# Pulsed capacitor calculation for required capacitance

#### SiCEngineer

Joined May 22, 2019
444
Hi All,

I am looking through the website here: https://www.microwaves101.com/encyclopedias/charge-storage-capacitors which goes throuugh requiredd capacitance for a certain level of droop for a given pulse width for pulsed power supply. The trouble I am having is for my -6kV output 10mA 5mS pulse, the required capacitance comes out at 10uF. This is quite high for a 12kV capacitor that I would need to buy for reinforced isolation.

I think the issue is that the equation does not take into account the ability of a switched mode power supply to increase the current supplied to the capacitor to raise its voltage and reduce the effect of the droop.

Does anyone have a better way of calculating the required amount of holdup capacitance for a power supply pulsed system, considering a minimum and maximum pulse duration/frequency?

SIC

#### KeithWalker

Joined Jul 10, 2017
3,178
You can reduce the capacitance needed by increasing the switching frequency.
Regards,
Keith

#### SiCEngineer

Joined May 22, 2019
444
You can reduce the capacitance needed by increasing the switching frequency.
Regards,
Keith
Yes I understand that. The thing I have sent in question shows that delta T is the pulse width of the pulsed output. There is no dependency on switching frequency there. My supply switches at 300kHz.

I need to know how to calculate the required hold up capacitance for the pulsed load to specify a maximum voltage droop on the capacitor.

#### Marc Sugrue

Joined Jan 19, 2018
222
Yes I understand that. The thing I have sent in question shows that delta T is the pulse width of the pulsed output. There is no dependency on switching frequency there. My supply switches at 300kHz.

I need to know how to calculate the required hold up capacitance for the pulsed load to specify a maximum voltage droop on the capacitor.
C = (I x dt) / dv

dt is required holdup duration
dv is the voltage change in the duration

#### SiCEngineer

Joined May 22, 2019
444
C = (I x dt) / dv

dt is required holdup duration
dv is the voltage change in the duration
Yes, like I say I understand that and it is in the equations that I attached to the question. The issue is that gives me 10uF for my 6kV output which is not at all realistic. The equations do not take into account the ability of the supply to provide current and charge to the capacitors during the pulse etc. My converter operating at 300kHz will probably not need 10uF of capacitance on the 6kV and it will be even more on the 3kV since the current is much higher on that output.
So although this is a basic calculation for hold up capacitance it is yielding values that I believe to be way too high for my application. The energy storage on a 10uF capacitance at this voltage will be very big.

#### Marc Sugrue

Joined Jan 19, 2018
222
The maths dictate if you don’t like the answer you will have choose a variable To change. As post number 2 Already suggests frequency being the obvious candidate or Consider relaxing your requirement to allow more ripple. Or add inductance to support the current in the form of an LC filter

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#### SiCEngineer

Joined May 22, 2019
444
The maths dictate if you don’t like the answer you will have choose a variable To change. As post number 2 Already suggests frequency being the obvious candidate or Consider relaxing your requirement to allow more ripple. Or add inductance to support the current.
yes, I understand that, but frequency is not a function in the equations that I have presented. The pulse duration is T and that equation would then dictate that the size of the capacitor will not be affected by changing the frequency, which as I have said, does not seem right.

#### Marc Sugrue

Joined Jan 19, 2018
222
yes, I understand that, but frequency is not a function in the equations that I have presented. The pulse duration is T and that equation would then dictate that the size of the capacitor will not be affected by changing the frequency, which as I have said, does not seem right.
Period T = 1/F

If you have to hold up for example 50% of your duty cycle the Tholdup would be 50% of T

#### SiCEngineer

Joined May 22, 2019
444
Period T = 1/F

If you have to hold up for example 50% of your duty cycle the Tholdup would be 50% of T
yes, so assuming a 5ms pulse and 200mA, the pulse capacitor should be 10uF. If I remember correctly. Again that is not taking into account the supplies ability to supply charge when the voltage drops at the output..?

#### Marc Sugrue

Joined Jan 19, 2018
222
yes, so assuming a 5ms pulse and 200mA, the pulse capacitor should be 10uF. If I remember correctly. Again that is not taking into account the supplies ability to supply charge when the voltage drops at the output..?
When you say pulse i assume you mean the hold up period.

So assuming you defined that you could allow 100v dip in your line

200mA•5e-3/100 would be 10uF

1 reduce holdup time
2 allow larger dip in voltage
3. Use the capacitance required to achieve your spec. Maybe paralleling parts if needed

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#### SiCEngineer

Joined May 22, 2019
444
When you say pulse i assume you mean the hold up period.

So assuming you defined that you could allow 100v dip in your line

200mA•5e-3/100 would be 10uF

1 reduce holdup time
2 allow larger dip in voltage
You are correct, assuming that the supply is turned off during the pulse and is unable to supply any voltage to account for the droop...

#### RPLaJeunesse

Joined Jul 29, 2018
255
I don't see a "simple" equation to handle this situation. The issue at hand is what is the behavior of the power supply during (and after!) the 5ms pulse load. That might be predictable from the loop response of the supply regulator, which in turn will be somewhat affected by the added capacitance. Unfortunately I'm not enough of a supply expert to provide a solid answer. It might be possible to simulate a theoretical supply having the response charecteristics of the actual supply, and observe the results with a selection of capacitors and various width pulse loads up the 5ms goal. Alternatively, build a scaled down version, perhaps at 50V, and play with that more safely on the bench.