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A zener diode has a breakdown voltage of 9.1 V with maximum power dissipation of 364mW. What the maximum current A that the diode can withstand?
Correct answer is '0.04'. Can you explain this answer?
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A zener diode has a breakdown voltage of 9.1 V with maximum power diss...
Maximum current 
The correct answer is: 0.04
The correct answer is: 0.04
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A zener diode has a breakdown voltage of 9.1 V with maximum power diss...
Given Information:
- Breakdown voltage (Vz) = 9.1 V
- Maximum power dissipation (Pmax) = 364 mW
Explanation:
1. Zener Diode:
A zener diode is a type of diode that operates in the reverse breakdown region. It is designed to have a specific breakdown voltage (Vz), at which it starts conducting in the reverse direction.
2. Maximum Power Dissipation:
The maximum power dissipation (Pmax) is the maximum amount of power that a diode can safely handle without getting damaged. It is given by the product of the maximum current (Imax) and the voltage across the diode (Vz).
Pmax = Imax * Vz
Given that Pmax = 364 mW and Vz = 9.1 V, we can rearrange the equation to find Imax.
3. Calculation:
Using the formula Pmax = Imax * Vz, we can rearrange it to find Imax:
Imax = Pmax / Vz
Substituting the given values:
Imax = 364 mW / 9.1 V
Imax = 0.04 A
4. Maximum Current:
Therefore, the maximum current (Imax) that the zener diode can withstand is 0.04 A.
Answer:
The maximum current (Imax) that the zener diode can withstand is 0.04 A.
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A zener diode has a breakdown voltage of 9.1 V with maximum power dissipation of 364mW. What the maximum current A that the diode can withstand?Correct answer is '0.04'. Can you explain this answer?
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