Introduction

Single Phase Half Wave Controlled Rectifier is a rectifier circuit which converts AC input into DC output only for positive half cycle of the AC input supply. The word “controlled” means that, we can change the starting point of load current by controlling the firing angle of SCR, the SCR turn ON at certain point of time when it is forward biased. A Single Phase Half Wave Controlled Rectifier circuit consists of SCR / thyristor, an AC voltage source and load. The load may be purely resistive, Inductive or a combination of resistance and inductance.

Following points must be kept in mind while discussing controlled rectifier:

1. The necessary condition for turn ON of SCR is that, it should be forward biased and gate signal must be applied. In other words, an SCR will only get turned ON when it is forward biased and fired or gated.
   
2. SCR will only turn off when current through it reaches below holding current and reverse voltage is applied for a time period more than the SCR turn off time.
   

1. Single Phase Half Wave Controlled Rectifier with Resistive Load

A logarithmic amplifier is an electronic circuit that produces an output that is proportional to the logarithm of the applied input. An op-amp based logarithmic amplifier produces a voltage at the output, which is proportional to the logarithm of the voltage applied to the resistor connected to its inverting terminal. The circuit diagram of op-amp based logarithmic amplifier is shown in figure 1.

Fig. 1 Single Phase Half Wave Controlled Rectifier Circuit with R load

The circuit is energized by the line voltage or transformer secondary voltage, V = V_m sin ωt as shown in Fig 1. Here, V₀ = Load output voltage, i₀ = Load current and V_T = Voltage across the thyristor T. It is assumed that the peak supply voltage never exceeds the forward and reverse-blocking ratings of the thyristor. The thyristor can be triggered at any angle α in the positive half cycle and thus the output voltage can be controlled. The thyristor blocks during the negative half cycle. The various voltage and current waveforms for the Single Phase Half Wave Controlled Rectifier with resistive load circuit are shown in Fig 2.

Fig. 2 Waveform of input voltage, firing pulse, output voltage, load current, voltage across thyristor

During the positive half cycle of the supply voltage, the thyristor anode is positive w.r.t, its cathode and until the thyristor is triggered by a proper gate pulse, it blocks the flow of load current in the forward direction. When the thyristor is fired at an angle α, full supply voltage (ignoring the thyristor drop V_T) is applied to the load. Thus, the load is directly connected to the ac supply. With a zero reactance source and a purely resistive load, the current waveform after the thyristor is triggered will be identical to the applied voltage wave and of a magnitude dependent on the amplitude of the applied voltage and the value of the load resistance R. At zero crossing, the thyristor is turned off by natural commutation thereby switching off the power to the load. The voltage and current are in same phase, as illustrated in Fig. 2. The angle (π – α) during which the thyristor conducts is called the conduction angle. By varying the firing angle α, the output voltage can be controlled. During the period of conduction, voltage drop across the thyristor is of the order of one volt. During the negative half cycle of the supply voltage, the thyristor blocks the flow of load current and no voltage is applied to the load R.

Analysis : Average load voltage is given by

$$V_{dc} = \frac {1}{2\pi} \left[\int_{\alpha}^{\pi} V_m ~ sin ~ \omega t ~ d(\omega t)+\int_{\pi}^{2\pi}(0)~d(\omega t)\right]$$

$$V_{dc} = \frac {1}{2\pi} \bigg[-V_m ~ cos\omega t \bigg]_{\alpha}^{\pi} = \frac {V_m}{2\pi}(1+cos \alpha)........(1)$$

Where, V_m is peak value of the ac input voltage. From the above Eq. (1) it is obvious that the load voltage varies with firing angle α having extremes values for α = 0 and α = π or 180°. The maximum output voltage is obtained when α = 0 and is given by

$$V_{dc ~ max} =\frac {V_m}{\pi} .......(2)$$

Average load current with resistive load is directly proportional to the average load voltage and is given by

$$I_{dc} = \frac {V_{dc}}{R} = \frac {V_m}{2\pi R}(1+cos \alpha ).......(3)$$

The rms value of load voltage is given by

$$V_{L~rms}=\frac {V_m}{2 \sqrt{\pi} } \left(\pi -\alpha + \frac {1}{2}sin ~ 2 \alpha\right)^\frac {1}{2}.......(4)$$

For firing angle α =0

$$V_{L ~ rms} = \frac {V_m}{2}.......(5)$$

The rms value of load current

$$I_{L ~ rms} = \frac {V_{L ~ rms}}{R} = \frac {V_{L ~ rms}}{2 R \sqrt{\pi}}\left(\pi -\alpha + \frac {1}{2}sin ~ 2 \alpha\right)^\frac {1}{2}......(6)$$

Average output power is given by

$$P_{dc} = V_{dc}.I_{dc} = \frac {V_m^2}{4 \pi^2 R}(1+ cos \alpha)^ \frac {1}{2}.......(7)$$

2. Single Phase Half Wave Controlled Rectifier with RL Load

Fig. 3 Single Phase Half Wave Controlled Rectifier Circuit with RL load

The circuit diagram of a single phase half wave controlled rectifier with RL load is shown above. This circuit consists of a thyristor T, source Vs and RL load. The output voltage is the voltage across the load and shown as V_o. Output current is the current through the load and shown as i_o.

