Series-Parallel DC RC circuit

bg 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.
bg
Click to expand...
No.

The time constant for both the charging and discharging cycles is the same.
I.e: tau = Rth * C, where Rth is R1 * R2 / (R1 + R2).

You are ignoring the fact that the series resistor R1 is providing current
to both R2 and the capacitor. As the capacitor voltage changes, the current
through R2 is also changing. This changing current affects the shape of
the capacitor voltage curve. With a low capacitor voltage, the current through
R2 is low and the capacitor rises quickly. At higher capacitor voltages, the
the current through R2 is higher, this gives a slower rate of increase in the
capacitor voltage. The net effect is a shorter time constant than would be
expected if R2 is ignored.

If your argument were correct then we would need to throw away Thevenin's
theorem. See: http://en.wikipedia.org/wiki/Th%C3%A9venin%27s_theorem
 
Jon Kirwan wrote:
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
Click to expand...
As I said: "I agree that the thevenin resistance is the correct value to use
for the time constant."

I also agree that bj's statement is wrong.

I disagreed with your statement that bg had not specified what resistance he
meant to use for the time constant calculation. In his post which I quoted
and later posts, he argues that only the series resistance should be used
for the time constant calculation. I do not agree with him (and I do not
think that you do either).
 
On Sun, 8 Mar 2009 17:14:30 -0700, "bg" <bg@nospam.com> wrote:


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
Click to expand...
The charging time constant is C*R' where the value of R' is R1
parallel R2. My numerical example expresses this; try it.

The discharge time constant depends on whether you discharge the cap
by jumping V1 to zero volts, or by disconnecting it. In the first
case, the discharge time constant is identical to the charge time
constant, 500 us in my example. In the second case, R1 is open so only
R2 determines tau. Your statement that the cap "discharges through the
parallel equivalent of the two resistors" is the first case. Try it.

It's better to do it in real life, but you can also use LTSpice to see
this stuff happening.

Interesting: if the cap charge and discharge curves indeed had
different time constants, an applied sine wave at V1 would produce
harmonics across the cap. That's not possible.

John
 
Jon Kirwan wrote in message <59o8r41aerroqeoui0ber6pu1nr7kusg9m@4ax.com>...
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
Click to expand...
My mistake !!
Yes I see what you guys are getting at, the time constant needs to use the
parallel eq of both resistors.
 
Bob Engelhardt 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.

Bob
Click to expand...
A little bit of nit picking: Actually the capacitor charges FASTER
to a lower voltage. The thevenin equivalent resistance gives a smaller
time constant which means a quicker initial increase in voltage. The
final voltage will be the thevenin voltage which is lower since the
two resistors form a voltage divider.
 
On Sun, 08 Mar 2009 17:56:32 -0700, Dan Coby <adcoby@earthlink.net>
wrote:

Jon Kirwan wrote:
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

As I said: "I agree that the thevenin resistance is the correct value to use
for the time constant."

I also agree that bj's statement is wrong.

I disagreed with your statement that bg had not specified what resistance he
meant to use for the time constant calculation. In his post which I quoted
and later posts, he argues that only the series resistance should be used
for the time constant calculation. I do not agree with him (and I do not
think that you do either).
Click to expand...
Ah! I couldn't tell from your last sentence. Thanks.

Jon
 
On Sun, 08 Mar 2009 18:11:24 -0700, Dan Coby <adcoby@earthlink.net>
wrote:

Bob Engelhardt 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.

Bob

A little bit of nit picking: Actually the capacitor charges FASTER
to a lower voltage. The thevenin equivalent resistance gives a smaller
time constant which means a quicker initial increase in voltage. The
final voltage will be the thevenin voltage which is lower since the
two resistors form a voltage divider.
Click to expand...
The initial slope of the capacitor charge, measured in volts/second,
is independent of R2 (for nonzero R2 of course). It's just

dv/dt = V1/(R1*C)

John
 
On Sun, 8 Mar 2009 17:39:58 -0700, "bg" <bg@nospam.com> wrote:

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.
Click to expand...
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.)

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.
Click to expand...
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
 
On Sun, 8 Mar 2009 18:07:24 -0700, "bg" <bg@nospam.com> wrote:

Jon Kirwan wrote in message <59o8r41aerroqeoui0ber6pu1nr7kusg9m@4ax.com>...
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

My mistake !!
Yes I see what you guys are getting at, the time constant needs to use the
parallel eq of both resistors.
Click to expand...
Okay. We are all good, now! We have achieved a shared reality!

Jon
 
Dan Coby wrote:
Bob Engelhardt 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.

Bob

A little bit of nit picking: Actually the capacitor charges FASTER
to a lower voltage. The thevenin equivalent resistance gives a smaller
time constant ...
Click to expand...
Right ... right ... right. I was focusing on the "All the shunt
resistor does is divide the source voltage ..." part & got sloppy. Bob
 
On Mon, 09 Mar 2009 00:19:17 GMT, Jon Kirwan
<jonk@infinitefactors.org> wrote:

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

Jon
 
John Larkin wrote:
On Sun, 08 Mar 2009 18:11:24 -0700, Dan Coby <adcoby@earthlink.net
wrote:

Bob Engelhardt 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.

Bob
A little bit of nit picking: Actually the capacitor charges FASTER
to a lower voltage. The thevenin equivalent resistance gives a smaller
time constant which means a quicker initial increase in voltage. The
final voltage will be the thevenin voltage which is lower since the
two resistors form a voltage divider.

The initial slope of the capacitor charge, measured in volts/second,
is independent of R2 (for nonzero R2 of course). It's just

dv/dt = V1/(R1*C)

John
Click to expand...
Yes, assuming that we are starting with a capacitor voltage of zero
(which has been true for all of the discussion so far on this thread).
In this case the initial current through R2 is zero which means that
for the initial instant we can ignore R2. Your equation then falls
out by simple inspection.
 
Dan Coby wrote:
Bob Engelhardt 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.

Bob

A little bit of nit picking: Actually the capacitor charges FASTER
to a lower voltage. The thevenin equivalent resistance gives a smaller
time constant which means a quicker initial increase in voltage. The
final voltage will be the thevenin voltage which is lower since the
two resistors form a voltage divider.
Click to expand...
A little more nit picking, this time with my own posting. (This is
what happens when I have dinner and have some time to think.)

I said that the capacitor charges 'FASTER'. I even capitalized the
word 'FASTER'. The change in capacitor voltage is only faster in the
circuit with the shunt resistor R2 (versus without R2) if you are looking
at change as a percentage of the final value. I.e. it will reach 63%
of its final value quicker in the version of the circuit with R2 then
in a circuit without it.

However, as Bob points out, the rate of change of the capacitor is
lower if the shunt resistor is present. As Bob says, the shunt resistor
is taking some of the current that would otherwise be charging the
capacitor. This gives us a lower rate of change of the capacitor
voltage (when measured in volts per second).

Both the capacitor voltage and the rate of change in the capacitor
voltage (in volts per second but not percentage) will always be lower
with the shunt resistor than without it. The only exception is at
t=0 when the capacitor voltage is zero. At that instant the capacitor
voltage is zero in either case and the rate of change is V1/(R1 * C)
(as John Larkin posted previously).
 
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