# Capacitor charge time question

Discussion in 'The Projects Forum' started by Fobio Design, May 31, 2016.

1. ### Fobio Design Thread Starter New Member

May 11, 2016
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Hello-

Thanks for taking the time to read my thread. I have a simple question (I think). My question is in regards to how long it will take for a 10uF capacitor to charge when being powered by a 9V 600mAH?

Thanks!

2. ### #12 Expert

Nov 30, 2010
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With no resistance involved, the answer is zero.
Assuming the 9v source has some internal resistance, the answer is:
Vcap = 9v e^(-time /RC)

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3. ### crutschow Expert

Mar 14, 2008
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3,362
The charge time of a capacitor depends upon the resistance in series with the capacitor (the RC time-constant).
So if you have no added resistor then it depends mainly on the resistance of the battery, the wire, and switch connecting the battery to the capacitor, as well as the capacitor ESR.

The 600 mAH is the battery capacity and has nothing directly to do with the capacitor charge time.

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4. ### dl324 Distinguished Member

Mar 30, 2015
3,376
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Using tex:
$\small V_f = V_ie^{\frac{-t}{RC}}$

The capacitor will be completely charged in 5 RC time constants.

EDIT: As has been pointed out, this is the formula for a capacitor discharging.

Last edited: Jun 1, 2016
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5. ### tcmtech Well-Known Member

Nov 4, 2013
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For the lazy people this is easier and more satisfying.

Calculators for everythign you cna think of and more. http://mustcalculate.com/

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6. ### dl324 Distinguished Member

Mar 30, 2015
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For the R and C in question, less than a microsecond.

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7. ### shortbus AAC Fanatic!

Sep 30, 2009
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Not to hijack the thread, but a similar question on the subject. How far away from the capacitor, in device terms not distance, does a resistor effect the RC time? Say a linear power supply with a current limiting resistor connected to a mosfet acting as a switch connected to the capacitor. Would the current resistor or the RDSon of the mosfet be considered the "R" in the RC formula, or both?

8. ### dl324 Distinguished Member

Mar 30, 2015
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It's the sum of all resistances in series with the cap.

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9. ### shortbus AAC Fanatic!

Sep 30, 2009
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Thank you for clearing that up for me.

10. ### crutschow Expert

Mar 14, 2008
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And if it's a low resistance you may need to include the capacitor ESR.

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11. ### Roderick Young Member

Feb 22, 2015
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Since the question says 600 mAh, it sounds like the source might be a rechargeable 9V battery? It will have an internal resistance, as others have mentioned. If it's a good NiMH, that could be as low as 0.1 ohm. As DL324 says, most engineers consider the capacitor to be fully charged after 5 time constants.

Example:

R = 0.1 ohm
C = 1 uF = 1 x 10^-6 F
Time constant = RC = 0.11 x 1 x 10^-6 = 0.1 uS
5 time constants = 5 uS.
So about 0.5 uS to fully charge.

If you use another kind of 9V battery, the time will likely be longer. This may help http://www.learningaboutelectronics.com/Articles/Battery-internal-resistance

12. ### crutschow Expert

Mar 14, 2008
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3,362
That equation is for discharging a capacitor through a resistor.
For charging a capacitor the equation is:

where Vc is the capacitor voltage and Vs is the source voltage.

As to when it the capacitor is fully charged, it depends on how you define "completely".
It will charge to 99.3% of the final value in 5 time constants.
In 10 time constants it charges to 99.95% of the final value.
Worst-case I suppose you could say it's completely charged when the capacitor voltage is within ±1 electron of equaling the source voltage.
I'll leave that time calculation as an exercise for the reader.
Hint: I think it something over 40 time constants.

Last edited: Jun 1, 2016
13. ### dl324 Distinguished Member

Mar 30, 2015
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Thanks for point that out. I wasn't thinking and just used information from an earlier post.

14. ### wayneh Expert

Sep 9, 2010
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For what it's worth, 3 time constants often is enough to get within error ranges of being "fully" charged or discharged. The closer you look, the longer it takes.