Power, Electronic Systems, Applications and Resources on Electrical and Electronic Project-Thesis: Single-Phase Full Converter with RL load

The output load current of the converter comprises of two components per cycle. One component flows when thyristors S1 and S2 are fired and connects the supply voltage to the load and the other component flows when thyristors S3 and S4 are turned on again connecting supply voltage to the load. Since both components of current are identical, only one component will be studied. For the interval α ≤ ωt ≤ (α + π) the load current is given by (at ωt = α, i_L = I_L0 )

$L\frac{{d{i_L}}}{{dt}} + R{i_L} + E = {V_m}\sin \omega t$

${i_L} = \frac{{{V_m}}}{Z}\sin (\omega t - \theta ) + \left[ {{I_{L0}} + \frac{E}{R} - \frac{{{V_m}}}{Z}\sin (\alpha - \theta )} \right]{e^{(R/L)(\alpha /\omega - t)}} - \frac{E}{R}$

Single-Phase Full Converter with RL load

The current magnitude at the end of the first component of load current is the same as that at the beginning of the second component of load current, that is at ωt = π+ α, i_L = I_L0 and the current I_L0 can be obtained by substituting this condition in the above equation which yields

${I_{L0}} = \frac{{{V_m}}}{Z}\frac{{\left[ { - \sin (\alpha - \theta ) - \sin (\alpha - \theta ){e^{ - (R/L)(\pi /\omega )}}} \right]}}{{1 - {e^{ - (R/L)(\pi /\omega )}}}} - \frac{E}{R}$

The value of firing angle α at which current I_L0 = 0 can be obtained for known values of the parameters in the above equation, using an iterative method. The rms value of thyristor current is given by

${I_R} = {\left[ {\frac{1}{{2\pi }}\int\limits_\alpha ^{\pi + \alpha } {i_L^2d(\omega t)} } \right]^{ - \frac{1}{2}}}$

The rms output current is given by

${I_{rms}} = {[I_R^2 + I_R^2]^{\frac{1}{2}}} = \sqrt 2 {I_R}$

The average output current is given by
I_dc = I_A + I_A = 2I_A
The average current of one thyristor is given by

${I_A} = \frac{1}{{2\pi }}\int\limits_\alpha ^{\pi + \alpha } {{i_L}d\left( {\omega t} \right)}$

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Single-Phase Full Converter with RL load

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