Phase-Controlled Three-Phase Full Converter DC Motor Drives

Three-Phase Full Converter Drives — Solved Problems

1. Key Formulae

Three-Phase Full Converter — Voltage and Current

Average dc output voltage \(\displaystyle V_{dc} = \frac{3\sqrt{6}}{\pi}\,V_\phi\cos\alpha = \frac{3\sqrt{2}}{\pi}\,V_{LL}\cos\alpha\) \(V_\phi\) = rms phase voltage; \(V_{LL}\) = line-to-line rms
Maximum voltage (\(\alpha = 0\)) \(\displaystyle V_{do} = \frac{3\sqrt{6}}{\pi}\,V_\phi\) For 480 V line: \(V_{do} = 648\,\text{V}\)
RMS line current (ripple-free) \(\displaystyle I_A = \sqrt{\tfrac{2}{3}}\,I_a\) 6-pulse converter; each thyristor conducts 120°
Supply power factor \(\displaystyle \text{PF} = \frac{V_{dc}\,I_a}{3\,V_\phi\,I_A} = \frac{3}{\pi}\cos\alpha \approx 0.9549\cos\alpha\) Ripple-free current assumption

Speed Regulation and Two-Region Speed Control

Speed regulation \(\displaystyle \text{SR} = \frac{N_0 - N_{FL}}{N_{FL}} \times 100\%\) \(N_0\) = no-load speed; \(N_{FL}\) = full-load speed
Armature-circuit efficiency \(\displaystyle \eta = \frac{E_g}{V_a} = 1 - \frac{I_a R_a}{V_a}\) Excludes field and rotational losses
Base speed \(N_b\) \(\displaystyle N_b = \frac{V_{a,\max} - I_{a,\text{rated}}\,R_a}{K_a\Phi_{\max}}\) Max speed at full armature voltage and full field
Region 1 — Armature Voltage Control (\(N \leq N_b\))
Field at maximum (\(\Phi = \Phi_{\max}\), \(\alpha_f = 0\)); \(V_a\) varied by adjusting \(\alpha_a\). Constant maximum torque capability. Speed increases with \(V_a\).
Region 2 — Field Weakening (\(N > N_b\))
\(V_a = V_{a,\max}\) (\(\alpha_a = 0\)); field flux \(\Phi\) reduced by increasing \(\alpha_f\). Speed increases at roughly constant power. Torque capability decreases with speed.

2. Three-Phase Full Converter — Motoring and Regeneration

Problem Statement & Given Data

Motor & Supply Ratings
Motor rating100 hp, 600 V, 1800 rpm
AC supply3-phase, 480 V (line), 60 Hz
Armature resistance \(R_a\)0.1 Ω
Armature inductance \(L_a\)5 mH
\(K_a\Phi\)0.3 V/rpm
Rated armature current130 A (ripple-free)
Objective
  1. Motoring at 1500 rpm, rated current:
    1. Firing angle \(\alpha\)
    2. Supply power factor
  2. Regenerative braking at 1000 rpm, rated current:
    1. Firing angle \(\alpha\)
    2. Power fed back and supply power factor

Converter Constant

\[ V_\phi = \frac{480}{\sqrt{3}} = 277\,\text{V}, \qquad V_{dc} = \frac{3\sqrt{6}}{\pi} \times 277\cos\alpha = 648\cos\alpha \]

Motoring at 1500 rpm

Firing Angle

\[ E_g = K_a\Phi \times N = 0.3 \times 1500 = 450\,\text{V} \] \[ V_{dc} = E_g + I_a R_a = 450 + 130 \times 0.1 = 463\,\text{V} \] \[ \cos\alpha = \frac{463}{648} = 0.7145 \implies \alpha = \mathbf{44.4°} \]

Supply Power Factor

\[ I_A = \sqrt{\tfrac{2}{3}} \times 130 = 106.1\,\text{A} \] \[ S = 3\,V_\phi\,I_A = 3 \times 277 \times 106.1 = 88{,}169\,\text{VA} \] \[ P_s = V_{dc}\,I_a = 463 \times 130 = 60{,}190\,\text{W} \] \[ \text{PF} = \frac{60{,}190}{88{,}169} = \mathbf{0.68} \]
Motoring Results
QuantityValue
Back-EMF \(E_g\)450 V
Converter voltage \(V_{dc}\)463 V
Firing angle \(\alpha\)44.4°
RMS line current \(I_A\)106.1 A
Power factor0.68

Regenerative Braking at 1000 rpm

Firing Angle (field polarity unchanged — generating)

\[ E_g = 0.3 \times 1000 = 300\,\text{V} \quad\text{(magnitude)} \] \[ V_{dc} = -E_g + I_a R_a = -300 + 130 \times 0.1 = -287\,\text{V} \] \[ \cos\alpha = \frac{-287}{648} = -0.4428 \implies \alpha = \mathbf{116.3°} \]

