Phase-Controlled DC Motor Drives

1. Key Formulae — Dual Converter Operation

Dual Converter — Structure and Constraints

Concept

A dual converter consists of two fully controlled converters (P and N) connected in anti-parallel across the motor armature. Selecting which converter is active achieves all four quadrants of motor operation without reversing the field current.

Voltage constraint	\(\alpha_P + \alpha_N = 180°\)	Prevents short-circuit in circulating-current mode
Converter P voltage	\(V_a = V_{do}\cos\alpha_P\)	Active for \(I_a > 0\) (forward current)
Converter N voltage	\(V_a = -V_{do}\cos\alpha_N\)	Active for \(I_a < 0\) (reverse current)

Four-Quadrant Operating Table

Quadrant	\(V_a\)	\(I_a\)	Active Conv.	\(\alpha\) Range
Q1 (Fwd. motoring)	+	+	P	0° – 90°
Q2 (Fwd. regen.)	+	−	N (inverter)	90° – 180°
Q3 (Rev. motoring)	−	−	N	0° – 90°
Q4 (Rev. regen.)	−	+	P (inverter)	90° – 180°

Continuous Conduction Condition

Continuity criterion	\(\dfrac{L_a}{R_a} \gg T_r\)	\(\tau = L_a/R_a\) must greatly exceed ripple period \(T_r\)
1-\(\phi\) full converter ripple period	\(T_r = T/2 = 1/(2f)\)	Two voltage pulses per supply cycle
3-\(\phi\) full converter ripple period	\(T_r = T/6 = 1/(6f)\)	Six voltage pulses per supply cycle
1-\(\phi\) practical rule	\(\omega_s L_a / R_a \geq 5\)	Continuous conduction likely if satisfied
3-\(\phi\) practical rule	\(\omega_s L_a / R_a \geq 2\)	Less stringent due to smaller ripple amplitude

2. Single-Phase Dual Converter — Forward Motoring and Forward Regenerative Braking

Problem Statement & Given Data

Motor & Supply Ratings

Motor rating	10 kW, 200 V, 1000 rpm
AC supply	260 V (rms), 50 Hz
Armature resistance \(R_a\)	0.4 Ω
\(K_a\Phi\)	0.18 V/rpm = 1.719 V·s/rad
Rated armature current \(I_{a,\text{rated}}\)	50 A

Objective (Circulating-Current-Free Mode)

Forward motoring at 800 rpm, rated current: find \(\alpha_P\) (active) and \(\alpha_N\) (standby)
Forward regenerative braking at 800 rpm, rated braking current: find new firing angles
Power returned to the AC supply during regenerative braking

Converter Constant and Back-EMF at 800 rpm

\[ V_m = \sqrt{2} \times 260 = 367.7\,\text{V}, \qquad V_{do} = \frac{2V_m}{\pi} = \frac{2 \times 367.7}{\pi} = 234.1\,\text{V} \] \[ E_g(800) = K_a\Phi \times N = 0.18 \times 800 = 144\,\text{V} \]

Case 1 — Forward Motoring at 800 rpm (Quadrant 1)

Converter P active; \(I_a = +50\,\text{A}\) (into motor).

\[ V_a = E_g + I_a R_a = 144 + 50 \times 0.4 = 164\,\text{V} \quad(> 0\;\checkmark) \] \[ \cos\alpha_P = \frac{V_a}{V_{do}} = \frac{164}{234.1} = 0.700 \] \[ \alpha_P = \mathbf{45.6°} \quad(\alpha_P < 90°,\;\text{motoring}\;\checkmark) \]

Standby Converter N

\[ \alpha_N = 180° - 45.6° = \mathbf{134.4°} \]

Q1 — Forward Motoring

Converter P supplies positive voltage and positive current. Converter N is armed at the complementary angle (\(134.4°\)) so it is ready to take over instantly if current direction reverses, eliminating switching delays during quadrant changes.

