# Understanding opamp feedback loop with parallel paths

I am trying to understand this feedback topology. C1 is probably there to form some kind of filter however I am not sure how to go about analyzing this network. The 500 ohms resistor is just a dummy load. I think the DC feedback factor is (100+2.5)/2.5, we can just ignore R8 at DC because it would be inside the FB loop. However, I am not sure. Thank you. • Not so much a filter as some kind of compensation to allow the opamp to drive a heavy capacitive load, I think. Not sure though, so no answer. – JRE Nov 8 '17 at 17:57
• More like (10+2.5)/2.5 at DC. – Chu Nov 8 '17 at 18:03
• @Chu there is also path through R10 – rsg1710 Nov 8 '17 at 18:03
• @rsg1710 yes, but that makes 110k//10k, which I approximated to 10k, hence 'more like'. – Chu Nov 8 '17 at 18:05
• can you post the full schematic? ...it might be helpful to help us understand the circuit – berto Nov 14 '17 at 23:55

Important parts of the circuit diagram are cropped. The 500 ohm resistor is a level-translator programming component, and current through it (to ground) modulates the collector currents of the MJE340 and MJE350 which are level translators.

Positive power supply and negative power supply current to the op amp ARE OUTPUTS here, and mismatch according to the current to ground through that 500 ohm resistor. The voltage difference across R8 is the difference of the direct (from op amp) output and the driven (from those pullup/pulldown lines that go off-page) secondary output.

Whatever current-boost the circuitry outside this part of the diagram shows, is important; if there is gain due to resistor ratios in current mirrors, the 100k "R10" component might be a dominant feedback signal. Current through R8 and R7 + C1 into the output pin is another feedback (which modulates the op amp power currents into the secondary output).

The way to analyze opamp circuits in general is to first you assume that the feedback loop works correctly and thus the non-inverted and the inverted input of the input are kept at the same voltage. Then you apply Kirchhof's laws to get the condition on the output of the opamp under which this assumption holds. This will if you turn this formula around, you will get the output voltage for a given input voltage. Lastly, but importantly: you need to check whether the initial assumption that the feedback loop works correctly has to be checked. Ie under all circumstances the amplification of the opamp will cause the difference voltage between the two inputs to become lower.

This has the look of something designed to drive a current, and control how it ramps up/down. Is this a homework problem?

Lets look at DC first. At low frequencies, pretend C1 and R7 are not there. Later we will put them back in, combining them in with R8. C1 will have complex resistance, look this up if you need to.

We will talk in terms of:

Vout as the output voltage of the opamp

Iout as the output current

VN as the negative feedback terminal

VP as the positive feedback terminal

Then you have to construct a bunch of equations using kirchoff voltage and current laws, and systematically solve them. It's going to be a pain for this little network.

I9 = Current thru R9 = Vout/R9

I6 = current thru R6 = (Vout - VN)/R6

I10 = current thru R10 = (Vload-VN)/R10

Vload = Vout - (Iout-I9-I6)*R8 -

I1 = current thru R1 = VN/R1

I1 = I6+I10

etc etc, solve, and you will eventually get there don't forget the opamp rule: VN = VP

that will give you a formula of what happens at low frequencies. For high frequencies, combine R8 with C1 and R7 ... the "resistance" of C1 is 1/(jomegaC), or to simplify a bit, 1/(2*pifrequencyC) at any given frequency.

So R8 is repaced by (R8 in parallel with (C1 in series with R7)).

Enjoy !