DC Motor Speed Control — Formula Sheet & Solved Problems

1. Formula Reference Sheet

Steady-State Speed & Torque (Separately Excited Motor)

Steady-State Speed	\(\displaystyle \omega_m = \frac{V_a}{K_b\phi} - \frac{R_a\,T_e}{(K_b\phi)^2}\)	Two control handles: \(V_a\) and \(\phi\)
Back-EMF	\(\displaystyle E_b = K_b\,\phi\,\omega_m = V_a - I_a R_a\)	Steady state; omit brush drop unless given
Electromagnetic Torque	\(\displaystyle T_e = K_b\,\phi\,I_a\)	\(K_b\phi = E_b/\omega_m = T_e/I_a\)
Rated Back-EMF	\(\displaystyle E_r = V_{a,\max} - I_{ar}\,R_a\)	Key reference for both control regions
Motor Constant (rated)	\(\displaystyle K_b\phi_r = \frac{E_r}{\omega_r} = \frac{T_{er}}{I_{ar}}\)	Identified from nameplate data

Region 1 — Armature Voltage Control (\(0 \leq \omega_m \leq \omega_r\))

Condition	\(\phi = \phi_r\) (rated, fixed); \(V_a\) varies from 0 to \(V_{a,\max}\)	Constant-torque region
Max torque	\(T_{e,\max} = K_b\phi_r\,I_{ar} = T_{er} = \text{const}\)	Torque limited by rated current
Speed proportionality	\(\displaystyle \frac{\omega_{m1}}{\omega_{m2}} = \frac{E_{b1}}{E_{b2}} = \frac{V_{a1} - I_a R_a}{V_{a2} - I_a R_a}\)	At constant \(\phi\) and \(I_a\)
Simplified (const. \(\phi\), \(T\))	\(\displaystyle \frac{E_{b1}}{E_{b2}} = \frac{N_1}{N_2}\) and \(V_a = E_b + I_a R_a\)	Back-EMF \(\propto\) speed at constant flux
Armature current	\(\displaystyle I_a = \frac{T_e}{K_b\phi_r}\)	Constant if torque demand constant

Region 2 — Field Weakening (\(\omega_r \leq \omega_m \leq \omega_{\max}\))

Condition	\(V_a = V_{a,\max}\) (fixed); \(\phi\) reduced by reducing \(V_f\) or \(I_f\)	Constant-power region
Constant back-EMF constraint	\(\displaystyle E_b = K_b\phi\,\omega_m = E_r = \text{const}\)	Back-EMF stays at rated value
Flux–speed relationship	\(\displaystyle \frac{\phi_2}{\phi_r} = \frac{\omega_r}{\omega_{m2}} = \frac{N_r}{N_2}\)	Flux falls inversely with speed
Flux via field current	\(\displaystyle \frac{\phi_2}{\phi_r} = \frac{I_{f2}}{I_{fr}} = \frac{V_{f2}}{V_{f,\max}}\)	Linear magnetic circuit assumed
Torque at field-weakened speed	\(\displaystyle T_e = K_b\phi_2\,I_{ar} = \frac{\phi_2}{\phi_r}\,T_{er}\)	Torque \(\propto\) flux; power = const
Maximum speed	\(\displaystyle \omega_{m,\max} = \frac{E_r}{K_b\phi_{\min}}\)	\(\phi_{\min}\) set by minimum safe field current

Variable Load Torque — Quadratic Solution Method

General approach	When \(T_L = f(N)\) and \(\phi\) may change, set up two simultaneous equations: (i) KVL and (ii) torque balance \(T_e = T_L\), then solve.
KVL equation	\(\displaystyle V_a = K_b\phi\,\omega_m + I_a\,R_a\)	One unknown: \(I_a\) or \(\omega_m\)
Torque balance	\(\displaystyle K_b\phi\,I_a = T_L(\omega_m)\)	Links \(I_a\) to \(\omega_m\)
Quadratic form (field weakening)	\(\displaystyle R_a\,I_a^2 - V_a\,I_a + T_L\,\omega_m = 0\)	Take the lower (physically feasible) root

