--- title: "Armature-controlled DC motor transfer function and block diagram" subject: "Control Systems" chapter: 2 key: "semester_3/control_systems:tf-armature-dc-motor" source: https://www.bctengineer.com/q/semester_3/control_systems/tf-armature-dc-motor site: BCT Engineering --- # Armature-controlled DC motor transfer function and block diagram Control Systems · Chapter 2: Component Modeling - Asked in: 2080 Bhadra, 6 marks - Video: [Transfer Function & Block Diagram of Armature Controlled D.C motor](https://www.youtube.com/watch?v=3qmtb7t_AUk) — Saibal Manna - Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14 - Page: https://www.bctengineer.com/q/semester_3/control_systems/tf-armature-dc-motor - Whole chapter: https://www.bctengineer.com/md/semester_3/control_systems/chapter-2 ## Question Find the transfer function of armature controlled dc motor and also draw the block diagram of same. [Figure: armature circuit with supply va, armature current ia, resistance Ra and inductance La feeding the motor; field winding Lf with constant field voltage vf (field current if); motor shaft (angle θ) drives a load with inertia J and friction B] ## Short answer The transfer function is θ(s)/Va(s) = Kt / [s{(Ra+sLa)(Js+B) + Kt Kb}]. With negligible armature inductance it simplifies to Km/[s(1+sτm)] where Km = Kt/(RaB+KtKb) and τm = RaJ/(RaB+KtKb). The block diagram includes an inner speed feedback loop from back emf. ## Answer In an **armature-controlled DC motor** the field current $i_f$ is kept constant (constant $v_f$) and the speed/position is controlled by changing the armature voltage $v_a$. #### Assumptions and symbols - $R_a, L_a$: armature resistance and inductance; $i_a$: armature current - $e_b$: back emf; $K_b$: back-emf constant - $T_m$: motor torque; $K_t$: torque constant - $J, B$: inertia and viscous friction of motor plus load; $\theta$: shaft angle - Flux $\phi \propto i_f$ is constant, so torque and back emf are linear. #### Governing equations 1. Armature circuit (KVL): $$ v_a = R_a i_a + L_a \frac{di_a}{dt} + e_b $$ 2. Back emf: $e_b = K_b \dfrac{d\theta}{dt}$ 3. Torque: $T_m = K_t\, i_a$ 4. Load (Newton's law): $T_m = J\dfrac{d^2\theta}{dt^2} + B\dfrac{d\theta}{dt}$ #### Laplace transform (zero initial conditions) $$ \begin{aligned} I_a(s) &= \frac{V_a(s) - E_b(s)}{R_a + sL_a} \\ T_m(s) &= K_t I_a(s) \\ \theta(s) &= \frac{T_m(s)}{s(Js + B)} \\ E_b(s) &= K_b\, s\,\theta(s) \end{aligned} $$ #### Block diagram ```text Va + ┌──────┐ Ia ┌──┐ Tm ┌─────┐ ω ┌───┐ θ ──►(Σ)─►│ 1 ├───►│Kt├───►│ 1 ├─┬─►│1/s├─► ▲ - │Ra+sLa│ └──┘ │Js+B │ │ └───┘ │ └──────┘ └─────┘ │ │ Eb ┌────┐ │ └───────────┤ Kb │◄──────────────┘ └────┘ ``` The back emf forms an internal negative feedback loop on speed $\omega = s\theta$. #### Transfer function Reducing the inner loop (forward path $G = \dfrac{K_t}{(R_a+sL_a)(Js+B)}$, feedback $K_b$): $$ \frac{\omega(s)}{V_a(s)} = \frac{K_t}{(R_a + sL_a)(Js + B) + K_t K_b} $$ Multiplying by the integrator $1/s$: $$ \frac{\theta(s)}{V_a(s)} = \frac{K_t}{s\left[(R_a + sL_a)(Js + B) + K_t K_b\right]} $$ If $L_a$ is negligible, this becomes $\dfrac{\theta(s)}{V_a(s)} = \dfrac{K_m}{s(1 + s\tau_m)}$ with $K_m = \dfrac{K_t}{R_aB + K_tK_b}$ and $\tau_m = \dfrac{R_aJ}{R_aB + K_tK_b}$, a type-1 second-order system. ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. What defines an armature-controlled DC motor? A. Field current is constant and speed/position is controlled by armature voltage B. Armature current is constant and field voltage is varied C. Both field and armature voltages are varied together D. The load torque is varied while the voltage is constant Answer: A. Field current is held constant so flux is constant, and the armature voltage v_a is the control input. (Key point: Armature control principle) 2. How does the back emf enter the block diagram? A. As a disturbance added to the load torque B. As negative feedback of speed (Kb s theta) to the armature summing point C. As positive feedback of angle theta D. As an extra forward gain Kt Answer: B. e_b = Kb dtheta/dt opposes v_a, so it forms an internal negative feedback loop on speed. (Key point: Back emf feedback) 3. What is theta(s)/Va(s) for the armature-controlled motor? A. Kt / (s[(Ra + sLa)(Js + B)]) B. Kt / ((Ra + sLa)(Js + B) + Kt Kb) C. Kt / (s[(Ra + sLa)(Js + B) + Kt Kb]) D. Kb / (s[(Ra + sLa)(Js + B) + Kt Kb]) Answer: C. The inner loop gives omega/Va; multiplying by the integrator 1/s gives the angle transfer function. (Key point: Overall transfer function) 4. If La is negligible, theta/Va reduces to Km/(s(1 + s tau_m)). This is: A. A type-1 second-order system B. A type-2 second-order system C. A type-0 first-order system D. A type-1 first-order system Answer: A. There is one pole at the origin (type 1) plus one more pole from 1 + s tau_m, giving second order. 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