Single Phase Fully Controlled Halfwave Rectifier

Single Phase Fully Controlled Halfwave Rectifier is obtained by replacing the diodes of an uncontrolled converter with thyristors. Thyristors can be turned ON by applying a current pulse at its gate terminal at a desired instance. However, they cannot be turned off from the gate terminals. Analysis and performance of Single Phase Fully Controlled Halfwave Rectifier supplying an R and R-L load will be studied in the blog.

Working principle of thyristors based single phase fully controlled rectifiers will be explained first in the case of a single thyristor halfwave rectifier circuit supplying an R or R-L load.

1. Single Phase Fully Controlled Halfwave RectifierwithResistive load

The circuit diagram of a single-phase fully controlled halfwave rectifier supplying a purely resistive load. At ωt = 0 when the input supply voltage becomes positive the thyristor T becomes forward biased. However, unlike a diode, it does not turn ON till a gate pulse is applied at ωt = α. During the period 0 < ωt ≤ α, the thyristor blocks the supply voltage and the load voltage remains zero as no-load current flows during this interval.

As soon as a gate pulse is applied to the thyristor at ωt = α it turns ON. The voltage across the thyristor collapses to almost zero and the full supply voltage appears across the load. From this point onwards the load voltage follows the supply voltage. The load being purely resistive the load current i0 is proportional to the load voltage. At ωt = π as the supply voltage passes through the negative-going zero crossing the load voltage and hence the load current becomes zero and tries to reverse direction.

In this process, the thyristor undergoes reverse recovery and starts blocking the negative supply voltage. Therefore, the load voltage and the load current remains clamped at zero till the thyristor is fired again at ωt = 2π + α. The same process repeats thereafter.

For \alpha <\omega t\le \Pi

vo  = vi= \sqrt { 2 } Vi sinωtdωt

vo  =i0 = 0 otherwise

VOAV = \frac { 1 }{ 2\pi } \int _{ 0 }^{ 2\pi } vodωt

=\frac { 1 }{ 2\pi } \int _{ 0 }^{ \pi }\sqrt { 2 } Visinωtdωt

VOAV = vi \frac { 1 }{ \sqrt { 2 } \pi } \left( 1+cos\alpha \right)

Form factor FFVO = VORMS /VOAV = \frac { \pi \sqrt { 1-\frac { \alpha }{ \pi } +\frac { \sin { 2\alpha } }{ 2\pi } } }{ 1+\cos { \alpha } }

2. Single Phase Fully Controlled Halfwave Rectifier with Resistive-Inductive load

Circuit is hardly used in practice its analysis does provide useful insight into the operation of fully controlled rectifiers which will help to appreciate the operation of single-phase bridge converters.

As in the case of a resistive load, the thyristor T becomes forward biased when the supply voltage becomes positive at ωt = 0. However, it does not start conduction until a gate pulse is applied at ωt = α. As the thyristor turns ON at ωt = α the input voltage appears across the load and the load current starts building up.

However, unlike a resistive load, the load current does not become zero at ωt = π, instead it continues to flow through the thyristor and the negative supply voltage appears across the load forcing the load current to decrease. Finally, at ωt = β (β > π) the load current becomes zero and the thyristor undergoes reverse recovery. From this point onwards the thyristor starts blocking the supply voltage and the load voltage remains zero until the thyristor is turned ON again in the next cycle.

It is to be noted that the value of β depends on the load parameters. Therefore the resistive load, the average, and RMS output voltage depends on the load parameters. Since the thyristors do not conduct over the entire input supply cycle this mode of operation is called the discontinuous conduction mode.

For \alpha \le \omega t\le \beta

vo  = vi= \sqrt { 2 } Vi sinωtdωt

vo = 0 otherwise

VOAV = \frac { 1 }{ 2\pi } \int _{ 0 }^{ 2\pi } vodωt

VOAV = \frac { 1 }{ 2\pi } \int _{ \alpha }^{ \beta }{ \sqrt { 2 } } Vi sinωtdωt

= Vi \frac { 1 }{ \sqrt { 2 } \pi } \left( \cos { \alpha } -\cos { \beta } \right)

IOAV = VOAV/R

=Vi \frac { 1 }{ \sqrt { 2 } \pi R } \left( \cos { \alpha } -\cos { \beta } \right)

The average voltage drop across the inductor is zero.