Field Coil Reconnection

Parallel coils → flux	Each coil carries \(V_f/r\); total MMF \(= 2N_f(V_f/r)\)	Reference (rated) configuration
Series coils → flux	Total resistance \(2r\); coil current \(= V_f/(2r)\); total MMF \(= N_f\,V_f/r\)	Half the MMF of parallel connection
Flux ratio	\(\displaystyle \frac{\phi_{\text{series}}}{\phi_{\text{parallel}}} = \frac{1}{2} \;\Rightarrow\; K_b\phi_{\text{series}} = \frac{K_b\phi_{\text{parallel}}}{2}\)	Doubling \(K_b\phi\) denominator raises speed
Speed scaling	\(\displaystyle \omega_{m,\text{series}} \approx 2\,\omega_{m,\text{parallel}}\) (same \(V_a\), small \(I_a R_a\))	Series connection roughly doubles no-load speed

Normalised (Per-Unit) Quantities

Normalised speed	\(\displaystyle \omega_n = \frac{\omega_m}{\omega_r}\)	Base = rated speed \(\omega_r\)
Normalised torque	\(\displaystyle T_n = \frac{T_e}{T_{er}}\)	Region 1: \(T_n = 1\); Region 2: \(T_n = 1/\omega_n\)
Normalised flux	\(\displaystyle \phi_n = \frac{\phi}{\phi_r}\)	Region 1: \(\phi_n = 1\); Region 2: \(\phi_n = 1/\omega_n\)
Normalised power	\(\displaystyle P_n = T_n\,\omega_n\)	Region 1: \(P_n = \omega_n\); Region 2: \(P_n = 1\)
Intermittent duty	\(I_{a,\max} = k\,I_{ar}\); base speed \(= E_m/K_b\phi_r\), \(E_m = V_{a,\max} - k\,I_{ar}R_a\)	Higher \(k\) reduces field-weakening range

2. Armature Voltage & Field Weakening Control

Problem Statement

Given Data

Rated voltage \(V_r\)	230 V
Rated speed \(N_r\)	500 rpm
Rated armature current \(I_{ar}\)	100 A
Armature resistance \(R_a\)	0.1 Ω

Load Condition

Load torque is constant and speed-independent. Rotational losses neglected. \(\Rightarrow I_a = I_{ar}\) throughout Region 1.

Find the armature terminal voltage to reduce speed to 400 rpm (armature voltage control, rated flux).
Find the flux reduction required to raise speed to 800 rpm (rated armature voltage, field weakening).

Preliminary: Rated Back-EMF

\[E_r = V_r - I_{ar}\,R_a = 230 - 100 \times 0.1 = 220\,\text{V}\]

Motor Constant at Rated Flux

\[K_b\phi_r = \frac{E_r}{\omega_r} = \frac{220}{500 \times \pi/30} = \frac{220}{52.36} = 4.202\,\text{V·s/rad}\]

Solution — Part (1): Armature Voltage Control at 400 rpm

Speed is below rated \(\Rightarrow\) Region 1: armature voltage control with \(\phi = \phi_r\) (rated).

Step 1: Back-EMF at 400 rpm

With constant flux, \(E_b \propto \omega_m \propto N\):

\[E_1 = E_r \times \frac{N_1}{N_r} = 220 \times \frac{400}{500} = 176\,\text{V}\]

Step 2: Required armature voltage

Constant torque load with constant flux \(\Rightarrow I_a = I_{ar} = 100\,\text{A}\):

\[\boxed{V_1 = E_1 + I_{ar}\,R_a = 176 + 100 \times 0.1 = 186\,\text{V}}\]

Result

To achieve 400 rpm with rated flux, the armature terminal voltage must be reduced from 230 V to 186 V. The armature current remains at 100 A.

Normalised Check

\[\omega_n = \frac{400}{500} = 0.8\,\text{p.u.} \qquad V_n = \frac{186}{230} = 0.809\,\text{p.u.}\] Speed \(\approx\) proportional to \(V_a\) at constant \(\phi\) and \(T\).

Solution — Part (2): Field Weakening at 800 rpm

Speed is above rated \(\Rightarrow\) Region 2: field weakening with \(V_a = V_r = 230\,\text{V}\), \(I_a = I_{ar} = 100\,\text{A}\).

