# capacitor current draw

Discussion in 'General Electronics Chat' started by lokeycmos, Jan 6, 2013.

1. ### lokeycmos Thread Starter Active Member

Apr 3, 2009
432
7
Given a source voltage, and a capacitor value, is there a formula that will show the maximum current that the capacitor will draw until it is fully charged?

a very large value capacitor will suck a lot of current and take some time to charge. whereas a very small capacitor will draw less current and charge much faster. this current draw is what im solving for. thank you!

2. ### StayatHomeElectronics Well-Known Member

Sep 25, 2008
864
40
Add some resistance between the voltage source and the capacitor to limit the current.

As you first connect the source to the circuit, the capacitor looks like a short so the maximum current will be determined by the source voltage and the resistor, I = V / R.

Current will then exponentially decay to zero as the capacitor charges.

3. ### lokeycmos Thread Starter Active Member

Apr 3, 2009
432
7
my issue is trying to find a proper gate resistor for a IRFP460 mosfet. the input capacitance on the mosfet is about 4200pf. the cmos ic that is driving it can only drive 8.8mA with 15v to VCC(although im using 12v vcc). if i do ohms law 12v/.0088= 1363 ohms. however mosfets are voltage controlled and usually have quite small gate resistors. i would imagine that 4200pf wouldnt draw very much at all and would work with a small resistor, but at the same time i dont want to overdraw on the IC and burn it out. btw, it is a 4047 multivibrator and im only working with about 60 hz on the gate. please help me, im lost!

4. ### Jaguarjoe Active Member

Apr 7, 2010
770
90
At those low currents and high gate capacitance it will take quite a bit of time to fully turn on the mosfet. It will spend lots of time in it's linear region and generate a lot of heat.
If you can find the gate "charge" (Q) probably defined in "nC" (nanocoulombs), in the data sheet, then use I = QT to determine the current it wants.

5. ### shortbus AAC Fanatic!

Sep 30, 2009
3,908
1,337
Most times the reason for a gate resistor is only to prevent 'ringing' (oscillation) of the gate. You want the gate to charge as fast as possible to prevent heat in the mosfet. Use a gate driver between gate and the 4047 for best results. A driver will give more amperage to switch the mosfet faster.

6. ### kubeek AAC Fanatic!

Sep 20, 2005
4,587
775
Well, U=I*t/C so with 4.2nF and theoretically 8.8mA all the way from 0 to 12V it takes 5.7us to get to 12V. With the IC you should have some resistor in series to limit the current so that the IC can take if for long time, and the turn-on time will be about 3 times longer. Use a dedicated mosfet driver if you want to switch it more than few times a second, for 60Hz you will probably be ok, but this depends on the load you´re switching.

7. ### WBahn Moderator

Mar 31, 2012
17,332
4,662
Check your units. QT does not yield units of current.

8. ### Ron H AAC Fanatic!

Apr 14, 2005
7,050
655
If you are worried about the power dissipated by the 4047, the power required to charge and discharge a capacitor is
P=f*C*V^2
P=60*4.2e-9*12*12
p=36.3μw
Since you are driving 2 MOSFETs, the total power dissipated by the IC (due to driving the MOSFETs) is 73uW.

You don't have to worry about the IC. However, as others have mentioned, the MOSFET dissipation will be reduced if you used high current drivers for the MOSFETs, but at 60Hz, the savings will probably not be worth it.

9. ### Jaguarjoe Active Member

Apr 7, 2010
770
90
Yep, you're 110% right. I multiplied instead of dividing: I = Q/T

10. ### Audioguru New Member

Dec 20, 2007
9,411
895
Texas Instruments show that the typical output current of a CD4047 with a 10V supply is 10mA into a short circuit when the capacitor is beginning to charge and is 7.5mA when the capacitor is charged to half the supply voltage. The currents will be a little higher with a 12V supply.
The current is very low so the current-limiting resistor can be a small value (10 ohms to 47 ohms) to keep the Mosfet from oscillating.