Half Wave Rectification with Thyristor using Inductive and Resistive Load

Half Wave Rectification with Thyristor using Resistive Load

During the positive half cycle, thyristor remains forward biased. If the gate signal is applied with delay angle α , the thyristor will start conducting from α to π degree. At pi degree, the input voltage will be zero and according to output voltage and current will be zero. Hence, thyristor will be naturally commutated at this point(wt = π) . During the negative half cycle, the thyristor remains reverse biased and the thyristor won’t conduct and the output voltage will be zero. Again, the thyristor can be turned on by providing the gate signal at the gate terminal in the next positive half cycle.And the waveform can be viewed on the figure given besides.

Fourier Analysis

The basis experssion of fourier Analysis is :
$$V0 = {{A0} \over 2} + \sum\limits_{n = 1}^\infty {(AnCos(n\omega t) + } BnSin(n\omega t))$$

Average value of output voltage is given by:

$$\eqalign{
& V0 = {{A0} \over 2} = {1 \over {2\pi }}\int\limits_\alpha ^\pi {{\rm{Vm}}} \;Sin(\omega t)\;d(\omega t) \cr
& \;\;\;\;\;\;\;\;\;\;\; = \;{{{\rm{Vm}}} \over {2\pi }}\int\limits_\alpha ^\pi {sin(\omega t)\;d(\omega t)} \cr
& \;\;\;\;\;\;\;\;\;\;\; = {{{\rm{Vm}}} \over {2\pi }}[ - cos(\omega t)]_\alpha ^\pi \cr
& \;\;\;\;\;\;\;\;\;\;\; = \;{{{\rm{Vm}}} \over {2\pi }}[1 + cos(\alpha )] \cr} $$

If α = 0, V₀ = V_m/ π and if α = π, V₀ = 0

For An,$$An = {1 \over {\pi }}\int\limits_\alpha ^\pi {{\rm{Vm}}} \;Sin(\omega t)\;Cos(n\omega t)\;d(\omega t)$$

For Bn,$$Bn = {1 \over {\pi }}\int\limits_\alpha ^\pi {{\rm{Vm}}} \;Sin(\omega t)\;Sin(n\omega t)\;d(\omega t)$$

$$\eqalign{
& Cn = \sqrt {A{n^2} + B{n^2}} \cr
& {\phi _n}\;\; = \;{\tan ^{ - 1}}({{An} \over {Bn}}) \cr
& V0(t)\; = \;{{A0} \over 2} + \;C1\,\,Sin(wt + {\phi _1}) + C2\,\,Sin(2wt + {\phi _2}) + ............... \cr} $$

So, in this way a complete fourier analysis of the output waveform of the thyristor controlled rectifier can be performed.

For the rms voltage,

$$Vrms = {\mkern 1mu} \sqrt {{1 \over {2\pi }}\int\limits_\alpha ^\pi {{{(Vm{\mkern 1mu} Sin(\omega t))}^2}} d\omega t} $$

For the ripple voltage,

$$Vripple = \sqrt {Vrm{s^2} - Vd{c^2}} $$

Half Wave Rectification With Thyristor Using Inductive Load

During the positive half cycle, thyristor remains forward biased. If the gate signal is applied with delay angle α , the thyristor will start conducting from α to π degree. At pi degree, the input voltage will be zero and according to load current will not be zero due to the inductive load.So, the input voltage appears across the output even though the input voltage is in a negative cycle. At 'β' degree, the output current will be zero and after that point, thyristor stops conducting the current and output voltage will be zero. Again, the thyristor can be turned on by providing the gate signal at the gate terminal in the next positive half cycle(i.e 2π + α). And the waveform can be viewed in the figure given below.

For rectifier applications, peak inverse voltage (PIV) or peak reverse voltage (PRV) is the maximum value of reverse voltage which occurs at the peak of the input cycle when the diode/thyristor is reverse-biased.And PIV loss in the thyristor can be viewed at an angle (ie 0 to alpha; Beta to 2*pi and 2pi to (2pi + alpha) where thyristor doesn't conduct the current and voltage doesn't flow through it.

For the Dc components,

$$\eqalign{
& An = {{A0} \over 2} = {1 \over {2\pi }}\int\limits_\alpha ^\beta {{\rm{Vm}}} \;Sin(\omega t)\;d(\omega t) \cr
& \;\;\;\;\;\;\;\;\;\;\; = \;{{{\rm{Vm}}} \over {2\pi }}\int\limits_\alpha ^\beta {sin(\omega t)\;d(\omega t)} \cr
& \;\;\;\;\;\;\;\;\;\;\; = {{{\rm{Vm}}} \over {2\pi }}[ - cos(\omega t)]_\alpha ^\beta \cr
& \;\;\;\;\;\;\;\;\;\;\; = \;{{{\rm{Vm}}} \over {2\pi }}[cos(\beta ) + cos(\alpha )] \cr} $$

For The An components,$$An = {1 \over {\pi }}\int\limits_\alpha ^\beta {{\rm{Vm}}} \;Sin(\omega t)\;Cos(n\omega t)\;d(\omega t)$$

For The Bn components,$$Bn = {1 \over {\pi }}\int\limits_\alpha ^\beta {{\rm{Vm}}} \;Sin(\omega t)\;Sin(n\omega t)\;d(\omega t)$$