Power Electronics Test 4

By referring to the waveforms, the freewheeling diode will remain off till ωt = π since the positive load voltage across the load will reverse bias the diode.

However, beyond this point as the load voltage tends to become negative the freewheeling diode comes into conduction. The load voltage is clamed to zero thereafter. As a result,

• Average load voltage increases.
• RMS load voltage reduces and hence the load voltage form factor reduces.
• Conduction angle of load current increases as does its average value.
• The load current ripple factor reduces.

Concept:

Calculation:

Given that, supply voltage (V_s) = 230 V

Resistance (R) = 10 Ω

Average output voltage \(\left( {{V_0}} \right) = \frac{{2{V_m}}}{\pi } = \frac{{2 \times 230 \times \sqrt 2 }}{\pi }\)

= 207.073 V

Average load current \(\left( {{I_0}} \right) = \frac{{{V_0}}}{R} = 20.707\;A\)

Average diode current \( = \frac{{{I_0}}}{2} = \frac{{20.707}}{2} = 10.35\;A\)

RMS diode current \( = \frac{{{I_0}}}{{\sqrt 2 }} = 14.64\;A\)

For constant load current

rms value of load current (I_or) = I_o = 10A

\({{V}_{s}}{{I}_{or}}\cos \phi =E{{I}_{1}}+I_{o}^{2}R\)

\(\Rightarrow \cos \phi =\frac{E{{I}_{o}}+I_{o}^{2}R}{{{V}_{s}}{{I}_{or}}}\)

\(=\frac{\left( 120 \right)\left( 10 \right)+{{\left( 10 \right)}^{2}}\left( 0.4 \right)}{\left( 230 \right)\left( 10 \right)}\)

Concept:

In a single-phase full converter with RLE load

\({{V}_{o}}=\frac{2{{V}_{m}}}{\pi }\cos \alpha \)

V_o = E = I_oR

Calculation:            

Given that,

Supply voltage (V_rms) = 230 V

Maximum voltage \(\left( {{V}_{m}} \right)=230\sqrt{2}~V\)

Average load current (I_o) = 10 A

R = 0.4 Ω, L = 2 mH

\(\frac{2{{V}_{m}}}{\pi }\cos \alpha =E+{{I}_{0}}R\)

\(\Rightarrow \frac{2\times 230\sqrt{2}}{\pi }\cos \alpha =120+10\left( 0.4 \right)\)

Concept:

The average output voltage of a single-phase full converter by considering the effect of source-inductance is

\({V_0} = \frac{{2{V_m}}}{\pi }\cos \alpha - \frac{{2\omega {L_s}}}{\pi }{I_0}\;\)       ---(1)

And

\(\cos \alpha - \cos \left( {\alpha + \mu } \right) = \frac{{2\omega {L_s}}}{\pi }{I_0}\)        ---(2)

Calculation:

Given that, load current (I₀) = 5 A

Firing angle (α) = 45°

Supply voltage (V_m) = 330 V

Load voltage (V₀) = 140 V

From equation (1)

\( \Rightarrow 140 = \frac{{2 \times 330}}{\pi }\cos 45^\circ - \frac{{2 \times 2\pi \times 50 \times {L_s}}}{\pi } \times 5\)

⇒ L_s = 8.55 × 10^-3

From equation (2)

\( \Rightarrow \cos \left( {45} \right) - \cos \left( {45 + \mu } \right) = \frac{{8.55 \times {{10}^{ - 3}} \times 2 \times 2\pi \times 50}}{{330}} \times 5\)

Concept:

In a single-phase half wave-controlled rectifier, the average output voltage is given by

\({V_0} = \frac{{{V_m}}}{\pi }\left( {1 + \cos \alpha } \right)\)

Calculation:

\({N_{NL}} \propto \frac{1}{{{\phi _f}}}\)

\( \Rightarrow {\phi _f} \propto \;\frac{1}{{{N_{NL}}}}\)

From the above relation, to increase the no-load speed by 30%, ϕ_f should be reduced by 23%.

Therefore, the applied field voltage must be reduced by 23%.

At a firing angle 0, \({V_f}\left( {\alpha = 0} \right) = \frac{{2{V_m}}}{\pi }\)

At a firing angle α, \({V_f}\left( \alpha \right) = \frac{{{V_m}}}{\pi }\left( {1 + \cos \alpha } \right)\)

From the above two equations,

\({V_f}\left( \alpha \right) = {V_f}\left( {\alpha = 0} \right) \cdot \frac{{1 + \cos \alpha }}{2}\)

\( \Rightarrow 1 - \frac{{1 + \cos \alpha }}{2} = 0.23\)

\( \Rightarrow \frac{{1 - \cos \alpha }}{2} = 0.23\)

Concept:

In a single-phase fully controlled bridge converter,

The average output voltage, \({V_0} = \frac{{2{V_m}}}{\pi }\cos \alpha \)

The average load current, \({I_0} = \frac{{{V_0}}}{R}\)

\({V_{an}} = \frac{{2\sqrt 2 }}{\pi }{V_i}\left[ {\frac{{\cos \left( {2n + 1} \right)\alpha }}{{2n + 1}} - \frac{{\cos \left( {2n - 1} \right)\alpha }}{{2n - 1}}} \right]\)

\({V_{bn}} = \frac{{2\sqrt 2 }}{\pi }{V_i}\left[ {\frac{{\sin \left( {2n + 1} \right)\alpha }}{{2n + 1}} - \frac{{\sin \left( {2n - 1} \right)\alpha }}{{2n - 1}}} \right]\)

The RMS value of the nth harmonic is,

\({V_{0n}} = \frac{1}{{\sqrt 2 }}\sqrt {V_{an}^2 + V_{bn}^2} \)

The impedance offered by the load at nth harmonic frequency is given by

\({Z_n} = \sqrt {{R^2} + {{\left( {2n\omega L} \right)}^2}} \)

Calculation:

The average dc output voltage,

\({V_{aV}} = \frac{{2\;{V_m}}}{\pi }\cos \alpha = 179.33\;V\)

Average output load current \( = \frac{{{V_0}}}{{{R_L}}} = 17.93\;A\)

The harmonic components, V_a3 = 10.25 V

V_b3 = 35.5 V

V_{03 RMS} = 26.126 V

\({Z_3} = \sqrt {{{\left( {{R_L}} \right)}^2} + {{\left( {6 \times 2\pi \times 50 \times 50 \times {{10}^{ - 3}}} \right)}^2}} = 94.18\;{\rm{\Omega }}\)

\({I_{3RMS}} = \frac{{{V_{03\;RMS}}}}{{{Z_3}}} = 0.2756\;A = 1.54\% \;of\;{I_0}\)