3
\$\begingroup\$

Was looking into capacitor placement in dual ideal op-amp circuits and came across this circuit (R1=R2=R3=1kOhm, R4=10kOhm, C1=1uF):

enter image description here

I attempted to determine the voltage gain (transfer function) of this circuit G=(vo/vi), to which I get the following expression:

$$\frac{v_o}{v_i}=\frac{R_2}{R_{eq}}\frac{R_4}{R_3}=\frac{R_2R_4}{R_3R_1}(1+sC_1R_1)$$

where Req=(R1||(1/sC1)) and s=jw=frequency variable.

I decided to plot a Bode plot for this transfer function and obtained an unstable result at high frequencies, as expected since G approaches infinity as s approaches infinity. However, when I simulate this circuit (I used CircuitLab) the Bode plot I obtain is similar in shape to that of a band-pass filter.

This has me thinking that my derivation of the transfer function G is incorrect, and that it should match the transfer function associated with a 1st order band-pass filter. Would someone be able to confirm my suspicion?

\$\endgroup\$
5
  • \$\begingroup\$ Making your post easily readable will help get good responses. Use MathJax to format equations nicely. \$\endgroup\$ Commented Nov 25, 2020 at 3:17
  • 2
    \$\begingroup\$ Did the simulator use an ideal opamp or a real opamp? \$\endgroup\$
    – Mattman944
    Commented Nov 25, 2020 at 3:24
  • 1
    \$\begingroup\$ As @Mattman944 hints, you may not have factored into the analysis, the limited bandwidth of real opamps. \$\endgroup\$
    – user16324
    Commented Nov 25, 2020 at 10:28
  • \$\begingroup\$ Thanks for that @relayman357 ! On CircuitLab, I tried to make the op-amps as ideal as possible (setting input impedance to a very large value, open-loop gain to a very large value, and output impedance to a very small value). It is possible that they are not truly ideal on the simulator, but having the correct transfer function should provide insight into this. \$\endgroup\$
    – JTaft121
    Commented Nov 25, 2020 at 14:54
  • \$\begingroup\$ Sorry @JTaft121, I haven’t done much modeling of op amps, I’m not much help. \$\endgroup\$ Commented Nov 25, 2020 at 16:26

3 Answers 3

5
\$\begingroup\$

This circuit is wonderfully evil, and if I were teaching a circuits class I would make it a homework problem, and then put some derivative of it in the final.

Forget the second amp, and R3 and R4. That's just a distraction. For many, many combinations of real-world parts, the first stage will oscillate. Where it doesn't oscillate, at some frequency it will show a strong resonance, with a gain much higher than the expected \$H_{fs}(s)=\frac{R_2}{R_1}\left(R_1 C_1 s + 1\right)\$.

The reason for this is because \$C_1\$ actually puts a pole in the feedback loop, and most op-amps these days are stabilized against zeros in the feedback loop (i.e. a cap in parallel with \$R_2\$), they aren't stabilized against poles.

If you go back to KVL, you find that you can write out $$v_- = \frac{G_2 v_o + (G_1 + C_1 s)v_i}{G_1 + G_2 + C_1 s} \tag 1$$ (where I'm using conductance instead of resistance, because I'm lazy -- just take \$G_1 = 1/R_1\$, and so on).

Now forget that ideal op-amp stuff, and let \$v_o = - H_a(s) v_-\$. Solve (1) for \$v_-\$ and you get $$V_-(s) = \frac{C_1 s + G_1}{C_1 s + G_2 H_a(s) + G_2 + G_1}V_i(s) \tag 2$$

In a typical op-amp, \$H_a\$ has the form $$H_a(s) = \frac{\omega_{GBW}}{(s + \omega_0)(\frac{s}{\omega_1} + 1)(\frac{s}{\omega_2} + 1)\cdots(\frac{s}{\omega_\infty} + 1)}\tag 3$$ Usually \$\omega_0\$ is around \$1\mathrm{Hz}\$ to \$100\mathrm{Hz}\$, and \$\omega_1\$ through \$\omega_\infty\$ will be greater than \$\omega_{GBW}\$, and high enough so that the phase shift of \$H_a\$ is no more than 120 degrees or so at unity gain, thus insuring stability if you don't mess around.

However, as soon as you put that capacitor in the forward path, you're introducing a pole into the loop gain. If you play around with (2), you'll find that the general tendency of the circuit with \$C_1\$ in there is to break into song. If the op-amp were a perfect integrator (\$H_a(s) = \frac{\omega_{GBW}}{s}\$), then you'd just get a super-big resonance roughly at the geometric mean of \$\omega_{GBW}\$ and \$\frac{1}{G_2 C_1}\$. With any real poles in the op-amp response, it'll oscillate -- probably near that same geometric mean, or maybe a bit lower.

I would suggest that you simulate this circuit with a real op-amp model in the time domain -- not just using a frequency sweep. I haven't tried it, but I think you'll see an oscillation.

Note that if you wanted to do something like this in the real world and have it actually work, you'd put a resistance in series with \$C_1\$. If someone is reading this and is fuming at me because they have a circuit like this and it does too work -- check to see if \$C_1\$ is an electrolytic, and look at the preceding stage. For a lot of combinations of op-amp and electrolytic capacitor, the ESR of the capacitor may well stabilize the circuit enough that it'll at least be stable (if not well-behaved). For that matter, if the preceding stage has a non-zero impedance at the right range of frequencies, that would also stabilize the circuit.

\$\endgroup\$
2
\$\begingroup\$

Way too much mystery is going into the answers. Simply put, the gain the the first stage is $$\frac{Z_f}{Z_{\text{in}}}$$

The denominator goes to zero at high frequency, as the cap behaves like a short.

\$\endgroup\$
1
  • 2
    \$\begingroup\$ Yes, and the OP understood this. The question was why then did the simulation appear to show a bandpass response? \$\endgroup\$
    – td127
    Commented Nov 26, 2020 at 3:30
2
\$\begingroup\$

This is a curious case – I simulated it and got the same sharp “bandpass” response.

Your transfer equation is correct.

It is a high pass filter, and the gain does explode to infinity at high frequencies.

This makes sense: the impedance of C1 goes to zero, so the first stage gain R2/0 goes to infinity.

But in real life or even simulation the opamp can only output so much. At some point the inverting input of the opamp can no longer be maintained at a virtual ground because the opamp has run out of voltage swing.

So the gain will climb quickly as C1’s impedance dips, hit some maximum, and thereafter the opamp ceases to behave, becoming an unruly comparator smashing against the rails. Frequency domain simulation results at this point will become nonsensical because things have gone nonlinear (distortion).

The way to make this circuit behave is to add some source resistance Rs in your voltage source. This avoids the divide by 0 and as long as the first stage gain of R2/Rs is within the opamp’s range you’ll get your expected high pass response.

There will be an additional rolloff above 100kHz due to general opamp wimpiness at high frequencies.

EDIT Here is a plot of the simulation the OP is talking about. A highpass response was expected given the transfer function, yet this apparent sharp bandpass was observed.

enter image description here

\$\endgroup\$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.