Step 1: Back-EMF at 800 rpm

\(V_a\) and \(I_a\) are unchanged from rated conditions, so back-EMF is also unchanged:

\[E_2 = V_r - I_{ar}\,R_a = 230 - 100 \times 0.1 = 220\,\text{V}\]

Step 2: Required flux at 800 rpm

Since \(E_b = K_b\phi\,\omega_m = \text{const}\):

\[K_b\phi_r\,\omega_r = K_b\phi_2\,\omega_2\] \[\frac{\phi_2}{\phi_r} = \frac{\omega_r}{\omega_2} = \frac{N_r}{N_2} = \frac{500}{800}\] \[\boxed{\phi_2 = 0.625\,\phi_r \quad (62.5\%\text{ of rated flux})}\]

Physical Interpretation

To maintain \(E_b = 220\,\text{V}\) at 60% higher speed, the flux (and hence field current and voltage) must be reduced to 62.5% of rated.

Torque at 800 rpm

\[T_{e2} = K_b\phi_2\,I_{ar} = 0.625\,K_b\phi_r\,I_{ar} = 0.625\,T_{er}\] Torque falls to 62.5%, but power \(= T_{e2}\,\omega_2 = 0.625\,T_{er} \times 1.6\,\omega_r = T_{er}\,\omega_r\) remains constant. \(\checkmark\)

3. Speed Control with Variable Load Torque

Problem Statement

Given Data

Rated voltage \(V_r\)	220 V
Rated armature current \(I_{ar}\)	500 A
Rated speed \(N_r\)	600 rpm
Armature resistance \(R_a\)	0.02 Ω
Field resistance \(R_f\)	10 Ω

Load Characteristic

\[T_L = (2000 - 2N)\,\text{N·m}\] where \(N\) is speed in rpm. Load torque decreases linearly with speed.

Find terminal voltage \(V_a\) and armature current \(I_a\) at 450 rpm (armature voltage control).
Find field winding voltage \(V_f'\) and armature current \(I_a\) at 750 rpm (field weakening).

Preliminary: Rated Operating Conditions

\[\omega_r = \frac{2\pi \times 600}{60} = 62.83\,\text{rad/s}\]

\[E_r = 220 - 500 \times 0.02 = 210\,\text{V}\]

\[T_r = \frac{E_r\,I_{ar}}{\omega_r} = \frac{210 \times 500}{62.83} = 1671\,\text{N·m}\]

Verify: \(T_L(600) = 2000 - 2\times600 = 800\,\text{N·m} \neq T_r\). The rated armature current of 500 A is the maximum permitted current. At rated speed the actual torque drawn equals the load torque.

Solution — Part (1): Armature Voltage Control at 450 rpm

450 rpm < 600 rpm \(\Rightarrow\) Region 1: \(\phi = \phi_r\) (constant), vary \(V_a\).

Step 1: Load torque at 450 rpm

\[T_L(450) = 2000 - 2 \times 450 = 1100\,\text{N·m}\]

Step 2: Armature current

Flux is constant \(\Rightarrow T_e \propto I_a\). Using the rated point as reference:

\[I_a = \frac{T_L(450)}{T_r} \times I_{ar} = \frac{1100}{1671} \times 500 = 329\,\text{A}\]

Step 3: Back-EMF at 450 rpm

Constant flux \(\Rightarrow E_b \propto N\):

\[E_2 = E_r \times \frac{450}{600} = 210 \times 0.75 = 157.5\,\text{V}\]

Step 4: Required terminal voltage

\[\boxed{V_2 = E_2 + I_a\,R_a = 157.5 + 329 \times 0.02 = 164.1\,\text{V}}\]

Result — Part (1)

At 450 rpm: \(V_a = 164.1\,\text{V}\), \(I_a = 329\,\text{A}\).

Verification

\[E_b = K_b\phi_r\,\omega_{450} = \frac{E_r}{\omega_r}\times\frac{450\pi}{30} = \frac{210}{62.83}\times47.12 = 157.5\,\text{V}\;\checkmark\] \[V_a = 157.5 + 329\times0.02 = 157.5 + 6.58 = 164.1\,\text{V}\;\checkmark\] \[T_e = T_L(450) = 1100\,\text{N·m}\;\checkmark\]

Solution — Part (2): Field Weakening at 750 rpm

750 rpm > 600 rpm \(\Rightarrow\) Region 2: \(V_a = 220\,\text{V}\) (fixed), flux reduced. \(\omega_{750} = 750\pi/30 = 78.54\,\text{rad/s}\).

