# Power delivered to capacitor in a pulse

#### Bhope691

Joined Oct 24, 2016
26
I've got a black box which can deliver high voltage DC pulses connected to a capacitor. As in the below schematic.

I know what the input power to the black box is as it is powered by a power supply unit which shows the voltage and current.

I know the peak voltage out of the box (measured across HV+ and HV-) the width of the pulse and the frequency of the pulse. The pulse is shaped like:

I also know the capacitance of the capacitor. How do I determine the power that the black box delivers to the capacitor in one pulse and the average power in a second? Do I even have enough information? Can I use the normal energy in a capacitor equation or is that only for a constant DC input?

#### dl324

Joined Mar 30, 2015
12,224
Is this school work?

#### Bhope691

Joined Oct 24, 2016
26
No.

I've worked out the energy in the capacitor for each pulse using E=1/2C(V^2) and used Vpeak. I've also worked it out for Vavg and Vrms, however I'm not sure if any of these is correct as it is a pulse rather than steady DC.

#### Just Another Sparky

Joined Dec 8, 2019
107
For power you will need to know the voltage of the power source, the series inductances & resistances of the power supply, wiring and capacitor, the rise time & profile of the pulse and the charge rating of the capacitor. In an ideal arrangement given a square wave, current will be infinite. In reality it could be several hundred or thousand amps. You're likely to see voltages above the input value due to ringing caused by the inductance of the circuit in the presence of very high currents.

Due to the precision required of the measurements needed to accurately figure this, it might be easier and simpler to just probe the circuit by inserting a low impedance shunt in series and measure the current using a storage oscilloscope fitted with suitable probes.

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#### Brown Pulliam

Joined Nov 26, 2014
2
I've got a black box which can deliver high voltage DC pulses connected to a capacitor. As in the below schematicView attachment 225274.

I know what the input power to the black box is as it is powered by a power supply unit which shows the voltage and current.

I know the peak voltage out of the box (measured across HV+ and HV-) the width of the pulse and the frequency of the pulse. The pulse is shaped like:
View attachment 225275

I also know the capacitance of the capacitor. How do I determine the power that the black box delivers to the capacitor in one pulse and the average power in a second? Do I even have enough information? Can I use the normal energy in a capacitor equation or is that only for a constant DC input?

#### Brown Pulliam

Joined Nov 26, 2014
2
I think we need to know more about source impedance of the pulser in the black box, and lacking further information, the pulse shape in 225275 would be as measured open circuit or into some standard impedance, such as 50 ohms. The instantaneous power you are delivering from the black box is E * I = E * C * dv/dt, which will be largest when charging a fully discharged capacitor, dropping as the capacitor voltage increases toward the peak black box voltage. So you have to mathematically integrate the product of the instantaneous voltage times current for the duration of each pulse, not exactly a trivial task.

#### AlbertHall

Joined Jun 4, 2014
10,776
If there is no resistance then where would this power be disipated?

#### Bhope691

Joined Oct 24, 2016
26
Currently measuring the current with a shunt is not possible due to the very small currents and high voltages and using a resistor with a higher resistance and measuring the drop across it effects the output of the black box.

The open circuit voltage (with no capacitor) is higher than with the capacitor. The pulse shape is the same.
Realistically I don't have enough information in order to work out the power delivered by the black box with the above.

If I was to take readings with a resistor in place of the capacitor and measure the power dissipated across that resistor, whilst recording the PSU input power, can I infer that at that particular PSU power the black box will always deliver the same power as if a resistor was the load even if the load was a capacitor? i.e

Can I say that whenever the input is 6W the output delivered by the black box is 0.5W and whenever the input is 8W the output delivered by the black box is 2W? So if the input was 6W with the capacitor the power delivered by the box is 0.5W?

#### LesJones

Joined Jan 8, 2017
2,837
If you know the peak voltage into the capacitor and the value of the capacity you can calculate the energy that was delivered into the capacitor before that energy was discharged back unto the black box. The energy delivered to the capacitor at the end of the pulse is zero. Is this another way of asking the question in another thread relating to some mysterious black box ?

Les.

