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.

 

[danadak]

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

Yes it will default to a value which can be overridden.

In the sim I am using I can, and do, set the timestep I want to force on the simulator.
Quite frequently I do this in fact, mainly on circuits who reveal what I think are in-
adequately involved circuit response with the default.

C1 = C2. C, R, Vi all ideal

So in summary can I say for the ideal circuit above that if Vi is a Heavyside step it produces
an infinite current, eg an impulse of infinite current (Dirac impulse from dVi/dt of Heavyside),
and that if I apply Norton to the output node, than Vc1 is <> Vc2 because the R removes
some current from the infinite current flowing out of C1 so C2 has less current thru it ?

 

[Vincf]

This is a very classical situation when a Heaviside step is applied to a capacitor, or to an arrangement of capacitors forming a closed circuit in series with the step. The voltage across the capacitor is obviously not continuous in this situation, and an infinite current spike is associated with this discontinuity.

Note that the usual “proof” of the continuity of the voltage across a capacitor assumes that there is no Dirac delta in the current, and is therefore not valid when dealing with a pure capacitor without resistance.

Also note that, in real life, there are always parasitic resistances, so the Dirac current actually becomes a finite spike of short duration. However, this argument is not really relevant here, since we are studying an idealized circuit model and must remain within the assumptions of that model. The same type of argument could be made when a Heaviside step is applied to an $(R,C)$ circuit: the current is discontinuous. But “in real life,” there are always parasitic inductances, so the current is not truly discontinuous. Hence, the “real life” argument is not convincing.

More precisely, the differential equation of the circuit is:
$$
(C_1 + C_2)\,\frac{d v_s}{dt} + \frac{1}{R}\,v_s = C_1\,\frac{d v_e}{dt}.
$$

For $t > 0$, $v_e = 0$, and we obtain:
$$
(C_1 + C_2)\,\frac{d v_s}{dt} + \frac{1}{R}\,v_s = 0,
$$
whose solution is:
$$
v_s = A\,\exp\!\left(-\frac{t}{R(C_1 + C_2)}\right).
$$

Clearly, one cannot impose $v_s(0^+) = 0$.

However, one can note that the two plates, the negative plate of $C_1$ and the positive plate of $C_2$ are connected only through a resistive circuit, so the current cannot be infinite. Therefore, the total charge on both plates, which was zero at $t = 0^-$, is still zero at $t = 0^+$. It follows that:
$$
-q_1(0^+) + q_2(0^+) = 0
$$

And since:

$$
E = \frac{q_1(0^+)}{C_1} + \frac{q_2(0^+)}{C_2}
$$
We have :
$$
v_s(0^+) = \frac{C_1}{C_1 + C_2}\,E
$$
The solution is therefore:
$$
v_s = \frac{C_1}{C_1 + C_2} E \exp\!\left(-\frac{t}{R(C_1 + C_2)}\right).
$$

Finally, note that, if we want to avoid the argument of continuous charge on both plates, the Laplace transform, which in its “modern” formulations uses initial values at $t = 0^-$, can handle this type of problem without difficulty.

 

[DaveE]

R removes some current from the infinite current flowing out of C1 so C2 has less current thru it ?

The infinite current flows in an infinitely short time. So how much charge is diverted through R with a finite current in an infinitely short time?

Also, a question: What's ∞ - 2 = ?

When you get these infinite solutions, just really can't proceed without fixing your model to remove them.

In the sim I am using

Every simulator I ever used would reject this circuit because of the voltage source/capacitor loop previously mentioned. Hence the pΩ resistors, which aren't just fixing the simulator, they're creating a different network.

 

[Vincf]

To be more precise, let me complete the analysis using the Laplace transform, which is very straightforward since:
$E(p) = \frac{E}{p}$ and $v_s(0^-) = 0$

We have:
$$
H(p) = \frac{C_1}{C_1 + C_2}\,\frac{p}{p + \frac{1}{R(C_1 + C_2)}}
$$

Hence:
$$
S(p) = H(p)\,\frac{E}{p} =E \frac{C_1}{C_1 + C_2}\,\frac{1}{p + \frac{1}{R(C_1 + C_2)}}
$$

We directly obtain the solution for $t>0$ :
$$
v_s(t) = \frac{C_1}{C_1 + C_2}\,E\,\exp\!\left(-\frac{t}{R(C_1 + C_2)}\right)
$$

 

[Dale]

The OP question is about $t=0$, not $t>0$. The Laplace transform doesn’t give you that. There is no solution for $t=0$

 

[danadak]

Every simulator I ever used would reject this circuit because of the voltage source/capacitor loop previously mentioned. Hence the pΩ resistors, which aren't just fixing the simulator, they're creating a different network.

