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I am using 50 W solar panel(20V @Open Circuit/ 3.5A @short circuit) to charge my 4V/100Ah battery(found from industrial scrap). I tried charge battery using buck converter but as it get pulse input current from source, i can charge battery at maximum of ~3.3A@4.8V. As this battery rated at 100Ah it can charge upto 10A. How can i increase the current to increase charging speed? Buck converter is not perfect selection for this application. Does any other DC-DC converter do this? Or i should modify buck converter or else?

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A buck converter can output almost all the power it's getting from its input (less typically 5-15% losses), which means if the output voltage is significantly lower than the input, the output current can be significantly higher.

As a buck converter draws input current in pulses, this can be a problem if the input is a current source, like a solar panel. A large capacitor across the input should be used to source the current during the large input pulses, it will recharge from the panel between the pulses.

It's difficult to design a value for this capacitor without knowing the switching frequency of the buck converter. You would want to aim for a ripple voltage that was small (say <1v ripple compared to your 20v panel) when the buck was consuming the panel's maximum output power. Failing tools to do that (an oscilloscope, or a meter that can read AC at the buck's frequency, or a diode+capacitor to make a simple peak detector) you could simply try a small value of capacitor, and increase it until the buck was able to consume enough of the panel's power that you were happy.

A good commercial buck converter will already have enough input capacitance for its power rating and switching frequency, which would mean it draws smooth current at its input terminals. However a cheap knock-off, or an amateur-built one, might not have.

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You have a 20V open / 3.5A short solar panal. Let's model that as an ideal 20V source with an interal series resistance of 20v/3.5A = 5.7 Ohm. The maximum external power you can draw from such a source is when the external load resistor is equal to the internal series resistance. This gives a current of 20V / ( 11.4Ohm ) = 1.75A. The external voltage is 10V, the external power is 17.5W.

The buck output you saw was 4.8V * 3.3A = 15.8W, so the converter was 90% effective. That is as good as you will get.

You problem isn't the wrong converter: you expect too much from your solar panel. Given the open voltage V and the short current I, the maximum power you can draw is 1/4 * V * I.

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  • \$\begingroup\$ It would be nice to read an explanation with it's 1/4 (and not, let's say, 1/2). \$\endgroup\$ – Janka Jan 17 '18 at 7:21
  • \$\begingroup\$ @Janka Internal resistance=R=U/I. Optimum load=R. Total power to the load and internal loss = (U^2)/2R. To the load we get 50% of it, it's =(U^2)/4R = (U*I)/4 \$\endgroup\$ – user287001 Jan 17 '18 at 8:30
  • \$\begingroup\$ I know that but it belongs into the answer, given the question. \$\endgroup\$ – Janka Jan 17 '18 at 9:59
  • \$\begingroup\$ Your answer doesn't make sense. This not how impedance match. Solar panle has different resistance at different voltage. This panle has 17.5Vmax and 2.8Amax at MPP. If you measure voltage at Open circuit it will be Infinite and 0 Ohm at short Circuit. And i don't ask about solar panle, i asked about Dc-Dc converter. As i read on internet, buck converter has discontinuous input current so the peak current is same as input current. So there are other converter that has continuous input and output current. Such as cuk converter but i am not sure what to do. Huge capacitor as a filter can solve it \$\endgroup\$ – user3785133 Jan 17 '18 at 11:58

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