DC Motor Drives Fundamentals — Formula Sheet & Solved Problems

1. Formula Reference Sheet

Separately Excited DC Motor

Back-EMF \(\displaystyle E_b = K_b\,\phi\,\omega_m\) Voltage opposing \(V_a\); grows with speed
Electromagnetic Torque \(\displaystyle T_e = K_t\,\phi\,I_a = K_b\,\phi\,I_a\) \(K_t = K_b\) in SI (energy balance)
Armature KVL (steady state) \(\displaystyle V_a = R_a\,I_a + V_{br} + E_b\) \(V_{br}\): brush drop (≈ 2 V for the pair); omit if not given
Steady-State Speed \(\displaystyle \omega_m = \frac{V_a - R_a\,I_a}{K_b\,\phi}\) Derived from KVL + back-EMF
Mechanical Equilibrium \(\displaystyle T_e = T_L + B_m\,\omega_m\) \(B_m\): viscous friction; \(T_L\): load torque
Shaft (output) Torque \(\displaystyle T_{sh} = T_e - T_f = T_e - B_m\,\omega_m\) Torque available at the shaft
Motor Constant \(\displaystyle K_b = \frac{T_e}{I_a} = \frac{E_b}{\omega_m}\) Identifies \(K_b\) from rated data (const. \(\phi\))
With explicit field current \(\displaystyle T_e = K_v\,I_f\,I_a,\quad E_b = K_v\,I_f\,\omega_m\) \(K_v\) [V/(A·rad/s)]; \(I_f = V_f/R_f\) in steady state
Efficiency \(\displaystyle \eta = \frac{P_{out}}{P_{in}} = \frac{T_{sh}\,\omega_m}{V_a\,I_a + V_f\,I_f}\) Include field power input if given

Series DC Motor

Field–Armature Link \(I_a = I_f\) (same current loop) Defines the series connection
Back-EMF \(\displaystyle E_b = M_{af}\,I_a\,\omega_m\) \(M_{af}\): mutual inductance [H]
Electromagnetic Torque \(\displaystyle T_e = M_{af}\,I_a^2\) Torque \(\propto I_a^2\) (nonlinear)
Armature Current \(\displaystyle I_a = \sqrt{\frac{T_e}{M_{af}}}\) Inverted torque equation
KVL (steady state) \(\displaystyle V = \bigl(R_s + R_{ar} + R_f\bigr)I_a + E_b\) \(= \bigl(R_{total} + M_{af}\,\omega_m\bigr)I_a\)

Thermal Criterion — RMS Current

RMS Definition \(\displaystyle I_{a,\text{rms}} = \sqrt{\frac{1}{T}\int_0^T i_a^2(t)\,dt}\) Determines average heating
Two-Level Duty Cycle \(\displaystyle I_{a,\text{rms}} = i_{\max}\sqrt{d}\) \(d\) = on-fraction of period; off-current = 0
Thermal Safety Criterion \(\displaystyle I_{a,\text{rms}} \leq I_{ar}\) \(I_{ar}\) = rated (continuous) armature current
Copper Loss (RMS) \(\displaystyle P_{\text{loss}} = I_{a,\text{rms}}^2\,R_a\) Must not exceed rated copper loss \(I_{ar}^2 R_a\)

Unit Conversions Used in Problems

Speed: rpm to rad/s \(\displaystyle \omega_m = N \times \frac{2\pi}{60} = N \times \frac{\pi}{30}\) \(N\) in rpm; \(\omega_m\) in rad/s
Power: horsepower to watts \(1\,\text{hp} = 745.6\,\text{W} \approx 746\,\text{W}\)
Output power \(\displaystyle P_{out} = T_{sh}\,\omega_m\)
Shaft torque from rating \(\displaystyle T_{sh} = \frac{P_{out}}{\omega_m}\)

2. Problem 1 — Thermal Suitability of Two Machines

Problem Statement

Given
Two electrical machines have identical torque and speed ratings. Thermal limits are set by the rated armature copper loss \(I_{ar}^2 R_a\). Their torque–current relationships differ: \[ \text{Machine 1:}\quad T_e \propto i_1^2 \qquad\qquad \text{Machine 2:}\quad T_e \propto i_2 \] Load duty cycle: Torque \(= 2T_r\) for half the period; torque \(= 0\) for the other half (\(d = 0.5\)).
Physical Interpretation
Machine 1 behaves like a series DC motor (\(T \propto I^2\)) — at rated conditions, \(T_r \propto i_r^2\).
Machine 2 behaves like a separately excited DC motor (\(T \propto I\)) — at rated conditions, \(T_r \propto i_r\).
Thermal Criterion to Check
\[I_{a,\text{rms}} = i_{\max}\sqrt{d} \leq I_{ar}\]

