capacitor logic for providing extra current on short period of time

Thread Starter

yef smith

Joined Aug 2, 2020
841
Hello,i need to drive 5 such mosfets as shown below.
Each mosfet draws 15A when the pulse opens the mosfet.
My power supply is gives the 48V but only 3.3A,If I understand correctly I need to put very high decoupling capacitors at the voltage source footprint so I will have in that short period of time the 5*15=75A I need.
Is there a theory which could explain me the logic of this trick?
What kind of capacitors do you recommend me to use?
Thanks.

https://www.meanwell-web.com/content/files/pdfs/productPdfs/MW/HRPG-150/HRPG-150-spec.pdf
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Last edited:

ericgibbs

Joined Jan 29, 2010
19,135
Hi yef,
The 48V supply discharge at:
"Secs to DisCharge to Vcap = (R*C) * log(Vsupply/(Vsupply-Vtimed))/log(e)"

What is the time period you want and what level of drop of the 48V is acceptable.?
E
 

Thread Starter

yef smith

Joined Aug 2, 2020
841
Hello , There is I=CdV/dt formula my pulse is width is 500ns rise and fall times are 30ns.
so at these 500ns the current that will flow must be 75A.

could you show some manual or the formula in a document that shows this formula?
Secs to DisCharge to Vcap = (R*C) * log(Vsupply/(Vsupply-Vtimed))/log(e)
Thanks.
 

Thread Starter

yef smith

Joined Aug 2, 2020
841
Hello Eric,sorry i made a mistake with the pulse width.
pulse width is 10us.
Ijust want that during the mosfet will be on that the voltage will be kept 45V and current is 75A.
What kind of discharge is it?
Thanks.
 

ericgibbs

Joined Jan 29, 2010
19,135
Hi yef.
This is using a 10,000uF with an theroretical ESR of 0R
The two small images show the Voltage and Current drop over 10uSec period.
E
EG 1466.pngEG 1467.pngEG 1468.png
 

Thread Starter

yef smith

Joined Aug 2, 2020
841

Thread Starter

yef smith

Joined Aug 2, 2020
841
Hello, I have created such current limited voltage source as shown below.
I have added a 10uF capacitor as shown below.
However i cant see that the capacitor creates a current source about the limited 4A.
My mosfets get zero current.
Where did i go wrong in the simulation?
LTspice files are attached,
Thanks.
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Attachments

Last edited:

dovo

Joined Dec 12, 2019
72
Requirements
Nominal voltage: 48V
Current: 75A
Period: 7 us
Voltage droop ? That is the missing parameter.

How to calculate the capacitance
The formula to use is i = Cdv/dt, i in amps, C in Farads and dv/dt in volts/second

Rearrange this and we have C = i/(dv/dt)

Let's say you are happy with 1 volt of capacitor discharge (droop) over the 7 us discharge period. Plugging these numbers into the formula we have C = 75A/(1V/7us) = 525 uF.

Add to this, as others here said, the voltage drop across the capacitor ESR (Equivalent Series Resistance). If 1V works for you here the ESR must be 1V/75A = 13 milliohms or less.

Provide the allowable voltage at the beginning and at the end of the 7 us current pulse and I will help you find a suitable capacitor, or capacitors. I will also explain how much circuit inductance can be tolerated.
 

ericgibbs

Joined Jan 29, 2010
19,135
Hi yef
Your .inc UCC5304.txt is required to run that asc file.

When choosing your capacitor value, consider the method of using several capacitors in parallel to give the total capacitance required.
By using very low ESR caps an overall low ESR can be achieved.

E
 

Thread Starter

yef smith

Joined Aug 2, 2020
841
Hello, I want to understand the math of the phenomena using this simple example .
First I have created a current limited voltage source which give me 45V and current limit of 4A at most.
Without the limitation i would have current and voltage on my 1.5Ohm load as shown below in the NO LIMITATION photo.
But then i plug the limited current source (45V 4A) I see both the current and voltage go down very rapidly.
The slope is too high.
Suppuse i want that over my 2u pulse( ON I'll have only 0.5V drop and current also drop by 0.1A.
I=C*dV/dt =10uF*(0.5V)/(2usec)=2.5A
I dont understand this formula result what is the meaning of 2.5A in my pulse current plot?
I want my slope tobe as small as possible.
LTspice file simulation is attached.
Thanks.

1715669273831.png.


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Attachments

Thread Starter

yef smith

Joined Aug 2, 2020
841
Hello , i put a resistor in parralel to the capacitor as shown below.
I get a pulse of viltage and current on the pulse .
The higher the ressistor the higher the higher voltage i get on my load resistor.
How the capacitor and resistor shown below play together .
I want to have a 45V and 30A on my load resistor over 2u pulse when its on.
Thanks.
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Thread Starter

yef smith

Joined Aug 2, 2020
841
Hello Eric, Yes exactly, I have a 45V source 4A current limit and i need some how during the 2u seconds when the pulse is on to have maximally stead 45V on the R6 and the more important is to have 30A threw R6.
I see that the capitor make the current jump much higher then the current limit of the supply.
But how do i control the slope?
Thanks.
 
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