Classification of DC Motor Speed Control Methods

DC motors operate based on Faraday's law of electromagnetic induction. When a current passes through the armature winding placed in a magnetic field, it generates torque causing the motor to rotate. The speed of the motor can be altered by changing either the applied voltage or the field flux.

Factors Controlling Motor Speed:

E = PØNZ/60A

• From above formula it is clear that, the speed of DC Motor can controlled by:
• By variation in flux (Φ) (Flux Control Method)
• By changing armature resistance (Ra) of armature circuit (Rheostatic Control) and
• Applied voltage control (Voltage Control Method)

Table of Contents

1. Speed Control of Shunt Motors
   1. (i) Flux Control Method (Field Control)
   2. (ii) Armature or Rheostatic Control Method
   3. (iii) Voltage Control Method
      1. (a) Multiple Voltage Control
      2. (b) Ward–Leonard System
2. Speed Control of Series Motors
   1. (i) Flux Control Method
      1. (a) Field Divertor
      2. (b) Armature Divertor
      3. (c) Tapped Field Control
      4. (d) Paralleling Field Coils
   2. (ii) Variable Resistance in Series with Motor
   3. (iii) Series–Parallel Control (Traction Use)
3. Speed Control Methods of DC Motors — Advantages & Disadvantages

Speed Control of Shunt Motors

(i) Variation of Flux (Field Control Method)

From the speed equation of a DC motor, speed is inversely proportional to flux: N ∝ E_b / (Φ). With a nearly constant supply, decreasing field flux increases speed and increasing flux reduces speed.

How flux is varied

The field flux (Φ) of a DC shunt motor is adjusted by changing the shunt field current (I_sh) using a shunt field rheostat. Since I_sh is relatively small, the rheostat carries small current and has low I²R loss, making the method efficient.

Practical limits

• Non-interpolar machines: typical safe speed increase up to about 2 : 1 by weakening flux.
• With interpoles: a maximum-to-minimum speed ratio of around 6 : 1 is fairly common.
• Too much flux weakening impairs commutation, limiting the maximum obtainable speed.

Example 1

Problem. A 500 V shunt motor runs at 250 r.p.m. when the armature current is 200 A. Armature resistance is 0.12 Ω. Find the speed when a field resistance reduces the shunt field to 80% of normal and armature current is 100 A.

Solution (click to expand)

E_b1 = 500 − 200 × 0.12 = 476 V
E_b2 = 500 − 100 × 0.12 = 488 V
Φ₂ = 0.8 Φ₁,  N₁ = 250 r.p.m.

Using  N &propto E_b/Φ:
\[ \frac{N_2}{N_1} \;=\; \frac{E_{b2}}{E_{b1}} \cdot \frac{\Phi_1}{\Phi_2} \;=\; \frac{488}{476} \cdot \frac{1}{0.8} \Rightarrow N_2 \approx \mathbf{320\ r.p.m.} \]

Example 2

Problem. A 250 V DC shunt motor has armature resistance 0.25 Ω. On load it takes I_a1=50 A and runs at 750 r.p.m. If flux is reduced by 10% without changing load torque, find the new speed.

Solution (click to expand)

Torque condition: T_a ∝ Φ I_a, and load torque is constant.

T_a1 = T_a2 ⇒ Φ₁ I_a1 = Φ₂ I_a2
Φ₂ = 0.9 Φ₁ ⇒ I_a2 = I_a1 (Φ₁/Φ₂) = 50 / 0.9 = 55.6 A

E_b1 = 250 − 50 × 0.25 = 237.5 V
E_b2 = 250 − 55.6 × 0.25 = 231.1 V

Using N &propto E_b/Φ:
\[ \frac{N_2}{N_1} \;=\; \frac{E_{b2}}{E_{b1}} \cdot \frac{\Phi_1}{\Phi_2} \;=\; \frac{231.1}{237.5} \cdot \frac{1}{0.9} \Rightarrow N_2 \approx \mathbf{811\ r.p.m.} \]

Example 3

Problem. A 220 V shunt motor has armature resistance 0.5 Ω and takes I_a1=40 A on full load. By how much must the main flux be reduced to raise the speed by 50% if the developed torque is constant?

