Capacitor charge time question

Thread Starter

Fobio Design

Joined May 11, 2016
17
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!
 

#12

Joined Nov 30, 2010
18,224
With no resistance involved, the answer is zero.
Assuming the 9v source has some internal resistance, the answer is:
Vcap = 9v e^(-time /RC)
 

crutschow

Joined Mar 14, 2008
34,285
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.
 

dl324

Joined Mar 30, 2015
16,846
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:

shortbus

Joined Sep 30, 2009
10,045
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.
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?
 

dl324

Joined Mar 30, 2015
16,846
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?
It's the sum of all resistances in series with the cap.
 

Roderick Young

Joined Feb 22, 2015
408
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
 

crutschow

Joined Mar 14, 2008
34,285
Using tex:
\( \small V_f = V_ie^{\frac{-t}{RC}} \)

The capacitor will be completely charged in 5 RC time constants.
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. :rolleyes:
Hint: I think it something over 40 time constants.
 
Last edited:

wayneh

Joined Sep 9, 2010
17,496
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.
 
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