Electronics-I Lecture 09

# Electronics-I Lecture 09 - 2 is off v o at cathode of D 2...

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Lecture 09 Electronics I 1 Half-Wave Rectifier v o = 0   for v s  < V DO v o = R(v – V DO )/(R+r D )      for v s  ≥ V DO Since r D  << R  v o 2245  v – V DO

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Lecture 09 Electronics I 2 Half-Wave Rectifier Transfer characteristics of  rectifier circuit Output voltage Important Design  parameters Current handling capability  of diode  Peak Inverse Voltage PIV PIV = V S Circuit can not be used  when input signal is small,  i.e. 100 mV amplitude
Lecture 09 Electronics I 3 Full-Wave Rectifier Utilizes both halves of  sinusoidal input Centre tapped  transformer D 1  conducts in first  half D 2  conducts in  second half

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Lecture 09 Electronics I 4 Full-Wave Rectifier Neglecting diode  resistance r D Transfer function of Full- Wave rectifier circuit PIV peak inverse voltage D 1  is On and D

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Unformatted text preview: 2 is off v o at cathode of D 2 -v s at anode of D 2 Voltage across D 2 is v o + v s PIV = 2V s - V D Lecture 09 Electronics I 5 Bridge Rectifier • Does not require centre-tapped transformer • Uses four diodes • In positive half D 1 and D 2 conduct • In negative half D 3 and D 4 conduct • v o 2245 V s – 2V D Lecture 09 Electronics I 6 Bridge Rectifier • PIV peak inverse voltage across D 3 can be determined from the loop D 3 , R and D 2 • V D3(reverse) = V o + V D2(forward) V o = V s – 2V DO • PIV = V s – 2V D + V D = V s – V D • PIV is half of centre-tapped based rectifier circuit...
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## This note was uploaded on 12/12/2010 for the course EE 1 taught by Professor Shahidanwar during the Spring '10 term at NUCES.

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Electronics-I Lecture 09 - 2 is off v o at cathode of D 2...

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