Solution — Machine 1 (\(T_e \propto i_1^2\), series type)

Step 1: Peak current when load torque \(= 2T_r\)

Using \(T_e \propto i_1^2\), the ratio of torques equals the ratio of currents squared:

\[ \frac{i_1^2}{i_r^2} = \frac{2T_r}{T_r} = 2 \qquad\Rightarrow\qquad i_1 = \sqrt{2}\;i_r \]

Step 2: RMS current over the duty cycle (\(d = 0.5\))

\[ I_{a1,\text{rms}} = i_1\sqrt{d} = \sqrt{2}\;i_r \times \sqrt{0.5} = \sqrt{2}\;i_r \times \frac{1}{\sqrt{2}} = i_r \]

Step 3: Compare with rated current

\[ P_{\text{loss},1} = I_{a1,\text{rms}}^2\,R_a = i_r^2\,R_a = I_{ar}^2\,R_a \quad (\text{rated copper loss}) \quad \checkmark \]
Result — Machine 1
\[I_{a1,\text{rms}} = i_r = I_{ar}\] Machine 1 operates exactly at its thermal limit. Thermally suitable.

Solution — Machine 2 (\(T_e \propto i_2\), sep. excited type)

Step 1: Peak current when load torque \(= 2T_r\)

Using \(T_e \propto i_2\), the ratio of torques equals the ratio of currents:

\[ \frac{i_2}{i_r} = \frac{2T_r}{T_r} = 2 \qquad\Rightarrow\qquad i_2 = 2\,i_r \]

Step 2: RMS current over the duty cycle (\(d = 0.5\))

\[ I_{a2,\text{rms}} = i_2\sqrt{d} = 2\,i_r \times \sqrt{0.5} = 2\,i_r \times \frac{1}{\sqrt{2}} = \sqrt{2}\;i_r \]

Step 3: Compare with rated current

\[ P_{\text{loss},2} = I_{a2,\text{rms}}^2\,R_a = \bigl(\sqrt{2}\;i_r\bigr)^2 R_a = 2\,i_r^2\,R_a = 2 \times I_{ar}^2\,R_a \quad \boldsymbol{\times} \]
Result — Machine 2
\[I_{a2,\text{rms}} = \sqrt{2}\;i_r > I_{ar}\] Machine 2 exceeds its thermal rating by a factor of 2 in copper loss. Not thermally suitable.
Conclusion
Machine 1 (series type, \(T \propto I^2\)) is thermally suitable because doubling the torque only requires \(\sqrt{2}\times\) the current, whereas the separately excited machine (Machine 2) requires \(2\times\) the current, causing four times the instantaneous copper loss.

3. Problem 2 — Efficiency of a Separately Excited DC Motor

Problem Statement

Given Data
Rated output power\(P_{out} = 1500\,\text{kW}\)
Rated armature voltage\(V_a = 600\,\text{V}\)
Rated armature current\(I_a = 2650\,\text{A}\)
Rated speed\(N = 600\,\text{rpm}\)
Brush voltage drop\(V_{br} = 2\,\text{V}\)
Field power input\(P_f = 50\,\text{kW}\)
Armature resistance\(R_a = 0.003645\,\Omega\)
Viscous friction\(B_m = 15\,\text{N·m·s/rad}\)

Find the efficiency at the rated operating point.

Note: Brush Voltage Drop
Real DC machines have carbon brushes. Each brush contributes \(\approx 1\,\text{V}\), giving a constant total drop \(V_{br} = 2\,\text{V}\). The armature KVL therefore reads: \[V_a = R_a\,I_a + V_{br} + E_b\]

Solution — Steps 1–5

Step 1: Rated angular speed

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

Step 2: Rated shaft torque

\[ T_{sh} = \frac{P_{out}}{\omega_m} = \frac{1\,500\,000}{62.83} = 23\,873\,\text{N·m} \]

Step 3: Friction torque

\[ T_f = B_m\,\omega_m = 15 \times 62.83 = 942.5\,\text{N·m} \]

Step 4: Electromagnetic torque

\[ T_e = T_{sh} + T_f = 23\,873 + 942.5 = 24\,815.5\,\text{N·m} \]

