RC Time Constant

One word: Thevinin.

no

Obviously the capacitor is only

yes, with the proviso you have to recalculate the impedance the cap sees

- in your specific case with r1 = r2 then ri = r1/2

I've been tossing this one around in my head all afternoon

afternoon

Replace the voltage source and resistor divider with its Thevenin equivalent. You now have a single voltage and resistor in series with the capacitor, for which you can easily work out the time constant as usual.

Given equal resistors of value R, the cap charges to V/2, and the time constant is R*C/2.

The capacitor "sees" a thevenin equivalent of V/2 at a resistance of R/2.

The initial voltage slope, the dV/dT at turnon, is unchanged by adding the bottom resistor.

John

Since the real world doesn't run on 1/e, I prefer to remember a generalized expression...

V = (VI-VF)e^(-t/TAU)+VF

VI = Vinitial (starting point), VF = Vfinal (charging toward)

Plug the threshold for your "tripping" device as "V", and you can calculate TAU needed for a given "t" trip point.

...Jim Thompson

Apologies! I didn't read your question thoroughly.

In your case, apply my equation, taking VI=0, VF=V+/2, and TAU=R*C/2

...Jim Thompson

"RST Engineering - JIm"

The devil's advocate has got right inside his head.

Poor fellow ......

...... Phil

Another word: Norton.

The others have given the relevant equations, but perhaps that's not really what you're asking.

If you posted your arguments, we could point to the flaw(s) in your reasoning.

Sylvia.

rnoon

It's getting the same amount of current as if the low side resistor was any value. Hence in both situations the current is the same. But the two resistors form a voltage divider which reduce the voltage. This is what is important because this is what effects the time constant.

A capacitor charges up with V(t) =3D (Vcc - V0)*(1 - e^(-t/R/C)) + V0

In the first case(no low side resistor and no charge), we have

V1(t) =3D Vcc*(1 - e^(-t/RC))

and the second we have

V2(t) =3D Vcc/2*(1 - e^(-t/rC))

In terms of current we have

I1(t) =3D -C*Vcc/RC*e^(-t/RC) I2(t) =3D -C*Vcc/2/rC*e^(-t/rC)

But I1 =3D I2, so

We can easily see that r =3D R/2

You can also see this by solving the system of DE's but it's not as intuitive. The main thing to realize is that the low side resistor does complicate matters as the current is the same but it does reduce the voltage by a factor of 2.

In fact can you see the general cause for arbitrary resistances?

" snipped-for-privacy@gmail.com"

It's getting the same amount of current as if the low side resistor was any value.

The ONLY time the current is the same is at the exact moment the supply is connected.

Why?

Cos the cap is effectively a short circuit at that *instant* and so the upper resistor is the only thing setting the current.

Obviouly, a short circuit by-passes all current from any resistor in parallel with it.

Yawnnnnnnnnn.....

...... Phil

Zero change; the R to be used for the calculation is the thevnin equivalent value across the capacitor.

Thanks; sorry i got the spelling rong.

On a sunny day (Mon, 13 Jul 2009 16:53:23 -0700) it happened "RST Engineering

- JIm" wrote in :

If the supply impedance is neglible low, then the resistors are in parallel, so use the parallel vale as R, and teh half voltage a U.