Series-Parallel DC RC circuit

Are you familiar with Thevenin's theorem?

No. But I will take a look. I believe I am fairly competent at understanding series / parallel circuits. Mostly, I am concerned with finding what I can guess would be called the "effective resistance" of C1 in any point in time.

Thanks!

You mean the instantaneous voltage across the capacitor divided by the instantaneous current flowing into it?

You don't mention, but it seems implied, that your driving voltage is DC and fixed (it could be AC and have a DC bias, as well, for example) and that you have a switch which you engage at some point in time, say at t=0.

The initial voltage across C1 is needed. I assume we can consider it zero, as an initial condition. So, obviously, at t=0, V=0. That's a start, if trivial. At that point, R1 experiences the entire source voltage -- call it Vs. So the initial current is Vs/R1. No current flows through R2 at that first moment, since R2 has 0V across it, so the current initially arrives into C1. Which raises it's voltage and begins some current into R2.

So, speaking in absolute magnitudes only and not necessarily getting the signs correct:

I(R1) = (Vs - V) / R1 I(R2) = V / R2 I(C1) = C * dV / dt

You also know that the current into node V, arriving from R1, must be equal to the other two currents going out. Knowing this, let's rewrite the above with this additional information and from the point of view of node V (positive incoming, negative leaving):

I(R1) = (Vs - V) / R1 incoming I(R2) = -V / R2 leaving I(C1) = -C1 * dV / dt leaving I(R1) + I(R2) + I(C1) = 0

From this, you can substitute into the last equation to arrive at:

(Vs - V) / R1 - V / R2 - C1 * dV / dt = 0

or, Vs/R1 - V/R1 - V/R2 - C1 * dV/dt = 0

This is a simple, linear differential equation. There are lots of approaches to solving them.

Before we go there, let's just sit back and take a qualitative look at your circuit and make some guesses based on other experience. Let's assume away R2 for a moment. In that case, it is a simpler RC charging circuit (assuming V starts at 0, of course.) The basic equation for that is:

V(t) = Vs*[1-e^(t/tau)], where tau = R2*C1

As you can see, when t=0 then V(t=0)=0. Which is expected. As t approaches infinity, then V(t=infinity)=Vs. Also expected.

What might your new circuit, with R1 added back into it? Well, if you are familiar with basic resistor dividers, you might guess that V will approach the divider voltage as t goes to infinity, instead of Vs itself. And you'd be right. Eventually, you'd expect the node voltage V to approach that value. At first, of course, it will start at 0V. So let's substitute that divider voltage in:

V(t) = [Vs*R2/(R1+R2)]*[1-e^(t/tau)]

Now, we don't know tau. It was R2*C1, but now we aren't sure. If you are familiar with Thevenin equivalents and have some working experience with them, you might guess that R2 should be replaced with a new equivalent that is the parallel resistance of R1 combined with R2 -- and you'd guess right if you did imagine that.

The full equation should be:

V(t) = [Vs*R2/(R1+R2)]*[1-e^(t/tau)], tau = [R1*R2/(R1+R2)]*C1

Okay, that was a bunch of intuition working. Is it right? Or just an argument that doesn't really mean anything and could just as well be wrong as right?

Using integrating factors, try and organize the earlier equation into this form: dV/dt + P*V = Q:

dV/dt + [(1/C)*(1/R1 + 1/R2)]*V = [Vs/(C*R1)]

thus, P = [(1/C)*(1/R1 + 1/R2)] Q = [Vs/(C*R1)]

Assuming the initial condition of V(t=0)=0, the solution (if you remember using integrating factors) is:

V = (Q/P)*[1 - e^(-P*t)]

The Q/P part works out like:

Q/P [Vs/(C*R1)] / [(1/C)*(1/R1 + 1/R2)] [Vs/R1] / [(1/R1 + 1/R2)] [Vs/R1] / [(R1 + R2)/(R1*R2)] Vs / [(R1 + R2)/R2] Vs*R2/(R1 + R2)

Now, that looks like what we earlier expected to see. So far, so good.