Solution (click to expand)

Given: N₂ / N₁ = 3/2, torque constant ⇒ Φ₁ I_a1 = Φ₂ I_a2.

Let x = Φ₁/Φ₂ ⇒ I_a2 = I_a1 × x = 40x.

E_b1 = 220 − 40 × 0.5 = 200 V
E_b2 = 220 − (40x) × 0.5 = (220 − 20x) V

Using N &propto E_b/Φ:
\[ \frac{N_2}{N_1} \;=\; \frac{E_{b2}}{E_{b1}} \cdot \frac{\Phi_1}{\Phi_2} \;=\; \frac{220-20x}{200}\cdot x \;=\; \frac{3}{2} \]

⇒ x² − 11x + 15 = 0 ⇒ x = 1.6 \text{ or } 9.4\overline{(rejected)}

Hence Φ₁/Φ₂ = 1.6 ⇒ Φ₂/Φ₁ = 1/1.6 = 0.625
Percentage reduction in flux: \[ \frac{\Phi_1 - \Phi_2}{\Phi_1}\times 100 \;=\; \left(1 - 0.625\right)\times 100 \;=\; \mathbf{37.5\%} \]

Example 4

Problem. A DC shunt motor supplied at 230 V runs at 990 r.p.m. Calculate the resistance required in series with the armature circuit to reduce the speed to 500 r.p.m., assuming armature current is 25 A.

Solution (click to expand)

Note on given text: The provided solution steps use 900 r.p.m. as the initial speed (not 990). Both cases are shown below.

Case A (as per solution steps using 900 r.p.m.)

E_b1 = 230 = K × 900   (assuming negligible armature resistance internally or included in constant K)
At 500 r.p.m.: E_b2 = K × 500 = E_b1 × (500/900) = 127.8 V
Drop across series resistor R: E_b1 − E_b2 = 25R
R = (230 − 127.8)/25 = 4.088 Ω

Case B (if initial speed is 990 r.p.m. as stated in the problem)

E_b2 = 230 × (500/990) = 116.16 V
R = (230 − 116.16)/25 = 4.554 &Omega (approx.)

Summary: In shunt motors, weakening the field flux via a field rheostat raises speed efficiently, but commutation limits set practical bounds (≈2:1 without interpoles; up to ≈6:1 with interpoles). Use the proportionality N &propto E_b/Φ and torque condition T_a &propto ΦI_a to solve numericals.

(ii) Armature or Rheostatic Control Method

This method is preferred when speeds below no-load speed are required. With a constant supply voltage V, a variable series resistance R (controller resistance) is inserted in the armature circuit to reduce the armature p.d. and hence the speed.

Working Principle

• As the controller resistance increases, the p.d. across the armature decreases → speed falls.
• For a constant torque load (I_a1 = I_a2), speed is approximately proportional to armature p.d.
• From the speed–armature current curves, more series resistance gives greater speed drop.

Relations

Let R_a = armature resistance, R = controller resistance, and R_t = R + R_a.

For two operating points with the same flux (Φ₁ = Φ₂):
N ∝ E_b,

hence N₁/N₂ = E_b1/E_b2, 

where E_b1 = V − I_a1R_a, 

E_b2 = V − I_a2R_t.

No-Load Line & Stalling Current

Considering the no-load reference (I_a0 small) and neglecting I_a0R_a versus V:

Speed–current line: N = N₀ − (I_a R_t) × (N₀/V)

Setting N = 0 gives the intercept: I_a,stall = V / R_t → stalling current.

Limitations & Improvement

• I²R loss in controller makes the method inefficient and costly in operation.
• Unsuitable for rapidly varying loads because speed changes with I_a for a fixed R_t.
• With an armature divertor (a shunt across the armature) in addition to series control resistance, changes in I_a have a smaller effect on armature p.d., giving more stable speed.

Example

Problem. A 200 V DC shunt motor running at 1000 r.p.m. takes I_a=17.5 A. Reduce speed to 600 r.p.m. Find series resistance to insert if original R_a=0.4 Ω. Assume I_a remains constant.

