Ch5. Problems Solution. 5-1. : IS = IZ = VS - VZ = RS Ω = 24.2㎃ Ω = 84.8㎃ 20V - 12V 330 5-2. : IS = IZ = VS - VZ = RS 40V - 12V 330 5-3. VS - VZ I S = I Z ( max ) = R = Ω 20V - 12V 297 S ( min ) = 26.9㎃ 5-4. VL = RL VS = R S+R L 1.5㏀ 330 Ω + 1.5㏀ 20V = 16.39V 5-5 VL = RL VS = R S+R L 1.5㏀ 330 Ω + 1.5㏀ 12V = 14.64V 5-6. IS = VS - VZ Ω = 24.24㎃ Ω = 26.94㎃ 20V - 12V = RS 330 5-7. IS= IL= VS - VZ RS VL R = = L ( max ) 20V - 12V 297 12V 1.65㏀ = 7.27㎃ I Z = I S - I L = 26.94㎃ - 7.27㎃= 19.67㎃ 5-8. IS = IL = VS - VZ RS VL RL = = 12V 1.5㏀ Ω 40V - 12V 330 = 84.85㎃ = 8㎃ I Z = I S - I L = 84.5㎃ -8㎃ = 76.85㎃ 5-9. VS - VZ IS= I L = 8㎃ RS Ω 20V - 12V = 330 I Z = 16.24㎃ P = I R = ( 24.2㎃ ) 330 2 2 P L = ( 8㎃ ) 1.5㏀ = 96㎽ 2 Ω = 24.24㎃ = 193.9㎽= P S P D = VI = ( 16.24㎃ )( 12V) = 194.9㎽ 5-10. IS = IL = VS - VZ VL = RL Ω 20V - 10V = RS 330 10V = 30.3㎃ = 6.67㎃ 1.5㏀ I Z = I S - I L = 30.3㎃ -6.67㎃ = 23.63㎃ 5-11. IS = IL = VS - VZ = RS VL = RL 15V Ω 25V - 15V 470 = 21.284㎃ = 15㎃ 1㏀ I Z = I S - I L = 21.28㎃ -15㎃ = 6.28㎃ 5-12. RZ V R ( out ) = V RS R ( in ) = Ω Ω 11.5 330 1V P - P = 34.85 ㎷ P - P 5-13. VS - VZ IS= IL= △ RS VL RL Ω 21V - 12V 330 12V = 27.3㎃ = 8㎃ 1.5㏀ I Z = I S - I L = 27.3㎃ -8㎃ = 19.3㎃ Ω Ω V L = I ZR Z = ( 19.3㎃ )( 1 1.5 VS - VZ IS= IL= △ = = RS VL RL = = ) = 0.222V 17.5V - 12V 12V 1.5㏀ 330 = 16.7㎃ = 8㎃ I Z = I S - I L = 16.7㎃ -8㎃ = 8.7㎃ V L = I ZR Z = ( 8.7㎃ )( 11 .5 Ω ) = 0.1V 5-14. R S ( max ) = S ( max ) = [ V S ( min ) VZ ] -1 R L ( min ) = V - 1 ]500Ω = 250Ω [ 18 12 V 5-15. R V S ( min ) - VZ I L ( max ) = 16V - 12V 25㎃ = 160 Ω 5-16. R S ( max ) = R L ( min ) = [ V ] S ( min ) -1 R VZ R V S ( max ) S ( min ) L ( min ) = -1 VZ 330 20V Ω = 495 -1 12V Ω 5-17. P Z = V ZI Z = ( 10 V)( 20㎃ ) = 0.2W 5-18. P Z = V ZI Z = ( 20 V)( 5㎃ ) = 0.1W 5-19. V Z = 12V±5% ( 12V)( 0.0 5 ) = 0.6V 12V + 0.6V = 12.6V 12V - 0.6V= 11.4V 5-20. I= PZ VZ = 400㎽ 10V I ZM1 + I ZM2 I= 2 = 40㎃ 35㎃ + 45㎃ = 2 = 40㎃ 5-21. 부하전압은 a. 0V (제너다이오드 단락시) b. 16.39V c. 0V ( d. 0V 5-22. Open Zener Diode 5-23. PIV = V P = 18.85V 5-24. a. 불은 켜진다. b. 전류가 흐르지 않아 불이 꺼진다 c. 리플이 커지고 불은 계속 켜진다. d. LED가 불이 꺼진다. 5-25. IS= VS - VD RS = 14V - 2V 2.2㏀ = 5.91㎃ 5-26. IS = VS - VD RS = 40V - 2V = 17.27㎃ 15V - 2V = 13㎃ 2.2㏀ 5-27. IS = VS - VD RS = 1㏀ 5-28. 1㏀ 5-29. VS - VZ IS= VL IL= △ = RS RL 330 12V = Ω 20V - 12V = 24.24㎃ = 8㎃ 1.5㏀ I Z = I S - I L = 24.24㎃ -8㎃ = 16.24㎃ V L = I ZR Z = ( 16.24㎃ )( 11.5 V L = 12.19V Ω ) = 0.19V 5-30. I I △ △ △ △ I VL L ( max ) = L ( min ) = Z ( min ) = IS- I R VL R = L ( max ) L ( min ) 12V 10㏀ = 12 ㎃ = 1.2㎃ = 24.24 ㎃ - 12 ㎃ = 12.24㎃ Ω Ω R Z = ( 12.24㎃ )( 11.5 Z ( min ) = I L ( min ) V = I L ( max ) V = 12.141V L ( min ) L ( max ) 1㏀ L ( min ) V V 12 V = ) = 0.141 V R Z = ( 23.04㎃ )( 11.5 Z ( max ) ) = 0.265V = 0.265 V 5-31. R S ( max ) R Smin ) = = V S ( min ) I - VZ = 20V - 6.8V VS - VZ = I ZM = 440 30 ㎃ L ( max ) 20 V - 6.8V 55㎃ = 240 Ω Ω 5-32. VS - V RS= I LED ( min ) = 5V - 1.5V 20㎃ LED = 175 Ω 5-33. V sec ( max ) = V sec + V sec ( 10%) = 12.6V a c + 12.6 V a c ( 10% ) = 13.86V a c V P = 1.414 V a c = 1.414 ( 13.86 V a c ) = 19.6 V V Z = 6.2 V±10% = 6.2 V - 6.2 V ( 10% ) = 5.58 V R 2 = 560 IS= Ω± 5% = 560 VS - VD R S+R Z = Ω - 560 Ω ( 5% ) = 532 Ω Ω 19.6V - 5.58 V 532 +7 Ω = 26 ㎃ 5-34. I= I 1 N5314 f out = 2f +I LED + I Z = 4.7㎃ + 15.6 ㎃ + 21.7㎃ = 42㎃ = 2 ( 60㎐ ) = 120 ㎐ in C = 1000 ㎌±20% = 1000㎌ - 1000㎌ ( 20% ) = 800㎌ I 42 ㎃ VR = = = 0.438V fC ( 120 ㎐ ) ( 800 ㎌ ) 5-35. V P ( in ) V = 1.414 V P ( o ut) = V r ms P ( in ) = 1.414 ( 6V a c ) = 8.48V - 0.5V = 7.98V 5-36. ① open RS ② B와 D or open E ③ Zener open ④ RS shorted 5-37. ⑤ open at A ⑥ open RL , B와 C사이 open, RL과 Groumd ⑦ open at E ⑧ Zener shorted, B와 C 사이 or D 와 ground 가 shorted