Determine the minimum value of load resistance that can be used in the...
The I=(VIN-VZ)/R=(10-6)/500=8mA. (IL)MAX=I-(IZ)MIN=8-3=5mA. (RL)MIN=VZ/(IL)MAX=6/5m=1.2KΩ.
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Determine the minimum value of load resistance that can be used in the...
Given data:
IZ(min) = 3mA
Input voltage = 10V
Resistance R = 500Ω
VZ = 6V
IZ(max) = 90mA
To find: Minimum value of load resistance (RL) that can be used in the circuit
Concepts used:
- Zener diode
- Zener diode as voltage regulator
- Zener breakdown voltage
- Zener current
Explanation:
- In the given circuit, a Zener diode is used as a voltage regulator.
- A Zener diode operates in the breakdown region, where a small increase in voltage causes a large increase in current.
- The breakdown voltage of a Zener diode is denoted by VZ, which is the voltage at which the diode enters the breakdown region and maintains a constant voltage drop across it.
- The Zener current is the maximum current that can flow through the diode without damaging it.
- In the given circuit, the Zener diode is connected in reverse bias, so it allows only a small amount of current to flow through it until the breakdown voltage is reached.
- Once the breakdown voltage is reached, the Zener diode maintains a constant voltage of VZ across it, regardless of the current flowing through it.
- The current flowing through the load resistance RL is given by:
IZ = (Vin - VZ) / RL
- To find the minimum value of RL that can be used, we need to consider the minimum value of IZ, which is given as 3mA.
- Substituting the given values in the above equation, we get:
3mA = (10V - 6V) / RL
RL = (10V - 6V) / 3mA
RL = 1.2KΩ
- Therefore, the minimum value of load resistance that can be used in the circuit is 1.2KΩ.
Conclusion:
- The minimum value of load resistance that can be used in the circuit with IZ(min) = 3mA is 1.2KΩ.
- This value is obtained by using the equation IZ = (Vin - VZ) / RL and substituting the given values.
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