1-Phase Semi Converter Solved Problems

1. Key Formulae

Single-Phase Semi-Converter vs. Full Converter — Voltage Equations

Semi-converter average voltage	\(\displaystyle V_{dc} = \frac{\sqrt{2}\,V_s}{\pi}(1 + \cos\alpha)\)	Single quadrant; \(V_{dc} \geq 0\) always
Full converter average voltage	\(\displaystyle V_{dc} = \frac{2\sqrt{2}\,V_s}{\pi}\cos\alpha\)	Two quadrant; \(V_{dc}\) can be negative
RMS supply current — semi-converter	\(\displaystyle I_{rms} = I_a\sqrt{\frac{\pi - \alpha}{\pi}}\)	Current flows only in positive half-cycle
RMS supply current — full converter	\(I_{rms} = I_a\)	Current flows in both half-cycles
Supply power factor (general)	\(\displaystyle \text{PF} = \frac{V_{dc}\,I_a}{V_s\,I_{rms}}\)	Ratio of real power to apparent power

Series DC Motor Equations

Developed torque	\(T_d = K_{af}\,I_a^2\)	\(K_{af}\) = field-armature coupling constant
Back-EMF	\(E_g = (K_{af}\,I_a + K_{res})\,\omega\)	\(K_{res}\) accounts for residual flux
Armature current	\(\displaystyle I_a = \frac{V_{dc} - K_{res}\,\omega}{R + K_{af}\,\omega}\)	\(R = R_a + R_f\) = total series resistance

Separately Excited Motor with Separate Field Converter

Developed torque	\(T_d = K_v\,I_f\,I_a\)	\(K_v\) in V/(A·rad/s)
Back-EMF	\(E_b = K_v\,I_f\,\omega\)	Proportional to both field current and speed
Motor speed	\(\displaystyle \omega = \frac{E_b}{K_v\,I_f} = \frac{V_a - I_a R_a}{K_v\,I_f}\)	Speed set by armature voltage and field flux
Field current (full converter, \(\alpha_f = 0\))	\(\displaystyle I_f = \frac{V_{f,\max}}{R_f},\quad V_{f,\max} = \frac{2\sqrt{2}\,V_s}{\pi}\)	Maximum field current at zero firing angle

Semi-Converter Advantage

The semi-converter draws supply current only during the positive half-cycle, giving a higher power factor than the full converter at the same firing angle. The trade-off is that \(V_{dc}\) cannot go negative — regenerative braking requires field reversal.

2. Semi-Converter Drive — Firing Angle and Power Factor

Problem Statement & Given Data

Motor & Supply Ratings

AC supply	208 V, 60 Hz
Armature resistance \(R_a\)	0.25 Ω
Field resistance \(R_f\)	147 Ω
Motor constant \(K_v\)	0.7032 V/(A·rad/s)
Load torque \(T_L\)	45 N·m
Operating speed \(N\)	1000 rpm

Drive Configuration

Armature fed by a single-phase semi-converter.
Field fed by a single-phase full converter at \(\alpha_f = 0\) (maximum field).

Find:

Field current \(I_f\)
Armature delay angle \(\alpha_a\)
Input power factor of the armature converter

Solution

Step 1 — Preliminary Values

\[ V_m = \sqrt{2} \times 208 = 294.2\,\text{V} \] \[ \omega = 1000 \times \frac{\pi}{30} = 104.7\,\text{rad/s} \]

Step 2 — Maximum Field Current (\(\alpha_f = 0\))

\[ V_f = \frac{2V_m}{\pi} = \frac{2 \times 294.2}{\pi} = 187.3\,\text{V} \] \[ I_f = \frac{V_f}{R_f} = \frac{187.3}{147} = \mathbf{1.274\,\text{A}} \]

Step 3 — Armature Current and Back-EMF

\[ I_a = \frac{T_L}{K_v I_f} = \frac{45}{0.7032 \times 1.274} = \mathbf{50.23\,\text{A}} \] \[ E_b = K_v\,I_f\,\omega = 0.7032 \times 1.274 \times 104.7 = \mathbf{93.82\,\text{V}} \] \[ V_a = E_b + I_a R_a = 93.82 + 50.23 \times 0.25 = \mathbf{106.4\,\text{V}} \]

Step 4 — Armature Delay Angle \(\alpha_a\) (Semi-Converter)

\[ V_a = \frac{V_m}{\pi}(1 + \cos\alpha_a) \] \[ 1 + \cos\alpha_a = \frac{V_a\,\pi}{V_m} = \frac{106.4\,\pi}{294.2} = 1.136 \] \[ \cos\alpha_a = 0.136 \implies \alpha_a = \mathbf{82.2°} \]

Step 5 — Input Power Factor

\[ P_o = V_a\,I_a = 106.4 \times 50.23 = 5344\,\text{W} \] \[ \begin{aligned} I_{s,rms} &= I_a \sqrt{\frac{\pi - \alpha_a}{\pi}} \\ &= 50.23 \sqrt{\frac{180^\circ - 82.2^\circ}{180^\circ}} \\ &= 50.23 \sqrt{0.543} \\ &= 37.03\,\text{A} \end{aligned} \] \[ \text{PF} = \frac{P_o}{V_s\,I_{s,rms}} = \frac{5344}{208 \times 37.03} = \mathbf{0.694} \]

Results Summary

Quantity	Value
Field current \(I_f\)	1.274 A
Armature current \(I_a\)	50.23 A
Back-EMF \(E_b\)	93.82 V
Armature voltage \(V_a\)	106.4 V
Armature delay angle \(\alpha_a\)	82.2°
RMS supply current \(I_{s,rms}\)	37.03 A
Power factor (armature conv.)	0.694

Why \(I_{rms} < I_a\)?