Step 1: Load torque at 750 rpm

\[T_L(750) = 2000 - 2 \times 750 = 500\,\text{N·m}\]

Step 2: Set up the voltage equation

Let \(K_b\phi' = T_L / I_a = 500/I_a\). Then back-EMF is:

\[E' = K_b\phi'\,\omega_{750} = \frac{500}{I_a} \times 78.54 = \frac{39{,}270}{I_a}\]

Substituting into KVL:

\[220 = I_a\,R_a + E' = 0.02\,I_a + \frac{39{,}270}{I_a}\]

Step 3: Solve the quadratic

\[0.02\,I_a^2 - 220\,I_a + 39{,}270 = 0\] \[I_a = \frac{220 \pm \sqrt{220^2 - 4 \times 0.02 \times 39{,}270}}{2 \times 0.02}\] \[= \frac{220 \pm \sqrt{48{,}400 - 3{,}141.6}}{0.04} = \frac{220 \pm \sqrt{45{,}258}}{0.04} = \frac{220 \pm 212.75}{0.04}\]

Feasible (lower) root: \(\quad\displaystyle I_a = \frac{220 - 212.75}{0.04} = \frac{7.25}{0.04}\)

\[\boxed{I_a \approx 181\,\text{A}}\]

Higher root gives \(\approx 10{,}819\,\text{A}\) — physically impossible.

Step 4: Required \(K_b\phi'\)

\[K_b\phi' = \frac{500}{181} = 2.762\,\text{V·s/rad}\] \[K_b\phi_r = \frac{E_r}{\omega_r} = \frac{210}{62.83} = 3.342\,\text{V·s/rad}\]

Step 5: Required field voltage

\[\frac{V_f'}{V_{f,r}} = \frac{K_b\phi'}{K_b\phi_r} = \frac{2.762}{3.342} = 0.826\] \[\boxed{V_f' = 220 \times 0.826 = 181.7\,\text{V}}\]

Verification

\[E' = 2.762 \times 78.54 = 216.9\,\text{V}\] \[V_a = 0.02 \times 181 + 216.9 = 3.62 + 216.9 = 220.5\,\text{V} \approx 220\,\text{V}\;\checkmark\] \[T_e = 2.762 \times 181 = 500\,\text{N·m} = T_L(750)\;\checkmark\]

Summary — Part (2)

At 750 rpm: \(I_a = 181\,\text{A}\), \(V_f' = 181.7\,\text{V}\) (82.6% of rated field voltage).

4. Field Coil Reconnection: Speed Control

Problem Statement

Given Data

Rated voltage \(V_r\)	220 V
Rated armature current \(I_{ar}\)	100 A
Rated speed \(N_r\)	750 rpm
Armature resistance \(R_a\)	0.1 Ω

Field Coil Configuration

The motor has two field coils normally connected in parallel.
Load: \(T_L = (500 - 0.3N)\,\text{N·m}\), \(N\) in rpm.

Find armature current and speed when \(V_a\) is reduced to 110 V (field coils remain in parallel).
Find speed and armature current when field coils are reconnected in series at rated armature voltage (220 V).

Preliminary: Rated Conditions (Parallel Coils)

\[\omega_r = \frac{2\pi \times 750}{60} = 78.54\,\text{rad/s}\]

\[E_r = 220 - 100 \times 0.1 = 210\,\text{V}\]

\[K_e\phi_1 = \frac{E_r}{\omega_r} = \frac{210}{78.54} = 2.673\,\text{V·s/rad}\]

Solution — Part (1): Reduced Armature Voltage (110 V, Parallel Coils)

Field coils remain in parallel \(\Rightarrow \phi = \phi_1\) unchanged \(\Rightarrow K_e\phi_1 = 2.673\,\text{V·s/rad}\) unchanged.

