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A transistor has a maximum power dissipation of 350 mW at an ambient temperature of 25ºC. If derating factor is 2 mW/ºC, the maximum power dissipation for 40ºC ambient temperature is
- a)300 mW
- b)330 mW
- c)350 mW
- d)380 mW
Correct answer is option 'C'. Can you explain this answer?
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A transistor has a maximum power dissipation of 350 mW at an ambient t...
Derating improves reliability but does not affect maximum power dissipation.
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A transistor has a maximum power dissipation of 350 mW at an ambient t...
°C. The transistor is operating with a collector current of 20 mA and a collector-emitter voltage of 5 V. Determine if the transistor is operating within its safe operating area (SOA) and if not, calculate the power dissipation and the temperature rise.
To determine if the transistor is operating within its safe operating area (SOA), we need to compare the actual power dissipation to the maximum power dissipation. The power dissipation can be calculated using the formula P = V * I, where P is the power dissipation, V is the voltage across the transistor, and I is the current through the transistor.
P = 5 V * 20 mA
P = 0.1 W = 100 mW
Since the calculated power dissipation (100 mW) is lower than the maximum power dissipation (350 mW), the transistor is operating within its safe operating area (SOA).
To calculate the temperature rise, we can use the formula ΔT = Pd / (Rth * A), where ΔT is the temperature rise, Pd is the power dissipation, Rth is the thermal resistance, and A is the surface area of the transistor.
Since the thermal resistance and surface area of the transistor are not provided, we cannot calculate the exact temperature rise without these values. However, we can determine the maximum allowable temperature rise by assuming a worst-case scenario.
Assuming a thermal resistance of 100 °C/W (common for small-signal transistors) and a surface area of 1 mm², we can calculate the maximum allowable temperature rise.
ΔT = 100 mW / (100 °C/W * 1 mm²)
ΔT = 1 °C
Therefore, the maximum allowable temperature rise is 1 °C. If the ambient temperature is 25 °C, the maximum temperature the transistor can reach is 26 °C.
To determine if the transistor is operating within its safe operating area (SOA), we need to compare the actual power dissipation to the maximum power dissipation. The power dissipation can be calculated using the formula P = V * I, where P is the power dissipation, V is the voltage across the transistor, and I is the current through the transistor.
P = 5 V * 20 mA
P = 0.1 W = 100 mW
Since the calculated power dissipation (100 mW) is lower than the maximum power dissipation (350 mW), the transistor is operating within its safe operating area (SOA).
To calculate the temperature rise, we can use the formula ΔT = Pd / (Rth * A), where ΔT is the temperature rise, Pd is the power dissipation, Rth is the thermal resistance, and A is the surface area of the transistor.
Since the thermal resistance and surface area of the transistor are not provided, we cannot calculate the exact temperature rise without these values. However, we can determine the maximum allowable temperature rise by assuming a worst-case scenario.
Assuming a thermal resistance of 100 °C/W (common for small-signal transistors) and a surface area of 1 mm², we can calculate the maximum allowable temperature rise.
ΔT = 100 mW / (100 °C/W * 1 mm²)
ΔT = 1 °C
Therefore, the maximum allowable temperature rise is 1 °C. If the ambient temperature is 25 °C, the maximum temperature the transistor can reach is 26 °C.
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A transistor has a maximum power dissipation of 350 mW at an ambient temperature of 25ºC. If derating factor is 2 mW/ºC, the maximum power dissipation for 40ºC ambient temperature isa)300 mWb)330 mWc)350 mWd)380 mWCorrect answer is option 'C'. Can you explain this answer?
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