--- title: "Feedback effect in DC motor speed control" subject: "Control Systems" chapter: 1 key: "semester_3/control_systems:feedback-dc-motor-speed" source: https://www.bctengineer.com/q/semester_3/control_systems/feedback-dc-motor-speed site: BCT Engineering --- # Feedback effect in DC motor speed control Control Systems · Chapter 1: Control System Background - Asked in: 2082 Baisakh, 6 marks - Video: [DC Motor Speed Control System | Speed Control of DC Motor](https://www.youtube.com/watch?v=SyGzqR67Yc8) — Sync With Me - Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14 - Page: https://www.bctengineer.com/q/semester_3/control_systems/feedback-dc-motor-speed - Whole chapter: https://www.bctengineer.com/md/semester_3/control_systems/chapter-1 ## Question How would you consider the effect of feedback in speed control of an armature-controlled DC motor in comparison to open loop? Elaborate analytically. ## Short answer In open-loop, the motor’s speed drops directly with load torque and changes with parameter variations, with time constant tau_m. Closed-loop (with tachogenerator feedback) reduces the time constant to tau_m/(1+K) and the speed drop due to load torque to K_L T_L/(1+K), improving speed regulation and making it less sensitive to parameter changes, where K = amplifier gain times motor and tacho constants. ## Answer Feedback (a tachogenerator measuring the speed) makes an armature-controlled DC motor respond faster and keeps its speed nearly constant despite load and parameter changes. #### Open-loop model Motor equations ($L_a$ neglected): $V_a = R_aI_a + K_b\omega$, $T = K_tI_a = J\dot\omega + B\omega + T_L$. Hence $$ \omega(s) = \frac{K_m}{\tau_m s+1}V_a(s) - \frac{K_L}{\tau_m s+1}T_L(s) $$ $$ K_m = \frac{K_t}{R_aB + K_tK_b}, \quad K_L = \frac{R_a}{R_aB + K_tK_b}, \quad \tau_m = \frac{R_aJ}{R_aB + K_tK_b} $$ Steady-state speed drop on load: $\Delta\omega_{OL} = K_LT_L$. Any change in $R_a$, $K_t$ or $B$ changes the speed directly. #### Closed-loop scheme ```text ωr + e +-----+ Va +-------+ ω --->(Σ)---->| KA |--->| Motor |---+---> ^ - +-----+ +-------+ | | +-------+ | +------| Kg |<-------------+ | tacho | +-------+ ``` With amplifier gain $K_A$ and tacho constant $K_g$, the loop gain is $K = K_AK_mK_g$: $$ \frac{\omega(s)}{\omega_r(s)} = \frac{K_AK_m}{\tau_m s + 1 + K}, \qquad \frac{\omega(s)}{T_L(s)} = \frac{-K_L}{\tau_m s + 1 + K} $$ #### Comparison | Quantity | Open loop | Closed loop | |---|---|---| | Time constant | $\tau_m$ | $\tau_m/(1+K)$ | | Speed drop for load $T_L$ | $K_LT_L$ | $K_LT_L/(1+K)$ | | Sensitivity to $K_m$ | 1 | $1/(1+K)$ | | Gain | $K_m$ | $K_AK_m/(1+K)$ | 1. **Faster response:** the time constant is reduced by $(1+K)$. 2. **Better speed regulation:** the load-torque disturbance causes a speed drop $(1+K)$ times smaller. 3. **Less sensitivity** to changes in $R_a$ (heating), flux or friction. 4. **Cost:** lower overall gain (made up by $K_A$), a tachogenerator is needed, and with armature inductance included, very high $K$ can cause oscillation. Example: with $K = 19$, a 100 rpm open-loop drop on load becomes only 5 rpm. ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. With tacho feedback and loop gain K = KA Km Kg, the motor time constant tau_m becomes A. tau_m (1+K) B. tau_m/(1+K) C. tau_m/K D. tau_m Answer: B. The denominator tau_m s + 1 + K gives a time constant tau_m/(1+K). (Key point: Closed-loop time constant) 2. If the open-loop speed drop on load is 100 rpm, what is it with K = 19? A. 100 rpm B. 19 rpm C. 10 rpm D. 5 rpm Answer: D. The drop is divided by (1+K) = 20, so 100/20 = 5 rpm. (Key point: Load regulation) 3. The sensitivity of closed-loop speed to Km (for example due to heating of Ra) is A. 1 B. 1+K C. 1/(1+K) D. K Answer: C. Feedback reduces sensitivity from 1 to 1/(1+K). (Key point: Sensitivity to motor parameters) 4. Which is a drawback of the closed-loop motor scheme? A. Larger load-speed drop B. Lower overall gain, needs a tachogenerator, and very high K can cause oscillation C. Slower response D. Higher sensitivity to Ra Answer: B. The gain falls to KA Km/(1+K) (recovered with KA), a sensor is needed, and high gain can destabilize. (Key point: Cost of feedback) ## Related in Control Systems - [Armature-controlled DC motor transfer function and block diagram](https://www.bctengineer.com/question.md?key=semester_3%2Fcontrol_systems%3Atf-armature-dc-motor) — Control Systems, chapter 2 - [Dynamic model of armature-controlled DC motor](https://www.bctengineer.com/question.md?key=semester_3%2Fcontrol_systems%3Adc-motor-de-dynamic-model) — Control Systems, chapter 2 - [Armature-controlled DC motor with gear and load](https://www.bctengineer.com/question.md?key=semester_3%2Fcontrol_systems%3Adc-motor-gear-load) — Control Systems, chapter 2 - [Closed vs open loop step response analytically](https://www.bctengineer.com/question.md?key=semester_3%2Fcontrol_systems%3Acl-vs-ol-step-response) — Control Systems, chapter 1, asked 2× - [DC motor with tachometer feedback and P-controller](https://www.bctengineer.com/question.md?key=semester_3%2Fcontrol_systems%3Abd-dc-motor-tacho-p) — Control Systems, chapter 2 - [Effect of feedback on gain and speed of response](https://www.bctengineer.com/question.md?key=semester_3%2Fcontrol_systems%3Afeedback-gain-speed) — Control Systems, chapter 1 ## Related in other subjects - [Two quadrant armature controlled DC drive; closed loop](https://www.bctengineer.com/question.md?key=semester_7%2Futilization_of_electrical_energy%3Atwo-quadrant-armature-control-loop) — Utilization of Electrical Energy, chapter 3 - [Single-phase fully controlled converter DC motor speed](https://www.bctengineer.com/question.md?key=semester_7%2Futilization_of_electrical_energy%3Afull-converter-dc-motor-speed) — Utilization of Electrical Energy, chapter 3 - [Speed control methods of DC shunt motor](https://www.bctengineer.com/question.md?key=semester_4%2Felectrical_machine%3Ashunt-motor-speed-control-methods) — Electrical Machine, chapter 4, asked 2× - [Speed control methods of shunt DC motor](https://www.bctengineer.com/question.md?key=semester_4%2Felectrical_machines%3Adc-motor-speed-control-methods) — Electrical Machines, chapter 4, asked 2×