Full Wave Rectifier With Diode And Thyristor Using Resistive And Inductive Load

Full Wave Rectifier With Diode Using ResistiveLoad

A bridge rectifier is an arrangement of four diodes in a bridge circuit configuration which provides the same output polarity for either input polarity. It is used for converting an alternating current (AC) input into a direct current (DC) output.For the positive half cycle, D1 and D2 remain forward biased, while D3 and D4 are reversed biased. So, current flows the path (Source - D1 - Load - D2- Source) and positive polarity appears across the load.

In the negative cycle, D3 and D4 remain forward biased while other two remains forward.And current flows the path(Source - D3 - Load - D4 - Source) and positive polarity appears across the load even the input is of negative polarity. And the output waveforms of it is shown below.

The voltage across the load is given by:

V_L = V_m sin(ωt) for 0 ≤ wt ≤ π

V_L = -V_m sin(ωt) for π ≤ wt ≤ 2 π

Average value of output voltage across the load is given by

$$\eqalign{
& Vdc = {1 \over {\pi }}\int\limits_0^\pi {Vm\,Sin(\omega t){\kern 1pt} {\kern 1pt} } {\kern 1pt} d\omega t \cr
& \,\,\,\,\,\,\,\, = \,\,\,\,{{2Vm} \over \pi } \cr} $$

Average value of output current across the load is given by

$$\eqalign{
& Idc = {1 \over {\pi }}\int\limits_0^\pi {Im\,Sin(\omega t){\kern 1pt} {\kern 1pt} } {\kern 1pt} d\omega t \cr
& \,\,\,\,\,\,\,\, = \,\,\,\,{{2Im} \over \pi } \cr} $$

The rms value of voltage is$$\eqalign{
& Vrms = \,\sqrt {{1 \over {2\pi }}\int\limits_0^\pi {{{(Vm\,Sin(\omega t))}^2}{\kern 1pt} {\kern 1pt} } {\kern 1pt} d\omega t} \cr
& \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = \,\,\,\,\,\,{{Vm} \over{\sqrt 2}} \cr} $$

$$\eqalign{
& Irms = \,\sqrt {{1 \over {2\pi }}\int\limits_0^\pi {{{(Im\,Sin(\omega t))}^2}{\kern 1pt} {\kern 1pt} } {\kern 1pt} d\omega t} \cr
& \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = \,\,\,\,\,\,{{Im} \over{\sqrt 2}} \cr} $$

Where I_m = V_m/R.

$$\eqalign{
& Efficiency = \,\,{{{P_{dc}}} \over {{P_{ac}}}} \cr
& \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\;\;\;\;\; = \,\,\,{{{V_{dc}}*{\rm{ }}{I_{dc}}} \over {{V_{rms}}*{\rm{ }}{I_{rms}}}} \cr
& \;\;\;\;\;\;\;\;\;\;\;\;\;\; = \;{8 \over {{\pi ^2}}} \cr} $$

Hence, the efficiency of full wave rectifer is 0.81 which is 2 times of half wave rectiifer.

[For the fourier Analysis

A₀/2 = V_dc = 2V_m/pi

And refer these two images for the calculation of An and Bn.

[For detailed study on Fourier Analysis Refer H.Rashid]

Full Wave Rectifier With Thyristor Using Inductive Load

The output load current of the converter comprises of two components per cycle. One component flows when thyristors T1 and T2 are fired and connects the supply voltage to the load and the other component flows when thyristors T3 and T4 are turned on again connecting supply voltage to the load. Since both components of current are identical,the output voltage are identical[OBSERVE FROMα to (π + α) AND (π + α) to (2π + α) ).

REFERENCES

Futureelectronics. (n.d.). Bridge Rectifier. Retrieved from futureelectronics: http://www.futureelectronics.com/