Basic wattage and volt conversion when using solar panel

Question

I have a basic electronic question or two I need to ask.

I want to use a solar panel which delivers 5.2W and is rated for "8V open voltage and 650mA short circuit". If I connect a 5V voltage regulator to lower the volts to 5V, would the amps go up to be 1A after the regulator? (W = V * A)

Another thing I'd like to know is, if I regulate the voltage to 5V and get approximately 1A out of it. How would using different components together work? I.e. I am using an arduino with a servo and a GPS unit and an XBee connected. Would these individual components' amp-usage just be added up (lets say to 850mA) and then the "extra" 150mA would just not be used?

Answer 1

I want to use a solar panel which delivers 5.2W and is rated for "8V open voltage and 650mA short circuit". If I connect a 5V voltage regulator to lower the volts to 5V, would the amps go up to be 1A after the regulator? (W = V * A)

The percentages used below are "out of my head" based on experience with PV panels. Your panel may vary somewhat from my assumptions.

Depending on technology used the ratio of Vmax_power to Voc will vary but using 0.8 is a good first guess. So V_max_power ~= 8V x 0.8 = 6.4V.

Imax_power will be somewhat less than Isc - probably about 90% or better, so Imax_power ~= 650 mA x 90% = 585 MA -> say 600 mA.

So actual max power in bright light will be V x I = ABOUT 6.4V x 600 mA = 3840 mW.
Say about 3.8W.

This will be at about 6.4V so to run with best efficiency you want to convert it to 5V with a switching regulator. A switching regulator will be typically 80% to 90% efficient if you use a good one - and rather less with lower quality designs.
3.8W x 90% =~ 3.4W
3.8W x 80% =~ 3W
3.8W x 70% =~ 2.6W

If output is at 5V then Iout = Pout/Vout =
680 mA at 90% conversion efficiency
600 mA at 80%
520 mA at 70%

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Building a buck converter with an MC34063 will produce an OK result but efficiency will be lower than cxan be achieved with modern IC.
An IC with synchronous rectification is required to maximise efficincy.
The Richtek RC8286 looks ideal
Datasheet here

Answer 2

If you short circuit the panel the output voltage is zero, so that's not a usable indication for maximum output. The 5.2 W is not correct, it's just the product of the open voltage and the short circuit current, but you can't have both at the same time. The output voltage will decrease as current increases, and you'll never have 5.2 W.

A linear regulator will draw the same current from the panel as its load. So if the panel gives you 7 V at 400 mA, using a linear 5 V regulator will still only give 400 mA output. The difference in power: 2.8W (7 V x 400 mA) - 2 W (5 V x 400 mA) = 0.8 W will be dissipated in the regulator. You would need an LDO regulator (Low Drop-Out) if the input voltage comes close to the 5 V out.

A switching regulator is different, and can indeed have a higher output current at a lower voltage. The conversion isn't 100 % efficient, so the 2.8 W in will not give you 2.8 W out, but probably 2.6 W. Then you have 520 mA out for 400 mA in.

Connecting your devices you have to add their resp. currents. Should be less than what the power supply can deliver, the rest will indeed remain unused.

Answer 3

No, the solar panel cannot supply any more than it's rated short circuit current on its own. You could use a buck regulator to do something like this though.
A typical switching regulator can convert at maybe 85-90% efficiency, so if your max power available at the original voltage is e.g. 10V * 0.5A = 5W, with a 90% efficient regulator, you can drop the voltage to 5V and get (5W * 0.9) / 5V = 0.9A.

The maximum power is also not as simple as Voc * Isc, as you can't have both at the same time. For your example, when the panel is short circuited, I = 650mA but V = 0V, and when V = 8V, I = 0A. In both cases you get 0W.
Maximum power comes at somewhere in between for both I and V, on the knee of the V-I curve. The point is called the maximum power point:

To answer the xBee, etc current question, the supply will only supply the components with the current they "request", so yes, the extra 150mA "will not be used". It's like your tap is capable of supplying e.g. 1 litre a second, but you can "request" 1 millilitre a second if you want (by only turning the tap on slightly) - the other 999ml is not wasted.