# Circuit analysis of active LP-filter with 2 OP amps

So this is our diagram: • We want to find C1, C2, R1, R2 so that our frequency cut-off is 1kHz.

• We also need to have R2C2 = 2 * R1C1.

So, my first instinct is to:

1. Use KCL nodal analysis

2. Set C1 = C2 and R2 = 2* R1

3. Somehow calculate the values of the components involved with the characteristics of the filter V_out/V_in and the given cut-off frequency of 1kHz.

Right now I'm stuck at the part with the nodal analysis. I've never tried nodal analysis on this kind of diagram before, so I don't know if I've done it correctly. I end up with a very messy result, especially when I replace my Z1, Z2, Z3,Z4 with the 1/jwc etc. I've set Z2 = z4. Have I done this analysis correctly, or is there something I missed? Where do I go from here so that I don't get a hairy equation for V_out/V_in?

• Looks straightforward enough. Your equations make sense. Substitute V2 with Vout in equation 2, solve for V1, and substitute that result into equation 1. May 8 '14 at 13:57

Have I done this analysis correctly?

The equations (1) - (3) look correct but the equations for the capacitor impedances are incorrect. They should be

$$Z_2 = Z_4 = \frac{1}{j\omega C_1}$$

Where do I go from here so that I don't get a hairy equation for V_out/V_in?

Part of the skill set you develop working through problems like this is taking a 'hairy' equation and putting it into standard form.

Here's how I would approach the problem using voltage division instead of KCL.

Since the op-amps are configured as non-inverting amplifiers with a voltage gain of 1 (unity gain buffer), this circuit is particularly straightforward to analyze.

At the non-inverting input of the 1st op-amp, the AC (phasor) voltage is, by voltage division,

$$V_{1+} = V_{in}\frac{Z_{C1}}{R_1 + Z_{C1}} + V_{out} \frac{R_1}{R_1 + Z_{C1}}$$

Since the 1st op-amp is configured as a unity gain buffer, this is also the output voltage for the 1st op-amp.

At the non-inverting input of the 2nt op-amp, the AC voltage is, by voltage division,

$$V_{2+} = V_{1+}\frac{Z_{C2}}{R_2 + Z_{C2}}$$

Again, the 2nd op-amp is configured as a unity gain buffer, this is also the output voltage for the 2nd op-amp and thus, the output voltage of the filter.

Combining the two equations above yields

$$V_{out} = V_{in}\frac{Z_{C1}}{R_1 + Z_{C1}}\frac{Z_{C2}}{R_2 + Z_{C2}} + V_{out} \frac{R_1}{R_1 + Z_{C1}}\frac{Z_{C2}}{R_2 + Z_{C2}}$$

Note that $V_{out}$ appears on both sides. Grouping terms, factoring, and solving for the transfer function yields

$$\frac{V_{out}}{V_{in}} = \frac{Z_{C1}}{R_1 + Z_{C1}}\frac{Z_{C2}}{R_2 + Z_{C2}}\frac{1}{1 - \frac{R_1}{R_1 + Z_{C1}}\frac{Z_{C2}}{R_2 + Z_{C2}}}$$

Now, this may look 'hairy' but in fact, there isn't much algebra required to put this into standard form.

For example, the first step is to multiply the denominators together which yields

$$\frac{V_{out}}{V_{in}} = \frac{Z_{C1}Z_{C2}}{(R_1 + Z_{C1})(R_2 + Z_{C2}) - R_1Z_{C2}}$$

That looks much better already. You should be able to take it from here.

I'd do it like this and sorry I can't make out the figures in your picture: - You have to remember that the op-amps are just unity gain buffers and there is in fact only one unknown voltage and that is what I have called Va in the picture above.

All that's left is group together the terms for Vout and Vin to make your transfer function. Can you do this?

As another alternative for computing the transfer function you can use the superposition theorem:

1.) calculate (voltage divider, lowpass) the voltage called va (pos. input of the 1st opamp) in the drawing (assuming vout=0).

2.) Calculate the second part of va caused by Vout (simple C-R highpass, assuming Vin=0).

3.) Add both parts to get the final voltage va.

4.) Caculate Vout asa function of Va (simple RC lowpass).

The clculation is very simple if - from the beginning - the mentioned part simplifications (R2=2R1 and C1=C2=C) are introduced.