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enter image description here

I am having some trouble understanding the background of this question.

I think they are saying that as you sweep the frequency of Vin, the impedance Zin seen will be different as you pass through the different pole location. For example, at DC, impedance should be purely resistive? Capacitors introduce a pole so a -20dB/dec and a -45degree phase shift at the cap pole location which eventually becomes a -90 degree shift.

Can someone just provide me some hints here to help me solve this?

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    \$\begingroup\$ An opamp alone has no phase margin. Such a margin is defined for a closed-loop system (with gain) and is measured under open-loop conditions. \$\endgroup\$
    – LvW
    Jun 5 at 13:33
  • \$\begingroup\$ @LxW I assume they mean the closed-loop system here. \$\endgroup\$ Jun 5 at 13:59
  • \$\begingroup\$ Opamp is not like a simple RC lowpass filter, it causes more phase lag due many points of slowness in the internal circuit. It causes total 135 degrees of phase lag - there's still room 45 degrees to instability. Consider the opamp as by voltage controlled voltage source which has voltage gain =1 with 135 degrees of phase lag. Think the input current of the opamp still as zero. Calculate Vin/the input current of the whole circuit. \$\endgroup\$
    – user287001
    Jun 5 at 15:12
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    \$\begingroup\$ I'll return and write more tomorrow when I have time assuming no-one hasn't already shown it. The thing is quite elementary if one knows phasor calculus and dependent sources. \$\endgroup\$
    – user287001
    Jun 5 at 17:37
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    \$\begingroup\$ There's no need to insert anything. The full story is already shown perfectly in an answer by user emnha. Accept the answer and study it carefully. \$\endgroup\$
    – user287001
    Jun 5 at 20:05

2 Answers 2

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The OpAmp has a phase margin of 45 degree at UGF so you can model it as follows:

enter image description here

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  • \$\begingroup\$ Ohh, I see it now. One more question, how do you know that this is a ressitive + inductive impedance as opposed to resistive + capacitive impedance? \$\endgroup\$ Jun 5 at 18:23
  • \$\begingroup\$ The impedance of inductor is ZL = jwL (positive) and for a capacitor ZC = -j/(wC) (negative). If you look at the imaginary of Zin, Im(Zin) = j*1.21R > 0 so it's inductive. \$\endgroup\$
    – emnha
    Jun 5 at 19:51
  • \$\begingroup\$ @alayoiskgfbfqhxjiw Inductors and LC circuits are often simulated with circuits which have an opamp, capacitors and resistors. This happens to lead to the same effect albeit not especially effectively nor designed for it, because the capacitances are in the slowness factors inside the opamp. \$\endgroup\$
    – user287001
    Jun 5 at 20:09
  • \$\begingroup\$ I believe both you and the OP test source are overlooking the R/(R+R)=1/2 in the feedback loop. If you say Av=exp(-j135°) you actually have loop gain modulus 0.5, we are past UGF in frequency. Av=2exp(-j135°) is the gain at UGF instead, yielding Zin=1.31+j0.18 ohm. BTW1 have you considered using Miller's Theorem? BTW2 you have a typo in the final result. \$\endgroup\$
    – carloc
    Jun 6 at 5:11
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    \$\begingroup\$ @carloc As long as I have hung around I have thought that UGF=the frequency where the opamp alone (=all external parts ignored) has open loop voltage gain =1. And that's used in the given answer. You suggest something else. Miller's theorem assumes certain cause for complex input impedance. We know nothing of the actual placements of the reactive parts, but we assume the inputs of the opamp still work ideally. Circuit analysis program Micro-Cap gave the same result as the answer. \$\endgroup\$
    – user287001
    Jun 6 at 8:44
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Problem wording isn't very clear, op amps don't have a phase margin, a system with a feedback has a phase margin.

I suppose that PM=45° refers to the complete circuit, including the two feedback resistors. At UGF of the loop gain \$T\$, we have \$|T|=1\$ and \$\arg(T)=-135°\$. \$T\$ includes also the two resistors, otherwise phase margin has no meaning!

Let's use the Blackman formula to find the input impedance.

$$Z = Z_d {1+T_{sc} \over 1+T_{oc}},$$

where \$Z_d\$ is the input impedance with the system dead, i.e. op amp gain = 0, \$T_{sc}\$ is the return ratio with the input (where we want the impedance) shorted, and \$T_{oc}\$ is the return ratio with the input in open circuit.

\$Z_d=2R\$ because op amp has very high input impedance and very low output impedance and its gain is nil.

\$T_{sc}\$ is the return ratio with shorted input, so it is equal to $$T = \exp(-j 135°) = -(1+j) {\sqrt{2} \over 2}. $$

\$T_{oc}\$ is the return ratio with open input. The two resistors don't divide by two the return signal anymore, so the return ratio is twice as much: $$ T_{oc} = 2 T_{sc} = -(1+j) \sqrt{2}.$$

Now let's apply the Blackman's formula

$$ Z = 2R { 1-(1+j) \sqrt{2}/ 2 \over 1-(1+j)\sqrt{2} } = (0.809+0.651 j)R. $$

This result is confirmed by spice, setting the op amp (alone) gain at system UGF to \$ -2 \exp (-j135°) \$, so that \$|T|=1\$.

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  • \$\begingroup\$ Interesting enough we can notice that input impedance real part (resistance) is less then R. This means that op amp alone exhibits a negative input resistance, this matches with the fairly underdamped behaviour of system with only 45° PM. \$\endgroup\$
    – carloc
    Jun 12 at 14:05

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