Problems 12-13 [4 pts. ea pg = 12 v The unregulated voltage source in the above

Question

Problems 12-13 [4 pts. ea pg = 12 v The unregulated voltage source in the above circuit ranges from 18 V to 24 V. 12. What should be the Zener diode's minimum power rating for surviving the "inadvertent" situations of the load either being short-circuited or open-circuited? (Your answer should be the smallest power rating that will satisfy the specification-for this calculation, do not add any additional safety margin.) Pz ratings: A 0.125 W B 0.25 W C 0.50 W D 1.0 VW E 2.0 W What should be the limiting resistor's minimum power rating to survive the "inadvertent" situations of the load either being short-circuited or open-circuited? (Your answer should be the smallest power rating that will satisfy the specification -for this calculation, do not add any additional safety margin.) 13. PR values A 0.125 W B 0.25 W C 0.50 W D 1.0 W E 2.0 VW

Explanation / Answer

12. For the minimal loss of zener diode the input voltage be 18V (minimal) and let load be open circuit.

Thus voltage drop across 1kohm = 18V -12V =6V

Thus current through it = 6V/1kOhms = 6mA

So all current passes through zenerdiode. Thus power dissipation of zener = 6mA * 12V = 72mW (ANs)

13. For this let input voltage be 24V.

Thus voltage drop across 1kohm = 24V -12V =12V

Thus current through it = 12V/1kOhms = 12mA

So 6mA current passes through zenerdiode and rest from Rl

Thus Rl = 12V / 6mA = 2 kiloOhms

Thus Rl power rating = 6mA *12V = 72mW(ANS)

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