When a 20 A current is initially applied to a Silicon diode @ 300K of a particul

Question

When a 20 A current is initially applied to a Silicon diode @ 300K of a particular characteristic, the voltage across the diode is VD = 0.69 V. With so much current flowing through the diode, the power dissipation raises the operating temperature of the dev ice. The increased temperature eventually causes the diode voltage to stabilize at VD = 0.58V. What is the temperature rise in the diode? How much power is dissipated in the diode at the two operating conditions mentioned? What is the temperature rise per watt of power dissipation?

Explanation / Answer

for a given diode, forward bias voltage decreases with temperature. for silicon, it is 2mV /C so, change in V= 0.69 - 0.58 = 0.11. change in temperature = 55 C degrees. power dissipiated = VI ist case= 0.69 X 20 = 13.8 W 2nd case = 0.58 X 20 = 11.6 W. temperature rise per watt= 55 / [13.8 - 11.6 ] = 25 .

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