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Power Electronic Riddle No.6 - Thyristor turn-off condition
During thyristor turn-off stage, after avalanche breakdown how depletion region starts reducing and returns to the original position, if possible give me the answer in diagramatic form so that I may be able to understand, what actually happens during avalanche and how depletion region is reduced. I will be very thankful to you!!!!!!!!!!!
Author : Abhishek Kumar - From: India
 
#1
Mon, January 10th, 2011 - 17:34
A thyristor is a controllable device, which can be switched from a blocking state (high voltage, low current) to a conducting state (low voltage, high current) by a suitable gate pulse. Forward conduction is blocked until an external positive pulse is applied to the gate terminal. A thyristor cannot be turned off from the gate. During forward conduction, its behavior resembles that of a power diode and it also exhibits a forward voltage drop of between 1 to 3 volts. Like the diode, conduction is blocked in the reverse biased direction.



The thyristor is turned off when it becomes reverse biased and/or the forward current falls below the holding current. This must be controlled externally in the power circuit.
The transitional period from blocking to conducting, and vice versa, is called commutation and the period during which a component turns on/off, is called the commutation period. During commutation, the component comes under electrical stress due to changes in the circuit conditions and the thermal stress due to losses. These losses produce heat in the component and also stress the insulation and current paths.



In power electronics, the operation of any converter is dependent on the switches being turned ON and OFF in a sequence. Current passes through a switch when it is ON and is blocked when it is OFF. As mentioned above, the word commutation is used to describe the transfer of current from one switch turning OFF to another turning ON.
In a diode rectifier circuit, a diode turns ON and starts to conduct current when there is a forward voltage across it, i.e. the forward voltage across it becomes positive. This process usually results in the forward voltage across another diode becoming negative, which then turns off which stops conducting current. In a thyristor rectifier circuit, the switches additionally need a gate signal to turn them on and off.
The factors affecting commutation may be illustrated in the idealized diode circuit in Figure below, which shows two circuit branches, each with its own variable DC voltage source and circuit inductance. Assume, initially, that a current I is flowing through the circuit and that the magnitude of the voltage V1 is larger than V2. Since V1 > V2, diode D1 has a positive forward voltage across it and it conducts a current I1 through its circuit inductance L1. Diode D2 has a negative forward voltage across it and is blocking and carries no current.



Consequently, at time t1
I1 = I
I2 = 0
Suppose that voltage V2 is increased to a value larger than V1, the forward voltage across diode D2 becomes positive and it then starts to turn on. However, the circuit inductance L1 prevents the current I1 from changing instantaneously and diode D1 will not immediately turn off. So, both diodes D1 and D2 remain ON for an overlap period called the commutation time tc.
With both diodes turned on, a closed circuit is established which involves both branches. The effective circuit voltage VC = (V2 – V1), called the commutation voltage, then drives a circulating current ic, called the commutation current, through the two branches which have a total circuit inductance of LC = (L1 + L2).
In this idealized circuit, the volt drop across the diodes and the circuit resistance have been ignored. From basic electrical theory of inductive circuits, the current ic increases with time at a rate dependent on the circuit inductance. The magnitude of the commutation current may be calculated from the following equations.



If the commutation starts at a time t1 and finishes at a time t2, the magnitude of the commutation current IC at any time t, during the commutation period, may be calculated by integrating the above equation from time t1 to t.



During the commutation period:

- It is assumed that the overall current through the circuit remains constant.
I = (I1 +I2) constant
As the circulating commutation current increases:
- Current (I2) through the diode that is turning on increases in value
I2 = Ic increasing
- Current (I1) through the diode that is turning off decreases in value
I1 = I – Ic decreasing



For this special example, it can be assumed that the commutation voltage VC is constant during the short period of the commutation. At time t the integration yields the following value of IC, which increases linearly with time.



When IC has increased to a value equal to the load current I at time t2, then all the current has been transferred from branch 1 to branch 2 and the current through the switch that is turning off has decreased to zero. The commutation is then over.
Consequently, at time t2

I1 = 0
I2 = Ic = I

At the end of commutation when t = t2, putting IC equal to I in the above equation, the time taken to transfer the current from one circuit branch to the other (commutation time), may be calculated.



It is clear from this equation that the commutation time tc depends on the overall circuit inductance (L1 + L2) and the commutation voltage.
From this we can conclude that:

- A large circuit inductance will result in a long commutation time.
- A large commutation voltage will result in a short commutation time.

In practice, a number of deviations from this idealized situation occur.
- The diodes are not ideal and do not turn off immediately when the forward voltage becomes negative. When a diode has been conducting and is then presented with a reverse voltage, some reverse current can still flow for a few microseconds as indicated in Figure above. The current I1 continues to decrease beyond zero to a negative value before returning to zero. This is due to the free charges that must be removed from the PN junction before blocking is achieved.
- Even if the commutation time is very short, the commutation voltage of an AC fed rectifier bridge does not remain constant but changes slightly during the commutation period. An increasing commutation voltage will tend to reduce the commutation time.
 
Author : Hamid - From: Iran- Firouzabad in Fars
 
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