Solution (click to expand)

E_b1 = 200 − 17.5 × 0.4 = 193 V,  N₁ = 1000 r.p.m.
Let R_t = R_a + R (total series resistance). At N₂=600 r.p.m.: E_b2 = 200 − 17.5 R_t.

Since Φ constant, N &propto E_b:
600/1000 = E_b2/E_b1 = (200 − 17.5 R_t)/193

Solving → R_t = 4.8 Ω → Additional R = R_t − R_a = 4.8 − 0.4 = 4.4 Ω.

Note: Brush drop neglected.

(iii) Voltage Control Method

(a) Multiple Voltage Control

The shunt field is connected to a fixed exciting voltage. The armature can be switched to one of several discrete supply voltages using suitable switchgear. Armature speed is approximately proportional to the selected armature voltage. Intermediate speeds can be obtained by small adjustments of the shunt field rheostat. This method is simple but not widely used.

(b) Ward–Leonard System

Employed where an unusually wide (up to ≈ 10:1) and very sensitive speed control with smooth variation is required—e.g., colliery winders, electric excavators, elevators, and major mill drives (steel, blooming, paper).

It uses a motor–generator set to furnish a controllable DC voltage to the DC motor armature, achieving fine control over speed and direction.

Speed Control of DC Series Motors

Methods, merits/demerits, and series-parallel control — clean, printer-friendly HTML notes.

(i) Flux Control Method

Speed of a series motor is influenced by flux (Φ). Since N ∝ E_b / Φ, changing the flux alters speed — reducing Φ tends to increase speed and vice versa. Variations in flux can be achieved by the following:

(a) Field Divertors

A variable resistance (called a field divertor) is connected in parallel with the series field winding. Adjusting the divertor resistance bypasses part of the series current, reducing field current and flux, which increases speed.

(b) Armature Divertor

A divertor across the armature allows part of the armature current to bypass the armature. For a given constant load torque, a reduction in I_a requires an increase in Φ (because T_a ∝ Φ I_a). Increased supply current then increases flux and causes speed to fall. The speed variation is controlled by changing the divertor resistance.

(c) Tapped Field Control

Common in electric traction. Field turns (series field) are provided with taps so turns can be cut out in steps. With full field the motor runs at minimum speed; removing turns raises speed in discrete steps.

(d) Paralleling Field Coils

Used for multi-speed fan motors: field coils can be regrouped (paralleled or series combinations) to alter effective turns/flux. For example, a 4-pole motor can yield three speeds by regrouping coils.

(ii) Variable Resistance in Series with Motor

Placing a resistance in series with the motor reduces the voltage across the motor terminals, decreasing speed. Because the full motor current flows through this resistance, the I²R losses are large and efficiency suffers.

(iii) Series–Parallel Control

This system is widely used in electric traction where two or more similar series motors are mechanically coupled. Motors are arranged in series at low speeds and in parallel at high speeds. It is often used together with variable series resistance control.

When in Series

With two identical motors in series:

• Same current I flows through each motor.
• Voltage across each motor = V/2 (supply voltage V).
• For series motors, flux Φ ∝ I, so speed ∝ E_b/Φ ≈ V / I.
• Substituting V/2 and I gives: speed (series) ∝ (V/2) / I = V / (2I).
• Compared to parallel case (below) this speed is one-fourth of the speed when motors are in parallel.
• Torque ∝ Φ I ∝ I² → torque (series) is four times torque when in parallel (for same supply conditions).

When in Parallel

• Each motor receives full supply voltage V.
• Line current splits: each motor draws I/2 (for the same total current I).
• Speed ∝ V / (I/2) = 2V / I — which is four times the series speed in the standard comparison above.
• Torque per motor ∝ (I/2)² = I²/4; combined torque in parallel is therefore less than the series case by a factor of four.

Practical Switching Sequence (Traction)

Typical sequence used for starting and accelerating traction motors:

1. Start motors in series with variable series resistance to limit starting current.
2. Gradually cut out the variable resistance to increase speed while still in series.
3. Switch motors to parallel configuration (at a prescribed speed point) and reinsert series resistance as required.
4. Reduce the series resistance in steps until full speed is reached in parallel connection.

The variable series controller is not continuously rated — it is cut out progressively and selected running positions (A, B, C, D) can be held for indefinite periods.

