amp voltage noise (En) is quite high at high frequencies just as the conventional gain would be when you factor-in the effect of Cin.

This effect, if problematic, can be alleviated by adding a capacitor (Cf) across Rf and, at high frequencies the amplification (or noise gain) tends towards 1 + Cin/Cf. If you know the approximate value of Cin you can scale Cf to match Rf/Rg and get decent performance up to high frequencies with unnoticeable noise gain.

But why exactly does C1 increase the noise gain exactly?

Can you see why this happens for the non-inverting amplifier? Can you take it from here and take the small leap to the inverting amplifier?

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  • \$\begingroup\$ Why does the gain tend toward 1 + Rf/XCin if the frequency w --> infinity ? Isn't the parallel resistor equal to1 / (1/R + 1/jwC). So if w-->infinity the resistor of R//C tends to R. Right? \$\endgroup\$
    – Gamdschiee
    Commented Jan 29, 2018 at 17:05
  • \$\begingroup\$ If the reactance of Cin is much lower than the resistance of Rg then the parallel combination of XCin and Rg tends to be dominated by XCin. If you had a 10 ohm resistor in parallel with a 1 ohm resistor, the net resistance looks more like 1 ohm than 10 ohm. \$\endgroup\$
    – Andy aka
    Commented Jan 29, 2018 at 17:09
  • \$\begingroup\$ If Rg is in parallel with XCin, the impedance is \$\dfrac{1}{\frac{1}{Rg}+j\omega C_{IN}}\$ and, as frequency rises it becomes... \$\dfrac{1}{j\omega C_{IN}}\$ \$\endgroup\$
    – Andy aka
    Commented Jan 29, 2018 at 17:11
  • \$\begingroup\$ The gain of the non-inverting amp is A = 1/k with $$k = 1 + \frac{C_N}{C_1}$$ for high frequencies. I.e. $$A = \frac{C_1}{C_1+C_N}$$ So you can clearly see that C_1 increases A, but when C_N = 0 would be A=1. Does that make sense? \$\endgroup\$
    – Gamdschiee
    Commented Jan 29, 2018 at 18:13
  • \$\begingroup\$ Well done you spotted an error that I have fixed. Originally I said gain is 1 + Cf/Cin but I have corrected it to 1 + Cin/Cf. \$\endgroup\$
    – Andy aka
    Commented Jan 29, 2018 at 18:17
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The answer is simple: The term "noise gain" is per definition nothing else than the inverse of the feedack factor. And - as you can see - the feedback factor is reduced due to C1 because the portion of the output voltage that is fed back is now developped across C1||R1. hence the inverse of this factor goes high for rising frequencies.

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