Discharge path for buck-boost converter

[BlackMelon]

Hi!

I'm interested in no load condition of the converter. As what's in a simplified circuit, with load removed, there is no discharge path for the capacitor in the case that I want to reduce the output voltage after I pumped it up.

Is my understanding right?

Thanks

 

[Colin Mitchell]

I don't know what you want. The diode is around the wrong way.

 

[(*steve*)]

Typically your feedback network would draw some current.

 

[KrisBlueNZ]

The diode is around the wrong way.

It's a buck-boost converter.

 

[Arouse1973]

I don't know what you want. The diode is around the wrong way.

No the diodes not around the wrong way, this is fine. It's designed to reverse the polarity of the input voltage. This is a positive to negative buck-boost configuration. Buck boost being the name given to this kind of group configuration.
Adam

 

[Arouse1973]

@kris snap

 

[BlackMelon]

I mean this:
At the first time, I boosted the capacitor voltage up to 100V in no load operation(no resistor in the circuit diagram which I showed you). Luckily, I've got a feedback loop and it stops the circuit from pushing more energy to the capacitor (which causes overvoltage to the capacitor). Now I want to reduce the capacitor's 100V down to 20V but I have no resistor to discharge the capacitor so what I want to ask you is "The voltage will not be able to decrease, right?"

 

[KrisBlueNZ]

I don't understand your question.

If you have no load on the capacitor at all, then it will only discharge due to its internal leakage.

If you have a feedback circuit using a voltage divider across the capacitor, that will discharge the capacitor slowly, as Steve pointed out in post #3.

If you change the control circuit so it stops switching when the capacitor voltage reaches -20V then the capacitor will charge to -20V.

If that doesn't answer your question, then you need to ask your question clearly.