The dc motor: two time constants and three limits

Electricity becomes motion

• The loop's valve is turned by a small motor through a gearbox. Its two time constants and its limits (voltage, current, speed) decide how fast the valve can move.
• A permanent-magnet dc motor joins its circuit and rotor with two laws: torque $T = Ki$ and back emf $e = K\omega$. In SI units the two constants are equal, because $ei = T\omega$.

Speed per volt

• Eliminating the current: $\Omega/V = K/[(Js + b)(Ls + R) + K^2]$.
• With small $L$ it is first order, with gain $K/(bR + K^2)$ and mechanical time constant $\tau_m = JR/(bR + K^2) \approx JR/K^2$; the electrical one is $\tau_e = L/R$.

The valve motor

Through the gearbox

• With $N = 50$ the valve sees $\mathbf{0.398}$ rad/s per volt, so at 12 V it turns at most 4.78 rad/s: a 90° move takes at least $\mathbf{0.33}$ s.
• Through the 80%-efficient gearbox a 5 N·m load reaches the motor as $5/(50 \times 0.8) = 0.125$ N·m and costs $\mathbf{3.0}$ V. A 5 A current limit allows 10 N·m at the valve: the linear model is valid only below it.

The torque–speed line

• In steady state $V = iR + K\omega$ and $T = Ki$, so $\omega = V/K - TR/K^2$: speed falls linearly with load. This torque–speed characteristic 转矩-转速特性 runs from 240 rad/s with no load to 0.5 N·m at stall.
• Output power $T\omega$ peaks at half speed and half torque: $0.25 \times 0.5 \times 240 = \mathbf{30}$ W. That is the most this 12 V motor can deliver, whatever the gearbox.

Worked example: the motor under load

• The valve's load reaches the motor as 0.125 N·m, and friction adds $b\omega$. Solving $Ki = 0.125 + b\omega$ with $12 = 1.2i + 0.05\omega$ gives $i = \mathbf{2.54}$ A and $\omega = 179$ rad/s: $\mathbf{3.58}$ rad/s at the valve.
• It draws $12 \times 2.54 = 30.4$ W and delivers $0.125 \times 179 = 22.4$ W to the gearbox: 74% efficient, with 7.7 W lost in the copper as $i^2R$.

Three motor slips

Recap