I fully understand this addition of the R changes the circuit. But I did what every proof of Dirac
shows, I kept reducing rise time of step, while also changing R, reducing its size to effect taking
a limit. I stopped at 1 pOhm because the simulator math precision came in question.
Interestingly the decrease of the R kept showing repeatable results, that indeed the presence
of the R forced Vc1 <> Vc2, even as the step rise time was forced closer and closer to 0 to get "closer" to a Dirac impulse, in turn getting closer to infinity current.

I am not a mathematician or a physicist, but I keep pursuing do Norton rules hold in infinite cases.

I now wonder am I abusing rules of infinity. Eg. if I have an infinite current and I take away from it a finite current (the R) have I indeed proved my belief Vc1 <> Vc2 at t= 0

 

[Dale]

I fully understand this addition of the R changes the circuit.

But do you fully understand that the addition of R changes the problem? In other words, everyone will agree about your solution to the changed circuit. But many people will (with good reason) not agree that this changed circuit addresses the problem posed in the OP.

Here is the issue, in the way that you have changed the circuit you get $v_{C1}(0)=v_{C2}(0)=0$ and $v_{C1}(0)+v_{C2}(0)\ne v_I(0)$. You claim that since this does not depend on the value of the inserted resistor you can take the limit as $R\rightarrow 0$, recover the original circuit, and then claim this is the answer to the original problem.

However, someone else can change the circuit differently. They can leave the components unchanged but change $v_I(t)$ such that instead of being a step function it is a logistic function $v_I(t)=(1+e^{-2kt})^{-1}$. With this voltage the original network is solvable since the derivative is defined at all points in time, and at each point in time $v_{C1}+v_{C2}=v_I$. Now they can take the limit as $k\rightarrow \infty$, recover the original waveform, and claim this is the answer to the original problem.

The original problem simply does not have an answer.

 

[Vincf]

I had not understood that the OP was specifically asking for the value at t = 0.

In the general case, if one modifies the circuit in any way such that the order of the differential equation on the output side becomes strictly higher than that on the input side, then the output is necessarily continuous at $t = 0$, and therefore $$s(0)=0$$.

For example, by adding a resistor at the input, one obtains an equation of the form:
$$
a_2 \frac{d^2 s}{dt^2} + a_1 \frac{ds}{dt} + a_0 s = b_1 \frac{de}{dt}
$$

If $e(t)$ is a step function, then $\frac{de}{dt}$ is a Dirac distribution. By balancing the singular terms, one immediately sees that the output must be continuous but not differentiable. Its first derivative contains a step, and the second derivative contains a Dirac distribution.

When the parameter $a_2$ tends towards 0, we return to the original equation, but we see that it's a singular perturbation. We will therefore have a temporal boundary layer to recover the discontinuity predicted by the calculation with the original circuit.

 

[danadak]

So for this circuit, ideal components, am I correct in observing the following :

Vi is a step applied V.

1) Vi(t) produces a Dirac impulse of infinite current into C1
2) Norton demands Ic1 = Ic2 + Ir
3) Hence, at t = 0 then Vc1 ≠ Vc2 and Vc1 > Vc2

When I did the sim, although having to insert another Rseries between Vin and C1 to obviate the
Dirac simulator problem, the sim showed repeatability in output as Rseries was reduced to
extremely small values.

So thanking you all I just want to confirm with my very basic math ability (compared to Physicists)
is correct.

 

[DaveE]

at t = 0 then Vc1 ≠ Vc2 and Vc1 > Vc2

Vc1 and Vc2 are initial conditions that have not been specified. You can't really say anything about them. You might choose to specify that the circuit has settled to a steady state before t=0, in which case Vc2 = 0 and Vc1 = Vi.

