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# capacitor logic for providing extra current on short period of time

#### yefj

Hello,i need to drive 5 such mosfets as shown in the simulation 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

How much do you care about the voltage value and pulse flat-top?

Given that you want this capacity from the power supply and caps and no added circuitry the acceptable loaded droop at current for time gives you the bulk C value by the basic equation.

Hello ,current supply is much more important.

Also I have simulated 5 mosfets as shown below my V4 puwer supply in real life can supply only 4A.
https://www.meanwell-web.com/content/files/pdfs/productPdfs/MW/HRPG-150/HRPG-150-spec.pdf
As you can see from the simulation i have 38A going from V4 source.
Is there a way in LTspice to block the amount of current V4 could give and test the amount of current my 10uF capacitor will give in the pulse time?
Thanks.

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?
Thanks.

UPDATE:

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.

--- Updated ---

UPDATE:

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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There is something strange in your simulation. The resistor R1 can just sink some of the current from C1 subtracting it from that going to the MOS. Could you post you LT project ?
In order to calculate the capacitor you shoud write down the equations of the circuit, in any case you could estimate the value using the condition that the drawn current is almos constant over the period you need. In this case we know that the energy stored in the capacitor is:

Es = 0.5*C*V^2

energy is power*time. Then under the simplified hypothesis that I=constant, the required energy over a time period "t" is:

Er = V*I*t

equating now Es = Er and solving with respect to C:

C = 2*I*t/V

in your case I=30A, V=45V, t=2us that is

C= 2*30*2u/45 = 2.7 uF

However in real life you must consider also the output impedance of the generator as well as the ESR of the capacitor.

There is something strange in your simulation. The resistor R1 can just sink some of the current from C1 subtracting it from that going to the MOS. Could you post you LT project ?
In order to calculate the capacitor you shoud write down the equations of the circuit, in any case you could estimate the value using the condition that the drawn current is almos constant over the period you need. In this case we know that the energy stored in the capacitor is:

Es = 0.5*C*V^2

energy is power*time. Then under the simplified hypothesis that I=constant, the required energy over a time period "t" is:

Er = V*I*t

equating now Es = Er and solving with respect to C:

C = 2*I*t/V

in your case I=30A, V=45V, t=2us that is

C= 2*30*2u/45 = 2.7 uF

However in real life you must consider also the output impedance of the generator as well as the ESR of the capacitor.
You are off by at least a factor of 100, try again. Look at energy stored after 2 us and before with 2A charge current and sufficient time to restore 48V will be what duty cycle?

You are off by at least a factor of 100, try again. Look at energy stored after 2 us and before with 2A charge current and sufficient time to restore 48V will be what duty cycle?
Not sure to have correctly understood your comment, however I'm referring to a one shot discharge of a fully charged capacitor. I didn't read about repetitive pulses.

The question was what C permits a 0.5V sag in 2 us from a 30A load (45V/1.5 ohm)?

C = Ic dt/dV = 30A *2e-6s/0.5V = 120 uF ideal

With a 2 Adc supply that requires dt=C*dV/Ic= 120e-6 * 0.5V/2A = 30 us if the current is constant to restore the 0.5V.

A good low ESR Cap is around Tau = 1 us , so 8 mOhm * 30A = 240 mV means you need two Caps. to meet the 500 mV spec. or 240 uF.

so you were off by ~ 100

In the back of my mind the topic seems to be a railgun, coilgun... Or could benefit from techniques used in building such devices. I understand enormous banks of capacitors are required to supply the brief power surge.

The question was what C permits a 0.5V sag in 2 us from a 30A load (45V/1.5 ohm)?

C = Ic dt/dV = 30A *2e-6s/0.5V = 120 uF ideal

With a 2 Adc supply that requires dt=C*dV/Ic= 120e-6 * 0.5V/2A = 30 us if the current is constant to restore the 0.5V.

A good low ESR Cap is around Tau = 1 us , so 8 mOhm * 30A = 240 mV means you need two Caps. to meet the 500 mV spec. or 240 uF.

so you were off by ~ 100
Ok I lost a maximum dV=0.5v was required. Where is written ?

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Ok I lost a maximum dV=0.5v was required. Where is written ?
Post #5

Hello, I want to understand the math of the phenomena using this simple example ....

Suppose I want that my 2u pulse( ON) I'll have only 0.5V drop....
--- Updated ---

Ok I lost a maximum dV=0.5v was required. Where is written ?
Post #5

Hello, I want to understand the math of the phenomena using this simple example .

Suppuse i want that over my 2u pulse( ON I'll have only 0.5V drop

The proper math would be Ic =C dV/dt + Vc/ESR with |dV + Vc|=< 0.5V and ESR * C = Tau for a low ESR e-cap rated for >=60 V then solve for C and examine cost/size and other specs vs ESR.
In this range of C anything 0.2 to 20 us is a low ESR cap

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I don't understand your math (possibly you are considering a different circuit from mine). In any case If you have:

Ic costant current drawn
V voltage across the ideal capacitor
Vc voltage across the series C+ESR

and we suppose negligible the current from the power supply (i.e.: considering just the discharge of the capacitor with constant current Ic) equations are:

Ic = C*dV/dt
Vc = V - Esr*Ic --> V = Vc + ESR*Ic

that is:

ESR*Ic + dV <= 0.5 or ESR*Ic + Ic*dt/C <=0.5

solving with respect to C:

C >= I*dt/(0.5 - ESR*Ic)

of course must be ESR*Ic < 0.5 that is: ESR < 0.5/Ic

I don't understand your math (possibly you are considering a different circuit from mine). In any case If you have:

