8. Design a 7.5V regulator circuit with 7.5V zener +10 V specified at 10mA. The

Question

8. Design a 7.5V regulator circuit with 7.5V zener +10 V specified at 10mA. The zener diode has an incremental resistance rz30ohms and the knee current of 0.5mA. The regulator operates from a 10V supply and has a 1.5kohm load. !ViR (Problem 4.63 from Sedra and Smith) a. What is the value of R you need to b. What wil the regulator output voltage c. What is the smallest possible load choose? be if the supply is 10% higher and if the load is removed? resistor that can be used while the zener operates at a current of 10mA while the supply is 10% low? Also, what is the load voltage in this case?

Explanation / Answer

a) Zener resistance rz = 30 Ohms , Load resistance RL = 1.5 Kohm

maximum zener current allowed Iz = 10 mA

Current through load IL = Vz / RL = 7.5 V / 1.5 K = 5 mA

Total current I = Iz + IL = 10 + 5 mA = 15 mA

R = 10 - Vz / I = 10 - 7.5 / 15 mA = 2.5 V / 15 mA = 0.16666 K ohm = 166.6 Ohm

R = 166.6 ohm                     .................1

This is the required valur of R

b) If supply voltage is increased by 10% i.e incresed to 11 V and load is removed.

Output voltage will be 7.5 V only as as per Zener diode characterstics.Rest of the 3.5 V will drop across Resistor R
c)

Now supply is 10% low i.e. 9 V

Iz = 10 mA

Now current I through resistor R

I = 9 V - Vz / R = 9 - 7.3 / 166.6 ohm = 1.7 / 0.1666 K = 10.20 mA

Iz = 10 mA & IL = 0.2 mA

Now voltage of zener dropped to 7.3 V

RL = Vz / IL = 7.3 V / 0.2 mA = 36.5 K ohm

RL = 36.5 K Ohm                  ...................2

This is the required minimum value of load resistance .

And load voltage VL =Vz = 7.3 Volts

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