RC circuit current, large signal

[John Larkin]

On Fri, 6 Sep 2019 12:35:01 -0400, bitrex <user@example.net> wrote:

On 9/6/19 12:08 PM, jlarkin@highlandsniptechnology.com wrote:
On Thu, 5 Sep 2019 18:52:53 -0700 (PDT), seagirt555@gmail.com wrote:

On Thursday, September 5, 2019 at 6:19:53 PM UTC-4, bitrex wrote:
On 9/5/19 5:56 PM, Tim Williams wrote:
I'd have to think about it a bit or write it out, but it seems wrong
(but may be right) that there would be anything 2nd order (omega^2).

The solution seems reasonable otherwise: there is an exponential term
for startup transient, and the driven function and its derivative, with
coefficients (this is a nonhomogeneous system).

Tim


The solution comes from here:

https://www.wolframalpha.com/input/?i=C*%28A*omega*cos%28omega*t%29+-+R*i%27%28t%29%29+%3D+i%28t%29&assumption=%22i%22+-%3E+%22Variable%22

Wasn't sure myself but can't see where I went wrong in the original
equation to make it that way. Total current thru the capacitor is the
series current, that's I(t) = C*dv/dt, where v is the voltage across the
capacitor. Voltage across cap is d/dt A*sin(omega*t) minus drop across
the resistor

no, the voltage across the cap is just v or A*sin(omega*t) - voltage drop
across resistor. voltage drop across resistor is R * I or RC dv/dt

A*sin(omega*t) = v +RC*dv/dt


which is V(t) = I(t)*R, dV/dt = dV/dI*dI/dt = R*I'(t)?

The omega squared only scales the amplitude so you can't get frequency
multiplication out of it or anything (something would definitely be
wrong, then.) If it's assumed C^2*R^2*omega^2 is large with respect to 1
the omega^2 term goes away

You certainly can't get freq multiplication out of it, it's a linear
system driven by a sine wave. The only freq there is omega.

It's essentially a sine burst, so there is a DC component that fades
out in time but complicates the math.

I see that a lot in my alternator-simulation Spice runs; the first
few, sometimes many, cycles of a sine-forced system are different from
the steady state.


I think it would be good to fully understand the large signal
time-domain behavior of some of simple circuits, mathematically. Maybe
make good interview question?

Perhaps once we understand RC circuits we can be engineers!!!

I Spice stuff like that. I prefer waveforms to equations. And
equations are pretty much useless for nonlinear circuits.

There's a visual learning feedback loop that Spice assists, to train
my instincts... as long as the sims don't run too slow. What's that
Skinner Box mouse training time constant? 30 seconds?

 

On Friday, September 6, 2019 at 1:07:07 PM UTC-4, Tim Williams wrote:

"bitrex" <user@example.net> wrote in message
news:WUvcF.380363$AR4.33491@fx39.iad...
I think it would be good to fully understand the large signal time-domain
behavior of some of simple circuits, mathematically. Maybe make good
interview question?

Perhaps once we understand RC circuits we can be engineers!!!

I mean, I got A's in my DE and signals classes. I have the tools to solve
this problem, I just don't particularly care to.

I hear you, especially when it's already been done for you:

https://ocw.mit.edu/courses/electrical-engineering-and-computer-science/6-071j-introduction-to-electronics-signals-and-measurement-spring-2006/lecture-notes/09_sss.pdf

 

On Friday, September 6, 2019 at 12:08:15 PM UTC-4, jla...@highlandsniptechnology.com wrote:

On Thu, 5 Sep 2019 18:52:53 -0700 (PDT), seagirt555@gmail.com wrote:

On Thursday, September 5, 2019 at 6:19:53 PM UTC-4, bitrex wrote:
On 9/5/19 5:56 PM, Tim Williams wrote:
I'd have to think about it a bit or write it out, but it seems wrong
(but may be right) that there would be anything 2nd order (omega^2).

The solution seems reasonable otherwise: there is an exponential term
for startup transient, and the driven function and its derivative, with
coefficients (this is a nonhomogeneous system).

