Buck regulator question

Discussion in 'General Electronics Chat' started by fabieville, Feb 23, 2011.

  1. fabieville

    Thread Starter Member

    May 25, 2009
    When you step down a high voltage DC to a lower dc output using a buck regulator does the output current always increase? Just for example I had 100VDC at the input @ 1amp and I step it down to 18VDC, would the output current increase from 1amp? And if it does about how much current would i recieve now?
  2. JDT

    Well-Known Member

    Feb 12, 2009
    Well, the output current will be what you demand from it.

    But switching converters like these are normally quite efficient - normally greater than 80-90% so if the output was say 18V at 5A = 90W then the input is is going to be in the region of 100W. In your case 100V at 1A.

    So the input current will be less than the output current.

    In series linear regulator , on the other hand, the input and output current will be roughly equal.
  3. SgtWookie


    Jul 17, 2007
    100v is pretty high for a buck-type regulator. Once you get over around 50v on the input side, you really need to have galvanic isolation from the output side using coupled inductors aka a broadband transformer. A reasonably simple type of DC-DC converter is a flyback converter; not as efficient as more sophisticated designs, but usually much more efficient than linear regulators.

    For an intro to such supplies, have a look at Ronald Dekkers' "Flyback Converters for Dummies" page:
    Lots of good information.
  4. nigelwright7557

    Senior Member

    May 10, 2008
    You should get about 80% out of the power in.
  5. JMac3108

    Active Member

    Aug 16, 2010

    You are not clear on if the input current you quote is the rated input current or the measured input current.

    In any case, in a switching power supply, assuming 100% efficiency, the ouput POWER is equal to the input power. For example in a Buck converter that has 10V input and a 5V output. If we put a 2A load on the output, how much current will we draw from the input?

    Pin = Pout
    Vin*Iin = Vout*Iout
    10V*Iin = (5V)(2A)
    Iin = (5V)(2A)/(10V)=1A

    So we have 10W at the input and 10W at the output!

    Now lets get real and admit that the power supply is not 100% efficient. We'll repeat the calculation assuming 80% efficiency of the power supply.

    Pin = Pout / efficiency
    Pin = Pout / 0.8
    Vin*Iin = Vout*Iout / 0.8
    10V*Iin = (5V)(2A)/0.8
    Iin = (5V)(2A)/(0.8)(10V)
    Iin = 1.25A

    Hope this helps!