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# Max current measuring

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#### Electronman

##### New Member
Hello,

How can I detect how much of current my solar cell is able to give?
What about detecting the MAX current of it?

Thanks

Short across the solar cell with an ammeter. You are basically short-circuiting the solar cell and measuring the current in such a condition.

Of course, the voltage output in this condition will also be zero. It might be more useful to check what is the current the solar cell outputs at the minimum voltage you can tolerate.

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Short circuit current measurement of a solar cell should be brief, and in the lighting conditions you're using it in. If you measure the current under direct full sunlight and you expect to get that current all the time you'll be sadly mistaken.

Isn't the total dissipation of the cell LESS into a short circuit then into an open circuit??

Total dissipation of...?

Whaa?? With an open circuit there is no power dissipation... With a short circuit the dissipation is 100% in the resistance of the solar cells themselves, less wiring losses.

I believe with an open circuit solar panel the voltage rises until the panel body diode starts to conduct which stops the voltage rising any further, so with open circuit the panel silicon is dissipating 10A at 22v. Into a short circuit the panel dissipates 10A at a very low voltage.

There are some good web sites on how solar cells work, and I could be wrong because it's been a few years since I looked it up.

Ohm's Law says that 10A flowing through an open will require a bit more voltage than I'm able to generate.

Mr RB can you explain this body diode a little? I've never heard of one in refrence to a solar cell. I thought no current flowed at all through a solar cell during open circuit conditions, otherwise the voltage would drop.

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Solar cell model is an illumination based current source with a parallel inherent diode that will forward bias if no load is applied. With no load, all generated current goes down the inherent diode.

Maximum available loading power point is when inherent diode just barely starts to conduct, taking about 0.3% of illumination generated current. The inherent diode obeys silicon diode temperature profile, meaning hot cell yields lower maximum voltage power point, like 0.4vdc, (less power output), cold cell yields higher maximum voltage power point, like 0.7 vdc (more power output). At room temp maximum power point voltage loading is about 0.45 to 0.5 vdc.

I understand the model, but that doesn't seem to make sense logically to me. Does that mean when a 10 watt solar panel is in full illumination, excluding all other sources of IR radiation would that mean that the solar panel is producing 10 watts of heat since it's not being used anywhere else?

It's probably producing more than 10W of heat. 10w standard panel output will probably be specced at about 17v, so it's 10W = 17v * 588mA.

Now if the 588mA is produced based on illumination (remaining constant) and the open circuit voltage of the panel rises to 22v, the 10W panel silicon is dissipating 22 * 588mA or 13W. Into a short circuit that dissipation is almost zero.

Depends.. If you're talking about a clasical 0 ohm short circuit. The dissipation is going to be almost entirely in the bulk resistance of the solar panel, as the short circuit itself will consume virtually no power because of it's low resistance. You have to have a large number of solar cells in series to get a decent voltage so I'm going to assume the bulk resistance is relatively high under short circuit conditions. Even just using decent heavy gauge wire should cause the same power dissipation.

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Wouldn't the dissipation equal volts times amps? The amps is known by the short circuit amps and the panel cells are in series with each other and with the short, so the amps are known in every cell and must be equal.

The volts can't exceed the short circut voltage, with maybe a little bit from the connector wires and solder joints on the cells. So the max voltage would be maybe 100mV for the whole panel from 20+ joints of a few milliohms each.

So to my mind if no point of the system can have more than some millivolts potential, and the current through every part of the system is known, then the TOTAL dissipation can't be much above zero.

As in all electrical systems, maximum power transfer occurs when the load impedance equals the source impedance. Driving an ammeter is a dead short, and while it may produce maximum current, the voltage will be so low that power will be considerably lower. The panel also won't produce that level of current in to any useful load.

The trick is to measure the current and voltage in to it's matched impedance, which is difficult as the source impedance changes with the amount of light on the panel.

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