Dynamic model of armature-controlled DC motor

In a separately excited DC motor under armature control, the field current $i_f$ is held constant, so the air-gap flux $\phi$ is constant, and the armature voltage $v_a$ is the control input.

Symbols

$R_a$, $L_a$: armature resistance, inductance; $i_a$: armature current; $e_b$: back emf; $T_m$: developed torque; $T_L$: load torque; $J$: total inertia; $B$: viscous friction; $\omega = \dot\theta$: speed.

Governing differential equations

1. Armature circuit (KVL):

$$v_a = R_ai_a + L_a\frac{di_a}{dt} + e_b$$

2. Back emf (proportional to flux and speed; flux constant):

$$e_b = K\phi\,\omega = K_b\,\omega$$

3. Developed torque:

$$T_m = K\phi\,i_a = K_t\,i_a$$

4. Mechanical (Newton's law for rotation):

$$T_m = J\frac{d\omega}{dt} + B\omega + T_L$$

5. Position: $\dfrac{d\theta}{dt} = \omega$

In SI units $K_t = K_b = K$.

Dynamic model in state-space form

State variables $x_1 = i_a$, $x_2 = \omega$, $x_3 = \theta$; inputs $v_a$ and $T_L$:

$$\begin{bmatrix}\dot i_a\\ \dot\omega\\ \dot\theta\end{bmatrix} = \begin{bmatrix} -\frac{R_a}{L_a} & -\frac{K_b}{L_a} & 0 \\ \frac{K_t}{J} & -\frac{B}{J} & 0 \\ 0 & 1 & 0 \end{bmatrix} \begin{bmatrix} i_a\\ \omega\\ \theta\end{bmatrix} + \begin{bmatrix} \frac{1}{L_a} & 0 \\ 0 & -\frac{1}{J} \\ 0 & 0\end{bmatrix} \begin{bmatrix} v_a \\ T_L\end{bmatrix}$$

Output: $y = \omega$ (speed) or $y = \theta$ (position).

Transfer-function model (with $T_L = 0$)

Taking Laplace transforms:

$$\begin{aligned} I_a(s) &= \frac{V_a(s) - K_b\Omega(s)}{R_a + sL_a}, \qquad \Omega(s) = \frac{K_tI_a(s)}{Js + B} \end{aligned}$$ $$\frac{\Omega(s)}{V_a(s)} = \frac{K_t}{(R_a + sL_a)(Js + B) + K_tK_b}, \qquad \frac{\theta(s)}{V_a(s)} = \frac{K_t}{s\left[(R_a + sL_a)(Js + B) + K_tK_b\right]}$$

Block diagram of the dynamic model

 Va +    ┌───────┐ Ia ┌────┐ Tm +   ┌──────┐ ω  ┌───┐ θ
 ──►(Σ)─►│  1    ├───►│ Kt ├───►(Σ)►│  1   ├─┬─►│1/s├─►
    ▲ -  │Ra+sLa │    └────┘    ▲-  │Js+B  │ │  └───┘
    │    └───────┘              │   └──────┘ │
    │ Eb         ┌────┐        TL            │
    └────────────┤ Kb ├◄─────────────────────┘
                 └────┘

Remarks

• Electrical time constant $\tau_a = L_a/R_a$; mechanical time constant $\tau_m = J/B$. Usually $\tau_a \ll \tau_m$, so $L_a$ may be neglected, giving $\dfrac{\Omega}{V_a} = \dfrac{K_m}{1 + s\tau}$ with $K_m = \dfrac{K_t}{R_aB + K_tK_b}$, $\tau = \dfrac{R_aJ}{R_aB + K_tK_b}$.
• The back emf provides inherent speed feedback, which makes the motor more stable than a field-controlled motor.