Electrical – Finding Loop Gain (Aβ) of Transimpedance Amplifier

amplifierfeedbackloop-gainshunt

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.

Best Answer

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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