Tim


The solution comes from here:

https://www.wolframalpha.com/input/?i=C*%28A*omega*cos%28omega*t%29+-+R*i%27%28t%29%29+%3D+i%28t%29&assumption=%22i%22+-%3E+%22Variable%22

Wasn't sure myself but can't see where I went wrong in the original
equation to make it that way. Total current thru the capacitor is the
series current, that's I(t) = C*dv/dt, where v is the voltage across the
capacitor. Voltage across cap is d/dt A*sin(omega*t) minus drop across
the resistor

no, the voltage across the cap is just v or A*sin(omega*t) - voltage drop
across resistor. voltage drop across resistor is R * I or RC dv/dt

A*sin(omega*t) = v +RC*dv/dt


which is V(t) = I(t)*R, dV/dt = dV/dI*dI/dt = R*I'(t)?

The omega squared only scales the amplitude so you can't get frequency
multiplication out of it or anything (something would definitely be
wrong, then.) If it's assumed C^2*R^2*omega^2 is large with respect to 1
the omega^2 term goes away

You certainly can't get freq multiplication out of it, it's a linear
system driven by a sine wave. The only freq there is omega.

It's essentially a sine burst, so there is a DC component that fades
out in time but complicates the math.

It's only a sine burst if you include the case where the voltage source
is turned on, ie it's not steady state, continuous. Usually when dealing
with a sine wave source it's the steady state that we assume we're talking
about.

 

[Helmut Sennewald]

Am 05.09.2019 um 22:18 schrieb bitrex:

So without the Laplace transform just from first principles, apply a
sine of amplitude A, A*sin(omega*t) to a series RC to ground:

I(t) = C*dv/dt,

I(t) = C*(A*omega*cos(omega*t) - I'(t)*R)

I(t) = K*e^(-t/RC) + [A*C^2*R*omega^2*sin(omega*t)]/[C^2*R^2*omega^2 +
1] + [A*C*omega*cos(omega*t)]/[C^2*R^2*omega^2 + 1]

does that look right?

Hello,

Here is the solution from a script found in the Internet.
http://www.dr-seifert-online.de/Downloads/LaborGETI3_0.pdf

T = R*C
w = 2*pi*f
A = 1/(1+(w*T)**2)
B = w*T/(1+(w*T)**2)

ue(t) = sin(w*t)

uc(t) = B*exp(-t/T) +A*sin(w*t) -B*cos(w*t)
-------------------------------------------

Because B/T is the same as A*w,

ic(t) = w*C*(-A*exp(-t/T) +A*cos(w*t) +B*sin(w*t))
--------------------------------------------------

I have checked the result with LTspice.

Helmut

 

[bitrex]

On 9/6/19 1:07 PM, Tim Williams wrote:

"bitrex" <user@example.net> wrote in message
news:WUvcF.380363$AR4.33491@fx39.iad...
I think it would be good to fully understand the large signal
time-domain behavior of some of simple circuits, mathematically. Maybe
make good interview question?

Perhaps once we understand RC circuits we can be engineers!!!

I mean, I got A's in my DE and signals classes.  I have the tools to
solve this problem, I just don't particularly care to.

JL might not be able to, but he went to school probably before DEs
existed, so it's not entirely his fault.

Anyway, there are better tutors than us, elsewhere on the internet.
This simple system has been solved time and again.

You can even find videos on it, probably if you don't mind that there's
a hundred different Indian guys in the most recent videos that turn up
in a YT or Google search...

Tim

Basically the problem is that the DEQs as-written don't describe
anything physical.

Either the system is steady state and you have to _assume_ a form of the
solution that describes a steady-state system. The solution as given
isn't that.

Or the system isn't steady-state and there must also be a step function
at 0 to "switch on" the sine, in addition to the driving sine that runs
for all time. The equations as written don't describe that situation,
either.

 

[bitrex]

On 9/6/19 1:28 PM, John Larkin wrote:

On Fri, 6 Sep 2019 10:17:33 -0700 (PDT), seagirt555@gmail.com wrote:

On Friday, September 6, 2019 at 12:08:15 PM UTC-4, jla...@highlandsniptechnology.com wrote:
On Thu, 5 Sep 2019 18:52:53 -0700 (PDT), seagirt555@gmail.com wrote:

On Thursday, September 5, 2019 at 6:19:53 PM UTC-4, bitrex wrote:
On 9/5/19 5:56 PM, Tim Williams wrote:
I'd have to think about it a bit or write it out, but it seems wrong
(but may be right) that there would be anything 2nd order (omega^2).

