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
http://www.nomad.ee/micros/mc34063a/
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?

 

Hi,

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.

 

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.

 

...How can I fix this problem?

Repeal the laws of Thermodynamics.