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Q&A discharge slope logic (fixed sloppinees question) [duplicate]

Hello, I want to understand the math of the phenomena of capacitor as temporarary means of increasing the current supply. First I have created a current limited voltage source which give me 45V a...

0 answers · posted 7mo ago by yefj‭ · closed as duplicate 7mo ago by Olin Lathrop‭

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#2: Question closed by

Olin Lathrop‭ · 2024-05-14T11:29:52Z (7 months ago)

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#1: Initial revision by

yefj‭ · 2024-05-14T07:26:48Z (7 months ago)

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discharge slope logic (fixed sloppinees question)

Hello, I want to understand the math of the phenomena of capacitor as temporarary means of increasing the current supply. \
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 to be as small as possible. \
LTspice file simulation is attached in the link
Thanks.
https://file.io/pyYz2mG0nCGa
![Image_alt_text](https://electrical.codidact.com/uploads/auesjarqy0yclvmawkjw15o16ic3)
![Image_alt_text](https://electrical.codidact.com/uploads/lvdj4lomv0f622xl4u5qljq2o689)

![Image_alt_text](https://electrical.codidact.com/uploads/hxhefo4t09rxyhvhs5pjl0ql901c)

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