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Understanding the RC circuit equation
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AndrewHodgson:
This is a pretty good explanation:


If you want take into account the initial voltage across the capacitor:

rearrange that equation so t is the subject
t=RC*ln( 1-Vc/Vs)

then say you want to know how long it takes you to get from 1V to 4V with a supply voltage of 5V. you say
t= RC(1-4/5)-RC*ln(1-1/5)

so you're subtracting the time it takes to get to 1V.

Make sense?
AndrewHodgson:
Also, if you want to take account of the initial capacitor voltage in your derivation you just integrate between t1 and t2, and q1 and q2
Azmarith:

--- Quote from: AndrewHodgson on November 16, 2019, 07:03:07 pm ---This is a pretty good explanation:


If you want take into account the initial voltage across the capacitor:

rearrange that equation so t is the subject
t=RC*ln( 1-Vc/Vs)

then say you want to know how long it takes you to get from 1V to 4V with a supply voltage of 5V. you say
t= RC(1-4/5)-RC*ln(1-1/5)

so you're subtracting the time it takes to get to 1V.

Make sense?

--- End quote ---


--- Quote from: AndrewHodgson on November 16, 2019, 07:44:23 pm ---Also, if you want to take account of the initial capacitor voltage in your derivation you just integrate between t1 and t2, and q1 and q2

--- End quote ---

Thank you so much guys! I see where I was going wrong now. I wasn't integrating between t1 and t2, and q1 and q2. I was integrating and just adding on a constant which I assumed was the starting voltage/charge.

Thank you all so much for you help  ;D
bson:
As has been pointed out, just subtract the initial voltage from the supply voltage.  Charging a capacitor from 2 to 5V with a 5V supply is exactly the same as charging it from 0 to 3V with a 3V supply.
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