All About Circuits Forum capacitor current draw
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#1
01-06-2013, 07:55 PM
 lokeycmos Senior Member Join Date: Apr 2009 Posts: 432
capacitor current draw

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
01-06-2013, 08:07 PM
 StayatHomeElectronics Senior Member Join Date: Sep 2008 Location: NE Ohio, USA (GMT - 5) Posts: 787

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.
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#3
01-06-2013, 08:24 PM
 lokeycmos Senior Member Join Date: Apr 2009 Posts: 432

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
01-06-2013, 10:15 PM
 Jaguarjoe Senior Member Join Date: Apr 2010 Posts: 770

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
01-06-2013, 10:25 PM
 shortbus Senior Member Join Date: Sep 2009 Location: ohio,usa Posts: 2,777

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.
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#6
01-06-2013, 11:42 PM
 kubeek Senior Member Join Date: Sep 2005 Location: Prague, Czech Rep Posts: 3,510

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
01-07-2013, 12:23 AM
 WBahn Senior Member Join Date: Apr 2012 Location: Larkspur, Colorado Posts: 8,077 Blog Entries: 9

Quote:
 Originally Posted by Jaguarjoe 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.
Check your units. QT does not yield units of current.
#8
01-07-2013, 02:03 AM
 Ron H E-book Developer Join Date: Apr 2005 Location: Idaho, USA (GMT-7) Posts: 7,050

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
01-07-2013, 03:37 AM
 Jaguarjoe Senior Member Join Date: Apr 2010 Posts: 770

Quote:
 Originally Posted by WBahn Check your units. QT does not yield units of current.
Yep, you're 110% right. I multiplied instead of dividing: I = Q/T
#10
01-07-2013, 03:47 PM
 Audioguru Banned Join Date: Dec 2007 Location: Ontario, Canada Posts: 9,411

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.

 Tags capacitor, current, draw

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