Re: AC to Cap Dump

Body

Hi Hob,

If you have access to a digital storage oscilloscope with mathematical functions it can be rigged up as a power meter to actually calculate the power. Only problem is that the memory depth isn't high enough to run for hours. But it can be used to confirm your measurements at different moments during the experiment or to see if excess energy comes out during the experiment. If the mean power going out of a component is greater than the mean power going into that component it extracts energy from somewhere (or uses up itself like a battery).

If you make curves for the current and power a curve representing the energy can be made too by integrating the power curve. I.e. it's done by summing up all the points representing power along the time axis and drawing a curve with the summed up numbers. This is the energy.

One thing to be aware of is to be sure that there is no phase shift between the voltage potential and the current as this represents energy changing direction (reactive power). This can be checked with an oscilloscope. Dumping a capacitor directly into a low impedance battery will make ringing (or spikes) in the circuit because of the parasitic inductance of the wires. I.e. they form loops which are actually parasitic inductors.

Charging a lead acid battery takes more power in the beginning (high voltage X high current) than later during charging (high voltage X low current). This is revealed if plotting the power curve or the energy curve.


Source: http://batteryuniversity.com/learn/article/charging_the_lead_acid_battery http://batteryuniversity.com/learn/article/charging_the_lead_acid_battery

I don't know if your formula is good enough but it ought to be confirmed with a proper power measurement.

Regards
Ole

 

---In [email protected], <hob.nilre@...> wrote :

 Hi Ole,

Power out is simply Vbat²/R where Vbat is the voltage across the battery and R is the dump-resistor.

Power in is Vbat*f*C*(Vcap-Vbat) where f is the AC frequency, 50Hz where I am, and Vcap is the voltage across the cap just before the dump to the battery and C is the capacitance of the cap. C*(Vcap-Vbat) is the charge Q being dumped from the cap as in Q = C*V and so f*Q is the total charge during a second which in turn is the average current. Work done is Vbat*Q which for one second is the same as the avg power.

Does it make sense or am I making a mistake somewhere?

/Hob