Let’s call A the knot for the inverting entry of the amp op. We apply
the KCL for knot A:
But we have, as the amp op. is ideal, V- = V+ = VA
= 0
Thus we get
So this is:
where p = 1/R1C1 = 4000
1+ jω/p can be replaced by 1 when ω < p and by jR1C1ω
when ω > p
So we get:
H(ω) ≈ jR1C2ω when ω < p
C2/C1 when ω > p
So we get
|H(ω)| ≈ R1C2ω when ω < p
C2/C1 when ω > p
Finally, as we are looking for the Bode graph, we need G = 20 log H(ω)
G = 20 log(R1C2ω) =20 log (0.05.10-2) + 20
logω when ω < 4000
20 log (C2/C1) when
ω > 4000
This is a line of slope 20 dB / dec up to ω = 4000 and then a
straight line.