**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 T_{1 }and T_{2} are fired together while T_{3} and T_{4} are fired 180º after T_{1 }and T_{2}. From the circuit, it is understood that for any load current to flow at the minimum one thyristor from the top group (T_{1}, T_{3} ) and one thyristor from the bottom group (T_{2}, T_{4}) must conduct. It can also maintain that neither T_{1}T_{3} nor T_{2}T_{4} can conduct simultaneously.

Let’s find the possible conduction modes when the current i_{0} can flow through T_{1}T_{2} and T_{3}T_{4}. 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 T_{1}T_{2} and T_{3}T_{4}. 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** T _{1} and T_{2} conducts**, the voltage developed across T

_{3}and T

_{4}is

**– V**. Therefore T

_{i}_{3}and T

_{4}can be triggered only when V

_{i}is negative i.e, over the negative half cycle of the input supply voltage. Similarly,

**T**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

_{1}and T_{2}**off-state impedance of each device**.

If i_{0} 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 T_{1}T_{2} and T_{3}T_{4} conducts for alternate half cycles of the input supply in the thyristor.

## Operation In Continuous Conduction Mode

For the continuous conduction mode of operation i_{0} never reaches zero, therefore, either T_{1}T_{2} or T_{3}T_{4} 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^{–} T_{3}T_{4} was conducting. As T_{1}T_{2} are fired at ωt = α they both get turn on commutating T_{3}T_{4} immediately and T3T4 is again triggered at ωt = π + α. Till this point T_{1}T_{2} 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 i_{0} continues to drop down to this point. However, as T_{1}T_{2} are triggered at this point V_{0} becomes greater than E, and i_{0} starts increasing through R-L and E. At ωt = π – θ V_{0} again equals E.

Depending upon the load circuit parameters current i_{0} reaches its maximum at around this point and starts falling afterward. Continuous conduction mode will be possible only if i_{0} remains greater than zero-till T_{3}T_{4} are triggered at ωt = π + α whereupon the same process repeats. The resulting i_{0} waveform is shown below V_{0}. The input ac current waveform i_{i} is obtained from i_{0} by observing that whenever T_{1}T_{2} conducts i_{i} =i_{0} and i_{i} = – i_{0} whenever T_{3}T_{4} conducts.

The last waveform of the figure shows the typical voltage waveform across the thyristor T_{1}. 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 n ^{th} harmonic **

V_{onRMS}\quad =\quad \frac { 1 }{ \sqrt { 2 } } \sqrt { v^2_{an}\quad +\quad v^2_{bn} }

**RMS value of v _{0}** 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 n ^{th} harmonic frequency **is given by

Z_n\quad=\quad \sqrt { R^2+\left( 2n\omega L \right)^2 }

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

To find out a closed-form expression of i_{0}. 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 i_{0} 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 i_{i} in relation to the v_{i} waveform

Load is highly inductive and the ripple on i_{0 }is negligible compared to I_{0}. Under this assumption, the idealized waveform of i_{i} becomes a square wave with transitions at ωt = α and ωt = α + π as shown in the above Fig where i_{i1} is the fundamental component of this idealized i_{i}.