Alternator output collected on a capacitor

[capthook]

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!

[gyulasun]

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

[capthook]

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%.