#### Bhope691

Joined Oct 24, 2016
26
So I can just use E=1/2CV^2 using Vpeak to calculate the energy into the capacitor, and this is what the black box is delivering?

#### LesJones

Joined Jan 8, 2017
2,837
The black box net output is zero as the energy that goes into the capacitor during the first half of the pulse is returned to the black box during the second half of the pulse. What happens if yo connect a diode between the black box and the capacitor ?
Is this another way of asking the question in another thread relating to some mysterious black box ?
Les.

#### Bhope691

Joined Oct 24, 2016
26
Ah OK thanks. I think it may be black box 2.0 but not sure what the question in the other thread was, it was a while ago.

#### BobTPH

Joined Jun 5, 2013
2,727
The black box net output is zero as the energy that goes into the capacitor during the first half of the pulse is returned to the black box during the second half of the pulse. What happens if yo connect a diode between the black box and the capacitor ?
Is this another way of asking the question in another thread relating to some mysterious black box ?
Les.
How can you know that? It is a black box. It might contain a diode in series with the output or a MOSFET that shuts off if current starts flowing back into it.

Bob

#### LesJones

Joined Jan 8, 2017
2,837
If that was the case then there would be no falling part of the pulse. The capacitor would remain charged at the peak value.

Les.

#### crutschow

Joined Mar 14, 2008
25,991
So I can just use E=1/2CV^2 using Vpeak to calculate the energy into the capacitor, and this is what the black box is delivering?
That will calculate the peak energy into the capacitor, but the net energy is zero.

#### Just Another Sparky

Joined Dec 8, 2019
107
Still not sure what the purpose of this circuit is and why determining the power to and/or from the capacitor is even relevant. Without further information it would seem that connecting a capacitor to the output of an unloaded pulsed DC power source is sort of pointless. What sort of system voltages are we even talking here? 100V? 1kV? 10kV+? Charge of the capacitor? Is there a rectifier diode present or are we dealing with alternating current? Is this a tiny little breadboard circuit or are we dealing with utility capacitors?

If the pulse time is sufficiently long and/or the current is sufficiently low given the inductance & resistance of the circuit to ignore the reactive & resistive components, and the HVDC source is suffuciently powerful given the size of the capacitor to negate drooping, you can figure the average power for the rising half of the pulse based on the charge of the capacitor, the peak voltage of the pulse and one half of the period of the pulse. Power = energy over time.

However with any sort of high frequency or current-limited system this is unlikely to yield an accurate result. You'll get impedance and/or droop - which make field measurements the only option without knowing intimate details regarding the HVDC source, capacitor and field wiring. Without knowing total circuit inductance, resistance, pulse period and whether this is an AC or DC system, an accurate model is infeasible. You need subjective field measurements.

Probe the output waveform both loaded and unloaded - that will at least tell you if your source is drooping and give you an accurate peak voltage and period to work with. From there you can calculate energy and power if an accurate capacitance value is known.

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#### BobTPH

Joined Jun 5, 2013
2,727
If that was the case then there would be no falling part of the pulse. The capacitor would remain charged at the peak value.

Les.
That is my point. I susoect the shape if the pulse is open circuit voltage. Who knows, it might even be current.

Bob

#### Bhope691

Joined Oct 24, 2016
26
That is my point. I susoect the shape if the pulse is open circuit voltage. Who knows, it might even be current.

Bob
The shape of the pulse is the same if open circuit but Vpeak is slightly larger by about 20%.

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#### Just Another Sparky

Joined Dec 8, 2019
107
So there is 20% power supply droop going on and we're dealing with non-sinusoidal AC.

If this is a continuous waveform, what is the period of the pulse and how much time is spent between pulses? If the period is between roughly 50-2.5ms (i.e. 20-400Hz equivalent) you can probably measure the RMS voltage using a true-RMS voltmeter fairly accurately. Determine average current based on the capacity rating of the capacitor, measured peak voltage and the frequency (1 ampere of current = 1 columb of charge per second). Multiply RMS voltage and average current to get apparent power within the tolerance rating of your capacitor.

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$I_{AVG}=\frac{C\cdot\ V_{peak}}{t}$