Set up two simultaneous equations

Let \(\omega_{m2} = N_2\pi/30\). The KVL equation (i) and torque balance (ii) are:

\[110 = 2.673 \times \frac{N_2\pi}{30} + 0.1\,I_{a2} = 0.2800\,N_2 + 0.1\,I_{a2} \tag{i}\] \[2.673\,I_{a2} = 500 - 0.3\,N_2 \tag{ii}\]

Solve by substitution

From (ii): \(\displaystyle I_{a2} = \frac{500 - 0.3\,N_2}{2.673}\). Substituting into (i):

\[0.2800\,N_2 + \frac{0.1(500 - 0.3\,N_2)}{2.673} = 110\] \[0.2800\,N_2 + 18.71 - 0.01122\,N_2 = 110\] \[0.26878\,N_2 = 91.29\] \[\boxed{N_2 \approx 340\,\text{rpm}}\] \[I_{a2} = \frac{500 - 0.3 \times 340}{2.673} = \frac{398}{2.673}\] \[\boxed{I_{a2} = 148.9\,\text{A}}\]

Result — Part (1)

At \(V_a = 110\,\text{V}\): \(N_2 = 340\,\text{rpm}\), \(I_{a2} = 148.9\,\text{A}\).

Why Does Current Increase?

Reducing \(V_a\) lowers the speed. At lower speed the load torque rises (\(T_L = 500 - 0.3N\) increases as \(N\) falls), requiring more armature current to meet the higher torque demand.

Verification (340 rpm)

\(\omega_{m2} = 340\times\pi/30 = 35.60\,\text{rad/s}\)
\(E_2 = 2.673\times35.60 = 95.2\,\text{V}\)
\(V_a = 95.2 + 148.9\times0.1 = 95.2 + 14.89 = 110.1\,\text{V}\;\checkmark\)
\(T_e = 2.673\times148.9 = 398\,\text{N·m}\)
\(T_L = 500 - 0.3\times340 = 398\,\text{N·m}\;\checkmark\)

Solution — Part (2): Field Coils in Series (220 V)

Step 1: Effect of series connection on flux

Configuration	Coil current	Total MMF
Parallel (original)	\(V_f/r\) each	\(2N_f(V_f/r)\)
Series (new)	\(V_f/(2r)\) each	\(N_f\,V_f/r\)

\[\frac{\text{MMF(series)}}{\text{MMF(parallel)}} = \frac{1}{2} \quad\Rightarrow\quad K_e\phi_2 = \frac{K_e\phi_1}{2} = \frac{2.673}{2} = 1.337\,\text{V·s/rad}\]

Step 2: Two equations with \(V_a = 220\,\text{V}\)

\[220 = 1.337 \times \frac{N_3\pi}{30} + 0.1\,I_{a3} = 0.14001\,N_3 + 0.1\,I_{a3} \tag{iii}\] \[1.337\,I_{a3} = 500 - 0.3\,N_3 \tag{iv}\]

Step 3: Solve by substitution

From (iv): \(\displaystyle I_{a3} = \frac{500 - 0.3\,N_3}{1.337}\). Substituting into (iii):

\[0.14001\,N_3 + \frac{0.1(500 - 0.3\,N_3)}{1.337} = 220\] \[0.14001\,N_3 + 37.40 - 0.02244\,N_3 = 220\] \[0.11757\,N_3 = 182.60\] \[\boxed{N_3 \approx 1553\,\text{rpm}}\]

\[I_{a3} = \frac{500 - 0.3 \times 1553}{1.337} = \frac{500 - 465.9}{1.337} = \frac{34.1}{1.337}\] \[\boxed{I_{a3} = 25.5\,\text{A}}\]

Verification (1553 rpm)

\(\omega_{m3} = 1553\times\pi/30 = 162.6\,\text{rad/s}\)
\(E_3 = 1.337\times162.6 = 217.4\,\text{V}\)
\(V_a = 217.4 + 25.5\times0.1 = 217.4 + 2.55 = 220.0\,\text{V}\;\checkmark\)
\(T_e = 1.337\times25.5 = 34.1\,\text{N·m}\)
\(T_L = 500 - 0.3\times1553 = 34.1\,\text{N·m}\;\checkmark\)

Summary — Problem 14

Condition	\(N\) (rpm)	\(I_a\) (A)
Rated (parallel, 220 V)	750	100
\(V_a = 110\,\text{V}\), parallel	340	148.9
\(V_a = 220\,\text{V}\), series	1553	25.5

Physical Insight

Halving the flux (by series connection) approximately doubles the no-load speed. The load torque at 1553 rpm is very low (\(34.1\,\text{N·m}\)), so only a small armature current (\(25.5\,\text{A}\)) is needed — well within the rated 100 A limit.