4 ms·
Nope, current scales more easily. Just add more batteries in parallel. The problem starts with voltage because now you need each battery or battery set connect
by ilyt 3y ago
Nope, current scales more easily. Just add more batteries in parallel.
The problem starts with voltage because now you need each battery or battery set connected to have very close capacity or else it will be "wasted", and you need cell balancer to keep them equally charged.
- asddubs 3y agoputting batteries in parallel comes with its own problems, namely that you need far higher current to charge them at a reasonable speed, which is more difficult to scale up effectively than voltage
- magicalhippo 3y ago> which is more difficult to scale up effectively than voltage For those who don't know, this is because losses scale as current squared, instead of just proportionally like it does with voltage. For example, charging a power bank with 20W of power at 5V you need to supply 4A of current. If you have a good USB cable with 0.1 Ohm resistance, that results in P = V*I = (I*R)*I = (4*0.1) * 4 = 1.6W of loss in the cable[1]. If instead you could use 20V to charge, then the current would only need to be 1A and hence the loss just (1*0.1) * 1 = 0.1W. [1]: https://en.wikipedia.org/wiki/Ohm%27s_law https://en.wikipedia.org/wiki/Ohm%27s_law