How to calculate the Resistor, when charging a Capacitor

Hello,
this might be a simple question for some of you, but I am stuck at this.

Imagine a simple circuit:

A 0.22 uF / 1000 V Capacitor is charged through a 1 k ohm Resistor and 1000 Volt 1 A Diode. The source voltage is 200 HZ AC Sine Wave with a peak to peak voltage of 400 Volts.

I want to use this circuit as a peak voltage detector.

At the moment of first charge, when the capacitor is completely empty, it is like a shortage to the circuit and theoretically the current is infinite, right?

How do I calculate the correct wattage for a 1 k Resistor?

Thanks

Tronoty

tronoty,
If you need to be ultra-conservative, procede as follows: Assume that power is applied when the voltage is at its 400V peak. The instantaneous current would be 400/1000 = 0.4 Amperes. The initial instantaneous power dissipated by the resistor is i^2R = .4^2 X 1000 = 160 Watts.
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To calculate the average power during the 1/2 cycle over which current flows, you would need to integrate the instantaneous power over 1/4 cycle. Let me know if you need help with this.
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A note of caution: The average power will be much lower thatn the initial instantaneous power. However, if you rate the resistor on the basis of average power, you may damage the resistor. Whether damage occurs dpends on the thermal time constant of the resistor. For example, you can't expect a 1Watt resistor to surve a 1 megawatt power surge, even if the average power is only 1Watt.
Regards,
Kral

Thank you Kral,

I started with a 5 Watt Resistor and that one died within seconds.
A 10 watt resistor got warm and eventually died.

When I went to a 50 watt wire wound resistor, I had no more problems.

I am still looking to find a way of calculating the wattage depending on the size of the capacitor. I know, if I use a 0.01 uF Cap, the load on the resistor is much smaller.

Or is this indeed a complicated calculation?

Thanks again

Tronoty

tronoty,
The dissipation is a function of the capcitor value. The smaller the capacitor, the smaller the duration of the current spike.
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Calculation of the actual dissipation as a function of C is not complicated, but does involve some integral calulus. I'll work it out when I have time. Probably some time today.
Regards,
Kral

Thank you, really appreciate this.

I have never read any paper/book about this case of resistor calculation.

This is an RC circuit. Every RC circuit has its time constant t=RC, and the capacitor is considered to be fully charged (from 0) or discharged (from full) in the duration of 4RC.

So, in my opinion, there are some matters here you have to take into accounts:

- For simple calculation, you can suppose the curve of voltage drop on R when C charging is linear, maximum = 400V when starting charging, and 0V when fully charged.
So, the average voltage drop on R: Vr = (400 + 0)/2 = 200V
And the average dissipation on R: Pr= 200V^2/1000 ohm = 40W.

- The charging time: = 4RC = 4*1000ohm*0.00000022Farad = 0.00088sec # 1ms.

Or: R will dissipate a average wattage of 40W in 1ms.

With the above calculation, any wattage of R should be satisfied since the time duration is fairly short.

- However, there is an important thing you have to consider: the maximum current allowed for the bonded contacts of the resistor/capacitor (component leads bond to resistor body or capacitor plates). This parameter is not informed by the component manufacturers for the common parts. In general, capacitor is made to survive with the current at its capacitance. And the resistor, the bigger size, the better conducting. That is why your 5W or 10W resistors died, they can only afford sqrt 5W/1000ohm = 0.07A max. or sqrt 10W/1000ohm = 0.1A max. while the maximum current in circuit = 400V/1000oym = 0.4A.

So, my opinion is you should use the resistor of 40W at least, 50W as good in your circuit and experience.

Just to discuss with you.
nguyennam

Thanks nguyennam, this was really informative and it goes well with my experience that a 50 watt wirewound resistor was working well. But now I am curios if may be the Capacitor is in any danger, because of the fast current rise or even the diode.

Since I am using a 1A 1000V Diode in series with the Capacitor, and the max current is 0.4 A, does that make the diode safe or will there be an issue with fast current rise times?

Thanks for any advise