Buck Converter Input current rquirement

[mshalash]

Hi All

I need some help on this one

I have DC DC Buck Converter 12v to 3V-5V 5A

I am using on a Resistive load of 2.5ohm at 4.04V, I measured the input current and it was around 0.63A using DC ameter.

I thought the input supply current should be more than the output current since on the SMPS on mode it is sourcing the current from the supply while in the output it is comming fom the energy stored in the inductor.

Here is the question can I run the circuit from 12v 1.5A for a load of say max 12W (3v to 5V) assuming 85% system effecincy?


Many thanks

 

[Jesse]

mshalash,
The power delivered by the source could never be less than the load power, therefore the input power is proportional to the load power, assuming there is no loss. In your case, for output voltage lower than the input voltage, the input current should be lower than the output current, your measurement is correct.

Pin = Pout + Ploss
= Pout/Efficiency

Pin*Iin = (Vout*Iout)/Efficiency

therefore,
Iout = 4.04V/2.5ohm = 1.616A
Iin = (Vout*Iout)/(Vin*Efficiency)
= (4.04*1.616)/(12*0.864) = 0.63A

Jesse

 

[mshalash]

mshalash,
The power delivered by the source could never be less than the load power, therefore the input power is proportional to the load power, assuming there is no loss. In your case, for output voltage lower than the input voltage, the input current should be lower than the output current, your measurement is correct.

Pin = Pout + Ploss
= Pout/Efficiency

Pin*Iin = (Vout*Iout)/Efficiency

therefore,
Iout = 4.04V/2.5ohm = 1.616A
Iin = (Vout*Iout)/(Vin*Efficiency)
= (4.04*1.616)/(12*0.864) = 0.63A

Jesse

Thanks Jesse