Capacitor Voltages at t = 0: Zero or Split?

[danadak]

TL;DR

Step Response R-RC Network

For this network :

For R finite, C1 = C2, vi a step function @ t = 0, what is the V on each Cap, more importantly is
Vc1 = Vc2 @ t = 0 and Vc1 + Vc2 = vi or ?

This is an ongoing discussion at EDAboard where one side states the Cap V's are 0 @ t = 0, the other the
caps have equal V totaling the step height @ t = 0. eg Vc1 + Vc2 = vi and that Vc1 = Vc2. The rule being used
by folks Vc's = 0 is cap rule that its V cannot change instantaneously, eg v(0-) = v(0+) = 0, and
Heaviside does not overrule constraint Vcap cannot change instantaneously.

Regards, Dana.

 

[Baluncore]

Welcome to PF.

An impossible situation has been hypothesised.
If V1 is an instant step function, then it must instantly charge the two capacitors in series. An instant charge would require an impossible infinite current.

To resolve the impossible situation, V1 must be defined as having an internal resistance, or a series inductance. The step function will then be a ramp. The two capacitors will start with zero voltage, at t₀, then ramp up to total V1 by t_rise. If C1 = C2, then they will approach half the V1 step voltage each, but R will exponentially discharge C2, while charging C1, to the full step voltage.

 

[Dale]

@Baluncore is right. If all of the components are ideal then the resulting system of equations has no solution. This is an “irresistible force meets immovable object” type of problem.

 

[danadak]

I did a sim and got following :

I also reasoned that R1 robs some of the charging current from Ic1 hence Vc2 will not
experience the same as VC1 at t = 0

I am looking for an exact mathematical analysis using Heaviside as source for Vc1 and Vc2
@ t = 0, and prove that Vc1(0) and Vc2(0) @ t = 0 because of charging rule of caps :

 

[Dale]

I did a sim and got following :

View attachment 368609

In your sim you added R2. This is perfectly reasonable to do, but it is no longer the circuit that the argument is about. Once you add R2 there is a solution, but I don't think that you have resolved the argument about the circuit in the OP. You have just resolved a circuit about which there was no argument.

I am looking for an exact mathematical analysis

This is easy enough to do directly. The only unknown node voltage (in the original circuit) is $v_0$. So we can use Kirchoff's current law to write a single node-voltage equation. The sum of the currents leaving the node is zero: $$0=\frac{v_0}{R}+C_2 \frac{dv_0}{dt} + C_1 \frac{d(v_0-v_I)}{dt}$$which simplifies slightly to$$0=\frac{v_0}{R}+(C_1+C_2)\frac{dv_0}{dt} - C_1 \frac{dv_I}{dt}$$

So this equation involves the derivative of $v_I$, which is the problem. The argument is about $t=0$, but the derivative of the Heaviside function does not exist at $t=0$. So, as I said earlier, mathematically there is no solution for real functions

 

[danadak]

R2 was added to get the sim to sim. Its 1 pico ohm. Its presence is why I want to
get at the equations to analytically solve without its presence.

I have a gent that did it in LaPlace and then transformed to time domain which I
am not sure is right.

 

[Dale]

R2 was added to get the sim to sim.

Yes. The equations for the original circuit have no real solution. So it makes sense that it wouldn’t simulate without R2.

I want to
get at the equations to analytically solve without its presence.

Yes, I showed that above.

 

[berkeman]

R2 was added to get the sim to sim. Its 1 pico ohm.

What time constant do you calculate with $R=1p\Omega$?

 

[Baluncore]

I did a sim and got following :

Step rise time, tr = 1 ps.
Total circuit resistance, R2 = 1 pΩ.
Initial charge current, I = 250 kA.
Those are quite unreal values for a real circuit.

You must consider the speed of light and the reality of copper wire. Transmission line impedances will probably be between 1Ω and 1kΩ.

Light will only travel the distance across a pin's head, 0.3 mm in 1ps. Maybe the rise-time should be closer to 100ps, since the circuit dimensions will be more than about 30 mm or 1 inch.

A real circuit, with real 500nF capacitors, and a voltage source, will have a series resistance closer to 100mΩ than 1pΩ.

250 kA cannot rise that quickly due to inductance of the conductors.
The inductance of a 25mm diameter circuit loop will be about 125 nH.
For a 1 volt step with rise-time of 100ps, V = L⋅di/dt ;
di = V⋅dt / L ; i = 0.8 mA, so I estimate a current of about 1mA.

From the definition of a capacitor, C = Q / V ;
and, Q = i ⋅ t ; then C = i ⋅ t / V ; C = i ⋅ dt/dv ; dv/dt = i / C .
So 1mA will charge what capacitance to 1V in 100ps ?
C = i ⋅ dt/dv ; C = 1e-3 * 100e-12 / 1 = 0.1pF .
Yet, you are assuming 500nF in series with 500nF = 250nF .

