Half Wave Rectification with Power Diode using Inductive and Resistive Load

Half-Wave Rectification with Power DiodeUsing Resistive Load

A rectifier is a circuit that converts an AC signal into a uni-directional DC signal. Power Diodes are extensively used in rectifier circuits

During the positive half cycle of V_s ,Diode D₁ conducts, and the input voltage appears across the load. During the negative cycle of V_S, Diode will be in reversed biased. Hence, the voltage loss takes place in the diode and no voltage appears across the load.

The voltage across the load is given by:

V_L = V_m sin(ωt) for 0 ≤ wt ≤ π

V_L = 0 for π ≤ wt ≤ 2 π

Average value of output voltage across the load is given by

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

Average value of output current across the load is given by

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

For the rms calculation of voltage,

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

For the rms calculation of current,

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

Where I_m = V_m/R.

Output D.c Power P_dc = V_dc * I_dc

Output A.c Power P_ac = V_rms * I_rms

Efficiency of rectification is given by:

$$\eqalign{
& Efficiency = \,\,{{{P_{dc}}} \over {{P_{ac}}}} \cr
& \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\;\;\;\;\; = \,\,\,{{{V_{dc}}*{\rm{ }}{I_{dc}}} \over {{V_{rms}}*{\rm{ }}{I_{rms}}}} \cr
& \;\;\;\;\;\;\;\;\;\;\;\;\;\; = \;{4 \over {{\pi ^2}}} \cr} $$

Ripple factor is the unwanted voltage which appears over and above the desired dc voltage. And it can be expressed as

$$\eqalign{
& Vr = \;\sqrt {Vrm{s^2} - Vd{c^2}} \cr
& R.F\;\;\; = \sqrt {{{Vrm{s^2} - Vd{c^2}} \over {Vd{c^2}}}} \cr
& \;\;\;\;\;\;\; = \;1.21(for\;half\;wave\;rectifier) \cr
& The\;transformer\;utilization\;factor\;is\;defined\;as \cr
& TUF\;\;\; = {{Pdc} \over {VsIs}} \cr
& where\;Vs\;and\;Is\;are\;rms\;values\;of\;\sec ondary\;voltage\;and\;current. \cr} $$

Half-Wave Rectification with Power Diode Using Inductive Load

In the diagram, below we will study two types of cases. In the case 1, we will study about the waveform of output voltage that appears across the load only using RL load and in the case 2, by the use of both RL Load And Freewheeling Diode.

Case-1:(Observe Waveform b)

A single phase diode rectifier feeding R-L load is shown in figure (a) [Ignore freewheeling Diode]and its output waveform is shown in figure(b). Current I_l continues to flow even after source voltage V_s become negative. This is because of the presence of an inductive load. After a positive half cycle of the source voltage, the diode still remains on. So, negative half cycle source voltage appears across the load until the load current decays to zero at (π + α = β).

Expression for V_dc can be calculated as:

$$\eqalign{
& Vdc = \;{1 \over {2\pi }}\int\limits_0^\beta {Vm{\mkern 1mu} Sin(\omega t)d\omega t} \cr
& \;\;\;\;\; = \;{{Vm} \over {2\pi }}(1 - \;\cos \beta ) \cr
& Idc = {{Vdc} \over R} \cr
& \;\;\;\;\; = \;{{Vm} \over {2\pi R}}(1 - \;\cos \beta ) \cr} $$

Case-2:(Observe Waveform c)

The performance of single phase half wave rectifier with inductive load can be improved by connecting a free wheeling diode across the load. The output voltage is Vs from 0 to pi. The source voltage is zero at wt = pi, but I₀ will not be zero at that point due to the inductive load. Just after wt = pi, the source voltage will be reversed in its polarity, the diode D_m gets forward biased. As a result, I₀ is immediately transferred from Diode D₁ to Diode D_m .And the energy stored in the inductive load is supplied to the resistive load and it slowly decays. In the case of highly inductive load, the I₀ can't fall to zero till 2*pi degree and such case are called as continuous conduction. If I₀ falls before 2*pi degree, then such a case is known as discontinuous conduction.