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Feedback effect in DC motor speed control

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.

How would you consider the effect of feedback in speed control of an armature-controlled DC motor in comparison to open loop? Elaborate analytically.

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 (LaL_a neglected): Va=RaIa+KbωV_a = R_aI_a + K_b\omega, T=KtIa=Jω˙+Bω+TLT = K_tI_a = J\dot\omega + B\omega + T_L. Hence

ω(s)=Kmτms+1Va(s)−KLτms+1TL(s)\omega(s) = \frac{K_m}{\tau_m s+1}V_a(s) - \frac{K_L}{\tau_m s+1}T_L(s) Km=KtRaB+KtKb,KL=RaRaB+KtKb,τm=RaJRaB+KtKbK_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: ΔωOL=KLTL\Delta\omega_{OL} = K_LT_L. Any change in RaR_a, KtK_t or BB changes the speed directly.

Closed-loop scheme

 ωr  +    e  +-----+ Va +-------+   ω
 --->(Σ)---->| KA  |--->| Motor |---+--->
      ^ -    +-----+    +-------+   |
      |      +-------+              |
      +------| Kg    |<-------------+
             | tacho |
             +-------+

With amplifier gain KAK_A and tacho constant KgK_g, the loop gain is K=KAKmKgK = K_AK_mK_g:

ω(s)ωr(s)=KAKmτms+1+K,ω(s)TL(s)=−KLτms+1+K\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

QuantityOpen loopClosed loop
Time constantτm\tau_mτm/(1+K)\tau_m/(1+K)
Speed drop for load TLT_LKLTLK_LT_LKLTL/(1+K)K_LT_L/(1+K)
Sensitivity to KmK_m11/(1+K)1/(1+K)
GainKmK_mKAKm/(1+K)K_AK_m/(1+K)
  1. Faster response: the time constant is reduced by (1+K)(1+K).
  2. Better speed regulation: the load-torque disturbance causes a speed drop (1+K)(1+K) times smaller.
  3. Less sensitivity to changes in RaR_a (heating), flux or friction.
  4. Cost: lower overall gain (made up by KAK_A), a tachogenerator is needed, and with armature inductance included, very high KK can cause oscillation.

Example: with K=19K = 19, a 100 rpm open-loop drop on load becomes only 5 rpm.

✓ Answer checked by Claude Opus 5.5 (high effort)

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