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I need to trick the buck converter feedback. The converter that I am going to use is TPS40192DRCT.

Instead of the regular resistor divider, I want to feed there voltage proportional to current somewhere else. So when somewhere the current rises, this voltage would go lower and make the converter increase the duty cycle. This will reduce the current I am measuring there.

I am not sure what to do with the compensation though. Can i simply connect the network to the GND instead the feedback?

enter image description here

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  • \$\begingroup\$ Increasing duty cycle would increase the current! Sounds like you want constant current as opposed to constant voltage. Yes, you can do as you propose. The devil is in the details though. \$\endgroup\$
    – Kartman
    Commented Mar 7, 2022 at 1:10
  • \$\begingroup\$ No, no. It will take current away from that other place, sort of a switching bypass \$\endgroup\$
    – TQQQ
    Commented Mar 7, 2022 at 1:10
  • \$\begingroup\$ Read the last two sentences. \$\endgroup\$
    – Kartman
    Commented Mar 7, 2022 at 1:17
  • \$\begingroup\$ I have, thank you \$\endgroup\$
    – TQQQ
    Commented Mar 7, 2022 at 1:34
  • \$\begingroup\$ Its just a test or do you have any spec ? Do you want to increase the voltage output or current output? \$\endgroup\$
    – user19579
    Commented Mar 7, 2022 at 5:02

2 Answers 2

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It will work in principle, but your polarity is inverted.

When FB rises, the converter will tend to reduce the duty cycle (in order to regulate FB at the required regulation point). Your as-drawn circuit has an inversion.

If you make your 1st amplifier into a fixed-gain difference amplifier with an offset, you should be able to set it to generate the 591 mV FV reference at your desired current.

Your compensation could be quite difficult because of the dynamics of what generates or affects the current you are sensing.

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  • \$\begingroup\$ Thank you. The dynamics in principle is very slow and adjustable - I don't hope to somply get it right. The polarity will be such that higher current will push the feedback lower, making the controller increase the duty cycle. \$\endgroup\$
    – TQQQ
    Commented Mar 7, 2022 at 2:23
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    \$\begingroup\$ So, higher current is the consequence of a higher duty cycle. And you're saying that this consequence of a higher duty cycle will push the duty cycle even higher. That's positive feedback, which is only good if you want the thing to tend to latch in a high current or low current state. \$\endgroup\$
    – TimWescott
    Commented Mar 7, 2022 at 2:43
  • \$\begingroup\$ I don't measure this converter current. It's the sum of the converter and some other current that is constant. \$\endgroup\$
    – TQQQ
    Commented Mar 7, 2022 at 4:12
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The part of your circuit that I've drawn a red trapezoid around is already there inside the TPS40192. Look at the functional block diagram in the datasheet -- the FB pin goes into an inverting input of an op-amp, there's an internal reference on that amplifier's non-inverting input, and the amplifier's output goes to the chip's "COMP" pin.

For example, the compensation network that they show is the classic feedback network for an op-amp proportional-integral stage.

As long as you can live with their reference voltage, you don't need to add an op-amp, with its delay and bandwidth limitations.

enter image description here

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  • \$\begingroup\$ It was thinking I would need an integrator stage. But I will prepare a bypass, thank you. \$\endgroup\$
    – TQQQ
    Commented Mar 7, 2022 at 4:18
  • \$\begingroup\$ Their compensation network provides proportional-integral action, with a lead-lag filter between Vout and the feedback pin. \$\endgroup\$
    – TimWescott
    Commented Mar 7, 2022 at 16:15
  • \$\begingroup\$ Yes, but then the output capacitor is taken out of the feedback loop. Frankly i don't know what will be the effect. \$\endgroup\$
    – TQQQ
    Commented Mar 7, 2022 at 16:47
  • \$\begingroup\$ Presumably you need to use compensation that's appropriate for whatever you're doing (you never said where your current is coming from). If you want info on that you should probably start a new question. \$\endgroup\$
    – TimWescott
    Commented Mar 7, 2022 at 20:17

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