Breaking the Loop in Circuit Design – Why Some Methods Are Unsuitable

circuit analysiscircuit-designsimulation

I've been working on simulating the open loop amplification of a series pass voltage regulator that I'm building as part of an assignment.

I've employed the following method for breaking the loop of the feedback circuit:

enter image description here

I deem it to be a good approach. It takes into account the need for the circuit to have a well defined operating point. And the approach of sticking a test signal in the non-inverting input of the diff pair resembles what I've seen being done in block diagrams in order to determine loop gain.

This is how the open loop gain looks like:
enter image description here

However, the feedback I received on this in class was unexpectedly harsh.

According to the person in charge, I should have cut the loop completely and paralleled the two resistors R3 and R4 at the non-inverting input of the diff pair, while adding an AC source at the positive terminal of the differential amplifier.

I tried to implement the suggested open loop circuit but I can't get it to work:

enter image description here

The resulting open loop amplification is a bit weird and slightly different from what I got with other methods.

enter image description here

Why are the values of the open loop amplification different, how should I change the second simulations so that it works?

Best Answer

I am not sure if the setup as shown with the 2nd figure works correctly. It is a rather uncommon approch because the test signal is not injected into the opening point. My doubt: The node between R3 and R4 is not AC-loaded at all (if compared with the closed loop where the base of Q4 is connected to this point).

Here is my approach: When you open the loop for finding the loop gain there are two important requirenments:

  • The DC operating point must be kept. This requirement can be fullfilled using the L-C combination which you have shown in the first figure.
  • Furthermore, the loading conditions at the opening must not be disturbed. That means: The node between R3 and R4 must still be loaded with the input resistance (base) of Q4.
  • For this purpose, you can create a mirror of Q4 (and the associated circuitry) at this point. For simulation purposes you could copy the relevant parts of the circuit. But if you have a good and sufficiently correct knowledge about this input resistance, you can of, course, insert a simple resistor of this value (if it is constant over the frequency range of interest).
  • As the most accurate method, one can also apply Middlebrooks "double-injection technique". This works for simulation only (not for measurements). To eliminate the loading problems at the opening, the simulation is divided into two parts: (a) injection of a test voltage and (b) injection of a test current. Then, both results are combined using a suitable formula.
Related Topic