# MC34063A DC-DC Step-Up Converter Power Consumption?

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

1. ### emaq

2
0
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 KappModeratorModerator

11,413
2,619
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.