Step 5: Motor constant \(K_b\) at rated conditions

\[ K_b = \frac{T_e}{I_a} = \frac{24\,815.5}{2650} = 9.364\,\text{N·m/A} \]
Verify Armature Voltage
\[\begin{align*} V_a &= R_a I_a + V_{br} + E_b \\ &= (0.003645 \times 2650) + 2 + (9.364 \times 62.83) \\ &= 9.66 + 2 + 588.4 \\ &= 600.1\,\text{V} \approx 600\,\text{V} \quad\checkmark \end{align*}\]

Solution — Efficiency Calculation

Step 6: Total input power

\[\begin{align*} P_{in} &= V_a\,I_a + P_f \\ &= (600 \times 2650) + 50\,000 \\ &= 1\,590\,000 + 50\,000 \\ &= 1\,640\,\text{kW} \end{align*}\]

Step 7: Efficiency

\[ \boxed{\eta = \frac{P_{out}}{P_{in}} \times 100 = \frac{1500}{1640} \times 100 = \mathbf{91.46\%}} \]
Loss Breakdown at Rated Load
Loss ComponentValue
Armature copper loss \(I_a^2 R_a\)25.6 kW
Brush contact loss \(V_{br}\,I_a\)5.3 kW
Friction & windage \(T_f\,\omega_m\)59.2 kW
Field power \(P_f\)50.0 kW
Total losses140 kW
Output power1500 kW
Input power1640 kW
Note on Armature Copper Loss
The armature copper loss \(I_a^2 R_a = 2650^2 \times 0.003645 = 25.6\,\text{kW}\) accounts for conduction losses in the winding only; the brush drop is treated separately as \(V_{br} I_a = 5.3\,\text{kW}\).

4. Problem 3 — Steady-State Speed of a Loaded Motor

Problem Statement & Solution

Given Data
Armature resistance \(R_a\)0.5 Ω
Motor constant \(K_b\)0.8 V·s/rad
Moment of inertia \(J\)0.0167 kg·m²
Viscous friction \(B_l\)0.01 N·m·s/rad
Load torque \(T_l\)100 N·m (constant)
Applied voltage \(V_a\)220 V

Find the steady-state speed \(\omega_m\).

In steady state \(d\omega_m/dt = 0\), so the inertia term \(J\,d\omega_m/dt\) vanishes.

Step 1: Torque balance (steady state)

\[ T_e = T_l + B_l\,\omega_m = 100 + 0.01\,\omega_m \]

Step 2: Armature current from torque

With constant flux, \(T_e = K_b\,I_a\):

\[ I_a = \frac{T_e}{K_b} = \frac{100 + 0.01\,\omega_m}{0.8} \]

Step 3: Back-EMF from KVL

\[ E_b = V_a - R_a\,I_a = 220 - \frac{0.5\,(100 + 0.01\,\omega_m)}{0.8} \] \[ E_b = 220 - 62.5 - 0.00625\,\omega_m = 157.5 - 0.00625\,\omega_m \]

Step 4: Equate \(E_b = K_b\,\omega_m\)

\[ 0.8\,\omega_m = 157.5 - 0.00625\,\omega_m \] \[ (0.8 + 0.00625)\,\omega_m = 157.5 \] \[ 0.80625\,\omega_m = 157.5 \] \[ \boxed{\omega_m = \frac{157.5}{0.80625} \approx 195.35\,\text{rad/s}} \]
Verification
\[T_e = 100 + 0.01 \times 195.35 = 101.95\,\text{N·m}\] \[I_a = 101.95/0.8 = 127.44\,\text{A}\] \[E_b = K_b\,\omega_m = 0.8 \times 195.35 = 156.3\,\text{V}\] \[V_a = R_a\,I_a + E_b = 0.5\times127.44 + 156.3 = 63.7 + 156.3 = 220\,\text{V}\;\checkmark\]

5. Problem 4 — Series DC Motor: Input Voltage and Efficiency

Problem Statement

Given Data
Rated output power\(P_r = 3\,\text{hp} = 2236.8\,\text{W}\)
Rated speed\(N_r = 2000\,\text{rpm}\)
Series resistance \(R_s\)1.5 Ω
Armature resistance \(R_{ar}\)0.7 Ω
Field resistance \(R_f\)0.0675 Ω
Mutual inductance \(M_{af}\)0.0675 H
Viscous friction \(B\)0.0025 N·m·s/rad

Find: (i) input voltage at rated torque and speed; (ii) efficiency.

