# Series-Parallel DC RC circuit

Discussion in 'Electronic Basics' started by kayvee, Mar 7, 2009.

1. ### kayveeGuest

+---------R1-------*-------*
| |
R2 C1
| |
GND------------------*-------*

Hi, I'm am interested in finding the voltage across the capacitor in
this DC circuit in time. Initially, when the capacitor is uncharged,
I know that the voltage across it is 0, and as time goes on the charge
will build and the voltage across it will as well. But, how do I
calculate the voltage across it as a function of time?

Thanks.

2. ### kayveeGuest

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!

3. ### Jon KirwanGuest

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

4. ### Jon KirwanGuest

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

5. ### Jon KirwanGuest

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

Jon

6. ### Jon KirwanGuest

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

7. ### Jon KirwanGuest

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

8. ### Bob EngelhardtGuest

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

9. ### Jon KirwanGuest

But that is incorrect to say.

Jon

10. ### Jon KirwanGuest

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

11. ### Jon KirwanGuest

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

12. ### Jon KirwanGuest

Ah! I couldn't tell from your last sentence. Thanks.

Jon

13. ### Jon KirwanGuest

Since the cap's voltage presents a voltage across the shunt resistor,
except at t=0, there is current present in it (except at t=0.)
That's not a very good way to look at it. But maybe I just can't read
your English and fit it with my mental models.

Since I already performed the complete calculations using various
techniques for solving differential equations, each with the same
result -- not to mention that I tested what I wrote with LTSpice
simulation to be absolutely certain that a spice program also agreed
with me -- Would you care to expose the calculations in detail,
starting from basic statements?

Jon

14. ### Jon KirwanGuest

Okay. We are all good, now! We have achieved a shared reality!

Jon

15. ### Bob EngelhardtGuest

Right ... right ... right. I was focusing on the "All the shunt
resistor does is divide the source voltage ..." part & got sloppy. Bob

16. ### Jon KirwanGuest

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

Jon  