Half Wave Rectifier

Half Wave Rectifier

When a diode is introduced between a sinusoidal voltage source and a linear load circuit, the excitation is no longer a pure sinusoid.

The forced response is no longer determined by the method of sinusoidal steady – state circuit analysis at source frequency. It should however be noted that the excitation provided by an alternating voltage source in series with a diode is however periodic and any such excitation can be represented by the sum of an infinite series of sinusoidal functions of related frequency.

Having done this, the forced response to each significant term of this series may then be determined by sinusoidal steady – state analysis of the appropriate frequency.

Given a linear load circuit, superposition theorem can be used to combine the forced responses to all significant terms.

Similarly, the current supplied by the source to a half-wave rectifier circuit is also non-sinusoidal but periodic. 

The fourier series is used to calculate and analyse the waveform of most converters to determine their suitability for the intended use.

If a non-ideal voltage source in the form of a single-phase transformer were employed to excite such a circuit, the direct component of the current would saturate the core of the transformer unless the core were extremely large.

Full – wave single phase and polyphase circuits are not subject to unrealistic limitations and the analysis of the halfwave can be extended to these

In the circuit,

Current will flow when

   v – v_o > 0  : V_AK = 0 [V]

The fourier series expansion of the rectified voltage is given by

 

 v_o = Vo +          a_n sin nωt +          b_n cos nωt  [V]

α : firing angle of the diode ie the angle at which it starts to conduct.

Β : extinction angle ie. Angle at which the diode stops conducting

ϒ = β – α (rad)

ϒ: conduction angle ie. Angle during which the diode is conducting ϒ

Active Load: A load which has a voltage source or excitation

Passive Load: A load with no voltage source or excitation.

If a diode is supplying a passive load circuit,

 α = 0 and  ϒ=β

The magnitude of β depends on the nature of the Load circuit.

The coefficients of the fourier expression is determined for the general case using the expressions  below.

V_o =(1/2π)∫      v_od(ωt)          =(1/2π)∫      v_od(ωt)                [ V ]

a_n =(1/π)∫      v_osinnωtd(ωt)                      =(1/π)∫      v_osinnωtd(ωt)   [ V ]

 

b_n =(1/π)∫      v_ocosnωtd(ωt) =(1/π)∫      v_ocosnωtd(ωt)                    [ V ]

V_o = average value of rectified voltage ie. Dc output voltage

The rms value of the nth harmonic voltage is given by.

v_nR = 1/√2[a²_n + b²_n ]^1/2  [V]

The rms value of the rectified voltage is

V_R = [(1/2π)  ∫   v²_od(ωt)]^1/2 

                                                                                                                                                    

= [V_o² + ΣV ²_{n  R} ]^1/2 [V]

So that the rms value of the harmonic component or ripple voltage is

V_RI  = [ΣV²_{n R} ]^1/2                [V]

   =   [V²_R  – V²_o ]^1/2 [V]

K_v : Voltage ripple factor

K_v = V_RI/V_o  

Similar equations can be written for the line and load current in the circuit.