Re: signal and capacitor
From: Terry Given (my_name_at_ieee.org)
Date: 03/04/05
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Date: Fri, 04 Mar 2005 18:31:36 +1300
jason wrote:
> Hello All
>
> I wish to know what will happen for the case of 1)DC and 2)AC input
> power or signal to the capacitor in a circuit.
> I heard that capcitor can act as a short or open circuit depending on
> DC or AC(in high or low frequency input power signal or signal itself).
>
> Can anyone help to explains this clearer in the point of view from
> physics and electronics?
>
> Kindly share
> Thank you
>
> rgds and thanks
> Jason
>
The maths behind a capacitor is simply i = C*dV/dt.
If the voltage truly is DC, then its rate-of-change (dV/dt) is zero, so
the current flowing into the cap is therefore zero. Given that Z=V/I,
the impedance at DC is infinite, ie the cap is an open circuit.
However that is only true once the cap has charged to V. Assuming it
started out uncharged (a valid assumption, as before it was fabricated
it wasnt a capacitor at all) then some current *must* flow into the cap
to charge it. How much? well, it depends on the rate-of-change of the
applied voltage. If you charge it with say a 12V supply, ramped from 0
to 12V in say 1ms then dV/dt = 12V/1ms = 12,000 V/s. If its a 1000uF
cap, then I = 1000uF * 12,000V/us = 12A.
OTOH if you ramped the voltage up to 12V in sa 1us, then dV/dt =
12,000,000V/us so I = 12,000A. To test this, get a 1000uF cap and slap a
car battery across it. See the big spark (and probably weld the damn
lead onto the cap).
in AC circuits the voltage is assumed to be sinusoidal. If its a
periodic signal of any arbitrarily stupid shape, it can be fabricated
from the sum of a whole bunch of sinusoids of differing amplitude, phase
and frequency. In this case, repeat the analysis for *each* sinusoid,
and sum the result.
if V=Vpk*sin*(wt) then dV/dt = w*Vpk*cos(wt)
so I = C*w*Vpk*cos(wt) = Ipk*cos(wt) where Ipk = C*w*Vpk
cos(x) is just sin(x) rotated (shifted) 90 degrees. mathematically this
is represented by the imaginary number j=sqrt(-1) (Euler gets the blame
for this) so we can write
I = C*j*w*Vpk*sin(wt)
and Z = V/I = 1/(C*j*w) = -j/(C*w)
the magnitude of this impedance is |Z| = 1/(w*C)
where w = 2*pi*f, f = frequency
so |Z| = 1/(2*pi*f*C), the familiar result.
our 1000uF cap has an impedance of 1/(2*3.14159*50*0.001) = 3.18 Ohms at
f = 50Hz. with a 230Vac 50Hz voltage, V = 325*sin(314*t)
so Ipk = 325V/3.18Ohms = 102.19A peak, or 72.26Arms
at f = 50kHz, V = 230Vrms the impedance is 3.18mOhms, so I = 72,260Arms
as you can see, the cap looks more and more like a short circuit as F
increases. You can also see why the line-to-line caps you find in
appliances tend to be < 1uF.
re-do the math for 50Hz, 1uF. |Z| = 3.18kOhms, Irms = 72.26mArms.
HTH
Cheers
Terry
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