Electromagnetic actuator: an incremental converter with simple excitation

An electromagnet with a movable armature is the core of every relay, contactor and solenoid actuator. Its inductance depends on the air gap, so the electrical and mechanical parts are coupled. The current pulls the armature, and the moving armature induces a voltage in the coil. The paper derives the coupled equations with an electromechanical Lagrangian, simulates eleven cases in Simulink, and concludes that the device has a "relay-type", unstable characteristic. The equations are correct and the results are numerically sound, except one plot. However, all runs stay far below the voltage at which the armature snaps closed. Several interpretations of the parameter study are also wrong. The interactive simulator shows the paper's cases, the static force balance and the snap-through ("pull-in") that the paper only mentions.

Problem statement

For the incremental converter in the figure, with simple excitation of the magnetic circuit intended for motor action, determine the dynamic equations of motion. The inductance changes with the air gap $\delta$ according to

$$L(\delta) = \mu_0\frac{N^2s}{\delta}$$

Here $N$ is the number of turns, $s$ the cross-section of the core and $\mu_0$ the permeability of vacuum.

The magnetic circuit consists of a fixed iron core with the excitation winding (resistance $R$) and a movable armature.
The armature (mass $m$) is held by a spring $k$ and has viscous friction $B$.
With the spring unstretched the gap is $x_0$. The armature displacement $x$ closes the gap, $\delta = x_0 - x$.
The coil is fed by a voltage $u(t)$, either sinusoidal or constant.

The iron's permeability is assumed much larger than $\mu_0$, so all magnetic reluctance lies in the air gap. The gap never closes completely.

Figure 1. Iron core with an excitation winding (N turns, resistance R) and a movable armature (mass m) held by a spring k and a damper B; the displacement x closes the air gap δ = x₀ − x

Energies of the electromechanical system

The generalized coordinates are the charge $q$ (with $\dot q = i$) and the armature displacement $x$. In the course's terminology:

(2.3.1)–(2.3.4)$$\mathcal{R}^* = \tfrac12m\dot x^2,\qquad \mathcal{A} = \tfrac12kx^2,\qquad W_m^* = \tfrac12L(x)\,\dot q^2 = \frac{\mu_0N^2s}{2(x_0-x)}\,\dot q^2,\qquad W_e = 0$$

The magnetic coenergy $W_m^*$ of the coil plays the role of a kinetic coenergy (it depends on $\dot q$). There is no capacitor, so the electric energy is zero. The Lagrangian and the Rayleigh dissipation function are:

(2.4.2), (2.5.4)$$\mathcal{L} = \tfrac12m\dot x^2 + \frac{\mu_0N^2s}{2(x_0-x)}\dot q^2 - \tfrac12kx^2,\qquad \mathcal{D} = \tfrac12B\dot x^2 + \tfrac12R\dot q^2$$

The voltage source $u(t)$ is the generalized force of the coordinate $q$. The paper writes the inductance in terms of the displacement, $L = \mu_0N^2s/(x_0-x)$, which is the problem's $L(\delta)$ with $\delta = x_0 - x$. The paper does not say this explicitly, but it is correct.

Equations of motion

Lagrange's equations with dissipation, $\frac{d}{dt}\frac{\partial\mathcal{L}}{\partial\dot q_i} - \frac{\partial\mathcal{L}}{\partial q_i} + \frac{\partial\mathcal{D}}{\partial\dot q_i} = Q_i$, give:

(2.7.1)$$\frac{\mu_0N^2s}{x_0-x}\,\frac{di}{dt} + \frac{\mu_0N^2s}{(x_0-x)^2}\,\dot x\,i + Ri = u(t)$$

(2.7.2)$$m\ddot x + B\dot x + kx = \frac{\mu_0N^2s}{2(x_0-x)^2}\,i^2$$

Electrical equation. The first term is the self-induced voltage with an inductance that depends on position. The second term is the motional EMF $i\,\frac{dL}{dx}\dot x$.
Mechanical equation. The right-hand side is the magnetic (ponderomotive) force $\tfrac12i^2\frac{dL}{dx}$. It always pulls the armature towards the core, whatever the sign of the current, because it is proportional to $i^2$.

The derivation is correct and well explained. The paper's remark on the energy balance (half of the electrical input power becomes mechanical, half is stored in the field) holds for motion at constant current. In general the split depends on how the current changes during the motion.

Time scales

For the base parameters ($N=200$, $s=0.01$ m², $x_0=0.01$ m, $R=20\ \Omega$, $m=0.2$ kg, $k=100$ N/m, $B=10$ N·s/m) the system has three very different time scales:

Electrical: $L_0 = \mu_0N^2s/x_0 = 0.050$ H, so $\tau_e = L_0/R = 2.5$ ms.
Mechanical: $\omega_n = \sqrt{k/m} = 22.4$ rad/s and $\zeta = B/(2\sqrt{km}) = 1.12$, i.e. slightly overdamped.
Excitation: the sine source has $\omega = 1$ rad/s, Simulink's default; the paper does not state the frequency. The force varies at $2\omega = 2$ rad/s.

