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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...
#2: Question closed
by
Olin Lathrop
·
2024-05-14T11:29:52Z (11 months ago)
#1: Initial revision
by
yefj
·
2024-05-14T07:26:48Z (11 months ago)
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   