Three Phase A.C to D.C Conversion using Diode and Fourier Analysis of the Waveforms

Three Phase A.C to D.C Conversion

In every D.C Load, the conversion of the single phase A.C to D.C doesn’t meet the requirement for that load to be operable. So, the use of three phase A.C to D.C come into practice. Along with it, the percentage of harmonics present in three phase A.C to D.C is less in comparable to the Single Phase A.C to D.C without any filtering or p.f correction scheme.

Three Phase A.C to D.C conversion using diode and Fourier Analysis of the Waveforms

Figure besides shows a simple three phase half wave rectifier. And it is fed from a three-phase voltage source of the sinusoidal waveform of ‘f’ Hz line of line voltage V_S via a delta-star transformer The star connected secondary provides the three phase output at anode terminals of three diodes and the cathodes of these three diodes are joined to form a positive pole of the dc output voltage.

Figure below shows the waveform of three phases of voltage whose equation are defined as;

V_A = V_max * Sin(wt); V_B = V_max * Sin(wt – 120^o); V_C = V_max * Sin(wt – 240^o)

NOTE: FOR THE CONDUCTION OF DIODE IN THIS CONFIGURATION, THAT DIODE CONDUCTS WHICH HAVE HIGHER ANODE VOLTAGE AMONG ALL OTHER THREE DIODES. SO, KEEP IT IN MIND AND TRACE THE OUTPUT WAVEFORM OF IT.

Again, back to the discussion, as we can see from the waveform above, V_B is equal to - V_max; V_A and V_C are equal at wt = π/6(i.e V_A = V_max/2 and V_C = V_max/2) but V_A is increasing while V_C is decreasing beyond that point. Therefore, the anode of the Diode D₁ is most positive and it conducts until 5π/6.

Again, at 5π/6, V_C is equal to - V_max; V_A and V_B are equal at wt = 5π/6(i.e V_A = V_max/2 and V_B = V_max/2) but V_B is increasing while V_A is decreasing beyond that point. Therefore, the anode of the Diode D₂ is most positive and it conducts until 3π/2.

Again, at 3π/2, V_A is equal to - V_max; V_C and V_B are equal at wt = 3π/2(i.e V_A = V_max/2 and V_B = V_max/2) but V_C is increasing while V_B is decreasing beyond that point. Therefore, the anode of the Diode D₃ is most positive and it conducts until 13π/6.

So, this is the conduction pattern of three diodes in three phase half wave rectifier using a diode.

Average value of output waveform is calculated as:

$$\eqalign{
& Vdc = {{A0} \over 2} = {3 \over {2\pi }}\int\limits_{\pi /3}^{5\pi /6} {{\rm{Vm}}} \;Sin(\omega t)\;d(\omega t) \cr
& \;\;\;\;\;\;\;\;\;\;\;\;\, = \;{{3\sqrt 3 } \over {2\pi }}{\rm{Vm}} \cr} $$

The rms value of output voltage can be calculated as:

$$\eqalign{
& Vrms = {\mkern 1mu} \sqrt {{3 \over {2\pi }}\int\limits_{\pi /3}^{5\pi /6} {{{(Vm{\mkern 1mu} Sin(\omega t))}^2}} d\omega t} \cr
& solving\,it,\;we\;get \cr
& Vrms = {\mkern 1mu} \sqrt 3 Vm{\mkern 1mu} \sqrt {({1 \over 6} + {{\sqrt 3 } \over {8\pi }})} \cr
& \,\,\,\,\,\,\,\,\,\,\,\, = \,0.84064Vm{\mkern 1mu} \cr} $$

The ripple voltage is

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

$$ = {\rm{ }}0.151{\rm{ }}Vm$$

For d.c current I_DC, we get

I_DC = V_dc/R = 0.7067 I_m

And r.m.s value of current I_rms , we get

I_rms = V_rms/R = 0.84064 I_m

Where I_m = V_m/R.

The efficiency of the system is
$$Efficiency = {\mkern 1mu} {\mkern 1mu} {{{P_{dc}}} \over {{P_{ac}}}}{\mkern 1mu} {\mkern 1mu} {\mkern 1mu} = {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {{{V_{dc}}*{I_{dc}}} \over {{V_{rms}}*{I_{rms}}}} = \;0.9676$$

For the fourier Analysis,

$$V0 = {{A0} \over 2} + \sum\limits_{n = 1}^\infty {(AnCos(n\omega t) + } BnSin(n\omega t))$$

A_{0 =} 2 V_dc

For the determination of An,

$$\eqalign{
& An = {1 \over \pi }[\int\limits_{\pi /6}^{5\pi /6} {{\rm{Vm}}} \;Sin(\omega t)\;Cos(n\omega t)\;d(\omega t) + \int\limits_{5\pi /6}^{3\pi /2} {{\rm{Vm}}} \;Sin(\omega t - 120)\;Cos(n\omega t)\;d(\omega t) \cr
& \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, + \int\limits_{3\pi /2}^{13\pi /6} {{\rm{Vm}}} \;Sin(\omega t - 240)\;Cos(n\omega t)\;d(\omega t)] \cr} $$

In the above calculation, as output voltage consists of these three waveforms at respective interval,the

V_A from (30^o to 150^o)

V_B from (150^o to 270^o)

V_c from (270^o to 390^o)

These three voltage waveforms are considered on the fourier calculation of A_n at the respective interval.

For the determination of Bn,

$$\eqalign{
& Bn = {1 \over \pi }[\int\limits_{\pi /6}^{5\pi /6} {{\rm{Vm}}} \;Sin(\omega t)\;Sin(n\omega t)\;d(\omega t) + \int\limits_{5\pi /6}^{3\pi /2} {{\rm{Vm}}} \;Sin(\omega t - 120)\;Sin(n\omega t)\;d(\omega t) \cr
& \,\,\,\,\,\,\,\,\,\,\,\,\, + \int\limits_{3\pi /2}^{13\pi /6} {{\rm{Vm}}} \;Sin(\omega t - 240)\;Sin(n\omega t)\;d(\omega t)] \cr} $$

References

1) Singh, Nitish Kumar. 3-phase-diode-rectifierspower-electronics. 2. 15 05 2016. <http://image.slidesharecdn.com/presentation1-140416012854-phpapp01/95/3-phase-diode-rectifierspower-electronics-3-638.jpg?cb=1443073572>.

Information from class notes and handouts provided by my subject teacher have also been used.