RC Time Constant

Anybody who slogged their way through freshman circuit design knows the well-worn formula for "time constant" (tau) being equal to the product of R and C (time in seconds). One tau gets you to 63% of V+, two tau to 86%, three tau to 95%, and so on. (1- (1/e^tau)).

Suppose, though, you had a resistive voltage divider comprised of equal value R. The voltage at the midpoint is, quite naturally, V+/2. Now suppose you put that capacitor across the "bottom" resistor that goes from the midpoint to the minus supply or ground.

Does the time constant equation change? Obviously the capacitor is only going to charge to V+/2, but does it do it in the same percentage as with a single resistor? I've been tossing this one around in my head all afternoon and can argue both sides of the discussion.

Comments appreciated.

Jim

Reply to
RST Engineering - JIm
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One word: Thevinin.

Reply to
krw

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

Reply to
David Eather

R
a

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.

Reply to
Greg Neill

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

Reply to
John Larkin

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

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| James E.Thompson, P.E.                           |    mens     |
| Analog Innovations, Inc.                         |     et      |
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Reply to
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

--
| James E.Thompson, P.E.                           |    mens     |
| Analog Innovations, Inc.                         |     et      |
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Reply to
Jim Thompson

"RST Engineering - JIm"

** Oh dear.

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

Poor fellow ......

...... Phil

Reply to
Phil Allison

Another word: Norton.

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Paul Hovnanian     mailto:Paul@Hovnanian.com
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Reply to
Paul Hovnanian P.E.

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.

Reply to
Sylvia Else

R
,
m
y
a

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?

Reply to
bob.jones5400

" snipped-for-privacy@gmail.com"

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

** This is very misleading cos the poster is a pedantic and illieterate idiot.

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

Reply to
Phil Allison

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

Reply to
Robert Baer

Thanks; sorry i got the spelling rong.

Reply to
Robert Baer

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.

Reply to
Jan Panteltje

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