Saturday, November 3, 2012

Power Electronics Graduate Trainee Interview [BGR Energy]

These questions were asked in Power Electronics Graduate Trainee Interview conducted by BGR Energy Systems Company at MIT Chennai.

[1] Draw and Explain the SCR V-I characteristic?
Vak = +ve & Vg = 0
(i) When a positive voltage is applied to anode with respect to cathode, the junctions J1 and J3 are         forward biased, but the junction J2 is reverse biased. 

(ii) The SCR is in its forward blocking state. At this time the Gate signal is not applied. 

(iii) As shown in figure a small amount of forward leakage current is flows through the device.

Vak = + & Vg = +ve
(i) When the small amount of positive voltage is applied, while positive voltage is applied to anode with respect to cathode, the junction J3 becomes forward biased. 

(ii) Thus the SCR conducts a large value of forward current with small voltage drop. With the application of gate signal the SCR changed from forward blocking state to forward conducting state. It is called as latching.Without gate signal it happens at forward breakdown voltage (Vfbd).

(iii) When the gate signal value is increased, the latching happens for low Vak voltages as mentioned in the figure. In the presence of forward current (i.e. after the thyristor is turned on by a suitable gate voltage) it will not turn off even after the gate voltage has been removed.  

(iv) The thyristor will only turn off when the forward current drops below holding current. The holding current is defined as the minimum current required to hold the SCR in the forward conduction state.

Vak = -ve
(i) When a negative voltage is applied to anode with respect to cathode, the junctions J1 and J3 are reversed biased, but the junction J2 is forward biased. The SCR is in its reverse blocking state. 

(ii) As shown in figure a small amount of reverse leakage current flows through the device. 
If the applied voltage is more than reverse breakdown voltage, the device will collapse and the large amount of current flows through it.

[2] Explain Buck topology
(i) When the switch is ON condition (ie, TON), the input provides energy to inductor and output. 
(ii) The difference between the input and output voltage is applied to inductor as shown in the below waveform.
(iii) When the switch is turned off, the inductor current will flow in the same direction to load. It is inductor property to maintain the current flow in the same direction.
(iv) The freewheeling diode D1 is forward biased at this time and completes the inductor current path.

The output of the buck converter with continuous output current is 

Vout = D*VIN

D = Duty cycle = TON / [TON + TOFF]

[3] Compare MOSFET and BJT
             BJT                                                                     MOSFET
More Power handling Capability                             Less Power handling capability
Low switching speed                                             Fast switching speed
High On state resistance                                        Low on state resistance
Has second breakdown voltage problem                 No second breakdown voltage problem

1 comment:

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