Fig. 4 Waveform of input voltage, firing pulse, output voltage, load current, voltage drop across thyristor
The waveform of source voltage, load voltage, load current and voltage across thyristor is shown above. It is assumed that the thyristor T is fired at an angle ωt = α. As soon as the thyristor T is fired at ωt = α, load voltage equal to the source voltage instantaneously appears across the load terminal. This is because, the thyristor is forward biased in between ωt = 0 to α. Hence, once the thyristor is gated, it starts conducting.

However, the current does not start at this instant of firing. This is just because of the nature of load. Since, the load is inductive, it will not allow any sudden change. Therefore, at ωt = α, the output current will be zero and will gradually increase. The output current will become maximum and then start decreasing. It should be noted here that, this behavior of load current i_o will not be observed for purely resistive load.

At ωt = π, the load voltage V_o reduces to zero. However, the load current will not be zero at this instant because of inductance L. Due to this, thyristor will not turn off, even though it is reversed biased. Rather it will continue to conduct till ωt = β. At ωt = β, the load current becomes zero and thyristor is reversed biased, hence it will turn off. This is a case of natural commutation.

After ωt = β, V_o = 0 and i_o = 0. At ωt = (2π+α), the SCR is triggered again, V_o is applied to the load and load current develops as discussed before. The angle β where the load current becomes zero is called extinction angle and the angle (β-α) for which thyristor is ON is called conduction angle.

Carefully observe the voltage across the thyristor. The SCR is reverse biased from ωt = β to ωt = 2π. During this period, the current through thyristor is also zero. Therefore, circuit turn off time is t_c = [(2π – β) / ω] second. This time must be greater than the thyristor turn-off time otherwise thyristor may turn on at undesired instant and will lead to commutation failure.

The Voltage equation is :

$$V_{m}~ sin \omega t = Ri_0+L \frac {dI_0}{dt}$$

The load current i_o consist of two component, one steady state component i_s and the transient component i_t. Here i_s is given by

$$i_{s} = \frac {V_m}{\sqrt{ (R^2+X^2)}}sin(\omega t- \phi)$$

The transient component i_t can be obtained from force free equation

$$Ri_t + L\frac{di_t}{dt} = 0$$

Its solution gives,

$$i_t = A e^{-\left(\frac{R}{L}\right)t}$$

$$i_o = i_s + i_t =\frac {V_m}{Z} sin(\omega t-\phi) + Ae^{-\left(\frac {R}{L}\right)t} .........(8)$$

Constant A can be obtained from the boundary condition at ωt = α. At time

$$t = \frac{\alpha}{\omega}$$

$$i_o = 0$$

Thus from equation (8)

$$A = -\frac {V_m}{Z} sin(\alpha - \phi )e^{\frac {R \alpha}{L \omega}}$$

Substitution of A in Eq. (8) gives

$$i_o = \frac {V_m}{Z} sin(\omega t- \phi) -\frac {V_m}{Z}sin(\alpha -\phi ) exp {-\frac{R}{\omega L}(\omega t-\alpha)}.......(9)$$

It is also seen from waveform of i_o in fig. 4 that when ωt = β, load current i_o = 0.

The average voltage V_o is given by

$$V_o = \frac {1}{2 \pi} \int_\alpha ^\beta V_m sin \omega t ~ d(\omega t) =\frac {V_m}{2 \pi}(cos \alpha -cos \beta).......(10)$$

The average load currrent I_o is given by

$$I_o = \frac {V_m}{2 \pi R}(cos \alpha -cos \beta)$$

Rms Load voltage is

$$V_{or} = \left[ \frac {1}{2 \pi } \int_\alpha ^\beta V_m^2 ~ sin^2 ~ \omega t.d(\omega t)\right]^{\frac {1}{2}}$$

$$V_{or} = \frac {V_m}{2 \pi} \sqrt {\left[(\beta - \alpha)- \frac {1}{2}(sin 2 \beta - sin 2 \alpha )\right]}........(11)$$

