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Author Topic: Alternator output collected on a capacitor  (Read 3514 times)

capthook

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Alternator output collected on a capacitor
« on: February 28, 2009, 11:17:28 AM »
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!
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gyulasun

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Re: Alternator output collected on a capacitor
« Reply #1 on: February 28, 2009, 03:49:27 PM »
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
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capthook

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Re: Alternator output collected on a capacitor
« Reply #2 on: February 28, 2009, 08:16:13 PM »
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%.

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