--- title: "Armature-controlled DC motor with gear and load" subject: "Control Systems" chapter: 2 key: "semester_3/control_systems:dc-motor-gear-load" source: https://www.bctengineer.com/q/semester_3/control_systems/dc-motor-gear-load site: BCT Engineering --- # Armature-controlled DC motor with gear and load Control Systems · Chapter 2: Component Modeling - Asked in: 2068 Chaitra, 10 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/dc-motor-gear-load - Whole chapter: https://www.bctengineer.com/md/semester_3/control_systems/chapter-2 ## Question For an electromechanical system shown below, derive an expression for VL(s) [?] considering it as armature controlled dc motor. Motor: i) Moment of inertia = Jm (ii) Frictional coefficient = Dm (iii) Torsional [rest not printed]. Load: i) Moment of inertia = Jm [?] (ii) Frictional coefficient = Dm [?] (iii) Torsional [rest not printed]. [Figure: armature circuit with supply ea, armature current Ia, resistance Ra, inductance La and back emf eb; field winding Lf, Rf with field current If; motor shaft (angle θm) drives the load (angle θL) through a gear pair N1:N2] ## Short answer Reflect load inertia and friction to motor shaft: Jeq = Jm + JL(N1/N2)², Deq = Dm + DL(N1/N2)². The load-angle transfer function is θL(s)/Ea(s) = (N1/N2)Kt / [s((Ra+sLa)(Jeq s+Deq) + KtKb)]. ## Answer The printed question is incomplete. It is read here as the standard problem: find the load-angle transfer function $\theta_L(s)/E_a(s)$ of an armature-controlled DC motor driving a load through a gear pair $N_1{:}N_2$. Motor data: inertia $J_m$, friction $D_m$; load data: inertia $J_L$, friction $D_L$; shafts are taken as rigid. #### Assumptions - Field current $I_f$ (through $L_f$, $R_f$) is constant, so flux is constant. - Torque $T_m = K_tI_a$; back emf $e_b = K_b\dot\theta_m$. - Gears are ideal (no loss, no backlash): $\theta_L = \dfrac{N_1}{N_2}\theta_m$. #### Step 1: Electrical equation $$ e_a = R_ai_a + L_a\frac{di_a}{dt} + e_b \;\Rightarrow\; I_a(s) = \frac{E_a(s) - K_bs\theta_m(s)}{R_a + sL_a} $$ #### Step 2: Reflect the load to the motor shaft Load impedances move to the motor side multiplied by $(N_1/N_2)^2$: $$ J_{eq} = J_m + J_L\left(\frac{N_1}{N_2}\right)^2, \qquad D_{eq} = D_m + D_L\left(\frac{N_1}{N_2}\right)^2 $$ #### Step 3: Mechanical equation at the motor shaft $$ T_m = K_tI_a = J_{eq}\ddot\theta_m + D_{eq}\dot\theta_m \;\Rightarrow\; K_tI_a(s) = s(J_{eq}s + D_{eq})\theta_m(s) $$ #### Step 4: Motor-angle transfer function Substituting $I_a$: $$ K_t\left[E_a - K_bs\theta_m\right] = s(R_a + sL_a)(J_{eq}s + D_{eq})\theta_m $$ $$ \frac{\theta_m(s)}{E_a(s)} = \frac{K_t}{s\left[(R_a + sL_a)(J_{eq}s + D_{eq}) + K_tK_b\right]} $$ #### Step 5: Load-angle transfer function $$ \frac{\theta_L(s)}{E_a(s)} = \frac{N_1}{N_2}\cdot\frac{K_t}{s\left[(R_a + sL_a)(J_{eq}s + D_{eq}) + K_tK_b\right]} $$ Expanded denominator: $s\left[L_aJ_{eq}s^2 + (R_aJ_{eq} + L_aD_{eq})s + R_aD_{eq} + K_tK_b\right]$. The load speed is $\Omega_L(s) = s\theta_L(s)$, so $\dfrac{\Omega_L(s)}{E_a(s)} = \dfrac{(N_1/N_2)K_t}{(R_a + sL_a)(J_{eq}s + D_{eq}) + K_tK_b}$. #### Block diagram ```text Ea + ┌──────┐Ia┌──┐Tm┌───────┐ωm┌───┐θm┌─────┐ θL ──►(Σ)►│ 1 ├─►│Kt├─►│ 1 ├┬►│1/s├─►│N1/N2├──► ▲- │Ra+sLa│ └──┘ │Jeq s+ ││ └───┘ └─────┘ │ └──────┘ │ Deq ││ │ Eb ┌────┐ └───────┘│ └─────────┤ Kb │◄────────────┘ └────┘ ``` #### Simplified form ($L_a \approx 0$) $$ \frac{\theta_L(s)}{E_a(s)} = \frac{K}{s(\tau s + 1)}, \quad K = \frac{(N_1/N_2)K_t}{R_aD_{eq} + K_tK_b}, \quad \tau = \frac{R_aJ_{eq}}{R_aD_{eq} + K_tK_b} $$ A large reduction ratio ($N_2 \gg N_1$) makes the load inertia and friction look much smaller to the motor, which is why gears are used for matching. ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. How is the load inertia reflected to the motor shaft? A. J_eq = J_m + J_L (N1/N2)^2 B. J_eq = J_m + J_L (N1/N2) C. J_eq = J_m + J_L (N2/N1) D. J_eq = J_m * J_L Answer: A. Load inertia and friction are multiplied by the square of the gear ratio N1/N2 when moved to the motor side. (Key point: Reflecting load) 2. Which equation holds at the motor shaft? A. K_t I_a = J_eq s^2 θm + D_eq s θm B. K_b I_a = J_eq s θm C. E_a = J_eq s^2 θm D. K_t I_a = D_eq θm Answer: A. Developed torque K_t I_a drives the equivalent inertia and friction: T_m = J_eq θm'' + D_eq θm'. (Key point: Mechanical equation) 3. How is θL(s)/Ea(s) obtained from θm(s)/Ea(s)? A. Multiply by N2/N1 B. Multiply by N1/N2 C. Divide by s D. Multiply by (N1/N2)^2 Answer: B. For ideal gears θL = (N1/N2) θm. (Key point: Load angle) 4. The denominator of θm/Ea is A. (R_a + sL_a)(J_eq s + D_eq) + K_t K_b B. s[(R_a + sL_a)(J_eq s + D_eq) + K_t K_b] C. s(R_a + sL_a) D. (J_eq s + D_eq) + K_t K_b Answer: B. The extra s comes from integrating speed to angle. 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