Electronic Question

[rova]

Just wondering if anyone knows the answers, or where i can learn more about
this,

1.why do you use a resistor when charing a capacitor?

2.How do you add tolerances in a circuit?


thanks
rova

 

[scada]

"rova" <rova@optonlineNOSPAM.net> wrote in message
news:esIhb.49355$Ri4.19594429@news4.srv.hcvlny.cv.net...

Just wondering if anyone knows the answers, or where i can learn more
about
this,

1.why do you use a resistor when charing a capacitor?

Time constant = R*C
IE: 10,000 ohms * .000001(1 micro farad) = 0.01 seconds

2.How do you add tolerances in a circuit?

Take your worse case condition that you will allow, caculate your

components for that condition, both high and low, then select values that
will put you somewhere in the middle.

thanks
rova

 

[Russell Powell]

Starting with the definition of capacitance as being the
proportionality constant between the magnitude of charge
on either conductor and the potential difference between them.

q = CV,

we get current as
dq/dt = C(dv/dt) or i = C dv/dt.

From this we see i will go to infinity as dt approaches zero.
Hence a capacitor is charged through a resistor to prevent this.

Adding a resistor in series and following the loop theorm, we
get
v - iR - q/C = 0 ==> v = R(dq/dt) + q/C

solving the diff eq. gives q = Cv(1-e^^(-t/R*C))
and differentiaiating w.r.t time gives dq/dt (= i) = (v/R)*e^^(-t/R*C).

Again, we see at t=0, i = (v/R). As R-->0, i-->gets big. Also note that
as t-->infinity, q-->CV (as we postulated earlier).


--
sincerely,

Russell Powell

Artisan Components
SR. FAE - U.S. Central
rpowell@artisan.com
469-438-6589





"rova" <rova@optonlineNOSPAM.net> wrote in message
news:esIhb.49355$Ri4.19594429@news4.srv.hcvlny.cv.net...

Just wondering if anyone knows the answers, or where i can learn more
about
this,

1.why do you use a resistor when charing a capacitor?

2.How do you add tolerances in a circuit?


thanks
rova

 

[Overlord]

Wow, lotta math.....
When I really want to char a capacitor, I go with the high voltage AC
and no resistor<G>.


On Sun, 12 Oct 2003 16:33:27 GMT, "Russell Powell"
<powell270@comcast.net> wrote:

Starting with the definition of capacitance as being the
proportionality constant between the magnitude of charge
on either conductor and the potential difference between them.

q = CV,

we get current as
dq/dt = C(dv/dt) or i = C dv/dt.

From this we see i will go to infinity as dt approaches zero.
Hence a capacitor is charged through a resistor to prevent this.

Adding a resistor in series and following the loop theorm, we
get
v - iR - q/C = 0 ==> v = R(dq/dt) + q/C

solving the diff eq. gives q = Cv(1-e^^(-t/R*C))
and differentiaiating w.r.t time gives dq/dt (= i) = (v/R)*e^^(-t/R*C).

Again, we see at t=0, i = (v/R). As R-->0, i-->gets big. Also note that
as t-->infinity, q-->CV (as we postulated earlier).


--
sincerely,

Russell Powell

Artisan Components
SR. FAE - U.S. Central
rpowell@artisan.com
469-438-6589





"rova" <rova@optonlineNOSPAM.net> wrote in message
news:esIhb.49355$Ri4.19594429@news4.srv.hcvlny.cv.net...
Just wondering if anyone knows the answers, or where i can learn more
about
this,

1.why do you use a resistor when charing a capacitor?

2.How do you add tolerances in a circuit?


thanks
rova

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