2
\$\begingroup\$

Based on my previous question about IC UC3845 I want to ask about feedback in these schematic:

enter image description here

Pin 2 on schematic above is VFB, Pin 1 is COMP, Pin3 is CURRENT SENSE.

enter image description here

I can't define negative feedback of Error Amplifier. As I presume it is P1+R10 and R8+C12. But why is C12 if it is DC. How it will works? As I understand voltage drop on R11 is 2.5V at steady-state because of Op-Amp features. And will voltage on out of Error Amp(COMP pin) be negative because current flows through P1+R10, C12+R8 and voltage on pin2 is 2.5V.

And this remaining part of schematic for full understanding:

enter image description here

\$\endgroup\$
1

1 Answer 1

1
\$\begingroup\$

The error amplifier will not always have a negative output and not always have a positive output. This is the point of the error amplifier.

Because of the negative-feedback characteristic of the op-amp it will try to keep 2.5V at its negative pin, as you say yourself. So if the incoming signal over the potentiometer P1 drops below that it will swing up the output to start trying to pull it up. When the signal incoming goes above the target 2.5V it will swing the output low.

There is your feedback, that signal gets propagated through the current sense amplifier, which is yet another kind of error amplifier, but for the primary current, as a protection, into the control circuitry.

The P1 potentiometer allows you to adjust "how violently" it reacts to errors in the output. The lower its value the more heavily the error amplifier will react to small errors on the red arrow.

The point of all the capacitance in the feedback, is that it is not at all a DC signal. The switching regulator switches the output on and off with a frequency, likely between 20kHz and 200kHz (I have not looked up this exact chip, but those are common fly-back outer frequencies). So, the capacitances in the feedback loop allow the error amplifier to filter out frequency responses that are too low or too high, by adjusting the feedback impedance depending on the frequency it tries to amplify.

\$\endgroup\$
3
  • \$\begingroup\$ The problem is that error amplifier is not comparator. so it hasn't only low and high output. According to onsemi.ru.com/pub_link/Collateral/UC3844-D.PDF page 8 Error Amp should have negative feedback. \$\endgroup\$
    – Robert
    Commented May 22, 2015 at 16:34
  • \$\begingroup\$ No, it's not a comparator. If P1 is high it will start working like a gradual amplification. As pointed out the behaviour will be more or less sharp depending on P1. As a point of interest, I did not say the output is switching, I said it swings up and down. That is what happens. Whether by 10mV or by 10V is up to the design and specific values, which I ignore in my explanation, because they were not relevant to your question. You wanted to know where the feedback is. There it is. \$\endgroup\$
    – Asmyldof
    Commented May 22, 2015 at 18:50
  • \$\begingroup\$ from this you can see that error amp has feedback from soloelectronica.net/PDF/AppNote03.pdf page 2 figure 3a and it is called direct duty control. So it's not only depend from P1, as I think P1 just should set Error Amp output. \$\endgroup\$
    – Robert
    Commented May 23, 2015 at 7:30

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

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