Ic costant current drawn
V voltage across the ideal capacitor
Vc voltage across the series C+ESR

and we suppose negligible the current from the power supply (i.e.: considering just the discharge of the capacitor with constant current Ic) equations are:

Ic = C*dV/dt
Vc = V - Esr*Ic --> V = Vc + ESR*Ic

that is:

ESR*Ic + dV <= 0.5 or ESR*Ic + Ic*dt/C <=0.5

solving with respect to C:

C >= I*dt/(0.5 - ESR*Ic)

of course must be ESR*Ic < 0.5 that is: ESR < 0.5/Ic
Not quite.
You might suggest 0.5V drop shared equally by ESR*I and I*dt/C=dV to be 0.25 V each though if possible from Tau = 1us my Vc was the same as your ESR*Ic so different assumptions

I would assume the current is constant from 45 to 44.5V

Then my C >= I*dt/(0.5 - ESR*Ic) becomes 0.5V*2e-6*I - ESR*2e-6*I^2= C
With 2 equations for C, ESR , you can now solve my assuming ESR = 1e-6/C approx.
--- Updated ---

It's a non-linear quadratic equation, so iterations of assumptions are needed. But my answer of two 120 uF caps should be correct. Copilot took me 8 attempts to correct its assumptions and could not solve it correctly or verify the results ( I just tried asking Bing.)

So maybe in GPT 6 or 7 will be as good as a graduate EE with a few years of experience.
The errors were in its assumptions and verification even when told what to assume and verify.

e.g.

Copilot
I apologize for the confusion. Let’s verify the assumptions and calculate the correct values for ESR and C, ensuring that the voltage drop does not exceed 0.5V with a -20% tolerance, meaning the actual voltage drop can be up to 0.4V.

Given:
• The load current ( I ) is 30A.
• The maximum allowable voltage drop ( \Delta V ) is 0.4V.
• The discharge time ( \Delta t ) is 2us (or ( 2 \times 10^{-6} ) seconds).
• The product of ESR and C is ( 1 \times 10^{-6} ) seconds.
We need to find the values of ESR and C that satisfy the equation:
ΔV=I×ESR+I×ΔtCΔV=I×ESR+CI×Δt
Substituting ( ESR \times C = 1 \times 10^{-6} ) seconds into the equation, we get:
0.4V=30A×ESR+30A×2×10−6sC0.4V=30A×ESR+C30A×2×10−6s
0.4V=30A×ESR+60A×10−6sC0.4V=30A×ESR+C60A×10−6s
0.4V=30A×ESR+60V×ESR0.4V=30A×ESR+60V×ESR
0.4V=ESR×(30A+60V)0.4V=ESR×(30A+60V)
Now, we solve for ESR:
ESR=0.4V30A+60VESR=30A+60V0.4V
ESR=0.4V30A+60×ACESR=30A+60×CA0.4V
ESR=0.4V30A+60×A1×10−6sESRESR=30A+60×ESR1×10−6sA0.4V
ESR=0.4V30A+60×A×ESR1×10−6sESR=30A+60×1×10−6sA×ESR0.4V
ESR=0.4V30A+60A×ESR1×10−6sESR=30A+1×10−6s60A×ESR0.4V
This is a nonlinear equation in ESR, and to solve for ESR, we would typically use numerical methods or iterative approaches. However, since we are looking for an analytical solution, let’s make an initial assumption that the voltage drop across the ESR is much smaller than the voltage drop across the capacitor. This would imply that ( 30A \times ESR) is much smaller than ( 0.4V ), allowing us to neglect the ( 30A \times ESR ) term in the denominator for the initial estimation:
ESR≈0.4V60×A×ESR1×10−6sESR≈60×1×10−6sA×ESR0.4V
ESR2≈0.4V×1×10−6s60AESR2≈60A0.4V×1×10−6s
ESR2≈0.4×10−6V⋅s60AESR2≈60A0.4×10−6V⋅s
ESR2≈0.4×10−660Ω2ESR2≈600.4×10−6Ω2
ESR≈0.4×10−660ΩESR≈600.4×10−6Ω
ESR≈0.460×106ΩESR≈60×1060.4Ω
ESR≈1150×106ΩESR≈150×1061Ω
ESR≈1150×106ΩESR≈150×1061Ω
ESR≈112,247ΩESR≈12,2471Ω
ESR≈81.6μΩESR≈81.6μΩ
With this estimated ESR, we can now calculate the capacitance ( C ):
C=1×10−6sESRC=ESR1×10−6s
C=1×10−6s81.6×10−6ΩC=81.6×10−6Ω1×10−6s

C≈12,247μC≈12,247μ
(end Copilot) 12 mF wow

This would be a very expensive solution with a low SRF and overkill on sag.

Two 120 uF caps with 8 or 9 mohms each should work.

1st find is close but no cigar. 12mOhm 63V 120 uF polymer elect.

Three 100 uF 10mohm 63V alum polymer caps are the best that Digikey has , but elsewhere there are better ones in DE.land for this hypothetical question.

To satisfy the equations, these are cheap and the current is only 28A with 2 from the supply.
But this low ESR might make the Banggood supply unstable for step response overshoot, so never make assumptions without testing.

Conclusion

1. one must know the properties of ultralow ESR caps 1~2 us in this range of alum caps
2. ΔV=I*ESR+C*I*Δt is the solution but this leads to a non-linear quadratic
3. Assuming 250 mV for I=CdV/dt and ESR*I each leads to problems as the ESR is harder to meet than C, so multiple caps tend to lead to better solutions, which is commonplace as you may have seen. But it might be better to assume 300 mV for ESR drop and 200 mV for sag.
Using Runga Cutta or other iterative solutions can also work, but experience beats Chat GPT4 after telling 8 times how to solve it before it gave up.

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