Speed Control Methods of DC Motors — Advantages & Disadvantages

Shunt Motor speed control Methods Merit and Demerit

Method	Advantages	Disadvantages
Flux Control (Field Control)	Simple & economical (small field rheostat) Wide speed range above rated speed Low power loss (field current small)	Only speeds above rated speed obtainable Poor commutation at weak field (high speeds) Not suitable for constant low-speed operation
Armature / Rheostatic Control	Simple, provides speeds below rated value Easy to implement for small drives	Large power loss in series resistance (poor efficiency) Poor speed regulation under varying load Suitable mostly for intermittent duty
Voltage Control — Multiple Voltage	More efficient than rheostatic control Can obtain speeds up to rated value and below	Requires special multi-tapped supply or converter (costly) Less flexible than electronic drives
Voltage Control — Ward‑Leonard System	Smooth, wide-range control above & below rated speed Excellent speed regulation and reversibility High efficiency (no wasted rheostat power)	High initial & maintenance cost (motor‑generator set) Large footprint and complexity

Notes

Field control is safe for shunt motors because they retain a field even at light load. Ward‑Leonard was historically popular for cranes, lifts and steel mill drives before power electronics took over.

DC Series Motor speed control Methods Merit and Demerit

Method	Advantages	Disadvantages
Flux Control	Simple to implement Can obtain speeds above normal	Field weakening may cause runaway at no-load Unstable operation if over-weakened
Variable Resistance in Series with Motor	Simple method to reduce speed Commonly used in traction for starting	Poor efficiency (high I^2R losses) Poor speed regulation Not ideal for continuous operation
Series‑Parallel Control (Traction)	Economical for starting heavy trains Provides stepped control: series (low) → parallel (high) More energy‑efficient than pure resistance control	Speed control is stepwise (not smooth) Requires switching gear and contactors Limited range compared to modern converters

Speed Control of DC Motor Important MCQ

DC Motor Speed Control Previous Year SSC JE Electrical MCQ Questions

Which field control method for a DC series motor involves shunting a portion of motor current around the series field to reduce the field flux and increase the speed of the motor?
A) Tapped field control
B) Armature control method
C) Field diverter method
D) Rheostatic control method
Correct Answer: Option 3

Which speed control methods offers below normal speed in DC shunt motor?
A) Field control method
B) Voltage control method
C) Armature control method and Voltage control method
D) Ward Leonard system of speed control
Correct Answer: Option 3

The starting torque of a DC motor is independent of
A) Speed
B) Magnetic flux density
C) Armature current
D) Armature current
Correct Answer: Option 1

For speed control of DC shunt motor, flux can be varied by
A) connecting a variable resistance in parallel with the shunt field winding
B) connecting a variable resistance in series with the armature
C) connecting a variable resistance in series with the shunt field winding
D) changing the number of coils in armature
Correct Answer: Option 3

​Which statement is correct for AC machine comparative to dc machine?
A) In this motor, carbon brushes and commutators are present.
B) Motor speed control can be done by changing the frequency.
C) Motor’s efficiency is high
D) It requires maintenance
Correct Answer: Option 2

Which of the following statements about the speed control of DC motors is correct?
A) The flux control method is otherwise called constant torque drive method.
B) In the speed control by varying field flux, the speed can be regulated at the base speed.
C) The field flux control is otherwise called constant power drive method.
D) The armature resistance method of speed control is employed to obtain speed above the base speed.
Correct Answer: Option 3

A shunt generator running at 1000 rpm has generated emf as 200 V. If the speed increases to 1200 rpm, the generated emf will be nearly
A) 150 V
B) 240 V
C) 175 V
D) 290 V
Correct Answer: Option 2

A separately excited, 200 V, DC motor runs with a speed of 1500 rpm at no load current of 5 A. When operated at full load current is found to be 50 A. Assuming constant flux operation with armature resistance of 0.2 Ω, calculate full load speed.
A) 1400 rpm
B) 1578 rpm
C) 1432 rpm
D) 1500 rpm
Correct Answer: Option 3

The speed of DC motor is proportional to the:
A) square of armature current
B) inverse of armature current
C) armature current
D) field current
Correct Answer: Option 2