Your model is almost as unrealistic as the original hypothesis, which is I guess, what you were trying to simulate.
For a 1V rise in 1ns, L1=125nH, R2=100mΩ, R1=1kΩ, C1=C2=1pF.
The circuit will ring at about 600MHz, with a circulating current of 1.5mA.

 

[DaveE]

I did a sim and got following

Because you added a series resistance R2. Yes, I know you know that, but others may need to have it explicitly stated. This is a different network.

I have a gent that did it in LaPlace and then transformed to time domain which I
am not sure is right.

If he didn't have R2, then it's BS. He probably ignored the initial conditions, which are pretty important for a transient response problem. $\mathcal L {f′(t)}=sF(s)−f(0)$

------

If you study network analysis, you will learn that there are two situations that simply can't be analyzed:
1) Any loop that contains only capacitors and/or voltage sources.
2) Any node that connects only inductors and/or current sources.
If the initial conditions match then it's a trivial solution at t=0 (and a miracle, IRL). Otherwise it's a contradictory set of assumptions. Then when the voltage (or current) changes instantaneously you get infinite solutions. The voltage on a capacitor can't change instantaneously. The current in an inductor can't change instantaneously.
Memorize this.

The broader point here is that circuit analysis is ALWAYS a simplified model of the real world. This is seldom explained to beginning EE students. There is no perfect capacitor, or ANY other circuit element. Given this it is the engineer's job to add the salient parts of their real world system to their model. This is simply a bad (incomputable) model. Experienced EEs will waste zero time on this question, they'll spot the problem right away. But, if you don't know this you can easily create circuits that can't be solved.

So really, what's the point? Why do you ask? Since you've posted a nice diagram, I suspect someone gave you this problem. If you have to, show them our responses, but try your very hardest not to call them an idiot. That may have additional consequences that aren't worth your time/effort/expense.

 

[danadak]

Here is the solution / thread I got out of ChatGPT. Attached.

The whole point of the exercise was to prove if Vc1 = Vc2 at t =0 for ideal circuit. This is not
a homework problem, its a discussion amongst several engineers at edaboard.

The 1 pico ohm was to get by the simulator having issue with a V source looking into a short,
eg. caps, at t=0.

The net from the ChatGPT is that a Heaviside step into a net with caps produces a Dirac impulse
of current, and that forces, for equal caps, equal voltages. The presence of the R does not respond
to a impulse of current of zero width. So the series caps experienced exactly the same current,
hence they spit the step V equally at t = 0.

Thanks all for the help, regards, Dana.

 

[A.T.]

Here is the solution / thread I got out of ChatGPT
...
its a discussion amongst several engineers at edaboard.

ChatGPT is just regurgitating what is written online. Using it to decide online debates is rather pointless. You can probably make it argue either side of the debate, by asking leading followup questions.

 

[Dale]

The whole point of the exercise was to prove if Vc1 = Vc2 at t =0 for ideal circuit.

The ideal circuit has no real solution. The voltage at $t=0$ is undefined.

The 1 pico ohm was to get by the simulator having issue with a V source looking into a short,
eg. caps, at t=0.

Yes. Does that not clue you into the fact that the original circuit is inconsistent?

The net from the ChatGPT is that a Heaviside step into a net with caps produces a Dirac impulse
of current

LLM output is not a reliable source. It hallucinates. A Dirac delta is undefined at $t=0$. You cannot make any definite statement about the value of an undefined quantity. What does it mean to say that two undefined quantities are equal?

Now, what you can do is change the problem into a consistent problem, like what you did by adding R2, and you can take a limit where the changed quantity goes to 0. That process can generate a definite result, but often other changes will generate other results in the limit. (Although I don’t immediately see another limit here)

 

[Baluncore]

This would not be the first time that ChatGPT followed a thread on a forum, quoting earlier posts back at a participant.

 

[Dale]

I have a gent that did it in LaPlace and then transformed to time domain which I
am not sure is right.

The Laplace transform also has the same issue. If we take the Laplace transform of the equation above we get $$0=\frac{V_0(s)}{R}+(C_1+C_2) \ s \ \left(V_0(s)-v_0(0)\right)- C_2 \ s \ \left( V_I(s)-v_I(0) \right)$$

This expression depends on $v_I(0)$ which is undefined and $v_0(0)$ which is the quantity that we want to determine.

 

[Baluncore]

SPICE simulates circuits with discrete time steps of dt, that it adaptively selects.

If two 500nF capacitors in series = 250nF, are charged to 1V, then
C = Q / V ; ∴ Q = C / V ; Q = 250n coulomb.
Since the SPICE simulation showed, i = 250kA, we can tell that,
dt = 250nC / 250kA ; ∴ dt = 1 ps .
So, SPICE selected the minimum step rise-time for dt .

However, the real world is continuous, not discrete, so the infinite impulse current that flows for zero time, actually has an integral of 250nC.