Summary: If I have a Capacitive Discharge through a coaxial cable to some load, how does the length of the cable effect the current felt by the load?

I believe that the schematic breakdown for this problem is a charged capacitor $C_1$ separated by a switch from a second capacitor $C_2$ (the capacitance of the coaxial cable), and the load $R_1$. All three loads are in parallel.

From this, I believe the differential equation comes to:

$$-C_1{dV/dt} = C_2{dV/dt} + {V/R_1}$$

which solves to $$V=\frac{Q}{C_1 + C_2}e^{-t/R(C_1+C_2)}$$

where the initial conditions would be at $t=0$, voltage is equal to the charge on the original capacitor. And at $t=\infty$, voltage goes to zero.

This makes sense to me as the two 'capacitors' are added together in parallel in the circuit simplification. Then the current through the resistor as a function of time is $V/R=\frac{Q}{R(C_1 + C_2)}e^{-t/\tau}$ where $\tau$ is the time constant $R(C_1+C_2)$

Then assuming a linear relationship between coaxial length and capacitance, you can easily substitute $C_2$ for some $LC_d$.

Here is where I am confused. It makes sense to me that the peak current is independent of the coax length. It makes sense to me that the time constant is dependent on coax length. But all the voltage, current, and charge curves are just simple exponential decays. When I shut the switch, doesn't the coaxial cable take some time to charge and then discharge? Shouldn't a plot of charge on $C_2$ be two competing exponentials? Theoretically, the coaxial cable should receive ALL the current initially and the resistor none, and then over time switch.

I think part of the issue is in the ideal circuit. Each capacitor should have it's own associated resistance $R_{C_1}, R_{C_2}$ in series. But how would you even begin to solve that circuit theoretically?


1 Answer 1


A coaxial cable is a transmission line. To describe the discharge process in complete detail, we need to deal with wave propagation along the transmission line, as modeled by the telegrapher's equations. In general, the process can be quite complicated, because there are signals propagating up and down the transmission line, getting partially reflected at both ends. If you are looking for a mathematical treatment of such circuits in time domain, this can be found in Chapter 3 of Foundations for Microwave Engineering, 2nd ed. by Robert E. Collin.

The time it takes for the transmission line to "charge" can be described as the delay time $T = \ell/v$ of the transmission line, which is the time it takes for the initial voltage wave to reach the load resistance. Here, $\ell$ is the length of the transmission line and $v$ is the propagation speed, usually a little lower than the speed of light. After the switch is closed, there is a "negotiating period" during which both the capacitor and the load voltages will have oscillations due to signals bouncing back and forth, which will eventually die down. The duration of this period depends on $T$, the characteristic impedance $Z_0$ of the transmission line, and the load resistance $R$.

During the first $2T$ after the switch is closed, the capacitor doesn't even "see" the load resistance: the transmission line just acts like a resistor with resistance $Z_0$, so the capacitor discharges with time constant $Z_0C$ during this time. Assuming $R > Z_0$, after the negotiation period, the voltage along the transmission line is roughly uniform, and the capacitor discharges as you describe. The effective capacitance of a "short" transmission line (meaning $T \ll RC_1$) is $$C_2 \approx \left(1 - \frac{Z_0^2}{R^2}\right)C'\ell$$ where $C'=(vZ_0)^{-1}$ is the capacitance per unit length of the transmission line. $R=Z_0$ is a special case where there is no reflection at the load, and the capacitor smoothly discharges through an effective resistance $Z_0$ the whole time.

I would encourage you to set up the circuit in a simulator and play with the various parameters. Here is a screenshot from the free circuit simulator LTspice, showing the capacitor and load voltages (up) and currents (down) as a function of time after the switch is closed at $t = 0.5\text{ ms}$. enter image description here

  • $\begingroup$ Ok this is a new subject for me, but I was able to find some resources to help me along and corroborate what you've said. How did you come about the 2T delay time though? Is that just the exponential decay/growth factor for ~75%. Or is it another physical property? $\endgroup$
    – Squatchis
    Dec 14, 2023 at 22:30
  • $\begingroup$ Disregard the clarification. I was confusing T for $\tau$. It takes two propagation times to go up and down the transmission line for the original capacitor to even be effected by the load resistance (i.e. current increasing/decreasing stepwise depending on the load size). $\endgroup$
    – Squatchis
    Dec 14, 2023 at 22:38
  • $\begingroup$ @Squatchis Yes, that's exactly right. $\endgroup$
    – Puk
    Dec 14, 2023 at 22:46

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