After performing some tests on a small sized solar panel, I have been able to determine that the panel outputs on average, around 6v (Open source voltage) 300mA (Short circuit current) with no load. Testing with variable loads the solar panel only managed to output around 5v 5mA.

In order to drive the circuit at the solar panels load, I need a current of up to 2A. Seeing as the increase is so high, is this at all possible using something like a current amplifier?

Alternatively, I can switch out the solar panel being used for a 'solar battery bank', the one I have is a solar panel mounted on a 15000mA battery. From here I can drive the circuit, although, ideally the output voltage should vary depending on the panels output rather than the battery. In this scenario I could use a potentiometer to vary the voltage. But I'd rather just amplify the current or the original panel of possible.

  • \$\begingroup\$ Can you provide a link to the datasheet for the panel? \$\endgroup\$ Commented Feb 29, 2016 at 14:54
  • \$\begingroup\$ Please be more descriptive of exzctly what you're wanting to use this panel + amplifies output for. FYI, PhotoVoltaic (PV) panels, like most batteries, are a bit of a "constant voltage" device. They inherently will produce a fairly 'flat' open-circuit voltage output under any illumination, with the available current being what varies with the intensity of illumination. - Are you wanting to use the PV cell as a power supply, "light sensor," or both in this application? If power supply or "both," then adding a battery & MPPT will probably be "step one." \$\endgroup\$ Commented Feb 29, 2016 at 14:55

2 Answers 2


You can't break physics.

5V @ 5mA is 25mW. Power must remain constant (less converter efficiency), so (25mW*90%)/2A = 11.25mV (assuming an optimistic 90% efficiency). So if whatever you are trying to power is fine with about a hundredth of a volt (which I very much doubt), that's fine. Otherwise, you're gonna need a lot more solar panels.

Edit: regarding a current amplifier:

OP, you can build a current amplifier, but think about this carefully: what will you power the current amplifier with? If you power it with itself, that's not going to work for reasons mentioned above: you can't get more power from something. Or you power it with an external supply (a nearby outlet, say), which completely defeats the purpose of having the panel.

  • \$\begingroup\$ The short circuit current is only 300mA. Given that, there's no way to get 2A even at 11.25mV. Solar panels do not put out constant power at all operating points, which is the reason for MPPT controllers and chargers. With the two data points given this panel likely can't even get near 1A at its maximum power point. \$\endgroup\$
    – John D
    Commented Feb 29, 2016 at 15:02
  • \$\begingroup\$ I was assuming a "dumb" step-down converter (of a very unusual flavour; 11.25mV output would be a strange converter). But yes, that's optimistic, although operating at the MPP might give more power. \$\endgroup\$
    – uint128_t
    Commented Feb 29, 2016 at 15:05
  • \$\begingroup\$ Ah, right, if you had a "dumb" buck that could put out 11.25mv you could get there from the 5V@5mA, and maybe a bit more power from the MPP. \$\endgroup\$
    – John D
    Commented Feb 29, 2016 at 15:10

In order to drive the circuit at the solar panels load, I need a current of up to 2A. Seeing as the increase is so high, is this at all possible using something like a current amplifier?

Yes, though you need to consider what that "current amplifier" might be.

Here's one kind of current amplifier:


simulate this circuit – Schematic created using CircuitLab

On the left we have I1, which represents the 300 mA you get out of the panels. This goes through an emitter follower to amplify the current.

The trouble with this setup is you need some kind of additional power source, drawn on the right as "???". But I'm guessing if you are using solar panels, the point is to not require any additional power source.

Alternately, there are devices like switched-mode power supplies which can increase the current available, at the cost of reducing voltage. These are essentially the electrical equivalent of mechanical gearboxes: you can increase speed and decrease torque, or you can decrease speed and increase torque, but you can't increase both.

The reason is conservation of energy. The electrical energy out of the solar panels can't exceed the solar energy (after inefficiencies) coming in to them. Electrical power is the product of current and voltage:

$$ \text{power} = \text{current} \cdot \text{voltage} \\ P = IE $$

You can alter I and E, as long as their product (P) is less than the maximum electrical power your panels can generate.

Another possibility: use the panels to charge a battery for a long time, then run your load from the batteries for a short time, once you've stored sufficient energy in the batteries.

  • 1
    \$\begingroup\$ Wouldn't it usually be recommendable to place the controlled load upstream of an NPN BJT, so the Vbe remains more predictable with the emitter held at GND potential? \$\endgroup\$ Commented Mar 8, 2016 at 21:52
  • \$\begingroup\$ @RobhercKV5ROB I'm not really sure what you mean by "upstream". \$\endgroup\$
    – Phil Frost
    Commented Mar 8, 2016 at 22:13
  • \$\begingroup\$ Sorry, between V+ & transistor's collector \$\endgroup\$ Commented Mar 8, 2016 at 23:01
  • \$\begingroup\$ @RobhercKV5ROB Sure, you could do that. The collector current is identical to the emitter current, neglecting the relatively small base current. The advantage to the circuit drawn is that the voltage across the load is almost the same as the voltage across I1. When I think "current amplifier", I think of a thing which amplifies current, and (ideally) does not change voltage at all. \$\endgroup\$
    – Phil Frost
    Commented Mar 9, 2016 at 1:57
  • \$\begingroup\$ Good point, I hadn't considered that as a potential benefit, but I can see it now. \$\endgroup\$ Commented Mar 9, 2016 at 5:21

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