Condenser microphone: an R–L–C circuit with a moving plate

Problem statement

Energies, Lagrangian and dissipation

The paper sets these up exactly like this. Because the charge \(q\) is the coordinate, the electric energy \(W_e(q,x)\) is the right function, and it enters with a minus sign, like a potential. The only dissipation is the resistor. The paper writes its Rayleigh function as \(\int_0^{q}Rq\,dq\), which is a slip of notation: the integration variable is the current, \(\mathcal{D} = \int_0^{\dot q}R\dot q\,d\dot q = \tfrac12R\dot q^2\). The result it gives is correct.

The generalized forces are \(f(t)\) for \(x\) and \(E\) for \(q\).

Equations of motion

• Equation (1) is Newton's law for the plate. The term \(q^2/(2C_0a) = q^2/(2\varepsilon S)\) is the familiar attractive force between the plates. It pulls the plate towards the fixed plate (\(+x\)) for either sign of the charge.
• Equation (2) is Kirchhoff's voltage law: \(u_L + u_R + q/C(x) = E\).
• The coupling is the \(x\)-dependent capacitance: the force depends on \(q\), and the capacitor voltage depends on \(x\).

Both equations agree with the paper's (2.14) and (2.16) and with its Simulink model (Figure 3.1).

Equilibrium, done properly

The paper finds the "steady state" by setting the derivatives to zero but keeps \(f(t)\). An equilibrium exists only for a constant force, so take \(f = 0\):

This exposes a small inconsistency in the problem statement. If the springs are relaxed at \(x = 0\), as the equations assume, then the charged capacitor already pulls the plate to \(x^* > 0\). The capacitance at equilibrium is then not \(C_0\) and the gap is not \(a\). For the paper's first case \(x^* \approx 0.020\) m, which is 4 % of the gap, so the error is small. Strictly, the springs carry a preload \(kx_p = q_0^2/(2C_0a)\) with \(q_0 = EC_0\). Then \(x = 0\) is the true equilibrium and (1) reads \(m\ddot x + kx = f(t) + (q^2 - q_0^2)/(2C_0a)\).

Simulink model and results

The Simulink model solves (1) for \(\ddot x\) and (2) for \(\ddot q\), with a sine block for \(f\), a step block for \(E\) and zero initial conditions. The text states \(E = 5\) V, \(F_0 = 5\) N and \(\omega = 5\) rad/s for all three cases. Our simulation of the paper's equations reproduces all six plots to the plotted accuracy, but only with \(F_0 = 1\) N:

The match with \(F_0 = 1\) N is exact in every extreme value, so the plots were almost certainly made with the sine block's default amplitude of 1. The paper's own explanation of the parameter study then says:

• the responses "did not change significantly" between cases 1 and 2;
• raising \(R\) made the charge amplitude larger;
• \(a\) is the parameter that most "stabilizes" the charge;
• the system "could only be modelled with very large parameter values", and the input values were not changed "so as not to cause instability".

The next section shows that none of these is the real explanation.

What the plots actually show

1. The mechanics is an ordinary forced oscillator

In case 1 the electric force is \(q^2/(2C_0a) \approx 0.011^2/(2\cdot0.0022\cdot0.5) = 0.055\) N, compared with a spring force of \(kx \approx 0.6\) N and a drive of 1 N. Neglecting it, equation (1) is an undamped oscillator started from rest:

With \(\omega_n = 1.83\) rad/s and \(\omega = 5\) rad/s, this is the beat-like sum of two sines seen in Figure 4.1. Nothing decays, because the plate has no damping. The free oscillation stays forever. This is why the shape of \(x(t)\) depends almost only on \(F_0\), \(m\), \(k\) and \(\omega\), and why the electrical parameters hardly change it.

