Connect with us

Discharge path for buck-boost converter

Discussion in 'General Electronics Discussion' started by BlackMelon, Oct 9, 2014.

Scroll to continue with content
  1. BlackMelon

    BlackMelon

    188
    5
    Aug 7, 2012
    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
     

    Attached Files:

  2. Colin Mitchell

    Colin Mitchell

    1,416
    314
    Aug 31, 2014
    I don't know what you want. The diode is around the wrong way.
     
  3. (*steve*)

    (*steve*) ¡sǝpodᴉʇuɐ ǝɥʇ ɹɐǝɥd Moderator

    25,501
    2,841
    Jan 21, 2010
    Typically your feedback network would draw some current.
     
  4. KrisBlueNZ

    KrisBlueNZ Sadly passed away in 2015

    8,393
    1,271
    Nov 28, 2011
    It's a buck-boost converter.
     
    Arouse1973 likes this.
  5. Arouse1973

    Arouse1973 Adam

    5,164
    1,087
    Dec 18, 2013
    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
     
  6. Arouse1973

    Arouse1973 Adam

    5,164
    1,087
    Dec 18, 2013
  7. BlackMelon

    BlackMelon

    188
    5
    Aug 7, 2012
    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?"
     
  8. KrisBlueNZ

    KrisBlueNZ Sadly passed away in 2015

    8,393
    1,271
    Nov 28, 2011
    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.
     
Ask a Question
Want to reply to this thread or ask your own question?
You'll need to choose a username for the site, which only take a couple of moments (here). After that, you can post your question and our members will help you out.
Electronics Point Logo
Continue to site
Quote of the day

-