Question

Two 1200V thyristor are series connected across them. The forward blocking characteristic of the first thyristor has a slope of 16.7uAN and the second thyristor slope is 12.5UAN. (a) How will they share the voltage (b) Voltage equalizing resistors of 6.8KN are connected across each thyristor. Calculate the voltage across each thyristor and the power dissipation in the equalizing resistors. (20 marks) QUESTION SEVEN (a) With the use of circuit diagrams and waveforms, explain the operation of a single phase to single phase cyclo-converter whose control signal is triggered at 60 degrees. (b) Draw a well labelled circuit diagram for a three-phase IGBT inverter which is P-type.

Answer 1

important to understand the slop

slop is given ampere/volts which gives us mho

but mho is the inverse of ohm,

for answer please see the images in sequence

Given :- o two which tyhen are thyristor - 12000 connected in each series The of the of forward blocking first thyristor 16.7 uAlu and characteristic has a slop 12.5 MAIN Now , we diagram like this Ti Too MWh R2 the is two thyrister consides voltage across not given let is V=3000 Volts volt Now Ampere we know R= Ī Ampere and slop given as I volts

(2 hence let's take inverse of given slop RT, = 16.7x06 = 59.88 ka w GOKR RT 2 = 12.5x10-6 = 80kn using voltage finding voltage is V.X Rt, (Rat Riz) - divide rule share across thyristors 3000 x Goxio3 ==1285.7V Thoxio3 i) ĈRait VX Rio 11) =1714.3 V 3000 X 80X10 - FRI, TR 12) = Thox 103 6) To find voltage thyristor first we their equivalent across need to Resistance each find Rt = 6.1lkre 60x6.8 X1OG (60+6.5] xios = 808 6.8x106 (GoT6. 8x183 a 6.27 KNL Riz =

their equi voltage across this equivalent Vi= URTI reiti toltz) = 3000X 6.1/X103 12.38x103 - 1480.6 V Viz= VRT2 (R'Tot RT2) 1519.38 3000x 6.27X103 - 12.38X103 - Power dissipation (1480.6). La 322.37 W - 6.8x103 - $6 (1319.38)2 - 339,99 w 6.8x103