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National Highways Authority of India (NHAI) has inked initial pacts with how many technical institutes to encourage them for voluntary adoption of highway stretches for research etc?
The correct answer is 200.
An SCR has ambient temperature of 25°C and Junction temperature of 50°C and Average power dissipated of 25 watt. If the junction temperature is reduced to 35°C, the percentage reduction in thermal resistance _______
Concept
Here,
Pav = Average rate of heat generated at a thyristor Junction
QJC = thermal resistance between the junction temperature TJ and case temperature TC
QCS = thermal resistance between the case temperature TC and sink temperature
QSA = thermal resistance between the sink temperature TS and ambient temperature TA
Where QJA = QJC + QCS + QSA
Calculations:
TA = 25°C
IJ = 50°C
Pav = 25 watt
∴ % change in thermal resistance
The thermal resistance for the thyristor – sin k combination are QJC = 0.18 and ACS = 0.09 °C/W for a heat sin k temperature of 75°C and maximum junction temperature is 130°C . In case the heat sin k temperature is brought down 70°C by forced coaling find the percentage increase in the device rating.
Thus total average power loss in the thyristor – sink combination is 203.70 W. with improved coaling
Thyristor rating is proportional to the square root of average power loss
∴ Percentage Increase in thyristor rating
For a thysistor maximum junction temperature is 200°C. The thermal resistance for the thyristor sink combination are θJC = 0.15 and θCS = 0.05∘c/w. Initially the heat sink temperature of 80° C and it is brought down to 70° C by force cooling, the percentage increase in the device rating is – (in%)
From the above equivalent circuit
TJ = Ts + Pav (θJC + θCS)
Thyristor rating is proportional to the square root of average power loss
% increase in thyristor rating
x 100 = 4.08%
The usual way to accomplish higher gate current for improved di/dt rating is by using
Pilot SCR is an SCR which is fired which activates the firing circuit and fires the main SCR.
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