At t = 0⁺, you will have a capacitive divider (based on conservation of charge) that will split the additional voltage after the initial readjustment. Since you haven't specified any values or initial conditions, you can't say which voltage is larger.

The resistor doesn't matter until you analyze t > 0. No charge will flow through the resistor in an infinitely short time.

However, analyzing this circuit doesn't make much sense, as I've said in every reply. Move on. Put a resistor in there. You're missing the bigger picture by ignoring unrealistic assumptions.

 

[Baluncore]

Before t=0, all voltages will be zero.
At t=0, the instant step in source voltage, will be attenuated by the capacitive voltage divider, to produce an output voltage step.
After t=0, the output voltage will begin to fall exponentially back to zero.

For the idealised circuit, the source and output voltage will be discontinuous, and indeterminate, at t=0.

For a real circuit, the rise time of the ramp, that starts with v=0, at t=0, will not yet have departed from zero. The rise will be limited by the impedance of the real circuit components. Depending on the circuit inductance, the output may overshoot, undershoot, or be critically damped.

For a simulation, the input voltage at t=0 will rise from zero to the step voltage while the output remains at zero. At the end of the first time step, t=dt, the voltages will have reached their step values. The current flow during the first time step, will be proportional to the voltage step and capacitance, and inversely proportional to the step time selected.

 

[Dale]

1) Vi(t) produces a Dirac impulse of infinite current into C1

Yes. In other words there is no real-valued solution to the problem.

2) Norton demands Ic1 = Ic2 + Ir

I am not sure why you say “Norton” here. It is Kirchoff’s current law.

3) Hence, at t = 0 then Vc1 ≠ Vc2 and Vc1 > Vc2

This doesn’t follow from the previous two statements. Whether it is true or not depends on the exact definition of the step function.

This idealized circuit simply does not have a solution at $t=0$ for a step input. Anyone who claims that the solution is one thing or another is wrong.

 

[Vincf]

It is possible to see what happens when we replace the Heaviside echelon with a more regular entry.
For simplicity, I restrict myself to the case where $C_1 = C_2$ and I consider the differential equation (with $\tau = RC$):
$$
2\tau \frac{ds}{dt} + s = \tau \frac{de}{dt}
$$

The impulse response is:
$$
s_{\text{imp}}(t) = \frac{1}{2}\,\delta(t) - \frac{1}{4\tau} \exp\!\left(-\frac{t}{2\tau}\right) H(t)
$$

At the input, let us replace the ideal step $E\,H(t)$ by a regular function $e_a(t)$, where $a$ is a parameter such that, as $a \to 0$, $e_a(t) \to E\,H(t)$.

If the capacitors are initially uncharged, the circuit response at any time is obtained by convolution:
$$
s(t) = (s_{\text{imp}} * e_a)(t)
$$

In particular, at $t=0$:
$$
s_a(0) = \int_{-\infty}^{0} s_{\text{imp}}(-u)\, e_a(u)\, du
$$

In this integral, only the Dirac term from the impulse response contributes, and therefore:
$$
s_a(0) = \frac{1}{2}\, e_a(0)
$$

We see that if the input is regularized, the output at $t=0$ depends entirely on the type of regularization chosen. If one selects a symmetric regularization around 0, one obtains $E/2$, but this is a very particular case.

For example, consider the input
$$
e_a(t) = E \,\frac{1}{2\,\exp(-t/a) + 1}, \quad \text{with } a>0.
$$

One readily checks that it converges to $E\,H(t)$ as $a \to 0$.

However,
$$
e_a(0) = \frac{E}{3},
$$

and therefore, with this regularization of the step, one obtains:
$$
s(0) = \frac{E}{3}.
$$

 

[Vincf]

There is a flaw in my previous argument.

Using the impulse response assumes zero initial conditions, in particular that the input vanishes for $t<0$. However, the regularization I introduced does not satisfy this, since $e_a(t)\neq 0$ for $t<0$.

Therefore the relation $s(0)=\frac{1}{2}e_a(0)$ does not apply here, and the conclusion $s(0)=E/3$ is incorrect.