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MC34063A DC-DC Step-Up Converter Power Consumption?

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

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  1. emaq


    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. Harald Kapp

    Harald Kapp Moderator Moderator

    Nov 17, 2011
    An ideal step-up (or ste-down) converter will require the same input power as it delivers output power.
    With 15mA*5V that is 75mW at the output.
    75mW at the input will require a higher currrent than at the output since the voltage is lower: Iin = 75mW/3V=25mA.
    Add the self consumption of the regulator (4mA without output current, probably a bit more with load), you arrive at Iin>=25mA+4mA=29mA, a value thats almost perfect in accordance with your measurements.

    Nothing is wrong with the circuit, it's the theory behind it. You can't get away with a 3V->5V step up converter and less input current than output current.
    Arouse1973 likes this.
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