Power Electronics

# Line-commutated single-phase half-wave rectifier

This post describes the topic: “What is line-commutated single-phase half-wave rectifiier”. Last posts described single-phase rectifier using diodes. Another configuration of line-commutated single-phase half-wave rectifier can be made with thyristor.

In case of line-commutated single-phase half-wave rectifier,  thyristor conducts when threshold voltage ${V}_{T}$ is positive and there is a current through the gate ${i}_{G}$. In order to control the load voltage delay of the firing angle can be used. Firing angle here is the phase angle of the AC supply voltage when the thyristor is on and and there is a current through the gate.

Let’s consider the case when the load is resistive, so current ${i}_{d}$ has the same waveform. When the load voltage and current are having negative value, the thyristor is off.

The load average voltage is ${V}_{d\left(av\right)}=\frac{1}{2\pi }{\int }_{\alpha }^{2\pi }{V}_{max}\mathrm{sin}\left(\omega t\right)d\omega t=\frac{{V}_{max}}{2\pi }\left(1+\mathrm{cos}\alpha \right)$, here $\alpha$ is s delay angle and ${V}_{max}$ is a peak supply voltage. This is the the way how to control the load voltage varying the delay angle of the firing current. The same technique can be used for controlling the transfer power. Load voltage waveform are depicted below.

Rectifier can also have inductive-resistive load (below), voltage across inductance ${v}_{L}={v}_{s}–R{i}_{d}=L\frac{di}{dt}$. If ${v}_{S}–R{i}_{d}>0$ then load current is raising, if ${v}_{S}–R{i}_{d}<0$ the load current is decreasing. The load current ${i}_{d}\left(t\right)=\frac{1}{\omega L}{\int }_{\alpha }^{\omega t}{v}_{L}d\theta$.

Despite the fact that these two controlled rectifiers configurations are simple, they are not often in used. Fully controlled bridge rectifiers are more popular and will be discussed in next post.

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