Design op-amp low-pass filter 1kHz gain of 10 SOLVED

Im supposed to show at least 2 different ways to make a low pass filter using op amps, capacitors, resistors. Corner 1000 hertz and gain of 10. The homework question is show how to do it with 1 op amp or with 2 op amps. Why are these different?

by CAMIE538
November 15, 2016

Does the gain have to be +10 or can it be inverting gain of -10?

by berniekorrie
November 15, 2016

either one

by CAMIE538
November 15, 2016

2 Answers

Answer by berniekorrie

Here's a way to do it with two. First opamp you can do the low-pass part, and second you can do the gain.

For the first one, say we pick $R_1 = R_2 = 10k\Omega$. We can just pick it as long as its reasonable for the rest of the circuit. Then for the capacitor we have

$$\tau = R_1 C_1 = \frac {1} {2 \pi f_c}$$

with $f_c = 1000$, so

$$C_1 = \frac {1} {2 \pi \times 1000 Hz \times 10k\Omega}$$

$$C_1 = 15.9 nF$$

For the second one we just do an inverting gain of 10, so maybe $R_3=10k\Omega$ and $R_4=100k\Omega$.

Here it is:

Click it, try the frequency domain simulation, look at the plot. It has +20dB gain at DC - that's a gain of 10 for all you non-decibel-speaking folks. It rolls off with the -3dB point (-3dB relative to the flat area, so the plot says +17dB) exactly at 1kHz.

Please accept my answer!

ACCEPTED +2 votes
by berniekorrie
November 15, 2016

Answer by CAMIE538

OK I copy+pasted berniekorrie's circuit and made it work with just one opamp.

  1. copy+paste berniekorrie 2 stage circuit
  2. deleted second stage
  3. changed $R_1$ from $10k\Omega$ to $1k\Omega$

Click it and run the bode plot, it works! Only difference is the phase: now it's 180 degrees at DC because it's inverting. Doesnt have 2nd stage to invert it back:

+1 vote
by CAMIE538
November 15, 2016

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