Time-varying capacitor generates higher than supply voltage? Capacitor power generators?

Question

I was playing around with electret capacitors, supposedly time-varying capacitors with accordance with sound waves, trying to understand them deeply when I found that I need to model it mathematically, so I set up a really simple circuit and it's differential equation

^{simulate this circuit – Schematic created using CircuitLab}

$$ V(t) = \frac{dq(t)}{dt} R + \frac{q(t)}{C(t)} $$ And for simplicity, I set $$V(t) = 5$$ and $$C(t)= Asin(\omega t)+B ~~,~~ \text{where $B$ $>$ $A$} $$

Note: Capacitance is inversely proportional to sound waves as noted by @glen-geek, this is just a model for a capacitor that changes with time in a sinusoidal manner.

Now although I thought this would be an easy differential equation, I couldn't solve it manually with any method I could know, Bernoulli, Laplace, etc.. so I had to simulate it on a computer.

Now unfortunately again I didn't find any tool that simulates electric circuits with time varying capacitors, so I just did my own crappy simulator using python (I simply solved for dq/dt and each nudge in time I multiply it by the nudge and add it to total charge)

import math
import matplotlib.pyplot as plt
import numpy as np


def voltage(time):
    return 5


def resistance(time):
    return 10000


def capacitance(time):
    change_in_capacitance = 1e-7
    base_capacitance = 1.01e-7
    return change_in_capacitance * math.sin((100 / 2 * math.pi) * t) + base_capacitance


q = 0
t = 0
dq_dt = 0  # Should be calculated first time
dt = 10e-6

t_points = np.array([])
q_points = np.array([])
dq_dt_points = np.array([])
output_v_points = np.array([])
output_v_r_points = np.array([])


def update_dq_dt(time):
    global dq_dt, dq_dt_points, output_v_points, output_v_r_points
    dq_dt = (voltage(time) - q / capacitance(time)) / resistance(time)
    output_v_points = np.append(output_v_points, voltage(t) - dq_dt * resistance(t))
    output_v_r_points = np.append(output_v_r_points, dq_dt * resistance(t))
    dq_dt_points = np.append(dq_dt_points, dq_dt)


def update_q():
    global q, q_points
    q_points = np.append(q_points, q)
    q += dq_dt * dt


def update_t():
    global t, t_points
    t_points = np.append(t_points, t)
    t += dt


for i in range(1, 100000):
    update_dq_dt(t)
    update_q()
    update_t()

print(t_points)
print(q_points)
print(dq_dt_points)

# plt.plot(t_points, q_points, label="charge")
# plt.plot(t_points, dq_dt_points, label="current")
plt.plot(t_points, output_v_points, label="voltage output")
plt.plot(t_points, output_v_r_points, label="voltage on resistor")
plt.show()

This is supposedly a circuit simulator with resolution of 10us and as a gut check that I put capacitance a constant value, and sure enough it generated the expected exponential graph so I moved on to simulate a time varying capacitance not expecting what's going to happen next.

(Blue is voltage on capacitor leads, and orange is voltage on resistor)

Voltage between the leads of capacitor exceeded the supply voltage (5 volts), because as capacitance changes instantaneously, sometimes charge has to move back into the source, aka current moving backwards, thus voltage on resistor is negative and as a compensation, voltage on capacitor is higher than the supply voltage.

I tried playing with the varying capacitor and could even reach double the supply voltage

Now of course practically this is I think is impossible as suggested by my practical experiment, I could barely sense any change in the voltage in the range of millivolts, actually all of this was because I was trying to find the best series resistance to put before the capacitor yet I fill into this rabbit hole.

My question in a nutshell is, Is this actually correct? and indeed the voltage on a capacitor can exceed supply voltage without the use of inductors or banks or the other regular means, and that my reasoning is correct, changing capacitance requires energy, sound waves bumping into it moving the diaphragm so indeed no laws broken here by achieving higher voltage.

And if this is actually correct, is there a possibility to construct a device that generates electricity by simply periodically igniting some fuel in such way that a piston hits hard a capacitor in the circuit to generate power?

Answer 1

Yes, this is correct. I mean, I don't have the exact data of your waveforms, and I'm not going to run sample code just yet, but you seem to have the right method at least, which is encouraging.

For an instantaneous change in capacitance, charge is conserved. (Charge won't be conserved in your circuit as the resistor bleeds charge in/out, of course, but "instantaneous" covers for that, in that no charge flows through a resistor in an infinitesimal unit of time.) Because \$Q = VC\$, voltage and capacitance must vary inversely with each other.

Energy is not conserved, within the circuit itself, because we are applying work to the capacitor to change its value. Try it! When separating charged insulators (say, dissimilar plastics, rubbed together), you apply a force to separate them; or they have an attractive force once so separated. This is the basis of many electrostatic generators, perhaps most iconically the Whimshurst machine (which uses counter-rotating glass discs to convey charge from one side to another; the voltage increases by cyclically decreasing capacitance and synchronous rectification).

It's easily demonstrated in SPICE:

("Easy" may be relative to one's familiarity with SPICE, however!)

We use a pair of ideal sources to act as an ideal (DC) transformer; we can set B2 as product and leave B1 alone (linear), or vice versa. As shown, voltage varies with control and therefore charge is conserved and work is added to the system; with B1 = I(B2)*V(CTRL) and B2 = V(VARCAP), we get variable loading on the system (capacitance) but without energy input, which can be useful for, say, tuning control systems.

This can also happen due to nonlinear capacitances, which is a frequent cause of questions on this Stack: type 2 ceramic capacitors have a significant C(V) dependency, which when combined with an inductive input connection and step voltage (such as hot-plugging a cable), can cause voltage to overshoot significantly, often to the detriment of a regulator or other circuitry connected to it.