Quote:
Originally Posted by Ken Maltby
So while 2x667=1334mAmps and the adapter is rated with an output of 1000 mAmps,
I am hoping that it can handle the input.
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This shows that you don't seem to have a proper understanding of basic electrical theory.
Of course the adapter will be able to handle the input. What do you think would happen if you plugged the adapter into the lighter socket in your car? A car battery can provide 60000mA (60 Amps).
You have to look at things in terms of power (watts) requirements, not current (amps). Power is calculated by the formula:
power (watts, W) = voltage (volts, V) X current (amps, A)
Note that power is usually signified by the letter P, current by I, and voltage by E or V, but I'll use the unit letters: W, A, V.
You can rearrange the formula to get:
V = W / A and A = W / V
The electronics in the adapter will do whatever it takes to provide a fixed 5V to the device connected to its USB output (in this case the Libre).
The Libre will draw whatever power it requires to run and charge its battery. You've stated in a previous post "If you look around at your USB powered devices you will find plenty that draw between 800mA and 900mA, especially if they are charging batteries", so let's assume that the Libre will draw a maximum of 900mA (0.9A). Since the adapter will always regulate the output voltage to 5V, we can use the formula to derive the maximum power that the Libre will draw: 5V X 0.9A = 4.5W
The job of the adapter is to convert the high input voltage (12V to 24V) to a fixed 5V output voltage. In a perfect world, it would do this 100% efficiently, so that the power that goes in would equal the power that goes out. Let's assume that the panel's output voltage will be 15V, as is stated in the little pamphlet (although this will vary depending on current draw and light available). Since the panel voltage is higher than 5V, less current will be required to provide the required power: 4.5W / 15V = 0.3A (or 300mA).
So, to charge the Libre, the panel need only be rated 300mA @ 15V. Any extra capacity that the panel could possibly provide will be wasted as heat; the panel itself will get hotter. Adding a second panel will just mean that this second panel's potential power will also be wasted as heat, with no difference in the Libre's charging speed.
However, as I've previously stated, the adapter won't be 100% efficient (even the red LED on it will use a little power) and the panel(s) won't normally be running, as the little pamphlet puts it, "under optimum conditions".
The adapter might be 70% to 90% efficient in power conversion but lets assume it's not very good, and therefore can only convert 60% of its input power to output power. So we need to multiply our output power by 1/60% = 167% to get input power requirements: 4.5W X 1.67 = 7.5W. Calculating input current: 7.5W / 15V = 0.5A (500mA). This 500mA is still within what a single panel can provide, even under slightly less than optimum conditions.
To summarize:
- The current that the panel needs to provide will be determined by the current that the Libre draws, but will be less (disregarding adapter inefficiencies) because the panel provides 15V and the Libre runs at 5V.
- As far as what the adapter can "handle", the only concern is that the panel(s) don't exceed the maximum 24V that the adapter specifies, and that the Libre doesn't try to draw more than the maximum 1000mA that the adapter specifies (both of these cases are probably true).
- One panel will be sufficient under optimum or somewhat less than optimum conditions. Two panels will allow charging under conditions that are less optimum than with one panel, but won't make any difference if conditions are sufficient for operation with one panel.