Alternator output collected on a capacitor:
Why so little charge collected on the capacitor?
I have a small 3-phase (star) alternator that at 60 RPMs puts out 7 watts after rectifying to DC:
16 Volts / 36 ohms / .4444 amps = 16V x .4444A = 7.11 watts
The output is then sent to a 9400uF 35V capacitor.
After 1 second, the voltage on the capacitor reads 12.1V.
.5(12.1 x 12.1 x .0094) = .688 J
1 watt = 1 joule per second
(7 watts output : .688 joules collected)
Why is 1/10th of the power being collected on the capacitor?
As the capacitors voltage climbs, the resistance increases.... is this part of it?
And if/is charging a capacitor REALLY only 50% efficient?
Is the capacitor charging going to be so lossy no matter what that it would be better to just charge a battery?
How do I increase the charge collected?
Thanks!
Hi,
I can give some hints on your setup but cannot answer all your questions. :) ::) :)
The very moment you connect the capacitor the voltage level across the 36 Ohm load should increase because you introduced an energy puffer (the 9400 uF will work as a normal puffer capacitor after a 3 phase rectification).
The observed resistor increase may happen due to the dissipation? 7W power heats your 36 Ohm and it may change to a higher value as it heats up. Maybe you can check it by measuring its resistance with a multimeter just after disconnecting it from the circuit and observe the change also as it cools down.
Here is good link for some calculation: http://hyperphysics.phy-astr.gsu.edu/hbase/electric/capchg.html#c1
If you substitute your data, you can see that under 1 second of charging time the charge is 142568.01 uC in the capacitor (1 second is 3*RC here, (RC time constant=0.3384). This amount of charge corresponds to a voltage: V=142568uC/9400uF=15.16V across the capacitor just after 1 second charging time. You measured 12.1V .
The difference may come from tolerance issue (you would like to measure or somehow check how much uF is really your 9400 uF capacitor), and/or may come from the fact that the calculation in the link is based on a stabil voltage source and your setup is not like that because the moment you connect the capacitor your source's output voltage increases a little? Also, there can be some uncertainty in the voltage measurement across the capacitor just after 1 second?
rgds, Gyula
Thanks for the reply Gyula!
If there were still karma points, you would be deserving of 1000+ as you are always so helpful.
Thanks for the thoughts on the RC, I hadn't been considering it.
So it could be thought of like this: ?
1. after 1 second 3RC's have passed
2. at 3RC the current has dropped to almost nothing
3. one could 'average' the current over the 3RC to be ~25% of initial
An initial of .4444 amps x 25% = 0.1111A
16V x .1111A = 1.7776 W (an approximation of actuall output vs. the 7 W rated)
1.776 x 50% (efficiency) = 0.8888
.8888 J is an approximation that is much closer to the actual of .688 J than the 7W.
Solutions: ?
1. increase capacitor size to increase RC thus keeping the current higher during full charge time and collecting more overall charge for a given time period.
2. implement a fly-back charge circuit in series just before the capacitor and increase the charge efficiency from 50% to maybe 85%
I have a modest understanding of #2 - flyback charger. It works on the idea that it is a constant current device. Since power lost in charging is I^2, limiting the high current in-rush at the start of the charging limits the IR losses, increasing efficiency beyond 50%.