# Power Transfer to Capacitor

1. Jan 7, 2010

### Nash78

Hi, Me again.

I have been tasked to replace a step up power transformer in a machine.
But i just can't get the answers i need after much thinking.

In a simplistic RC circuit, there are 4 items
DC Power source (Normally voltage driven)
Switch (On/Off the circuit)
Resistor
Capacitor

All are in series right?

In my simplified circuit.

1) Step up transformer (Pri230V:Sec2000V, 16A:1.84A, Single Phase, 50Hz)
2) Full Bridge Rectifer
3) Charging Resistor, 33 ohms
4) Capacitor Bank, 2000Vdc@ 2000uF, total energy rated 4000J

An SCR will turn on power to the transformer which in turn transfers power to the capacitors for charging till 2000Vdc.

My doubt is, this setup is able to charge the capacitor up at 600J/sec as stated by the manufacturer. The specifications are quite accurate. Normally it finishes before 6 seconds on my stopwatch.

I tried the few calculations but it didn't work out,

1) 2000V source with 33 ohms, RC constant does not work out right.
2) Using a source of 1.84A with 33 ohms, it does not work out right either.

Is my direction correct?

2. Jan 7, 2010

### vk6kro

That transformer would give a peak voltage of 2828 volts and the capacitor would charge up to this voltage. So, the capacitor should be rated to at least this voltage and probably 3500 volts for a safety factor.

The secondary resistance of the transformer is not given but it might be as high as 1086 ohms (2000 V / 1.84 A). You would have to consider this in series with the 33 ohm charging resistor.
Using those figures, the time constant would be about 2.2 seconds. ( 2000 uF * 0.001119 Mohms)

Also, the capacitor is not being charged continuously. It is getting a series of bursts of charge with each rectified mains cycle.
So, it would take longer than if the source was constant DC.

3. Jan 8, 2010

### Nash78

Hi,

I assume 230V would already be the rms value going into the primary side, so output should still be 2000V instead of 2828V, yah? There is a feedback circuit which cuts off the SCR signal, so overcharging is well taken of.

So, the secondary resistance plays a big part in the calculations? I will measure it and calculate again.

Thanks,
Nash78

4. Jan 8, 2010

### vk6kro

The 230 volts and 2000 volts are RMS values so you have to multiply them by 1.414 (square root of 2) to get the peak values.
So, the 2000 volts RMS will rectify and give pulsing DC with peaks at 2828 volts.

Actually, the 2000 volts is AT 1.84 amps, so it may be higher off load.

Yes, it would be worth measuring the resistance of the secondary. It is probably fairly high compared with your 33 ohm charging resistor.

This graph is for 1000 ohms series resistance and 2000 uF but 60 Hz input frequency. I tried 50 Hz and it didn't make any difference.

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5. Jan 8, 2010

### Nash78

Any idea on what relation this has with the charging power?

I can't relate these two things together.

Let's say i measure the secondary resistance and get a calculated answer close to what i see in practical. The charging of the bank completes 2000Vdc within 6 secs. I did the test many times to make sure of it.

Manufacturer states it charges 600Joules per sec, is it equal to 600W?
Bank = 4000J
So calculations show ~5 secs.

The stepup transformer has a power of 230V x 16A = 3680VA
Effective power = 3680 x Power Factor = 600W
That is a very low power factor is my equations are correct.

Am i missing something???

6. Jan 8, 2010

### vk6kro

600 joules / second is the same as 600 watts.

However, you can't take the rating of the transformer and say this is the current being used.

The actual current decreases as the capacitor charges up. My simulation shows it as starting at 2.8 amps and dropping to about 0.7 amps at the 6 second point. These are peak values.
Eventually, the current would drop to zero.

The capacitor has 4 coulombs of charge in it. (2000 V * 0.002 Farads). (Q = C * V)
So the average current to charge it was 4 coulombs / 6 seconds = 0.666 amps (Q = I * t)

So loss in 1000 ohms series resistor = I * I * R = .66 * .66 * 1000 = 443 watts or 443 joules / second.

Charge in capacitor = 0.5 * V * V * C = 0.5 * 2000 V * 2000 V * 0.002 F = 4000 joules

Charging rate = 4000 joules / 6 seconds = 666.6 joules / sec

Efficiency = (666.6 / 666.66 + 443) *100 = (666.6 / 1109.6) *100 = 60 %

7. Jan 9, 2010

### Nash78

Wow . . . . that's the BIG difference between just "knowing" and "knowing how to apply what u learnt"

I guess my constant snoozing during classes proved 1 thing.
10% knowledge and 90% crap

Thanks man.