# differentiating circuit voltage gain Hi! I came across this question in lab, and it's been several years since I last done analog circuits. The voltage gain here in this op-amp should be 0.707 (-3db cut-off), and it asks us to find the capacitance. The way I see it is

V_in/(1000 + 1/jwC) = -V_out/10000

And after performing some simple arithmetic, we will be able to find the capacitance. However, the answer that I'm getting is wrong, so could anyone please show me the workings of how this should be done.

Much thanks fellas!

• Are you remembering tthat w=2*pi*f? – BeB00 Mar 20 '18 at 13:04
• 1000 = |-j 1/ωC| did you use a slide rule? – Tony Stewart Sunnyskyguy EE75 Mar 20 '18 at 13:25
• @BeB00 yes I did – A. Garrod Mar 20 '18 at 13:27
• @TonyStewart.EEsince'75, no, sorry whats a slide rule? – A. Garrod Mar 20 '18 at 13:27
• @A.Garrod if you show us the steps in your calculation, we can show where you went wrong. We need to know the frequency – BeB00 Mar 20 '18 at 13:31

It doesn't matter here that 1/j=-j -meaning -ve phase shift , but it does matter that 1/|Zin|=1/√(1+1) = 1/√2= 1/ 1.414= 0.707

meaning $A_V= - \dfrac{R_f}{\sqrt{(R_{in}+2πfC)}}= -5/√2$ But if you do, You may find the above handy in future simulate this circuit – Schematic created using CircuitLab

f = 50 Hz and Xc = 1000 ohms so C = $\dfrac{1}{2\pi\times 50\times 1000}$ = 3.183 uF.

• but if Xc=1k, isnt Rin=2k and gain=-5? – BeB00 Mar 20 '18 at 13:38
• -5 at f >= 2 decades when Xc goes towards 0 but -3dB or 0.707 of -5 when R=|Xc| – Tony Stewart Sunnyskyguy EE75 Mar 20 '18 at 13:44
• are we assuming that the reactor of the capacitance is 1000 ohms? if so, why? – A. Garrod Mar 20 '18 at 13:45
• @BeB00 no, Rin isn't 2k it's 1000 - j1000 and gain is 7.071 – Andy aka Mar 20 '18 at 14:01
• @A.Garrod yes, that produces the 3 dB point in the spectrum of this high pass filter circuit. – Andy aka Mar 20 '18 at 14:01