With the semi-converter, current is drawn from the supply only during the positive half-cycle (conduction angle = \(180° - \alpha_a = 97.8°\)). The effective RMS current is therefore reduced by the factor \(\sqrt{(\pi - \alpha)/\pi} = 0.737\), improving the apparent power and power factor compared to a full converter carrying the same \(I_a\).

3. Series Motor — Semi-Converter vs. Full Converter

Problem Statement & Given Data

Motor & Supply Ratings

Motor rating	20 hp, 230 V, 900 rpm
Total series resistance \(R\)	0.15 Ω (field + armature)
Field-armature constant \(K_{af}\)	0.03 N·m/A²
Residual flux constant \(K_{res}\)	0.075 V·s/rad
AC supply	260 V, 60 Hz
Firing angle \(\alpha\)	30°
Operating speed \(N\)	900 rpm

Objective

Compare motor torque, armature current, and supply power factor when the series motor is supplied by:

A single-phase semi-converter
A single-phase full converter

Both at \(\alpha = 30°\), \(N = 900\,\text{rpm}\).

Common Preliminary Quantities

\[ \omega = 900 \times \frac{\pi}{30} = 94.2\,\text{rad/s} \] \[ K_{res}\,\omega = 0.075 \times 94.2 = 7.07\,\text{V} \] \[ R + K_{af}\,\omega = 0.15 + 0.03 \times 94.2 = 2.976\,\Omega \]

Case 1 — Single-Phase Semi-Converter

Average Voltage

\[ V_{dc} = \frac{\sqrt{2} \times 260}{\pi}(1 + \cos 30°) = 117.0 \times 1.866 = 218.3\,\text{V} \]

Motor Current and Torque

\[ I_a = \frac{V_{dc} - K_{res}\,\omega}{R + K_{af}\,\omega} = \frac{218.3 - 7.07}{2.976} = 70.97\,\text{A} \] \[ T_d = K_{af}\,I_a^2 = 0.03 \times 70.97^2 = \mathbf{151.1\,\text{N·m}} \]

Supply Power Factor

\[ P_s = V_{dc}\,I_a = 218.3 \times 70.97 = 15{,}494\,\text{W} \] \[ I_{rms} = I_a\sqrt{\frac{\pi - \alpha}{\pi}} = 70.97\sqrt{\frac{150°}{180°}} = 70.97 \times 0.9129 = 64.77\,\text{A} \] \[ \text{PF} = \frac{15{,}494}{260 \times 64.77} = \mathbf{0.92} \]

Semi-Converter Results

Quantity	Value
Average voltage \(V_{dc}\)	218.3 V
Armature current \(I_a\)	70.97 A
Developed torque \(T_d\)	151.1 N·m
Power factor	0.92

Case 2 — Single-Phase Full Converter

Average Voltage

\[ V_{dc} = \frac{2\sqrt{2} \times 260}{\pi}\cos 30° = 234.1 \times 0.866 = 202.7\,\text{V} \]

Motor Current and Torque

\[ I_a = \frac{202.7 - 7.07}{2.976} = 65.73\,\text{A} \] \[ T_d = 0.03 \times 65.73^2 = \mathbf{129.6\,\text{N·m}} \]

Supply Power Factor

\[ P_s = 202.7 \times 65.73 = 13{,}325\,\text{W} \] \[ I_{rms} = I_a = 65.73\,\text{A} \] \[ \text{PF} = \frac{13{,}325}{260 \times 65.73} = \mathbf{0.78} \]

Full Converter Results

Quantity	Value
Average voltage \(V_{dc}\)	202.7 V
Armature current \(I_a\)	65.73 A
Developed torque \(T_d\)	129.6 N·m
Power factor	0.78

Comparison Summary

Parameter	Semi-Converter	Full Converter
\(V_{dc}\)	218.3 V	202.7 V
\(I_a\)	70.97 A	65.73 A
\(T_d\)	151.1 N·m	129.6 N·m
Power factor	0.92	0.78
Quadrants	1 (no regen.)	2 (regen. possible)

Key Comparison Insight

At the same \(\alpha = 30°\), the semi-converter yields a higher power factor (0.92 vs. 0.78) because it draws supply current only during the positive half-cycle. The trade-off is a higher average voltage and current, producing a larger torque (151.1 vs. 129.6 N·m). The full converter is preferred when regenerative braking is required.