Power Fed Back and Supply Power Factor

\[ P_{dc} = E_g \cdot I_a = 300 \times 130 = 39{,}000\,\text{W} \] \[ P_R = I_a^2 R_a = 130^2 \times 0.1 = 1690\,\text{W} \] \[ P_s = P_{dc} - P_R = 39{,}000 - 1690 = \mathbf{37{,}310\,\text{W}} \] \[ \text{PF} = \frac{P_s}{S} = \frac{37{,}310}{88{,}169} = \mathbf{0.423} \]
Regeneration Results
QuantityValue
Converter voltage \(V_{dc}\)−287 V
Firing angle \(\alpha\)116.3°
Power generated \(P_{dc}\)39,000 W
Armature loss \(P_R\)1,690 W
Power returned \(P_s\)37,310 W
Power factor0.423
PF Degradation in Inverter Mode
Power factor drops from 0.68 (motoring) to 0.423 (braking) because the firing angle increases from 44.4° to 116.3°. PF \(\approx 0.9549\cos\alpha\): at 116.3°, \(\cos\alpha = -0.443\), so the converter absorbs reactive power while returning real power to the supply.

3. Three-Phase Full Converter — No-Load Speed and Speed Regulation

Problem Statement & Given Data

Motor & Supply Ratings
Motor rating125 hp, 600 V, 1800 rpm
AC supply3-phase, 480 V (line), 60 Hz
Armature resistance \(R_a\)0.0874 Ω
Armature inductance \(L_a\)6.5 mH
\(K_a\Phi\)0.33 V/rpm
Rated current \(I_{a,\text{rated}}\)165 A
No-load current \(I_{a,0}\)16.5 A (10% of rated)
Objective
  1. No-load speeds at \(\alpha = 0°\) and \(\alpha = 30°\)
  2. Firing angle \(\alpha\) and supply PF at rated speed and rated current
  3. Speed regulation at the firing angle found in Part 2

Converter Constant & No-Load Speeds

\[ V_\phi = \frac{480}{\sqrt{3}} = 277\,\text{V}, \qquad V_{dc} = 648\cos\alpha \]

No-Load Speed at \(\alpha = 0°\)

\[ V_{dc} = 648\,\text{V} \] \[ E_g = V_{dc} - I_{a,0}\,R_a = 648 - 16.5 \times 0.0874 = 646.6\,\text{V} \] \[ N_0 = \frac{E_g}{K_a\Phi} = \frac{646.6}{0.33} = \mathbf{1959\,\text{rpm}} \]

No-Load Speed at \(\alpha = 30°\)

\[ V_{dc} = 648\cos 30° = 561.2\,\text{V} \] \[ E_g = 561.2 - 16.5 \times 0.0874 = 559.8\,\text{V} \] \[ N_0 = \frac{559.8}{0.33} = \mathbf{1696\,\text{rpm}} \]
No-Load Speed Results
\(\alpha\)No-Load Speed
0°1959 rpm
30°1696 rpm
Insight
Even at no load, the motor draws \(I_{a,0} = 16.5\,\text{A}\) to overcome friction, windage, and core losses. The small voltage drop \(I_{a,0}R_a = 1.44\,\text{V}\) slightly reduces the no-load speed below the ideal \(V_{dc}/K_a\Phi\).

Rated Load: Firing Angle and Power Factor

Required Converter Voltage at Rated Load

\[ E_g = K_a\Phi \times N = 0.33 \times 1800 = 594\,\text{V} \] \[ V_{dc} = 594 + 165 \times 0.0874 = 608.4\,\text{V} \]

Firing Angle

\[ \cos\alpha = \frac{608.4}{648} = 0.9389 \implies \alpha = \mathbf{20.1°} \]

Supply Power Factor

\[ I_A = \sqrt{\tfrac{2}{3}} \times 165 = 134.6\,\text{A} \] \[ S = 3 \times 277 \times 134.6 = 111{,}881\,\text{VA} \] \[ P_s = 608.4 \times 165 = 100{,}386\,\text{W} \] \[ \text{PF} = \frac{100{,}386}{111{,}881} = \mathbf{0.90} \]

Cross-check: \(\text{PF} = 0.9549\cos(20.1°) = 0.9549 \times 0.939 = 0.897 \approx 0.90\) ✓

Rated Load Results
QuantityValue
Back-EMF \(E_g\)594 V
Converter voltage \(V_{dc}\)608.4 V
Firing angle \(\alpha\)20.1°
RMS line current \(I_A\)134.6 A
Power factor0.90

Speed Regulation at \(\alpha = 20.1°\)