Case 2 — Forward Regenerative Braking at 800 rpm (Quadrant 2)

Braking torque requires \(I_a = -50\,\text{A}\) (current reverses; out of armature).
Motor still spinning forward: \(E_g = +144\,\text{V}\).
Converter N now active.

\[ V_a = E_g + I_a R_a = 144 + (-50) \times 0.4 = 124\,\text{V} \quad(V_a > 0\;\checkmark) \]

For converter N in anti-parallel, \(V_a = -V_{do}\cos\alpha_N\):

\[ \cos\alpha_N = \frac{-V_a}{V_{do}} = \frac{-124}{234.1} = -0.530 \] \[ \alpha_N = \mathbf{122.0°} \quad(\alpha_N > 90°,\;\text{inverter mode}\;\checkmark) \]

Standby Converter P

\[ \alpha_P = 180° - 122.0° = \mathbf{58.0°} \]

Verification: \(V_{do}\cos(58.0°) = 234.1 \times 0.530 = 124\,\text{V}\;\checkmark\)

Q2 — Forward Regenerative Braking

Converter N operates in inverter mode (\(\alpha_N > 90°\)), returning energy to the supply. The motor terminal voltage remains positive (same direction as back-EMF), but current direction has reversed, creating a braking torque. No field reversal is needed.

Case 3 — Power Returned to the Supply

\[ P_g = E_g \times |I_a| = 144 \times 50 = 7200\,\text{W} \quad\text{(generated by motor)} \] \[ P_R = I_a^2\,R_a = 50^2 \times 0.4 = 1000\,\text{W} \quad\text{(copper loss)} \] \[ P_s = P_g - P_R = 7200 - 1000 = \mathbf{6200\,\text{W}} \]

Quadrant Transitions Summary

Operation	\(\alpha_P\)	\(\alpha_N\)	\(I_a\)	\(V_a\)
Forward motoring (Q1)	45.6°	134.4°	+50 A	+164 V
Fwd. regen. braking (Q2)	58.0°	122.0°	−50 A	+124 V

Converter N operates in inverter mode (\(\alpha_N > 90°\)) during forward regenerative braking, returning energy without any field reversal.

3. Three-Phase Dual Converter — All Four Quadrants

Problem Statement & Given Data

Motor & Supply Ratings

Motor rating	50 hp, 500 V, 1200 rpm
AC supply	3-phase, 415 V (line), 50 Hz
Armature resistance \(R_a\)	0.15 Ω
\(K_a\Phi\)	0.38 V/rpm
Rated armature current \(I_{a,\text{rated}}\)	80 A

Objective (Circulating-Current-Free Mode)

Forward motoring at 1000 rpm, rated current: \(\alpha_P\) and standby \(\alpha_N\)
Reverse motoring at 800 rpm, rated current: \(\alpha_N\) (active) and standby \(\alpha_P\)
Forward regenerative braking at 1000 rpm, rated braking current: \(\alpha_N\) and standby \(\alpha_P\); also power returned to supply

Converter Constant

\[ V_\phi = \frac{415}{\sqrt{3}} = 239.6\,\text{V}, \qquad V_{do} = \frac{3\sqrt{6}}{\pi} \times 239.6 = 560.5\,\text{V} \]

Case 1 — Forward Motoring at 1000 rpm (Quadrant 1)

Converter P active; \(I_a = +80\,\text{A}\).

\[ E_g = 0.38 \times 1000 = 380\,\text{V} \] \[ V_a = 380 + 80 \times 0.15 = 392\,\text{V} \] \[ \cos\alpha_P = \frac{392}{560.5} = 0.6994 \implies \alpha_P = \mathbf{45.6°} \] \[ \alpha_N = 180° - 45.6° = \mathbf{134.4°} \quad\text{(standby)} \]

Q1 Result

Back-EMF \(E_g\)	380 V
Terminal voltage \(V_a\)	392 V
Active \(\alpha_P\)	45.6°
Standby \(\alpha_N\)	134.4°

Case 2 — Reverse Motoring at 800 rpm (Quadrant 3)

Converter N active; \(I_a = -80\,\text{A}\) (reversed); \(E_g = -0.38 \times 800 = -304\,\text{V}\) (reversed speed).

\[ V_a = E_g + I_a R_a = -304 + (-80) \times 0.15 = -316\,\text{V} \quad(V_a < 0\;\checkmark) \]

Converter N provides the negative terminal voltage \((V_a = -V_{do}\cos\alpha_N)\):

\[ \cos\alpha_N = \frac{-V_a}{V_{do}} = \frac{316}{560.5} = 0.5638 \] \[ \alpha_N = \mathbf{55.7°} \quad(\alpha_N < 90°,\;\text{motoring}\;\checkmark) \] \[ \alpha_P = 180° - 55.7° = \mathbf{124.3°} \quad\text{(standby)} \]

Q3 — Reverse Motoring

Both \(V_a\) and \(I_a\) are negative: the motor drives a load in the reverse direction. Converter N acts as a rectifier (\(\alpha_N < 90°\)), supplying negative voltage. The standby converter P is armed at \(124.3°\) — in its inverter region — ready for Q4 (reverse regeneration).

