Converter theory hal bridge

It means that, the gate pulse is generated for this area only. The phase delay is set 0 sec, means we are not giving any delay to the gate pulse.

If there is any phase delay, it means gate pulse will be generated after this time. For example, if phase delay is 1e-3 then gate pulse will be generated after 10msec. In simulation, we will use logical NOT gate. The NOT gate inverse the output means it will convert 1 to 0 and 0 to 1.

This is how, we can exactly get opposite gate pulse so that DC source will never be short circuited. This time period is known as Dead Time. This screenshot is for the output voltage across the load. In this image, we can see that, the peak value of load voltage is 50V, which is half of DC supply and frequency is 50Hz. For complete one cycle, required time is 20 msec. If you get output of half bridge inverter, then it is easy to implement the full bridge inverter, because most of all things remain the same.

In full bridge inverter also, we need only two gate pulses which is same as half bridge inverter. This screenshot is for output voltage across the load. However, it does have a meaning with reference to output.

The output of single-phase bridge inverter is a single-phase output. The output frequency of this type of inverter may be controlled by controlling the switch ON and switching OFF time of thyristors. Figure below shows the power circuit diagram of a single-phase half bridge inverter. The circuit for turning ON and turning OFF the thyristor is not shown in the above circuit to maintain simplicity.

While analyzing the circuit, it is assumed that each thyristor conducts for the duration its gate pulse is present and is commutated as soon as this pulse is removed. The root-mean-square RMS value of the output voltage can be calculated by. Fourier transform can be used to express the instantaneous voltage. As there is no DC offset so a o is zero and due to quarter-wave symmetry, all the components in are a n zero. By putting the value of b n in Fourier series equation, we get.

The even harmonics of the output voltage are not present due to quarter-wave symmetry. Hence the result is. From here, the output voltage is approximately equal to half of the applied voltage. The current through the resistive load can be easily calculated out by just dividing the RMS voltage by its resistance. The working operation of half-bridge for both L and R-L load is the same. A pure inductor has some value of ohmic resistance, so R-L load is more common in practice.

Here we will consider the working operation with RL load in a half H Bridge inverter. The working operation can be understood in 4 modes where 2 of the modes are used for controlling switches and 2 for the feeding back of the stored energy to the sources.

Terminal A and B are considered for sign convention across the load. Time interval when S1 triggered, A is more positive while B is more positive when S2 triggered. The shape of the current waveform and voltage waveform are both the same if the load is resistive. In case of RL load, both the current and voltage waveform will not rise at its peak points at the same time. The current will lag when the load is inductive dominant while voltage will lag in case of a capacitive dominant load.

The damping current will pass through feedback diodes when the polarity of both current and voltage is not the same. Diode D1 will conduct when the current is positive, and the voltage is negative.

D2 will conduct when the voltage is positive while the current is negative. The waveform is a bit different because the inductor resists the change of flow of current.

For this reason, when switches are triggered to change the flow of current, the inductor will show resistance to the change. Hence the waveform differs. These modes are discussed by considering that all switches are initially off and zero energy is stored in storage components. In this mode, switch S1 will conduct from time interval t1 to t2. The current will start flowing from DC source to the load through s1 as soon as it is triggered.

The current will enter through terminal B, showing that B is more positive w. The inductor will store the energy as the polarity of both current and voltage is the same.

In other words, the inductor will provide the stored energy, and D2 will provide the path for the current to flow form load to the source.