4. Full-Converter Drive with Field Control — Torque, Speed, Power Factor

Problem Statement & Given Data

Motor & Supply Ratings

AC supply	440 V, 60 Hz
Armature resistance \(R_a\)	0.25 Ω
Field resistance \(R_f\)	175 Ω
Motor constant \(K_v\)	1.4 V/(A·rad/s)
Armature firing angle \(\alpha_a\)	60°
Field firing angle \(\alpha_f\)	0° (maximum field)
Armature current \(I_a\)	45 A

Drive Configuration & Objective

Both armature and field are supplied by single-phase full converters from the same 440 V AC supply.

Find:

Developed torque \(T_d\)
Motor speed \(\omega\)
Input power factor (combined)

Solution

Step 1 — Field Circuit (\(\alpha_f = 0\))

\[ V_m = \sqrt{2} \times 440 = 622.3\,\text{V} \] \[ V_f = \frac{2V_m}{\pi} = \frac{2 \times 622.3}{\pi} = 396.1\,\text{V} \] \[ I_f = \frac{V_f}{R_f} = \frac{396.1}{175} = \mathbf{2.26\,\text{A}} \]

Step 2 — Developed Torque

\[ T_d = K_v\,I_f\,I_a = 1.4 \times 2.26 \times 45 = \mathbf{142.4\,\text{N·m}} \]

Step 3 — Motor Speed

\[ \begin{aligned} V_a &= \frac{2V_m}{\pi} \cos \alpha_a \\ &= 396.1 \times \cos 60^\circ \\ &= 198.1\,\text{V} \\ \\ E_b &= V_a - I_a R_a \\ &= 198.1 - 45 \times 0.25 \\ &= 186.9\,\text{V} \\ \\ \omega &= \frac{E_b}{K_v\,I_f} \\ &= \frac{186.9}{1.4 \times 2.26} \\ &= \mathbf{59.1\,\text{rad/s}} \end{aligned} \]

Step 4 — Combined Input Power Factor

\[ \begin{aligned} P_{\text{in}} &= V_a\,I_a + V_f\,I_f \\ &= 198.1 \times 45 + 396.1 \times 2.26 \\ &= 8914 + 895 \\ &= 9809\,\text{W} \\ \\ I_s &\approx \sqrt{I_a^2 + I_f^2} \\ &= \sqrt{45^2 + 2.26^2} \\ &= 45.06\,\text{A} \\ \\ \text{PF} &= \frac{P_{\text{in}}}{V_s\,I_s} \\ &= \frac{9809}{440 \times 45.06} \\ &= \mathbf{0.495} \end{aligned} \]

Results Summary

Quantity	Value
Field current \(I_f\)	2.26 A
Developed torque \(T_d\)	142.4 N·m
Back-EMF \(E_b\)	186.9 V
Motor speed \(\omega\)	59.1 rad/s
Combined power factor	0.495

Low Power Factor Note

The combined power factor of 0.495 is low because the armature converter fires at \(\alpha_a = 60°\) (significantly delayed). Even though the field converter operates at \(\alpha_f = 0°\), it draws very little current (2.26 A vs. 45 A), so the armature converter's poor power factor dominates the total. Power factor improves as \(\alpha_a\) is reduced.

5. Regenerative Braking with Full-Converter Field Control

Problem Statement & Given Data

Continuation of Previous Problem

Back-EMF (from previous)	\(E_b = 186.9\,\text{V}\)
Armature current	\(I_a = 45\,\text{A}\)
Armature resistance \(R_a\)	0.25 Ω
\(V_m\)	622.3 V

New Condition & Objective

The field polarity is reversed, making \(E_b = -186.9\,\text{V}\). Armature current \(I_a = 45\,\text{A}\) is maintained.

Find:

Required delay angle \(\alpha_a\)
Power fed back to supply

Solution

Step 1 — New Armature Terminal Voltage

With reversed back-EMF \((E_b = -186.9\,\text{V})\):

\[ V_a = E_b + I_a R_a = -186.9 + 45 \times 0.25 = \mathbf{-175.7\,\text{V}} \]

Step 2 — Required Firing Angle

\[ \cos\alpha_a = \frac{V_a\,\pi}{2V_m} = \frac{-175.7\,\pi}{2 \times 622.3} = -0.443 \] \[ \alpha_a = \mathbf{116.3°} \]

Note: \(\alpha_a > 90°\) confirms converter operates in inverter mode.

Step 3 — Power Fed Back to Supply

\[ P_{\text{regen}} = |V_a| \cdot I_a = 175.7 \times 45 = \mathbf{7907\,\text{W}} \]

Results Summary

Quantity	Value
New terminal voltage \(V_a\)	−175.7 V
Firing angle \(\alpha_a\)	116.3°
Power fed back	7907 W

Motoring vs. Braking Comparison

Mode	\(\alpha_a\)	\(V_a\)	Power
Motoring	60°	+198.1 V	8914 W in
Regenerating	116.3°	−175.7 V	7907 W out

Field reversal shifts \(E_b\) negative, forcing \(V_a < 0\). The single full-converter can handle this directly by increasing \(\alpha_a\) beyond 90°.