Case 3 — Forward Regenerative Braking at 1000 rpm (Quadrant 2)

\(I_a = -80\,\text{A}\) (braking, reversed); \(E_g = +380\,\text{V}\) (forward spin, field unchanged).

\[ V_a = 380 + (-80) \times 0.15 = 368\,\text{V} \quad(V_a > 0\;\checkmark) \]

Converter N active in inverter mode:

\[ \cos\alpha_N = \frac{-368}{560.5} = -0.6566 \] \[ \alpha_N = \mathbf{131.0°} \quad(\alpha_N > 90°\;\checkmark) \] \[ \alpha_P = 180° - 131.0° = \mathbf{49.0°} \quad\text{(standby)} \]

Power Returned to Supply

\[ \begin{aligned} P_s &= E_g \times |I_a| - I_a^2\,R_a \\ &= 380 \times 80 - 80^2 \times 0.15 \\ &= 30{,}400 - 960 \\ &= \mathbf{29{,}440\,\text{W}} \end{aligned} \]

Q2 Result

Terminal voltage \(V_a\)	368 V
Active \(\alpha_N\)	131.0°
Standby \(\alpha_P\)	49.0°
Power returned	29,440 W

All Four Quadrants — Complete Summary

Quadrant	\(\alpha_P\)	\(\alpha_N\)	\(V_a\)	\(I_a\)
Q1 — Fwd. motoring	45.6°	134.4°	+392 V	+80 A
Q3 — Rev. motoring	124.3°	55.7°	−316 V	−80 A
Q2 — Fwd. regen.	49.0°	131.0°	+368 V	−80 A

4. Condition for Continuous Armature Current

Key Concept — Continuous vs. Discontinuous Conduction

Why It Matters

Converter–motor drive analyses assume continuous (ripple-free) current. This holds when the armature time constant \(\tau = L_a/R_a\) is large relative to the voltage ripple period \(T_r\). If current goes to zero during each supply cycle (discontinuous), the simple average-voltage equations no longer apply and speed control becomes nonlinear.

Practical Check Rules

1-\(\phi\) full converter: continuous if \(\omega_s L_a / R_a \geq 5\)
3-\(\phi\) full converter: continuous if \(\omega_s L_a / R_a \geq 2\)
Equivalently: \(\tau / T_r > 1\) (time constant exceeds ripple period)

Problem Statement & Given Data

Drive Parameters

Converter type	1-\(\phi\) full converter
AC supply	230 V, 50 Hz
Armature resistance \(R_a\)	1.5 Ω
Armature inductance \(L_a\)	25 mH
\(K_a\Phi\)	0.22 V/rpm

Objective

Check whether the ripple-free current assumption is justified
At \(\alpha = 20°\), \(N = 800\,\text{rpm}\): calculate \(I_a\) and armature-circuit efficiency

Solution

Part 1 — Continuity Check

\[ \omega_s = 2\pi \times 50 = 314.2\,\text{rad/s} \] \[ \frac{\omega_s L_a}{R_a} = \frac{314.2 \times 0.025}{1.5} = 5.24 \geq 5 \;\checkmark \]

Also check via time constants:

\[ \tau = \frac{L_a}{R_a} = \frac{25}{1.5} = 16.7\,\text{ms} \] \[ T_r = \frac{1}{2 \times 50} = 10\,\text{ms}, \qquad \frac{\tau}{T_r} = 1.67 > 1 \;\checkmark \]

Part 2 — Armature Current and Efficiency at \(\alpha = 20°\)

\[ V_{do} = \frac{2\sqrt{2} \times 230}{\pi} = 207.1\,\text{V} \] \[ V_a = 207.1\cos 20° = 207.1 \times 0.9397 = 194.6\,\text{V} \] \[ E_g = K_a\Phi \times N = 0.22 \times 800 = 176\,\text{V} \] \[ I_a = \frac{V_a - E_g}{R_a} = \frac{194.6 - 176}{1.5} = \mathbf{12.4\,\text{A}} \] \[ \eta = \frac{E_g}{V_a} = \frac{176}{194.6} = \mathbf{90.4\%} \]

Continuity Verdict

\(\omega_s L_a/R_a = 5.24 \geq 5\): the ripple-free (continuous conduction) assumption is justified for this drive.

