Purpose of a cap in a diff amp

[timlein]

What is the purpose of C6 in this diff amp circuit?

I have been told it's an RF filter consisting of R6 and C6.

If true, what formula do I use to determine the turnover frequency of that filter?

Reason I'm asking is that the amp had too much gain and I changed R6 to 4.2K to reduce the gain. Since I changed R6, do I now need to change C6? If yes, how do I determine what value to change C6 to?

[pebe]

R6 and C6 form a low-pass filter. The turnover frequency is the frequency where the reactance (Z) of C6 = R6. (Z of a cap = 1/[6.28 x f x C]). So with those values the frequency is about 106KHz. Above that frequency the reactance of C6 reduces and effectively forms a shunt across R6 and lowers the gain of the stage.

If you reduce R6 to 4.2K then you need to increase C6 to 350pF to keep the turnover frequency the same.

[timlein]

[quote="pebe"]R6 and C6 form a low-pass filter. The turnover frequency is the frequency where the reactance (Z) of C6 = R6. (Z of a cap = 1/[6.28 x f x C]). So with those values the frequency is about 106KHz. Above that frequency the reactance of C6 reduces and effectively forms a shunt across R6 and lowers the gain of the stage.

If you reduce R6 to 4.2K then you need to increase C6 to 350pF to keep the turnover frequency the same.[/quote]

Thanks for the explanation and the calculations! Will change C6 to 350pF.