Solution

Step 1: Rated angular speed and output power

\[ \omega_r = \frac{2\pi \times 2000}{60} = 209.44\,\text{rad/s} \] \[ P_{out} = 3 \times 745.6 = 2236.8\,\text{W} \]

Step 2: Shaft torque and friction torque

\[ T_{sh} = \frac{P_{out}}{\omega_r} = \frac{2236.8}{209.44} = 10.68\,\text{N·m} \] \[ T_f = B\,\omega_r = 0.0025 \times 209.44 = 0.524\,\text{N·m} \]

Step 3: Electromagnetic torque

\[ T_e = T_{sh} + T_f = 10.68 + 0.524 = 11.20\,\text{N·m} \]

Step 4: Armature current

Series motor: \(T_e = M_{af}\,I_a^2\)

\[ I_a = \sqrt{\frac{T_e}{M_{af}}} = \sqrt{\frac{11.20}{0.0675}} = \sqrt{165.9} = 12.88\,\text{A} \]

Step 5: (i) Input voltage

Total series resistance: \(R_{tot} = R_s + R_{ar} + R_f = 1.5 + 0.7 + 0.0675 = 2.2675\,\Omega\)

Back-EMF: \(E_b = M_{af}\,I_a\,\omega_r = 0.0675 \times 12.88 \times 209.44 = 181.9\,\text{V}\)

\[ V_{in} = R_{tot}\,I_a + E_b = 2.2675 \times 12.88 + 181.9 \] \[ V_{in} = 29.2 + 181.9 \] \[ \boxed{V_{in} \approx 211.1\,\text{V}} \]
Equivalent Compact Form
The same result follows from factoring: \(V_{in} = (R_{tot} + M_{af}\,\omega_r)\,I_a\) since \(E_b = M_{af}\,I_a\,\omega_r\): \[V_{in} = (2.2675 + 14.14)\times12.88 = 16.41\times12.88 = 211.4\,\text{V}\]

Step 6: (ii) Efficiency

\[ P_{in} = V_{in}\cdot I_a = 211.4 \times 12.88 = 2723\,\text{W} \] \[ \boxed{\eta = \frac{P_{out}}{P_{in}}\times100 = \frac{2236.8}{2723}\times100 \approx \mathbf{82.1\%}} \]

6. Problem 5 — Sep. Excited Motor at Partial Load and Speed

Problem Statement

Given Data
Rating15 hp, 220 V, 2000 rpm
Field resistance \(R_f\)147 Ω
Armature resistance \(R_a\)0.25 Ω
Motor constant \(K_v\)0.7032 V/(A·rad/s)
Field voltage \(V_f\)220 V
Operating load torque \(T_L\)45 N·m at \(N = 1200\) rpm

Find: (a) Back-EMF \(E_b\);  (b) Armature voltage \(V_a\);  (c) Rated armature current \(I_{a,r}\).

Model Used
With explicit field current: \(\;T_e = K_v\,I_f\,I_a\;\) and \(\;E_b = K_v\,I_f\,\omega_m\). Here friction is neglected (not given), so \(T_e \approx T_L = 45\,\text{N·m}\).

Solution

Step 1: Field current (steady state)

\[ I_f = \frac{V_f}{R_f} = \frac{220}{147} = 1.497\,\text{A} \]

Step 2: Operating angular speed

\[ \omega_m = 1200 \times \frac{\pi}{30} = 125.66\,\text{rad/s} \]

Step 3: Armature current from torque

\[ I_a = \frac{T_L}{K_v\,I_f} = \frac{45}{0.7032 \times 1.497} = \frac{45}{1.0527} \] \[ \boxed{I_a = 42.75\,\text{A}} \]

Step 4: (a) Back-EMF

\[ E_b = K_v\,I_f\,\omega_m = 0.7032 \times 1.497 \times 125.66 \] \[ E_b = 1.0527 \times 125.66 \] \[ \boxed{E_b = 132.3\,\text{V}} \]

Step 5: (b) Armature terminal voltage

\[ V_a = E_b + I_a\,R_a = 132.3 + 42.75 \times 0.25 = 132.3 + 10.69 \] \[ \boxed{V_a = 143.0\,\text{V}} \]

Step 6: (c) Rated armature current

\[ I_{a,r} = \frac{P_r}{V_r} = \frac{15 \times 745.6}{220} = \frac{11\,184}{220} \] \[ \boxed{I_{a,r} = 50.84\,\text{A}} \]
Sanity Check
\(I_a = 42.75\,\text{A} < I_{a,r} = 50.84\,\text{A}\;\checkmark\)
Motor is operating at 84% of rated current, confirming it is below full load.