Volts and Amps
- Thread starter trevornewkirk
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N017RW
Premium Pilot
You can determine this from the supply/charger that came with your aircraft.
The listed wattage divided by the listed output voltage will reveal the current.
The listed wattage divided by the listed output voltage will reveal the current.
The other number you need to look at is watts. I have done some extensive searching for those power bank batts with a plug that they say will change a laptop. The highest I found is 80 watts. You need at least 100 to charge your P3/4 batts. Realistically more than 100, because most all of those ratings are stretched a bit, and rated under "ideal lab conditions"
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100 W is not really another number here - it's the product of the 17.5 V and 5.7 A mentioned above. If the source meets those requirements then, by definition, it has enough power.
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Simple calc is P=V/I. Or power (watts) = voltage ÷ current (Amps). Careful with the formula as the AC calc will not reflect true power output due to heat losses across secondary coil during step down and AC to DC conversion. I'm not sure if the batteries are charged via DC or AC but my gut says DC.
sar104 is correct, 17.5 x 5.7 = 99.75 watts, and on the other side 10,000mah (10 ah) is probably double what 1 battery will take depending how low you run them (something less than the 5870mah the battery is rated. While the only way to know is to try it, you may get 2 batteries recharged.
Yes - but the 17.5 V, 5.7 A is the DC current delivered to the batteries, so the charger simply needs to supply that. Now if you are thinking about the power rating of whatever is driving the charger then yes - there may be losses to consider, since the charger will consume more power than it delivers to the battery.
100w is not another number, it is the only number. Have fun trying to charge with less. Make sure you can get a refund when it doesn't work
I think you misunderstood my point. 100 W is the power delivered by a charger supplying 17.5 V and 5.7 A. That is the way that chargers are specified, since the power alone does not mean it will charge the battery. A 12 V, 8 A charger will deliver 100 W, but it will not charge these batteries.
alokbhargava
Premium Pilot
No, not correct. 10,000 mah power bank supplies power at 5volts resulting into 50Whr energy
Whereas each battery is 5870mah at 15.2 v resulting into 89.2 Whr.
Hence battery pack doesn't have enough juice to charge phantom battery. There are 30,000 mah power packs available that with a D.C. Voltage booster may be enough to charge this battery.
That might work, as you say, but it will only manage one charge and it seems like it would be much more convenient just to buy one more battery.
alokbhargava
Premium Pilot
OP had asked if 10,000mah power pack can charge the battery and answer is NO as shown in my calculations whereas 30,000mah pack can.
Yes - I agree. I was just pointing out that it would be simpler and more effective just to get another flight battery rather than carry around a different battery that can only manage one recharge.
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The main issue is whether the power bank can sustain its voltage at the required current level all the way through. @HighTechPauper pointed out "17.5 volts @ 5.7 amps is what the charger is rated" as the guzzoutta capacity of the charger. That is supported by the AC mains going in at up to 100W.
A power bank is being depleted and its own batteries are falling in voltage. If the power bank doesn't have a DC-DC converter to hold up the voltage with the current, then it can't charge the P4P battery completely.
Next, capacity. 10,000 mAh at what voltage? What you need is WH in excess of each P4P battery you want to charge as @alokbhargava points out in a round about way. Also factor in conversion inefficiencies.
10,000 mAh @ 5V = 50 WH. The P4P batt is 90 WH. So you'd need in excess of 2 of your 10 AH power banks to charge a single P4P battery (conversion inefficiencies need to be factored in).
Finally of course, how are you going to get your 5V supply up to 17.5 V? Can the power bank output enough current?
17.5V/5 = 3.5 more current so the power bank has to output over 20 A @ 5V into a DC-DC converter that outputs 17.5V (5.7A). (Over 20A because the DC-DC will have conversion inefficiencies).
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