3. Single Phase Half Wave Controlled Rectifier with RLE Load

Fig. 5 Single Phase Half Wave Controlled Rectifier Circuit with RLE load

A Single Phase Half Wave Controlled Rectifier Circuit with RLE load is shown in Fig. 5. The counter emf E in the load may be due to a battery or a DC motor. The minimum value of firing angle is obtained from the relation V_msin ωt = E, this is shown to occur at an angle θ₁ where

$$\theta_1 = sin^{-1}\left(\frac{E}{V_m}\right).......(12)$$

In case thyristor T is fired at an angle α <θ₁, then E > V_s SCR is reversed biased and therefore it will not turn on. Similarly, maximum value of firing angle is θ₂ = π - θ₁ in Fig. 6. During the interval load current i_o is zero, Load voltage v_o= E and during the time i_o is not zero, v_o follows v_s curve, KVL gives the voltage differential equation as

$$V_m ~ sin \omega t = Ri_o + L\frac{di_o}{dt} + E.........(13)$$

Fig. 6 Voltage and current waveforms

The solution of this equation is made up of two components: namely steady-state current component i_s and the transient current component i_t. For conveninence, i_s may be thought of as the sum of i_s1 and i_s2, Where i_s1 is the steady state current due to ac source voltage acting alone and i_s2 is that due to dc counter emf E acting alone due to source voltage V_msin ωt is given by

$$i_{s1} = \frac{V_m}{Z} sin(\omega t-\phi)$$

If only E were present, then steady state current i_s2 would be given by

$$i_{s2} = -\left(\frac{E}{R}\right)$$

The transient current i_t is given by

$$i_t = Ae^{-\left(\frac{R}{L}\right)^t}$$

At ωt = α, i_o = 0, i.e., at t = α/ω, i_o = 0.

$$A = \left[\frac {E}{R}-\frac {V_m}{Z} sin(\alpha - \phi )\right]e^{-\frac{Ra}{L \omega}}$$

$$i_o =\frac {V_m}{Z}\left[sin(\omega t- \phi )-sin( \alpha - \phi)exp{-\frac {R}{\omega L}(\omega t- \alpha )}\right]-\frac {E}{R}\left[1-exp{-\frac{R}{ \omega L}(\omega t- \alpha)}\right].......(14)$$

Equation 11 is applicable for α ≤ ωt ≥ β. The extinction angle β depends upon load emf E, firing angle α and the load impedence angle φ the average load current i_o is given by

$$I_o = \frac{1}{2 \pi R} \bigg[V_m(cos \alpha -cos \beta )-E(\beta-\alpha) \bigg]..........(15)$$

Average load voltage V_o is given by

$$V_o= E+I_oR=E+ \frac {1}{2 \pi}\bigg[2V_m sin\left(\alpha + \frac{\gamma}{2}\right)sin\left(\frac {\gamma}{2}\right)-\gamma.E\bigg]$$

$$V_o=E\left(1-\frac {\gamma} {2 \pi}\right)+\frac {V_m} {\pi} sin\left(\alpha +\frac {\gamma}{2}\right)sin \left(\frac {\gamma}{2}\right)..........(16)$$

In case β is made equal to (γ+α) in the above expression, Equation (15) can be obtained. If load inductance L is zero in Fig. 5, then extinction angle β = θ₂ and γ = β - α = θ₂ - α but θ₂ = π - θ₁.

This gives β = θ₂ = π - θ₁ and γ = π - θ₁ - α. Putting this in Equation (15), we get average load current I_o as under :

$$I_o = \frac {1}{ 2 \pi R} \bigg[V_m (cos \alpha +cos \theta_1)-E( \pi - \theta_1- \alpha)\bigg].......(17)$$

For no inductance in the load circuit, β = π - α. The RMS value of load current for L = 0 is obtained as under:

$$I_{or} = \bigg[\frac {1}{2\pi R^2}\int_\alpha^{\pi-\alpha}(V_m sin \omega t-E)^2.d(\omega t)\bigg]^\frac {1}{2}.........(18)$$

Power delivered to load

$$P=I^2R+I_oE.......(19)$$

Supply power factor

$$= I_or^2R+I_o \frac {E}{V_sI_{or}}$$

Advantages of Half-wave controlled rectifier

1. Half wave controlled rectifier is a simple circuit, We can easily construct.
   
2. Low cost and Simple circuit with less number of components.
   
3. Economical at initial state. Although there is a higher cost over time due to more power losses.

Disadvantages of Half-wave controlled rectifier

1. The transformer utilization factor is low.
   
2. They produce a low output voltage.
   
3. DC saturation of transformer core resulting in magnetizing current and also some hysteresis losses and generation of harmonics.
   
4. The power is delivered only during the one-half cycle of the input alternating voltage.
   
5. Ripple factor is high, therefore required to give steady dc output.
   

Applications of Half-wave controlled rectifier

1. AM radio
   
2. Pulse generated circuits
   
3. Single demodulation
   
4. Voltage multiplier