Series-Parallel DC RC circuit

[kayvee]

+---------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.

 

[Greg Neill]

kayvee wrote:

+---------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.

Are you familiar with Thevenin's theorem?

http://en.wikipedia.org/wiki/Th%C3%A9venin's_theorem

 

[Greg Neill]

kayvee wrote:

On Mar 7, 5:14 pm, "Greg Neill" <gneil...@MOVEsympatico.ca> wrote:
kayvee wrote:
+---------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.

Are you familiar with Thevenin's theorem?

http://en.wikipedia.org/wiki/Th%C3%A9venin'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.

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

 

[kayvee]

On Mar 7, 5:14 pm, "Greg Neill" <gneil...@MOVEsympatico.ca> wrote:

kayvee wrote:
   +---------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.

Are you familiar with Thevenin's theorem?

http://en.wikipedia.org/wiki/Th%C3%A9venin'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!

 

[Jon Kirwan]

On Sat, 7 Mar 2009 14:08:56 -0800 (PST), kayvee <kmvgroups@gmail.com>
wrote:

+---------R1-------*-------* <----- V
| |
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?

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

 

[Jon Kirwan]

On Sun, 08 Mar 2009 01:03:35 GMT, Jon Kirwan
<jonk@infinitefactors.org> wrote:

, where tau = R2*C1

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

 

[Jon Kirwan]

On Sun, 08 Mar 2009 01:03:35 GMT, Jon Kirwan
<jonk@infinitefactors.org> wrote:

What might your new circuit, with R1 added back into it?

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

Jon

 

[John Larkin]

On Sat, 7 Mar 2009 14:08:56 -0800 (PST), kayvee <kmvgroups@gmail.com>
wrote:

+---------R1-------*-------*
| |
V1 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.

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

+---------R3--------------*
|
V2 C1
|
GND-------------------------*

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.

http://faraday.physics.utoronto.ca/IYearLab/capacitor.pdf


John

 

[bg]

kayvee wrote in message ...

+---------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.

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.
2 RC units = 86%
3 RC units = 95%
etc

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.

 

[John Larkin]

On Sun, 8 Mar 2009 14:08:03 -0700, "bg" <bg@nospam.com> wrote:

kayvee wrote in message ...

+---------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.

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.
2 RC units = 86%
3 RC units = 95%
etc

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.

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

 

[Jon Kirwan]

On Sun, 08 Mar 2009 14:25:54 -0700, John Larkin
<jjlarkin@highNOTlandTHIStechnologyPART.com> wrote:

On Sun, 8 Mar 2009 14:08:03 -0700, "bg" <bg@nospam.com> wrote:


kayvee wrote in message ...

+---------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.

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.
2 RC units = 86%
3 RC units = 95%
etc

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.

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.

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

 

[bg]

John Larkin wrote in message ...

On Sun, 8 Mar 2009 14:08:03 -0700, "bg" <bg@nospam.com> wrote:


kayvee wrote in message ...

+---------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.

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.
2 RC units = 86%
3 RC units = 95%
etc

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.

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

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

 

[Dan Coby]

Jon Kirwan wrote:

On Sun, 08 Mar 2009 14:25:54 -0700, John Larkin
jjlarkin@highNOTlandTHIStechnologyPART.com> wrote:

On Sun, 8 Mar 2009 14:08:03 -0700, "bg" <bg@nospam.com> wrote:

kayvee wrote in message ...
+---------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.
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.
2 RC units = 86%
3 RC units = 95%
etc

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.
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.

Well, he didn't specify what R was, as I read it (I may have missed
something, of course.)

I agree that the thevenin resistance is the correct value to use for the
time constant. However my reading of bg posting agrees with JL. Bg says:

"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."

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

 

[Jon Kirwan]

On Sun, 8 Mar 2009 17:14:30 -0700, "bg" <bg@nospam.com> wrote:

John Larkin wrote in message ...
On Sun, 8 Mar 2009 14:08:03 -0700, "bg" <bg@nospam.com> wrote:


kayvee wrote in message ...

+---------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.

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.
2 RC units = 86%
3 RC units = 95%
etc

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.

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


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.

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

 

[Bob Engelhardt]

bg wrote:

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. ...

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

 

[Jon Kirwan]

On Sun, 08 Mar 2009 17:15:41 -0700, Dan Coby <adcoby@earthlink.net>
wrote:

Jon Kirwan wrote:
On Sun, 08 Mar 2009 14:25:54 -0700, John Larkin
jjlarkin@highNOTlandTHIStechnologyPART.com> wrote:

On Sun, 8 Mar 2009 14:08:03 -0700, "bg" <bg@nospam.com> wrote:

kayvee wrote in message ...
+---------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.
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.
2 RC units = 86%
3 RC units = 95%
etc

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.
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.

Well, he didn't specify what R was, as I read it (I may have missed
something, of course.)

I agree that the thevenin resistance is the correct value to use for the
time constant. However my reading of bg posting agrees with JL. Bg says:

"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."

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

But that is incorrect to say.

Jon

 

[Jon Kirwan]

On Sun, 08 Mar 2009 20:20:26 -0400, Bob Engelhardt
<bobengelhardt@comcast.net> wrote:

bg wrote:
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. ...

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

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

 

[bg]

Big Snip----

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.

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

 

[Jon Kirwan]

On Sun, 8 Mar 2009 17:26:58 -0700, "bg" <bg@nospam.com> wrote:

Big Snip----

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.

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

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

 

[bg]

Jon Kirwan wrote in message ...

On Sun, 8 Mar 2009 17:14:30 -0700, "bg" <bg@nospam.com> wrote:

John Larkin wrote in message ...
On Sun, 8 Mar 2009 14:08:03 -0700, "bg" <bg@nospam.com> wrote:


kayvee wrote in message ...

+---------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.

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.
2 RC units = 86%
3 RC units = 95%
etc

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.

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


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.

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

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