# Solar power wattage requirement

I want to understand the Math behind the Solar Power Panel Wattage requirement in order to charge 10 mobile phones (use 0.8A each) at the same time. No battery is required. Solar panel output is 12V DC.

• What voltage do the mobile phones need when they are taking 4.7 amps? Commented Apr 25, 2013 at 17:04
• Are you planning to apply 12V to the phones? Just in case you are, this is a bad idea - you will need to provide the 0.8A at the voltage they expect - 5.7V according to your comment. For efficiency, you will be best with a switch-mode (buck) converter to drop the 12V down to this voltage. Commented Apr 25, 2013 at 19:36
• @Oli Glaser I am trying to determine the solar panel requirements (output in W and 12V) and any other hardware (converters) required to charge 10 cellular phones at the same time using the std chargers (marked DC 5.7V 800mA). Commented Apr 26, 2013 at 17:30

Okay, you have 10 chargers that output 5.7V at 0.8A. Ideally we need to know the input specs, as the conversion from 100V-240VAC to the 5.7VDC will not be 100% efficent (usually on the plate there should also be the input ratings also)

Anyway let's run some rough estimate numbers. Let's assume the chargers are 85% efficient for now.

So the wattage requirement for the 10 chargers input will be:

(1/0.85) * 5.7V * 0.8A * 10 = 53.7W

Now you have to decide how you are supplying the phones charging voltage - whether you bypass the chargers and drop the 12V solar panel down to 5.7V, or use an inverter to create say 120VAC to supply the chargers. The second option will include the inverters inefficiency since you will be going:

12VDC -> 120VAC -> 5.7VDC

If you just step the 12V down, you are going:

12VDC -> 5.7VDC

Inverter Option

There are plenty of cheap 12V inverters available on eBay, so lets take a look at the first option. We already assumed the chargers need 53.7W at their input. Now lets assume the 12VDC->120VAC inverter is 80% efficient, then we need:

(1/0.8) * 5.37W * 10 = 67.2W from the 12V panel.

So current wise this equates to:

67.2W / 12V = 5.6A

Bypass Chargers

For the other option:

12VDC -> 5.7VDC

If we assume the step down converter is 85% efficient, then we get the same result as above with the charger calculation:

(1/0.85) * 5.7V * 0.8A * 10 = 53.7W

And then the panel current would need to be:

53.7W / 12V = 4.48A

• Wow, all of the answers and comments have really helped me. I have a much better understanding. Great site (still learning how it works). Thanks Oli & all! Commented Apr 27, 2013 at 7:55

I'm not sure what you're question exactly is, but I think $P=V \cdot I$ is what you're looking for. $P$ is power in Watts, $V$ is voltage in Volts, and $I$ is current in Amperes.

You want $10\cdot0.8=8A$ with $12V$, so that gives you $P=8\cdot12=96W$.

• Got the same number, simple multiplication :) Which charger uses 4.7Amps? Serious current. My oldschool Nokia takes a max of 890mA, but I guess with modern power hungry devices this is a reasonable value Commented Apr 25, 2013 at 17:04
• @MartinH I believe my phone (Sony XPeria Sola, model of March 2012) has like 1.3Ah, it charges in somewhat more than an hour. 4.7A really is a lot.
– user17592
Commented Apr 25, 2013 at 17:08
• My mistake on the numbers. The phone charger is a std Nokia Output: DC 5.7V/800mA. P=V.I 10.0.8=8A with 12V, so that gives me P=8.12=96W? Commented Apr 25, 2013 at 18:45
• @Derek but can that charger handle a 12V input? I'll update my answer with the 0.8A, you please do the same with your question.
– user17592
Commented Apr 25, 2013 at 18:47
• @CamilStraps - I'm picking up what you're putting down.. The charger is manufactured to an input of:AC 100-240V(RSA). Seems like it would be a waste to regulate up then back down. I'll use a TSR15 Regulator Solar 15A (12V) direct from the panel. Commented Apr 25, 2013 at 19:04