2. The charge simply follows the plate

The circuit is fast: \(\omega_e = 1/\sqrt{LC_0} = 213\) rad/s and the decay rate is \(R/(2L) = 50\) s⁻¹ in case 1, and 500 s⁻¹ in cases 2 and 3. After a few tens of milliseconds the capacitor is always charged to the source voltage:

In case 1 the maximum \(x = 0.219\) m gives \(q = 0.011/(1-0.438) = 0.0196\) C, exactly the peak in Figure 4.2. The minimum \(x = -0.177\) m gives 0.0081 C. Equation (4) explains every trend the paper reports:

• The charge does not grow with \(R\). In case 2 the charge varies less because \(x\) varies less (stiffer spring, heavier plate). \(R\) only sets how fast the charge catches up.
• What "stabilizes" \(q\) is the ratio \(x/a\). In case 3, \(x/a\) is only ±0.19 and \(C_0\) is ten times smaller, so the charge is small and nearly constant.
• The asymmetry of the \(q\) plot (sharp peaks, flat valleys) comes from \(1/(1 - x/a)\).

3. The "instability" is a collision

The model is only valid while \(x < a\). If the drive pushes the plate further, it hits the fixed plate and the simulation diverges, because \(C(x)\to\infty\). With the stated \(F_0 = 5\) N, case 1 has a static deflection \(F_0/k = 1.67\) m, more than three times the gap. Even at \(\omega = 5\) rad/s the plate reaches the fixed plate after 0.54 s. At a low frequency such as \(\omega = 1\) rad/s even 1 N is too much (preset in the simulator). The real design rule is not "large parameters" but

A real microphone satisfies it with micrometre gaps and millipascal pressures (section 7).

Extension: electrostatic pull-in

The second way to lose the gap is electrical. The spring force grows linearly with \(x\), but the electrostatic force from (3) grows like \(1/(a-x)^2\). Equilibrium (3) exists only up to

Above \(E_{PI}\) the plate snaps onto the fixed plate whatever the force. For the paper's three cases \(E_{PI}\) = 10.1 V, 18.3 V and 60.3 V, so the chosen \(E = 5\) V is safely below, but not by much in case 1.

Two further effects follow from the same equations:

• Electrostatic softening. Below pull-in the charged capacitor acts as a negative spring. For slow motion, where \(q \approx EC(x)\), the effective stiffness is \(k_{ef} = k - E^2C_0a/(a-x^*)^3\). It falls to zero at \(E_{PI}\), and the resonance \(\sqrt{k_{ef}/m}\) drops as \(E\) rises.
• Dynamic pull-in. Switching on the voltage as a step (the paper's Simulink model) makes the undamped plate overshoot to twice its static deflection. It then collapses already at about \(0.92\,E_{PI}\). In case 1, \(E = 9.5\) V switched on from rest collapses after about 3 s, while the same voltage applied to a plate already at equilibrium is stable (presets "E = 9.5 V").

Extension: how a real condenser microphone works

The paper never asks the question that makes this system a microphone: what is the output signal? It is the voltage across the resistor, \(u_R = R\dot q\) (equivalently the AC part of the capacitor voltage, since \(u_C + u_R = E\) when \(L\) is negligible). Linearizing (1)–(2) about the equilibrium, with \(c = E/(a-x^*)\) (the field between the plates) and \(C^* = C(x^*)\), gives in the Laplace domain:

Two limits show the two operating modes:

• Small \(R\) (the paper's case, \(RC^* \ll\) period): the source keeps the voltage constant, \(q \approx EC(x)\), and \(u_R \approx (REC_0/a)\,\dot x\). The output is proportional to plate velocity and small. In case 1 it peaks at about 18 mV.
• Large \(R\) (real microphones, \(R \sim 1\) GΩ): the charge cannot follow, so \(q \approx\) constant \(= EC_0\). The capacitor voltage changes as \(u_C = q(a-x)/(C_0a)\), so the AC output is

A sound pressure \(p\) gives \(f = pS\). The constant-charge mode gives a flat response from the corner \(1/(2\pi RC_0)\) up to the diaphragm resonance. That is why studio microphones use a gigaohm bias resistor and a membrane tuned near the top of the audio band. The calculator below uses the linearized model with \(L = 0\). The default values are typical for a microphone with a 12 mm capsule and give a sensitivity of tens of mV/Pa, matching real data sheets.

Critical review

Interactive simulator

The simulator integrates (1)–(2) with 4th-order Runge–Kutta, with a time step adapted to the fast electrical circuit. An optional plate damper \(B\) is added to (1). The simulation stops if the plate touches the fixed plate (\(x \geq 0.98a\)). The dashed line in the charge plot is the quasi-static approximation (4).

Review questions