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24th July 2019, 18:18 #1
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charging capacitor partially
I got a silly question to ask.
I am quite new to electronics and these kind of stuff.
I am trying to calculate the value of resistor use for charging 4700uF cap
from 1/3 to 2/3 of the supply voltage(6 volts) I had tried to figure out how to, but I can't find the way.
Is there any equations or ways to solve this problem?
thx for your reply.

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24th July 2019, 20:45 #2
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Re: charging capacitor partially
Vc=Vapplied(1e^(t/RC))
Vc is voltage across the capacitor
t is time.

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25th July 2019, 00:07 #3
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Re: charging capacitor partially
Hi,
There are missing informations.
Simply saying: Any resistor may charge a capacitor from 1/3 to 2/3 of the supply voltage.
Varying the resistor value will vary the time ... and the current.
KlausPlease donīt contact me via PM, because there is no time to respond to them. No friend requests. Thank you.

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25th July 2019, 01:14 #4
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Re: charging capacitor partially
if you want to stop the charging at 2/3 of the applied volts you will need a resistive divider, e.g. 10 ohm and 20 ohm with the 20 ohm across the grounded cap
this will charge from zero to 2/3 of applied V,
Are you trying to design a long time timer ckt for a 555 by any chance ...?
The time to charge from 1/3 to 2/3 is very roughly half a time constant, which is Rcharge x C ( then divide by 2 )

25th July 2019, 06:15 #5
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Re: charging capacitor partially
Hi,
Ln(1/3) = 1.099
Ln(2/3) = 0.405
The difference is 0.694.
Thus ideal charging from 1/3 to 2/3 takes 0.694 x R x C.
Discharging from 2/3 to 1/3 also takes 0.694 x R x C.
KlausPlease donīt contact me via PM, because there is no time to respond to them. No friend requests. Thank you.

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29th July 2019, 09:31 #6
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Re: charging capacitor partially
Thank you for all
"Are you trying to design a long time timer ckt for a 555 by any chance ...?"
I am working on different circuits I can do with 555 but I dont really understand the calculation for capacitor
so I searched for some information in my language but there is not much I could find so I try to search in english
but I can't find a good keyword for it
now I come up with this equation "ln(1V1/Vs)ln(1V2/Vs) = t"
not sure if it works but it still work in all of my case

29th July 2019, 12:49 #7
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Re: charging capacitor partially
from 1/3 to 2/3 of the supply voltage(6 volts) I had tried to figure out how to, but I can't find the way....
If you want to know the time, you must specify the resistor and the capacitor; they decide together the time constant.
If you want to know the resistance value, you must specify the time.
Now for the details. RC (product of R and C) has the dimension of time. In this story, you will notice R and C will always appear together.
During charging, the voltage increases from 0 to V (the final value) but that takes infinite time. Both charging and discharging are exponential process.
33% charging will take RC*log(0.66) s and 66% charging will take RC*log(0.33) s. So the time to charge is RC*0.415s to RC*1.109s.
So you must say the R value or the time.
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