My bode plot does not mach with calculated cut-off frequencies

Question

I have modelled a band-pass filter, but when I calculate the cut off-frequencies and amplitude it does not match with the bode-plot. I am not able to find out what is wrong, hope that someone might help me out:

Circuit:

^{simulate this circuit – Schematic created using CircuitLab}

(It has am opposite filter on the other side)

• R1 = 1000

• C1 = 100e-9

• R2 = 22000

• C2 = 33e-12

• fc_low = 1/(2*np.pi*R1*C1)

• fc_high = 1/(2*np.pi*R2*C2)

• mag = -R2/R1

Fc_low = 1,6kHz, Fc_high = 219kHz, Magnitude= -22dB

When I plot the bode-diagram (using python control.bode, see code below) the cut-off frequencies and the magnitude becomes slightly higher than the calculated ones:

• tau_1 = R1*C1
• tau_2 = R2*C2
• tau_3 = R2*C1

---

G = control.tf([tau_3, 0],[tau_1*tau_2, (tau_1+tau_2),  1])


           0.0022 s
------------------------------
7.26e-11 s^2 + 0.0001007 s + 1

mag, phase, omega = control.bode(G, dB=True, Hz=True, deg=True)

Bode-plot attached, but approx values are as follows: fc_low = 1,05 kHz, fc_high = 250 kHz, magnitude = 28 dB

Why does this not match?

Answer 1

The gain is 22000/1000 = 22

Gain in dB = 20 log(22) = 26.8 dB

For the frequency cutoffs on the graph, you aren't reading them carefully. Drawing intersecting lines with the correct slope and using the phase where the shift is half (45 deg in this case), is often more accurate.

Answer 2

@Mattman944 Is already correct, I'll just thrown in the numbers that you could have used to verify, yourself, without the error-prone visual inspection:

$$\begin{align} H(s)&=\dfrac{0.0022s}{7.26\cdot 10^{-11}s^2+1.00726\cdot 10^{-4}s+1} \\ {}&=\dfrac{3.03030303030303\cdot 10^7s}{s^2+1387410.468319559s+1.377410468319559\cdot 10^{10}} \\ {}&=K\dfrac{\dfrac{\omega_0}{Q}s}{s^2+\dfrac{\omega_0}{Q}s+\omega_0^2} \\ {}&=K\dfrac{a_1s}{s^2+b_1s+b_0} \\ \Rightarrow \\ \omega_0&=\sqrt{b_0}=117363.132\;\text{kHz} \\ f_0&=\dfrac{\omega_0}{2\pi}=18678.923\;\text{Hz} \\ Q&=\dfrac{\omega_0}{b_1}=0.08459 \\ BW&=\dfrac{b_1}{2\pi}=220813.234\;\text{Hz} \\ K&=\dfrac{a_1}{b_1}=21.841 \\ K_{dB}&=20log_{10}(K)=26.786\;\text{dB} \\ \Rightarrow \\ f_{H,L}&=\dfrac12\left(\sqrt{4f_0^2+BW^2}\pm BW\right)=[22382.165,\;1568.93]\;\text{Hz} \end{align}$$

It's easier to do it when you convert the transfer function into its standard format (there's also the \$\zeta\$ version, but here \$Q\$ works better). Also, the numbers imply an ideal opamp, which is never the case in real life (also tolerances, temperature, parasitic capacitances, especially for that 33 pF, etc).