Now, what about tau? Well, tau=1/P. So that's not so hard:

tau = 1 / P = 1 / [(1/C)*(1/R1 + 1/R2)] = 1 / [(1/C)*[(R1 + R2)/(R1*R2)]] = 1 / [(R1 + R2)/(C*R1*R2)] = (C*R1*R2) / (R1 + R2)

Wow! Looks great and matches the earlier hand-waving expectations.

There are other ways, some far more powerful to learn. Laplace transforms can readily handle these kinds of differential equation solutions. But they can also handle more complex ones, like when your driving voltage isn't constant but is instead a sinusoidal driver or other complex waveform. The Laplace form of the earlier equation looks like:

s*L(V) + P*L(V) = Q/s L(V)*(s + P) = Q/s L(V) = Q/[s*(s + P)] = (Q/P)/s + (-Q/P)/(s+P) L(V) = Q/[s*(s + P)] = (Q/P)*(1/s) + (-Q/P)*(1/(s+P))

The rules aren't too complex. The constant value 1 in the time domain is just 1/s in Laplace. And the Laplace of a constant times a function is just the constant times the Laplace of the function (like derivatives, etc), so the Laplace of k is k/s. The reverse also works, so the inverse Laplace of k/s is k.

Substituting back (inverse), we get:

V = (Q/P) + (-Q/P)*e^(-P*t)

which, if you remember from earlier when we got:

V = (Q/P)*[1 - e^(-P*t)]

is the same thing.

(The inverse Laplace of 1/(s+P) is e^(-P*t), which I didn't mention earlier but where you can just look that up in a table.)

Laplace is actually something that is a lot of fun to learn about, if you get a chance. But so are integrating factors and simple, ordinary differential equations.

Jon

Sorry, when assuming away R2, this should actually read:

V(t) = Vs*[1-e^(t/tau)], where tau = R1*C1

I had mentally changed the designations.

Jon

And this should be R2 above, not R1. Oh, well.

Jon

As far as the cap sees life, the circuit is...

where V2 is V1 * R2/(R1+R2). That treats R1 and R2 as a simple voltage divider.

The value of R3 is the value of R1 in parallel with R2, namely 1/(1/R1

• 1/R2) or the more traditional product-over-the-sum.

Yup, it starts out a zero volts, and the cap charges up towards V2, getting very close after a long while. The shape of the curve is an exponential. If Tau = R3 * C1, it reaches 0.63 of V2 in the first Tau of time; it goes 0.63 of the remaining distance in another Tau. Like that forever.

The actual equation for voltage versus time T is

V = V2 * ( 1-e^-(T/Tau) )

It starts off pretty linear, reaching 1% of V2 in the first 1% of time Tau, but eventually it starts to curve, to flatten out.

John

When you charge a capacitor from a DC source, through a resistor, the voltage across the capacitor rises. The rise of the capacitor voltage with time plots according to the equation -

Vc = E (1 - e^ (-t/RC))

Vc = voltage across the cap E = applied DC voltage e = epsilon = base of natural logarithms = 2.71828 t = elapsed time R = resistance in ohms c = capacitance in farads

If you make your applied voltage equal to 1, then the capacitor voltage Vc is equal to a fraction or percentage of the applied voltage for any given time that has elapsed. This is called a universal time constant chart. What makes it universal, is that the shape of the curve never changes, only the values for time and percent change. RC is called a time constant. RC is actually equal to time in this instance. So you can mark the X axis in units of time or RC units. The y axis is marked in percent or as a decimal part of E applied

Now if you plot this equation - Vc = 1 - e^ (-t/RC) you get a universal time constant chart of the capacitor voltage expressed as a percentage or decimal part of the applied voltage for the elapsed time in RC units.

Examples ---- If you allowed the capacitor to charge for 1RC unit then the capacitor voltage will rise to 63% of the applied voltage.

