MC34063A DC-DC Step-Up Converter Power Consumption?

Discussion in 'General Electronics Chat' started by emaq, Sep 17, 2015.

  1. emaq

    Thread Starter New Member

    Sep 17, 2015
    I have made a DC-DC step-up converter using MC34063A with following components
    Ct=31 pF
    Rsc=0.682 Ohm
    Lmin=4 uH
    Co=701 uF
    R=180 Ohm
    R1=1k R2=3k

    and I used
    for the design with following input parameters
    Vin: 3V
    Vout: 5V
    Iout: 100mA
    Vripple: 1mV(pp)
    Fmin: 700kHz

    The problem is the power consumption... at no load the current is about 4mA but as I start drawing 15mA @ 5V the total current reaches approx. 31mA.
    I intend to use this converter for a low power application using a 3.6V lithium battery, but the converter current consumption of 16mA is beyond the budget.

    How can I fix this problem?
  2. MrAl

    Well-Known Member

    Jun 17, 2014

    Even with 100 percent efficiency PowerOut=PowerIn.

    For example, if you have a 3v input and 6v output with 10ma on the output then you must have 20ma on the input, and there is no way whatsoever around this. In fact, you are lucky if you see that because in the real world there is always inefficiencies, so you might see that with only 5v out.

    To get the best, you can try a larger inductance, but keep the series R of that inductor down as low as possible. If you have a large resistor for current sense that wont help either, but under 1 ohm sounds ok for this application.

    So the short story is you cant input 3v at 10ma and expect to get 5v at 10ma out.
  3. crutschow


    Mar 14, 2008
    You are seeing a conversion efficiency of about 81% [(15x5)/(31x3)], which is probably typical for that circuit.
    There are some newer circuits using synchronous flyback diodes (MOSFETs) that can approach 90% or better efficiency, but that's about as good as you can do.
  4. MikeML

    AAC Fanatic!

    Oct 2, 2009
    Repeal the laws of Thermodynamics.