Inductor spec question - solar boost circuit

Thread Starter

russpatterson

Joined Feb 1, 2010
353
Hi There,

I'm building a USB charging station that needs to boost voltage on a solar panel from .5V to ~9 or 10V in order to charge a 6V AGM battery. I'll use the AGM to charge the cell phone / iPod whatever.

I've bread boarded a boost circuit that works (not yet tested at the .5V to 10V boost). My question is what specs do I need for the inductor. The panel I'll be using will output .5V at 3.8 amps. It will out put ~10V at .152 amps (given an 80% efficiency).

So do I need the inductor to be rated for 4 amps since that's what I'm putting in to it? Or 1 amp or less since that's what I'm ultimately moving through it?

Thanks!
 

SgtWookie

Joined Jul 17, 2007
22,230
Well, where's your circuit?

You'll need an inductor rated for around 8A, and a low ESR cap (or a bank of them) on the input side.

If you got it to work with an input of 0.5V, you must've used Germanium transistors, as standard silicon transistors would not do much of anything for you.

Since Germanium transistors have been obsolete for awhile, it will be interesting to see what your circuit looks like.
 

Thread Starter

russpatterson

Joined Feb 1, 2010
353
Ahhh, reality rears it's ugly head. As I said in my post I haven't tried boosting the .5 volts yet. I boosted 5V up to around 100. I forgot about the .7V drop in the diode. What if I changed the diode out for another switching transistor? I've read that's a way to boost efficiency anyway. Is it as simple as toggling them on and off or is the timing more complex? It might be worth a try.

Another question. Isn't the diode on the high voltage side of the inductor? Wouldn't the high voltage coming off the inductor overcome the diode voltage drop? What am I missing?



So I need to spec the inductor (array) to double the input current? I think that if I score the cells I may be able to break one into multiple strips which would each put out .5V at lower amps. Then connect them in series. The other issue is that I need to generate enough voltage to turn on my circuit in the first place if the battery is dead.

Thanks for the response.
 

DickCappels

Joined Aug 21, 2008
10,169
Yes, that topology would do what you want, but to operate with such a low input voltage, you would have to add a start-up circuit, such as a momentary pushbutton to create an initial pulse to get things going, then you can power your logic and drive from the output of the power supply.

An input of 0.6 volts makes the converter very difficult to design. The main problem being voltage drops because of the high current involved. A MOSFET or array of MOSFETs would probably give you much better performance at these voltages than would a bipolar. Whichever way you go, good luck...
 

SgtWookie

Joined Jul 17, 2007
22,230
It's not just the diode; it's the Vf of the BE junction of a standard BJT, which is also like that of a diode - a transistor needs roughly 0.63Vbe in order to start current flow in the base at even a low level. A MOSFET, even a low threshold version (lower than typical logic level MOSFETs) would likely require over a volt to start conducting.

Dick Cappels' idea about a push-button start might work, but would also require some additional components to keep the inductor from saturating - which would simply act like a short to the solar panel.

However, if the circuit could be powered from the battery, and the battery were not allowed to discharge below, say, 6.2v, then it could simply be operated from the battery voltage.
 

retched

Joined Dec 5, 2009
5,207
I agree with the Sarge. Use the solar cell to keep the battery topped up. That may require a few extra components, but with the size of surface mount stuff, it shouldn't be too much of a problem.
 
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