No-Load Speed at Same Firing Angle

\[ E_{g,0} = 608.4 - 16.5 \times 0.0874 = 606.96\,\text{V} \] \[ N_0 = \frac{606.96}{0.33} = 1839.3\,\text{rpm} \]

Speed Regulation

\[ \text{SR} = \frac{N_0 - N_{\text{rated}}}{N_{\text{rated}}} \times 100\% = \frac{1839.3 - 1800}{1800} \times 100\% = \mathbf{2.18\%} \]
Excellent Speed Regulation
A speed regulation of 2.18% is very good for a phase-controlled drive. The low value arises from the small armature resistance (0.0874 Ω) relative to the high rated current, meaning the voltage drop \(I_a R_a = 14.4\,\text{V}\) is only 2.4% of the rated converter voltage. Three-phase drives naturally offer better regulation than single-phase drives due to lower ripple.

4. Three-Phase Full Converter with Field Weakening (Two-Region Speed Control)

Problem Statement & Given Data

Motor & Supply Ratings
Motor rating20 hp, 300 V, 900 rpm
AC supply3-phase Y, 208 V (line), 60 Hz
Armature resistance \(R_a\)0.25 Ω
Field resistance \(R_f\)145 Ω
Motor constant \(K_v\)1.2 V/(A·rad/s)
Objective
Both armature and field fed by 3-phase full converters.
  1. At \(T_d = 116\,\text{N·m}\), \(N = 900\,\text{rpm}\), field at maximum (\(\alpha_f = 0\)): find \(\alpha_a\)
  2. Same torque, \(\alpha_a = 0\): find speed \(N\) (base speed)
  3. Same torque, \(N = 1800\,\text{rpm}\), \(\alpha_a = 0\) (field weakening): find \(\alpha_f\)

Preliminary Values

\[ V_\phi = \frac{208}{\sqrt{3}} \approx 120\,\text{V}, \qquad V_{a,\max} = V_{f,\max} = \frac{3\sqrt{6}}{\pi} \times 120 = 280.7\,\text{V} \quad(\alpha = 0) \] \[ I_{f,\max} = \frac{V_{f,\max}}{R_f} = \frac{280.7}{145} = 1.936\,\text{A} \]

Part 1 — Armature Firing Angle at 900 rpm, Full Field (Region 1)

\[ \omega = 900 \times \frac{\pi}{30} = 94.25\,\text{rad/s} \] \[ I_a = \frac{T_d}{K_v\,I_{f,\max}} = \frac{116}{1.2 \times 1.936} = 49.93\,\text{A} \] \[ E_b = K_v\,I_{f,\max}\,\omega = 1.2 \times 1.936 \times 94.25 = 219.0\,\text{V} \] \[ V_a = E_b + I_a R_a = 219.0 + 49.93 \times 0.25 = 231.5\,\text{V} \] \[ \cos\alpha_a = \frac{231.5}{280.7} = 0.8248 \implies \alpha_a = \mathbf{34.5°} \]
Region 1 — Result
At 900 rpm with full field, the armature converter fires at \(\alpha_a = 34.5°\). This is Region 1 (armature voltage control): field flux is maximum and speed is controlled by adjusting \(V_a\) through \(\alpha_a\).

Part 2 — Base Speed at \(\alpha_a = 0\) (Maximum Armature Voltage)

\[ V_a = V_{a,\max} = 280.7\,\text{V} \] \[ E_b = 280.7 - 49.93 \times 0.25 = 268.2\,\text{V} \] \[ \omega = \frac{268.2}{K_v\,I_{f,\max}} = \frac{268.2}{1.2 \times 1.936} = 115.4\,\text{rad/s} \] \[ N_b = 115.4 \times \frac{30}{\pi} = \mathbf{1102\,\text{rpm}} \]
Base Speed Significance
\(N_b = 1102\,\text{rpm}\) is the boundary between Region 1 and Region 2. At this point, \(V_a\) has reached its maximum and field flux is still at its rated value. Further speed increase requires field weakening (Region 2).

Part 3 — Field Weakening at 1800 rpm (Region 2)

\(V_a = V_{a,\max} = 280.7\,\text{V}\) (armature at maximum); flux reduced to raise speed.

\[ \omega = 1800 \times \frac{\pi}{30} = 188.5\,\text{rad/s} \] \[ E_b = 280.7 - 49.93 \times 0.25 = 268.2\,\text{V} \] \[ I_f = \frac{E_b}{K_v\,\omega} = \frac{268.2}{1.2 \times 188.5} = 1.186\,\text{A} \] \[ V_f = I_f\,R_f = 1.186 \times 145 = 172.0\,\text{V} \] \[ \cos\alpha_f = \frac{V_f}{V_{f,\max}} = \frac{172.0}{280.7} = 0.6128 \implies \alpha_f = \mathbf{52.2°} \]
Field Weakening Result
QuantityValue
Required field current \(I_f\)1.186 A
Required field voltage \(V_f\)172.0 V
Field firing angle \(\alpha_f\)52.2°
Two-Region Control Summary
RegionSpeed Range\(\alpha_a\)\(\alpha_f\)
Region 1 (const. torque)0 – 1102 rpmvaries0°
Region 2 (const. power)1102 – 1800+ rpm0°increases

