0
\$\begingroup\$

Today I was learning about the negative feedback in the amplifier circuits and some quick ways of discovering whether a negative or a positive feedback exist in the circuit by finding a feedback loop and counting the number of pi-s (phase changes) that exist in that loop.

If we look at the general scheme of feedback in a system we can see that the loop of feedback goes from Output->B->+->A->Output and the total phase change is 0 meaning it's a positive feedback. enter image description here

Going to the actual circuit in question here is an example from class

schematic

simulate this circuit – Schematic created using CircuitLab

Clearly, there is feedback in the circuit. I know in advance that phase shift of a bjt transistor are: BE=0, BC=pi, EC=0.

Now i need to determine the loop and count the number of pi-s that exist inside the loop to determine which feedback it is. To me a logical way of determining this was going from Vout to the input and back and i did it like this:

LOOP: Vout -> BJT 2 CE (0) -> R5 (0) -> BJT 1 BC (pi) -> BJT 2 BC (pi) -> Vout which makes the total equal 2pi which would make it a positive feedback.

This is however not the way they showed the loop in class as the went like this:

LOOP: BJT 2 C -> R5 (0) -> BJT 1 BC (pi) -> BJT 2 BE (0) -> BJT 2 C and the result is pi which makes it a positive feedback.

Why is my way of finding a loop wrong and theirs right? Isn't the loop always from Vout to Vin or is there something I'm not understanding here?

I'm primarily in understanding this method rather than calculating and figuring out the sign.

\$\endgroup\$
  • \$\begingroup\$ the loop of feedback goes from Output->B->+->A->Output and the total phase change is 0 meaning it's a positive feedback - not necessarily. It depends on what function B implements. If B inverts or applies a 180-deg phase shift then it's negative feedback. \$\endgroup\$ – brhans May 7 at 18:37
3
\$\begingroup\$

Why is my way of finding a loop wrong and theirs right? Isn't the loop always from Vout to Vin or is there something I'm not understanding here?

No, the loop isn't always from \$V_{out}\$ to \$V_{in}\$. A loop happens if you can trace the signal path in a loop. In this case the loop that your prof is talking about is the common-emitter stage around Q1, which is applied to the base of Q2, where the emitter of Q2 follows Q2's base voltage. Then that emitter voltage is applied to R5, which feeds back to the base of Q1. Yes, \$V_{out}\$ is being picked off of Q2's collector -- but that's not part of the loop in question.

\$\endgroup\$
  • \$\begingroup\$ I agree to TimWescotts answer - nevertheless, I think it is important to add that the node where the feedback signal is sampled carries a signal that is proportional to the real output. In detail: The signal current through the transistor Q2 determines the output signal (across R4) as well as the feedback signal (across R1). \$\endgroup\$ – LvW May 8 at 8:42
  • \$\begingroup\$ @LvW Hey! I left that out on purpose. But yes. And it has it's own feedback loop, except the feedback signal is carried in the current through Q2, rather than a voltage. \$\endgroup\$ – TimWescott May 8 at 18:50
1
\$\begingroup\$

examine these polarity indicators

schematic

simulate this circuit – Schematic created using CircuitLab

\$\endgroup\$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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