Electronic – voltage buck converter which also boosts output current

efficiencysolar cellvoltage-regulator

Application: solar USB charger.

I have a 9V solar panel. I use a buck converter circuit to bring it down to 5V before going into the USB battery charging circuit.

Exact components:

http://www.amazon.co.uk/gp/product/B00I0MOVBO
http://www.amazon.co.uk/gp/product/B00EYT1DWW

This link has more information about the buck converter circuit (I think it's the same one I bought from amazon above):

http://www.minikits.com.au/LM2596-PSU-01

I was hoping that the buck converter would also increase the output current (say, for example, from 333mA to 450 mA, for a ~75% efficiency), but after measuring it, it seems to track exactly the input current (Ioutput = Iinput). That's why I didn't simply use a 7805 regulator, which I understand dissipates extra power as heat: I was hoping that the buck converter circuit would convert more efficiently, and translate in increased output current.

Are there any alternative buck converters (PCB or a design that I could build myself) which would reduce from 9V to 5V while also increasing the current somewhat? If yes, in practice, would their efficiency be worth it? (I think that anything below ~75% efficiency might not be worth it for such low power).

Best Answer

I measure by shorting the converter with the multimeter. So to measure input current/voltage, with the whole circuit connected:I place multimeter black on in-, multimeter red on in+. For output current/voltage: multimeter black on out-, multimeter red on in+ (shorting the battery)

This means you are not measuring the current that the solar cell is delivering to the switching converter. You are short circuiting the cell with the ammeter, and measuring what the cell can provide into a short circuit.

Then you short circuit the converter output and measure what it can provide into a short circuit. The way a buck converter works, when you short its output, it will essentially just short its input to its output. And again you will measure what the cell can drive into a short circuit.

Like Peter Bennett says in comments, in order to know what the currents in and out of the converter are in the operating circuit, you must break the circuit and insert the ammeter between the cell and the converter input; then insert the meter between the converter and the load.