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A 700 mW maximum power dissipation diode at 25°C has 5 mW/°C de-rating factor. If the forward voltage drop remains constant at 0.7 V, the maximum forward current at 65°C is
1. 700 mA
2. 714 mA
3. 1 A
4. 1 mA

1 Answer

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Correct Answer - Option 2 : 714 mA

Concept:

  • The derating factor of a transistor is the amount by which the power dissipating rating of a transistor falls when the transistor junction temperature increases.
  • The power dissipation rating of a transistor is usually given at 25°C.
  • At higher temperatures, the power dissipation rating is less. For most silicon transistors, the maximum allowable junction temperature is between 150° and 200°C.
  • Temperatures higher than this will destroy the transistor. This is why a manufacturer must specify a maximum power rating for the transistor.

ΔTJ = θ.PD     ----(1)

Where,

ΔTJ = Change in Temperature

θ = Derating factor and

PD = Power Dissipated.

Calculation:

Given:

θ = -5 mW/0C

ΔTJ = 65 - 25 = 400C

Power available after derating at 650C

From equation (1):

P = 700 - 5 × 40

P = 500 mW

As,

P = V.I

500 = 0.7 × I

I = 714 mA.

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