Since $\omega L_0 = 0.05\ \Omega \ll R$, the coil is practically a resistor: $i \approx u/R$. Since $2\omega \ll \omega_n$, the armature follows the force almost statically: $x \approx F/k$.

Steady state

The paper sets the mechanical derivatives to zero but keeps $\ddot q$ in the electrical equation:

(2.8.1)–(2.8.2)$$\frac{\mu_0N^2s}{x_0-x}\ddot q + R\dot q = u,\qquad kx = \frac{\mu_0N^2s}{2(x_0-x)^2}\dot q^2$$

This mixes a transient electrical equation with a static mechanical one. For a constant voltage $U$, the true steady state has $di/dt = 0$ as well:

$$i^* = \frac UR,\qquad kx^*(x_0-x^*)^2 = \frac{\mu_0N^2s}{2}\,\frac{U^2}{R^2}$$

This is a cubic equation for $x^*$. Depending on $U$ it has two solutions in the gap (one stable, one unstable) or none. In the latter case the armature snaps shut (section 7). For a sinusoidal voltage there is no static steady state, only a periodic one.

Simulink model and results

The model integrates

$$\ddot q = \frac{u(x_0-x)}{\mu_0N^2s} - \left(\frac{\dot x}{x_0-x} + \frac{R(x_0-x)}{\mu_0N^2s}\right)\dot q,\qquad \ddot x = \frac{\mu_0N^2s}{2m(x_0-x)^2}\dot q^2 - \frac kmx - \frac Bm\dot x$$

It uses five constant gains, product and square blocks, and a reciprocal block for $1/(x_0-x)$. The outputs are $q$, $i$ and $x$. Every case starts with $x(0) = 0.5$ mm (except case 9), $q = i = 0$, $U_m = 1$ V.

Case	Change	Paper's observation	Simulated steady x
1	base (m = 0.2)	reference	0 … 63 µm
2	B = 1	B does not affect q, i; larger B damps x faster	0 … 64 µm
3	N = 50	smaller N: smaller oscillation of x	0 … 3.9 µm
4	k = 20	"smaller k gives a smaller amplitude"	46 … 284 µm
5	m = 1	m has little effect	0 … 64 µm
6	m = 1, R = 100	smaller current and amplitude	0 … 2.5 µm
7	m = 1, s = 0.002	smaller amplitude	0 … 13 µm
8	m = 1, x₀ = 0.09	smaller amplitude, "less path to move"	0 … 0.8 µm
9	m = 1, x(0) = 1 mm	"smaller initial condition gives smaller amplitude"	0 … 64 µm
10	m = 1, DC E = 1 V	classic RL circuit; x settles faster	64 µm (constant)
11	m = 1, DC, B = 1	x rings, then settles	64 µm (constant)

The current (amplitude 50 mA = U/R) and the charge (0 to 0.1 C) are reproduced exactly in all cases, and so is $x$ in every case except case 4.

Critical review

The relay behaviour was never reached

The conclusion states that the system is unstable and has a relay-type characteristic. In all eleven simulations, however, the armature moves at most 64 µm, i.e. 0.6 % of the 10 mm gap, and returns smoothly. The snap-through that makes a relay a relay requires the armature to pass one third of the gap. For the base parameters that needs $U \gtrsim 4.9$ V (section 7). The paper uses 1 V throughout. The claim may be true, but the paper does not demonstrate it. The transfer function mentioned in the conclusion is never shown. A transfer function exists only for a linearized model around an operating point.

Stiffness: the conclusion has the wrong sign, and the plot does not match

Static deflection is $x = F/k$, so a softer spring ($k = 20$) must give a larger displacement, five times larger than the base case. The paper writes that a smaller $k$ gives a smaller amplitude "because a spring of higher stiffness stretches less", which contradicts itself. Its plot for $k = 20$ shows the slow decay expected for $k = 20$ (time constant ≈ $B/k = 0.5$ s). The steady ripple, however, stays below about 20 µm, while the model gives 46–284 µm, roughly 14 times more. That plot was apparently produced with different settings.

Initial condition: same amplitude, different axis

The steady oscillation is forced by $i^2$ and does not depend on $x(0)$. The initial condition only changes the decaying transient. The paper's own plot for $x(0)=1$ mm shows the same 0–64 µm ripple as the base case. Its vertical axis is scaled $\times10^{-3}$ instead of $\times10^{-4}$, which makes the ripple look smaller. The conclusion "smaller initial condition, smaller amplitude" is a misreading of the plot.

Gap and mass: right observation, wrong reason

Larger gap: the amplitude drops because the force falls as $1/(x_0-x)^2$, which is 81 times weaker for $x_0 = 0.09$ m. It is not because the armature has "less path to move".
Mass: it has no effect because the force changes slowly ($2\omega = 2$ rad/s) compared with $\omega_n$ (10–22 rad/s). The motion is quasi-static. Only near $2\omega \approx \omega_n$ would the mass matter.