5. Speed Regulation and Armature Efficiency

Problem Statement & Given Data

Motor & Supply Ratings
Motor rating20 hp, 300 V, 1800 rpm
AC supply3-phase Y, 208 V (line), 60 Hz
Armature resistance \(R_a\)0.25 Ω
Field resistance \(R_f\)245 Ω
Motor constant \(K_v\)1.2 V/(A·rad/s)
Rated current \(I_{a,\text{rated}}\)≈ 49.7 A
No-load current \(I_{a,0}\)10% of rated = 4.97 A
Drive Configuration & Objective

Both armature and field at maximum field (\(\alpha_f = 0\)).

  1. Armature delay angle \(\alpha_a\) at rated load and speed
  2. No-load speed and speed regulation
  3. Armature-circuit efficiency at rated load

Preliminary Quantities

\[ V_\phi = \frac{208}{\sqrt{3}} \approx 120\,\text{V}, \qquad V_{do} = \frac{3\sqrt{6}}{\pi} \times 120 = 280.7\,\text{V} \] \[ \omega_{\text{rated}} = 1800 \times \frac{\pi}{30} = 188.5\,\text{rad/s}, \qquad I_{a,\text{rated}} = \frac{20 \times 746}{300} = 49.73\,\text{A} \]

Field at maximum (\(\alpha_f = 0\)):

\[ I_f = \frac{V_{do}}{R_f} = \frac{280.7}{245} = 1.146\,\text{A} \]

Part 1 — Armature Delay Angle at Rated Load

\[ E_b = K_v\,I_f\,\omega_{\text{rated}} = 1.2 \times 1.146 \times 188.5 = 259.2\,\text{V} \] \[ V_a = E_b + I_{a,\text{rated}}\,R_a = 259.2 + 49.73 \times 0.25 = 271.6\,\text{V} \] \[ \cos\alpha_a = \frac{271.6}{280.7} = 0.9676 \implies \alpha_a = \mathbf{14.6°} \]
Part 1 Result
The small firing angle \(\alpha_a = 14.6°\) means the converter operates close to its maximum voltage. This is typical for large motors running near their rated conditions.

Part 2 — No-Load Speed and Speed Regulation

\(I_{a,0} = 4.97\,\text{A}\); same \(V_a = 271.6\,\text{V}\):

\[ E_{b,0} = V_a - I_{a,0}\,R_a = 271.6 - 4.97 \times 0.25 = 270.4\,\text{V} \] \[ \omega_0 = \frac{270.4}{K_v\,I_f} = \frac{270.4}{1.2 \times 1.146} = 196.6\,\text{rad/s} \] \[ N_0 = 196.6 \times \frac{30}{\pi} = 1877.5\,\text{rpm} \] \[ \text{SR} = \frac{1877.5 - 1800}{1800} \times 100\% = \mathbf{4.31\%} \]
Speed Regulation Result
A speed regulation of 4.31% is acceptable for industrial drives. The regulation is slightly higher than the previous example (4.31% vs. 2.18%) because this motor has a higher \(R_a / V_a\) ratio (0.25 Ω vs. 0.0874 Ω), causing a larger speed drop under load.

Part 3 — Armature-Circuit Efficiency

\[ \eta = \frac{E_b}{V_a} = \frac{259.2}{271.6} = \mathbf{95.4\%} \]

Verification via Power Balance

\[ P_{\text{in,arm}} = V_a\,I_a = 271.6 \times 49.73 = 13{,}505\,\text{W} \] \[ P_{\text{copper}} = I_a^2\,R_a = 49.73^2 \times 0.25 = 618\,\text{W} \] \[ P_{\text{mech}} = 13{,}505 - 618 = 12{,}887\,\text{W} \] \[ \eta = \frac{12{,}887}{13{,}505} = 95.4\% \;\checkmark \]
Complete Results Summary
QuantityValue
Field current \(I_f\)1.146 A
Back-EMF \(E_b\)259.2 V
Armature voltage \(V_a\)271.6 V
Firing angle \(\alpha_a\)14.6°
No-load speed \(N_0\)1877.5 rpm
Speed regulation SR4.31%
Armature efficiency \(\eta\)95.4%
Copper loss \(P_{\text{copper}}\)618 W