Capacitive discharge

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camino75080
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Capacitive discharge

Post by camino75080 » Sun Sep 05, 2004 9:55 pm

Man, am I on a roll tonight. ;) Anyway a am looking for a formula calculate the current of a capacitive dischare, varibles: voltage, current (duh), and resistance. So far I've got this I=(((C*C)*V*R)/2). Please help me.

bodgy
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Re: Capacitive discharge

Post by bodgy » Mon Sep 06, 2004 2:47 am

From your formula I think you are talking about Joules.<p>W = (C * (V^2)) * 0.5<p>For (dis)charging current the formula is<p>I = C * V /t
OR more correctly<p>Q = C * V<p>I = Q/t<p>Where 't' is the time at a particular instant and Q is the charge in Coulombs.<p>So a 100uF capacitor with 12v across it <p>Q = 1.2mC<p>After 100mS of charging time the current will be<p>I = 1.2mC /100mS = 12mA<p>Colin<p>[ September 06, 2004: Message edited by: bodgy ]</p>
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camino75080
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Re: Capacitive discharge

Post by camino75080 » Mon Sep 06, 2004 7:30 am

Colin, thanks for the formula, i was way off, but I don't understand how that works without having the resistance. If you can, would you please explain that. Thanks again for the help.

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Chris Smith
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Re: Capacitive discharge

Post by Chris Smith » Mon Sep 06, 2004 11:34 am

Resistance is everything. Think of a water tower and many different size pipes either feeding it or discharging it. The bigger the pipe size the less resistance, [the longer it is the more resistance] the faster it fills or empties and Vice Versa. <p>The tank size is the capacity, or capacitance, and the height of the tower is pressure [psi], or volts, while the pipe size is a division of the perfect zero ohms, [which really isn’t possible with out super cooling], so everything is more than Zero. One inch is half of two, one ohm is half of two ohms, etc.

camino75080
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Re: Capacitive discharge

Post by camino75080 » Mon Sep 06, 2004 12:31 pm

Okay, thanks. So if I plugin time time constant for a given resistor/capacitor combination, for S then I get the current :cool: . Thanks again for the help

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MrAl
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Re: Capacitive discharge

Post by MrAl » Mon Sep 06, 2004 6:08 pm

Hi there,<p>If you are using a cap being discharged by
a resistor you can use this 'formula' which
by the way includes the resistance and is exact:<p>i=i0*e^(-t/RC)<p>where
i is the instantaneous current at time t,
i0 is the initial current at time t=0,
t is the time in seconds,
RC is the RC time constant = R*C with
R in ohms and C in farads,<p>To get the initial current i0 knowing the
voltage across the cap Vc simply divide
Vc by the resistance R:
i0=Vc/R
where Vc is the voltage across the cap at time
t=0.<p>Also, the voltage across the cap is:
v=v0*e^(-t/RC)
where v0 is the voltage at time t=0.<p>
Take care,
Al<p>PS. The formulas like dv=i*dt/C are only
accurate for very small time intervals--
'small' relative to the time constant.
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camino75080
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Re: Capacitive discharge

Post by camino75080 » Mon Sep 06, 2004 6:22 pm

Thanks Al, I dint realize there were there were other formulas, I looked on the internet and only for PHD research papers, that sucked, I could make any sense of them!

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jwax
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Re: Capacitive discharge

Post by jwax » Mon Sep 06, 2004 7:46 pm

You'll find that the resistance of an arc discharge in air is a very low value. A thousand volts at a thousand amperes is an ohm, although it's over in a microsecond or so.

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Re: Capacitive discharge

Post by bodgy » Tue Sep 07, 2004 2:12 am

<blockquote><font size="1" face="Verdana, Helvetica, sans-serif">quote:</font><hr>
<p>
Also, the voltage across the cap is:
v=v0*e^(-t/RC)
where v0 is the voltage at time t=0.<p>
<hr></blockquote><p>Of course I shouldn't have been lazy, and should have posted the full formula.<p>But Mr Al shouldn't the voltage across C be -<p>Vc = V0 * [1- e ^(-t/CR)] ?<p>where e is the nat exponent 2.71828<p>Just to add voltage discharge is normally measured at Vc * 0.37 and Current at I * 0.63<p>For charging it is the other way around Vc * 0.63 and I * 0.37<p>Colin
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camino75080
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Re: Capacitive discharge

Post by camino75080 » Tue Sep 07, 2004 10:51 am

Buy chance, I am calculating the current though an arc. What would be a good aproximation of the arc's resistance?

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jwax
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Re: Capacitive discharge

Post by jwax » Tue Sep 07, 2004 3:51 pm

Take a look at N & V magazine, Sept 2004, p.54-"The Anatomy of a Spark".
There can be a leakage path through air from high voltage of milliamps, or a catastrophic breakdown that carries thousands of amps. The energy available determines the arc properties, along with temperature, humidity, electrode shape, etc.
So the "resistance" of the discharge depends on all those factors.

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Chris Smith
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Re: Capacitive discharge

Post by Chris Smith » Tue Sep 07, 2004 6:48 pm

The [over] general value for an arc is 25k/v per inch. This does not include many factors, but is just a figure for Dry air, general over simplistic conditions such as leakage through a break in insulation to ground.

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MrAl
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Re: Capacitive discharge

Post by MrAl » Wed Sep 08, 2004 1:21 am

Hi again,<p>bodgy:<p>The equation:
v=v0*e^(-t/RC)<p>is actually correct for a *discharging* cap,
where
v is the voltage across the cap and
v0 is the initial voltage across the cap
and the resistor is connected right across
the cap so as to discharge it.<p>
If the cap were *charging* then it would be
v=vp*[1-e^(-t/RC)]
where
v is the voltage across the cap and
vp is the power supply voltage (resistor
connects between it and the cap).<p>
Take care,
Al
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camino75080
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Re: Capacitive discharge

Post by camino75080 » Wed Sep 08, 2004 10:22 am

Is the formula for a cap's charging current, I=Io*(1-e^(-t/RC))? Thanks or all the help.<p>[ September 08, 2004: Message edited by: camino75080 ]</p>

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MrAl
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Re: Capacitive discharge

Post by MrAl » Wed Sep 08, 2004 7:04 pm

Hi there,<p>Here are all the formula's with the
following resistor connections:<p>When discharging the cap the resistor is connected across the cap.
When charging the cap the resistor is connected
one lead to the cap and the other lead to the
power supply voltage (vp) and the other end
of the cap is grounded (zero volts).<p>Also, the voltage v is across the cap and
the current i is through the cap.<p>CHARGING VOLTS:
v=vp*[1-e^(-t/RC)]<p>CHARGING AMPS:
i=i0*e^(-t/RC), i0=(vp-vc)/R<p>DISCHARGING VOLTS:
v=v0*e^(-t/RC), v0=initial v across cap<p>DISCHARGING AMPS:
i=i0*e^(-t/RC), i0=initial i through cap<p>Note there is only one formula with the "1" in
it and that is for the voltage when charging.<p>If i get a chance i'll post some graphs if you
think that will help?<p>Take care,
Al
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