Damping: forced, not "undamped" oscillations

With larger $B$ the transient decays faster. The remaining oscillation of $x$ is the forced response to the force $\propto i^2 \propto \sin^2\omega t$, at twice the supply frequency and never negative. Calling it "undamped oscillations" and a "stable position" mixes up two different things. That N, B and m barely affect the current follows from $\omega L \ll R$, which the paper does not note.

Smaller points

The steady-state equations keep $\ddot q$ (section 4). The sine frequency is not given. The DC current plot shows a straight line at 50 mA without the 2.5 ms rise, which is simply too fast to see on a 20 s axis.

What is goodThe electromechanical Lagrangian, with the magnetic coenergy and the Rayleigh function for both domains, is set up correctly and clearly. The physical interpretation of the motional EMF and the magnetic force is good. The Simulink model is correct, and the parameter study is systematic.

Extension: static characteristic and pull-in

For a constant current $I$, the armature is in equilibrium where the spring force equals the magnetic force:

$$kx = \frac{\mu_0N^2s\,I^2}{2(x_0-x)^2}$$

The magnetic force grows faster than the spring force as the gap closes. Its slope is a "negative stiffness" $dF/dx = \mu_0N^2sI^2/(x_0-x)^3$. An equilibrium is stable only while the spring is stiffer than this, which holds for $x < x_0/3$. The largest current with an equilibrium is reached at $x = x_0/3$:

$$I_\text{pi} = \sqrt{\frac{8kx_0^3}{27\,\mu_0N^2s}},\qquad U_\text{pi} = R\,I_\text{pi}$$

For the base parameters, $I_\text{pi} = 0.243$ A and $U_\text{pi} = 4.86$ V. Above this voltage no equilibrium exists in the gap. The armature accelerates ever faster and slams into the core (in practice onto a thin non-magnetic stop). When the voltage is then reduced, the armature stays closed until the current drops far below $I_\text{pi}$, because the force is much larger with a small gap. This hysteresis is exactly the relay characteristic the paper refers to.

The simulator includes a mechanical stop at a residual gap of 5 % of $x_0$. It plots the static characteristic for the present DC current, showing both curves and the stable and unstable equilibria. The presets "Below pull-in" and "Pull-in" show the two regimes. During pull-in the current visibly dips: the fast-moving armature induces a back EMF $i\,\dot x\,dL/dx$, and the inductance itself grows as the gap closes. This dip is used in practice to detect that a solenoid has actually moved.

Interactive simulator

The simulator solves (2.7.1)–(2.7.2) with 4th-order Runge–Kutta (step 0.05 ms, small enough for the 2.5 ms electrical time constant). The armature is stopped by an inelastic stop at $x = 0.95\,x_0$.

Cases

Voltage

Parameters

U [V]

ω [rad/s]

t_off [s]

R [Ω]

N [turns]

s [m²]

x₀ [m]

m [kg]

k [N/m]

B [N·s/m]

x(0) [m]

Duration [s]

Current i(t)

Armature displacement x(t)

Magnetic force F(t)

Static characteristic

Spring force kx and magnetic force at I = U/R over the gap.

Animation

Results table

Review questions

Why does the magnetic coenergy appear in the Lagrangian with a plus sign, like kinetic energy?

In the charge formulation, $\dot q$ plays the role of a velocity, and the inductor's coenergy $\tfrac12L\dot q^2$ is the analogue of $\tfrac12m\dot x^2$. A capacitor's energy $q^2/2C$ would be the analogue of a spring's potential energy.

Why does the armature oscillate at twice the supply frequency?

The force is proportional to $i^2 \propto \sin^2\omega t = \tfrac12(1-\cos2\omega t)$: a constant pull plus a component at $2\omega$. That is also why AC relays hum at 100 Hz on a 50 Hz grid.

Why do N, B and m hardly affect the current in the paper's cases?

The reactance $\omega L$ (about 0.05 Ω) is tiny compared with $R = 20\ \Omega$, and the motional EMF is tiny because the armature barely moves. The current is set by Ohm's law, $i \approx u/R$.

What is the pull-in voltage, and why is the critical position $x_0/3$?

The largest current for which an equilibrium exists comes from maximizing $kx(x_0-x)^2$. That maximum is at $x = x_0/3$, giving $U_\text{pi} = R\sqrt{8kx_0^3/(27\mu_0N^2s)}$. Beyond it the magnetic force grows faster than the spring force and the armature snaps shut.

Why does a closed relay stay closed when the voltage is reduced somewhat?

With a small gap, the force $\propto I^2/\delta^2$ is much larger than at the open position. The current must drop far below the pull-in value before the spring can open the armature again. This gives hysteresis between pull-in and drop-out.

When would the mass of the armature matter?

When the force varies at a frequency comparable with $\omega_n = \sqrt{k/m}$, e.g. $2\omega\approx\omega_n$, or during fast switching. In the paper's cases ($2\omega = 2$ rad/s) the motion is quasi-static.