The solution seems reasonable otherwise: there is an exponential term
for startup transient, and the driven function and its derivative, with
coefficients (this is a nonhomogeneous system).

Tim


The solution comes from here:

https://www.wolframalpha.com/input/?i=C*%28A*omega*cos%28omega*t%29+-+R*i%27%28t%29%29+%3D+i%28t%29&assumption=%22i%22+-%3E+%22Variable%22

Wasn't sure myself but can't see where I went wrong in the original
equation to make it that way. Total current thru the capacitor is the
series current, that's I(t) = C*dv/dt, where v is the voltage across the
capacitor. Voltage across cap is d/dt A*sin(omega*t) minus drop across
the resistor

no, the voltage across the cap is just v or A*sin(omega*t) - voltage drop
across resistor. voltage drop across resistor is R * I or RC dv/dt

A*sin(omega*t) = v +RC*dv/dt


which is V(t) = I(t)*R, dV/dt = dV/dI*dI/dt = R*I'(t)?

The omega squared only scales the amplitude so you can't get frequency
multiplication out of it or anything (something would definitely be
wrong, then.) If it's assumed C^2*R^2*omega^2 is large with respect to 1
the omega^2 term goes away

You certainly can't get freq multiplication out of it, it's a linear
system driven by a sine wave. The only freq there is omega.

It's essentially a sine burst, so there is a DC component that fades
out in time but complicates the math.




It's only a sine burst if you include the case where the voltage source
is turned on, ie it's not steady state, continuous. Usually when dealing
with a sine wave source it's the steady state that we assume we're talking
about.

That's the dilemma of a differential equation solution: the sine has
to be turned on.

You see the same thing in Spice.

Basically the problem is that the DEQs as-written don't describe
anything physical.

Either the system is steady state and you have to _assume_ a form of the
solution that describes a steady-state system. The solution as given
isn't that.

Or the system isn't steady-state and there must also be a step function
at 0 to "switch on" the sine, in addition to the driving sine that runs
for all time. The equations as written don't describe that situation,
either.

 

[John Larkin]

On Fri, 6 Sep 2019 12:07:03 -0500, "Tim Williams"
<tiwill@seventransistorlabs.com> wrote:

"bitrex" <user@example.net> wrote in message
news:WUvcF.380363$AR4.33491@fx39.iad...
I think it would be good to fully understand the large signal time-domain
behavior of some of simple circuits, mathematically. Maybe make good
interview question?

Perhaps once we understand RC circuits we can be engineers!!!

I mean, I got A's in my DE and signals classes. I have the tools to solve
this problem, I just don't particularly care to.

JL might not be able to, but he went to school probably before DEs existed,
so it's not entirely his fault.

The issue is that I got my EE degree before Spice existed. I had to
write my own simulation code. Sometimes I still do.

 

[bitrex]

On 9/6/19 2:14 PM, Helmut Sennewald wrote:

Am 05.09.2019 um 22:18 schrieb bitrex:
So without the Laplace transform just from first principles, apply a
sine of amplitude A, A*sin(omega*t) to a series RC to ground:

I(t) = C*dv/dt,

I(t) = C*(A*omega*cos(omega*t) - I'(t)*R)

I(t) = K*e^(-t/RC) + [A*C^2*R*omega^2*sin(omega*t)]/[C^2*R^2*omega^2 +
1] + [A*C*omega*cos(omega*t)]/[C^2*R^2*omega^2 + 1]

does that look right?

Hello,

Here is the solution from a script found in the Internet.
http://www.dr-seifert-online.de/Downloads/LaborGETI3_0.pdf

T = R*C
w = 2*pi*f
A = 1/(1+(w*T)**2)
B = w*T/(1+(w*T)**2)

ue(t) = sin(w*t)

uc(t) = B*exp(-t/T) +A*sin(w*t) -B*cos(w*t)
-------------------------------------------

Because B/T is the same as A*w,

ic(t) = w*C*(-A*exp(-t/T) +A*cos(w*t) +B*sin(w*t))
--------------------------------------------------

I have checked the result with LTspice.

Helmut

Okay, so there is an omega^2 term in the generalized solution. ¯\_(ツ)_/¯