# Single Phase Fully Controlled Bridge Converter

Single Phase Fully Controlled Bridge Converter is one of the suitable converter circuits and is mostly used in the speed control of separately excited dc machines. Indeed, the R–L–E load shown in this figure may represent the electrical equivalent circuit of a separately excited dc motor. Single-phase fully controlled bridge converter is got by replacing all the diode of the corresponding uncontrolled converter by thyristors.

Thyristors T1 and T2 are fired together while T3 and T4 are fired 180º after T1 and T2. From the circuit, it is understood that for any load current to flow at the minimum one thyristor from the top group (T1, T3 ) and one thyristor from the bottom group (T2, T4) must conduct. It can also maintain that neither T1T3 nor T2T4 can conduct simultaneously.

Let’s find the possible conduction modes when the current i0 can flow through T1T2 and T3T4. It is workable that at a given moment none of the thyristors conduct. This situation will mostly occur when the load current becomes zero in between the firings of T1T2 and T3T4. When the load current becomes zero all thyristors remain off. In this mode, the load current remains zero. However, this operation of the converter is called in the discontinuous conduction mode.

Whenever T1 and T2 conducts, the voltage developed across T3 and T4 is – Vi. Therefore T3 and T4 can be triggered only when Vi is negative i.e, over the negative half cycle of the input supply voltage. Similarly, T1 and T2 can be triggered only over the positive half cycle of the input supply. The voltage across the devices when none of the thyristors conduct depends on the off-state impedance of each device.

If i0 is more than zero then the converter is called to be operating in the continuous conduction mode. In this mode of operation of the converter T1T2 and T3T4 conducts for alternate half cycles of the input supply in the thyristor.

## Operation In Continuous Conduction Mode

For the continuous conduction mode of operation i0 never reaches zero, therefore, either T1T2 or T3T4 conducts. Below fig shows the waveforms of different variables in the steady-state. The firing angle of the converter is α. The angle θ is given by

\sin { \theta } =\frac { E }{ \sqrt { 2 } V_1 }

It is supposed that at t = 0 T3T4 was conducting. As T1T2 are fired at ωt = α they both get turn on commutating T3T4 immediately and T3T4 is again triggered at ωt = π + α. Till this point T1T2 conducts. The period of conduction of different thyristors is expressed in depicted in the second waveform.

It is seen that the emf source E is greater than the dc-link voltage till ωt = α. Therefore, the load current i0 continues to drop down to this point. However, as T1T2 are triggered at this point V0 becomes greater than E, and i0 starts increasing through R-L and E. At ωt = π – θ V0 again equals E.

Depending upon the load circuit parameters current i0 reaches its maximum at around this point and starts falling afterward. Continuous conduction mode will be possible only if i0 remains greater than zero-till T3T4 are triggered at ωt = π + α whereupon the same process repeats. The resulting i0 waveform is shown below V0. The input ac current waveform ii is obtained from i0 by observing that whenever T1T2 conducts ii =i0 and ii = – i0 whenever T3T4 conducts.

The last waveform of the figure shows the typical voltage waveform across the thyristor T1. It is to be observed that when the thyristor turns off at ωt = π + α a negative voltage is applied across it for a duration of π – α. The thyristor must turn off during this interval for the successful operation of the converter.

It can be expressed in a Fourier series as follows

v_0\quad =\quad V_{OAV}+\sum _{ n=1 }^{ \alpha }{ \left[ v_{an}\cos { 2n\omega t+v_{bn}\sin { 2n\omega t } } \right] }

V_{OAV}\quad =\quad \frac { 1 }{ \pi } \int _{ \alpha }^{ \pi +\alpha }{ v_0d\omega t\quad =\quad \frac { 2\sqrt { 2 } }{ \pi } } V_i\cos { \alpha }

RMS value of the  nth harmonic

RMS value of v0 can of course be completed directly from

V_{ORMS}\quad =\quad \sqrt { \frac { 1 }{ \pi } \int _{ \alpha }^{ \alpha +\pi }{ v^2_0d\omega t } } =\quad V_i

The impedance offered by the load at nth harmonic frequency is given by

I_{onRMS}=\frac { V_{onRMS} }{ Z_n }

To find out a closed-form expression of i0. This will also help to establish the condition under which the converter will operate in the continuous conduction mode.

The current waveform is periodic over an interval π. Therefore, finding out an expression for i0 over any interval of length π will be sufficient. We choose the interval

For α ≤ ωt ≤ π + α.

L\frac { di_0 }{ dt } +Ri_0+E=\sqrt { 2 } V_i\sin { \omega t }

Z=\sqrt { R^2+\omega^2 L^2 } ;\quad \tan { \varphi =\frac { \omega L }{ R } } ;\quad E=\sqrt { 2 } V_i\sin { \theta } ; \quad R=Z\cos { \varphi }

Using this boundary condition we obtain

The waveform of ii in relation to the vi waveform

Load is highly inductive and the ripple on i0  is negligible compared to I0. Under this assumption, the idealized waveform of ii becomes a square wave with transitions at ωt = α and ωt = α + π as shown in the above Fig where ii1 is the fundamental component of this idealized ii.