Your circuit has a shunt resistor and a series resistor. The capacitor charges through the series resistor, and it is the value of the series resistor that you should use for the RC units. The capacitor does not charge through the shunt resistor, so it has no effect on the capacitor voltage other than to modify the value of E applied. For example, if both resistors were equal, the applied voltage is cut in half. You can modify the equation to include the attenuation of the voltage divider to show that-

Vc as a fraction of E applied = (Rshunt/Rseries+Rshunt) - e^ (-t/RC)

You might want to google for universal time constant charts or something like that and see if you can find an actual plot to make this clearer.

No. The Thevenin equivalent resistor value sets the time constant, and that's the equivalent value of the two resistors in parallel.

If V=1, R1=R2=1K and C=1uF, C will charge to 0.5 volts with a time constant of 500 microseconds.

John

Well, he didn't specify what R was, as I read it (I may have missed something, of course.) If it was R_th, and if you read his use of "applied voltage" as being V_th, then it could be read as equivalent. I didn't like the "Vc = 1 - e^ (-t/RC)" part, as the right hand expression is usually meant to be taken to be unitless when written like that. Otherwise, it's only true if the asymptote voltage at t=infinity is 1V. Further, he writes "RC is actually equal to time __in this instance__" when, in fact, the dimensions of RC is _time_ no matter the instance. (Joule-second/Coulomb^2) * (Coulomb^2/Joule) = seconds This is a good thing, because e^(t/RC) then means e is raised to a unitless power, as should be.

For anyone wondering where the 63%, 86%, and 95% values come from, just imagine (1-1/e^1), (1-1/e^2), and (1-1/e^3).

Jon

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time

The capacitor charges through the series resistor but discharges through the parallel equivalent of the two resistors. All the shunt resistor does is divide the source voltage, so that the cap voltage will never equal the source voltage. The op asked about the cap voltage during the charge cycle. If he needs the cap voltage during the discharge cycle , we have a different set of conditions. bg

To me, this says that only the series resistance is used for the time constant calculation.

The full equation is:

V(t) = V_source * [R2/(R1+R2)] * ( 1 - e^( t/[C1*R1*R2/(R1+R2)] ) )

Note that both resistor values are needed to compute the rate while charging. Are you suggesting that R2 does not factor into the tau value during charging? (It does, so I hope not.) Maybe I'm misunderstanding what you write, though.

Jon

No no no. The shunt resistor also divides the charge /current/, so the capacitor charges more slowly. The Thevenin equivalent is the appropriate approach.

Bob

But that is incorrect to say.

Jon

Yes. I wonder why it is so hard to get right. Particularly after John Larkin clearly pointed out why Thevenin works (looking at things from the point of view of the capacitor makes it pretty clear) and I also took the calculations and laid them out several ways to Sunday.

If someone wants to disagree, it would be appropriate to show the errors in the calculations or approach. Not just hand-wave about it.

In any case, I guess I'm a little bit bothered seeing Dan Coby and bg both appearing to get this wrong. It's just an RC problem, after all.

Jon

Big Snip----

Exactly, I should have converted the whole thing to represent a decimal or percentage of the applied voltage, and in this case the applied voltage would be the voltage at the output of the voltage divider. In which case you would need to replace the 1 with that voltage. But eventually I did get around to -- Vc as a fraction of E applied = (Rshunt/Rseries+Rshunt) - e^ (-t/RC) and I could have taken it one step further Vc = (E applied * (Rshunt/Rseries+Rshunt)) - e^ (-t/(Rseries*C)) for the actual voltage bg

But that is wrong. The time constant is NOT based only on Rseries (R1 in the diagram by the OP.) If you think so, you've got it wrong.

Jon

with

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The current through the shunt resistor does not flow through the cap during the charge cycle. So it does not contribute to charging the cap. On the other hand, the shunt resistor will draw current from the cap as it discharges and so there is a different time constant for the charge and discharge cycles. bg