Figure 2-11 39.(a) A sine wave with a positive peak at 0.7 V, a negative peak at ±7.3 V, and a dc value of ±3.3 V. (b) A sine wave with a positive peak at 29.3 V, a negative peak at ±0.7 V, and a dc value of +14.3 V. (c) A square wave varying from +0.7 V to ±15.3 V with a dc value of ±7.3 V. (d) A square wave varying from +1.3 V to ±0.7 V with a dc value of +0.3 V. 40.(a) A sine wave varying from ±0.7 V to +7.3 V with a dc value of +3.3 V. (b) A sine wave varying from ±29.3 V to +7.3 V with a dc value of +14.3 V. (c) A square wave varying from ±0.7 V to +15.3 V with a dc value of +7.3 V. (d) A square wave varying from ±1.3 V to +0.7 V with a dc value of ±0.3 V. Section 2-8 Voltage Multipliers 41. VOUT= 2Vp(in)= 2(1.414)(20 V) = 56.6 VSee Figure 2-12. Figure 2-12
Chapter 2 10 42.VOUT(trip)= 3Vp(in)= 3(1.414)(20 V) = 84.8 VVOUT(quad)= 4Vp(in)= 4(1.414)(20 V) = 113 V See Figure 2-13. Figure 2-13 Section 2-9 The Diode Datasheet 43.The PIV is specified as the peak repetitive reverse voltage = 100 V. 44. The PIV is specified as the peak repetitive reverse voltage = 1000 V. 45.IF(AVG)= 1.0 A RL(min)= 50 V1.0 A= 50 :Section 2-10 Troubleshooting 46.(a) Since VD= 25 V = 0.5VS, the diode is open. (b) The diode is forward-biased but since VD= 15 V = VS, the diode is open. (c) The diode is reverse-biased but since VR= 2.5 V = 0.5VS, the diode is shorted. (d) The diode is reverse-biased and VR= 0 V. The diode is operating properly. 47.VA= VS1= +25 V VB= VS1±0.7 V = 25 V ±0.7 V = +24.3 VVC= VS2+ 0.7 V = 8 V + 0.7 V = +8.7 V VD= VS2= +8.0 V 48.If a bridge rectifier diode opens, the output becomes a half-wave voltage resulting in an increased ripple at 60 Hz.
Chapter 2 11 49.Vavg= #SSV)(1.414)115(22pV104 V The output of the bridge is correct. However, the 0 V output from the filter indicates that the surge resistoris openor that the capacitor is shorted.50.(a) Correct (b) Incorrect. Open diode. (c) Correct (d) Incorrect. Open diode. 51.Vsec= 120 V5= 24 V rms Vp(sec)= 1.414(24 V) = 33.9 V The peak voltage for each half of the secondary is ()33.9 V22p secV= 17 V The peak inverse voltage for each diode is PIV = 2(17 V) + 0.7 V = 34.7 V The peak current through each diode is ()0.7 V17.0 V0.7 V2330p secpLVIR±±:= 49.4 mA The diode ratings exceed the actual PIV and peak current. The circuit should not fail. Application Activity Problems 52.(a) Not plugged into ac outlet or no ac available at outlet. Check plug and/or breaker. (b) Open transformer winding or open fuse. Check transformer and/or fuse. (c) Incorrect transformer installed. Replace. (d) Leaky filter capacitor. Replace. (e) Rectifier faulty. Replace. (f) Rectifier faulty. Replace. 53.The rectifier must be connected backwards. 54.±16 V with 60 Hz ripple Advanced Problems 55.Vr= )(1inpLVCfR¸¸¹·¨¨©§C= V35V)5.0)(kHz)(3.3(12011)(¸¸¹·¨¨©§:¸¸¹·¨¨©§inprLVVfR= 177 PF
Chapter 2 12 56. VDC= )(211inpLVCfR¸¸¹·¨¨©§±¸¸¹·¨¨©§±CfRVVLinp211)(DC)(DC121inpLVVCfR±¸¸¹·¨¨©§±)(DC121inpLVVfR= CC= )067.0)(kHz)(1.0240(1)933.01)(kHz)(1.0240(1:±:= 62.2 PF Then Vr= ()1115 V(120 Hz)(1.0 k)(62.2F)p inLVfR CP§·§·¨¸¨¸:©¹©¹= 2 V57.The capacitor input voltage is Vp(in) = (1.414)(24 V) ±1.4 V = 32.5 V R
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