Results Summary

Quantity	Value
\(\omega_s L_a / R_a\)	5.24 (≥ 5 ✓)
Time constant \(\tau\)	16.7 ms
Ripple period \(T_r\)	10 ms
\(\tau / T_r\)	1.67 (> 1 ✓)
Converter voltage \(V_a\)	194.6 V
Back-EMF \(E_g\)	176 V
Armature current \(I_a\)	12.4 A
Efficiency \(\eta\)	90.4%

Design Insight — Operating Range

At \(\alpha = 45°\), the converter would output \(V_a = 207.1\cos 45° = 146.4\,\text{V}\), which is less than \(E_g = 176\,\text{V}\). The converter cannot sustain motoring at 800 rpm with \(\alpha = 45°\) — the firing angle must be reduced (or speed lowered) to maintain positive current. Always verify \(0 \leq V_a \leq V_{do}\) for motoring.

5. Single-Phase Semi-Converter — Torque–Speed Characteristic

Problem Statement & Given Data

Motor & Supply Ratings

Motor rating	10 hp, 220 V, 1200 rpm
AC supply	240 V, 50 Hz
Armature resistance \(R_a\)	0.5 Ω
\(K_a\Phi\)	0.17 V/rpm
Firing angle \(\alpha\)	60°
Load torque \(T_L\)	45 N·m

Objective

At \(\alpha = 60°\) and load torque \(T_L = 45\,\text{N·m}\):

Armature current \(I_a\)
Motor speed \(N\)
Supply power factor

Solution

Step 1 — Converter Output Voltage

\[ V_m = \sqrt{2} \times 240 = 339.4\,\text{V} \] \[ V_a = \frac{V_m}{\pi}(1 + \cos 60°) = \frac{339.4}{\pi}(1 + 0.5) = 108.0 \times 1.5 = 162.0\,\text{V} \]

Step 2 — Armature Current from Torque

Convert \(K_a\Phi\) to SI units first:

\[ K_a\Phi\,\text{(SI)} = 0.17\,\text{V/rpm} \times \frac{60}{2\pi} = 1.623\,\text{V·s/rad} \] \[ I_a = \frac{T_L}{K_a\Phi\,\text{(SI)}} = \frac{45}{1.623} = \mathbf{27.73\,\text{A}} \]

Step 3 — Motor Speed

\[ E_g = V_a - I_a R_a = 162.0 - 27.73 \times 0.5 = 148.1\,\text{V} \] \[ N = \frac{E_g}{K_a\Phi} = \frac{148.1}{0.17} = \mathbf{871\,\text{rpm}} \]

Step 4 — Supply Power Factor

\[ P_s = V_a\,I_a = 162.0 \times 27.73 = 4492\,\text{W} \] \[ I_{rms} = 27.73\sqrt{\frac{180° - 60°}{180°}} = 27.73\sqrt{0.667} = 22.65\,\text{A} \] \[ S = V_s\,I_{rms} = 240 \times 22.65 = 5436\,\text{VA} \] \[ \text{PF} = \frac{4492}{5436} = \mathbf{0.826} \]

Results Summary

Quantity	Value
Converter voltage \(V_a\)	162.0 V
\(K_a\Phi\) (SI)	1.623 V·s/rad
Armature current \(I_a\)	27.73 A
Back-EMF \(E_g\)	148.1 V
Motor speed \(N\)	871 rpm
RMS supply current	22.65 A
Apparent power \(S\)	5436 VA
Power factor	0.826

Unit Conversion Note

When using \(T_d = K_a\Phi \cdot I_a\), both quantities must be in consistent SI units. If \(K_a\Phi\) is given in V/rpm (as is common in nameplate data), convert to V·s/rad by multiplying by \(60/2\pi = 9.549\). Using V/rpm directly gives torque in incorrect units and leads to errors.