Single-Phase Full Converter DC Motor Drives

Step 1 — Motor Torque

\[ T_d = K_a\Phi \cdot I_a = 1.74 \times 38 = \mathbf{66.12\,\text{N·m}} \]

Step 2 — Average Converter Voltage

\[ V_a = \frac{2\sqrt{2} \times 260}{\pi}\cos 30° = 234.1 \times 0.866 = 202.8\,\text{V} \]

Step 3 — Back-EMF and Motor Speed

\[ E_g = V_a - I_a R_a = 202.8 - 38 \times 0.3 = 191.4\,\text{V} \] \[ N = \frac{E_g}{K_a\Phi} = \frac{191.4}{0.182} = \mathbf{1051.6\,\text{rpm}} \]

Step 4 — Supply Power Factor

\[ I_{rms} = I_a = 38\,\text{A} \] \[ P_s = V_a \cdot I_a = 202.8 \times 38 = 7706\,\text{W} \] \[ S = V_s \cdot I_{rms} = 260 \times 38 = 9880\,\text{VA} \] \[ \text{PF} = \frac{P_s}{S} = \frac{7706}{9880} = \mathbf{0.780} \]

Step 5 — Armature-Circuit Efficiency

\[ \eta = \frac{E_g}{V_a} = \frac{191.4}{202.8} = \mathbf{94.4\%} \]

Step 1 — New Back-EMF (field reversed)

Reversing the field polarity reverses the back-EMF: \(E_g = -191.4\,\text{V}\).
Armature current direction is unchanged: \(I_a = +38\,\text{A}\).

Step 2 — Required Terminal Voltage

\[ V_a = E_g + I_a R_a = -191.4 + 38 \times 0.3 = -180.0\,\text{V} \]

Step 3 — Firing Angle for Inverter Mode

\[ \cos\alpha = \frac{V_a \cdot \pi}{2\sqrt{2}\,V_s} = \frac{-180.0\,\pi}{2\sqrt{2} \times 260} = -0.769 \] \[ \alpha = \cos^{-1}(-0.769) = \mathbf{140.3°} \]

Note: \(\alpha > 90°\) confirms inverter (regenerating) operation.

Step 4 — Power Fed Back to Supply

\[ P_g = |E_g| \cdot I_a = 191.4 \times 38 = 7273\,\text{W} \] \[ P_R = I_a^2 R_a = 38^2 \times 0.3 = 433\,\text{W} \] \[ P_s = P_g - P_R = 7273 - 433 = \mathbf{6840\,\text{W}} \]

Converter Maximum Voltage

\[ V_{do} = \frac{2\sqrt{2} \times 120}{\pi} = 108.04\,\text{V} \]

Back-EMF and Required Terminal Voltage

\[ E_g = K_a\Phi \times N = 0.09 \times 1000 = 90\,\text{V} \] \[ V_a = E_g + I_a R_a = 90 + 30 \times 0.4 = 102\,\text{V} \]

Firing Angle

\[ \cos\alpha = \frac{V_a}{V_{do}} = \frac{102}{108.04} = 0.9441 \] \[ \alpha = \mathbf{19.3°} \]

Power Delivered to Motor

\[ P = V_a \cdot I_a = 102 \times 30 = \mathbf{3060\,\text{W}} \]

Supply Power Factor

\[ S = V_s \cdot I_{rms} = 120 \times 30 = 3600\,\text{VA} \] \[ \text{PF} = \frac{P}{S} = \frac{3060}{3600} = \mathbf{0.85} \]

New Terminal Voltage (field reversed)

\(E_g = -90\,\text{V}\); \(I_a = +30\,\text{A}\) (direction unchanged).

\[ V_a = -90 + 30 \times 0.4 = -78\,\text{V} \]

Firing Angle for Inverter Mode

\[ \cos\alpha = \frac{-78}{108.04} = -0.7219 \] \[ \alpha = \mathbf{136.2°} \]

Power Fed Back to Supply

\[ P_{dc} = |E_g| \cdot I_a = 90 \times 30 = 2700\,\text{W} \] \[ P_R = I_a^2 R_a = 900 \times 0.4 = 360\,\text{W} \] \[ P_s = P_{dc} - P_R = 2700 - 360 = \mathbf{2340\,\text{W}} \]

Back-EMF Constant

\[ E_{\text{rated}} = V_{\text{rated}} - I_{\text{rated}}\,R_a = 220 - 11.6 \times 2 = 196.8\,\text{V} \] \[ \omega_{\text{rated}} = \frac{1500 \times 2\pi}{60} = 157.1\,\text{rad/s} \] \[ K = \frac{E_{\text{rated}}}{\omega_{\text{rated}}} = \frac{196.8}{157.1} = 1.253\,\text{V·s/rad} \]

Maximum Converter Voltage

\[ V_{do} = \frac{2\sqrt{2} \times 230}{\pi} = 207.1\,\text{V} \]

\[ \begin{aligned} \omega_{1000} &= \frac{1000 \times 2\pi}{60} \\ &= 104.7\,\text{rad/s} \\ \\ E_{1000} &= K\,\omega_{1000} \\ &= 1.253 \times 104.7 \\ &= 131.2\,\text{V} \\ \\ V_{dc} &= E_{1000} + I_{\text{rated}}\,R_a \\ &= 131.2 + 11.6 \times 2 \\ &= 154.4\,\text{V} \\ \\ \cos\alpha &= \frac{154.4}{207.1} \\ &= 0.7455 \implies \alpha = \mathbf{41.7^\circ} \end{aligned} \]

Rated braking torque requires \(I_a = +11.6\,\text{A}\) (same direction).
Reverse speed means \(E_g < 0\):

\[ \begin{aligned} E_{-1500} &= 1.253 \times (-157.1) \\ &= -196.8\,\text{V} \\ \\ V_{dc} &= -196.8 + 11.6 \times 2 \\ &= -173.6\,\text{V} \\ \\ \cos\alpha &= \frac{-173.6}{207.1} \\ &= -0.8384 \\ \\ \alpha &= \mathbf{147.0^\circ} \end{aligned} \]

\[ \begin{aligned} V_{dc} &= 207.1 \cos(160^\circ) \\ &= 207.1 \times (-0.9397) \\ &= -194.6\,\text{V} \\ \\ E &= V_{dc} - I_{\text{rated}}\,R_a \\ &= -194.6 - 11.6 \times 2 \\ &= -217.8\,\text{V} \\ \\ \omega &= \frac{E}{K} \\ &= \frac{-217.8}{1.253} \\ &= -173.8\,\text{rad/s} \\ \\ N &= -173.8 \times \frac{60}{2\pi} \\ &= \mathbf{-1660\,\text{rpm}} \end{aligned} \]