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I have been having trouble finding the loop gain (Aβ) of the following transimpedance amplifier:

schematic

simulate this circuit – Schematic created using CircuitLab

To find Aβ, the books I have looked at (Sedra&Smith as well as Razavi CMOS ICs), the following method is used:

  1. Break the feedback loop at any point (preferably at a point of infinite impedance)
  2. Place a test voltage at one side
  3. At the other side reconstruct impedance seen from the test voltage
  4. Eliminate the input source
  5. Loop gain Aβ will be the ratio Vreturn/Vtest where Vreturn is the voltage at the other side of the break point

I have tried the break point above for this circuit and as you can see above it doesn't give the right answer. For example, using that break point Aβ would be found to be:

\$ Vreturn/Vtest = \mu \$

However using method of finding separate A and β circuits:

\$ Vreturn/Vtest = \mu (R_f/(R_f+r_o)) \$

I find it easy to find loop gain Aβ using method of finding seperate A and β circuits, but would like to know how to apply the described method here. I have found the method successful in other amplifiers such as voltage, and transconductance amplifiers, but can't seem to figure out how to apply it here.

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There are two important conditions for finding a "correct" breakpoint:

  • The dc operating point of the circuit must not be influenced. That means: For circuits with REAL opamps there must be still a DC stabilizing negative feedback loop. During simulation and for an IDEAL opamp this requirement can be neglected.

  • The loading at the opening must not change (in comparison to closed-loop operations). Hence, the opening must therefore be directly at the opamp output (zero output resistance assumed). Otherwise one must apply alternative methods, which are available (mirroring of the load,...). However, in your case, this condition is not relevant (because of the next point).

  • In the present case - assuming an ideal voltage-in and voltage-out amplifier with gain A - the loop gain is identical to the open-loop gain "-A" of the amplifier (100% voltage feedback) because there is no current or voltage drop caused by the test voltage.

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  • \$\begingroup\$ For this question I was trying to figure out the loop gain for a non-ideal Op amp with output resistance comparable to the feedback resistor (like 20k to 10k in the diagram). I do what you describe in the second point (mirroring the load seen by the test voltage) but since that load is infinite because it's a CMOS op amp I don't see any effect. Through other analysis I have found the loop gain to be what I describe in the second equation. But, I was wondering if this "breaking the loop" method can come up with that answer somehow? \$\endgroup\$ – matthewd49 Apr 29 '16 at 18:51
  • \$\begingroup\$ When the input impedance of the amplifier is very large and we neglect the input current, r,out does not matter. There will be no voltage drop across the resistors. The situation changes for finite input impedances. \$\endgroup\$ – LvW Apr 30 '16 at 7:56
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Here's a video tutorial on this that I used for general op amp configurations: http://www.linear.com/solutions/4449

schematic

simulate this circuit – Schematic created using CircuitLab

It works because:

  • loop gain is A * B
  • B = ZI / (ZI + ZF)
  • FB = (-VN * A) * B
  • FB/VN = -A * B

The tricky part for a transimpedance amplifier is that ZI may not seem to be part of your amplifier at all: e.g. it could be the shunt capacitance and resistance of a photodiode. It may help your mental model to think of the thevenin equivalent of your current/signal source so that things are then setup like a regular voltage amplifier.

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