diode circuits: part 2sequel/ee101/mc_diodes_2.pdfclamper circuits 0v vi clamper vo 0v 0v vi clamper...

$Page 1: Diode Circuits: Part 2sequel/ee101/mc_diodes_2.pdfClamper circuits 0V Vi Clamper Vo 0V 0V Vi Clamper Vo 0V * A clamper circuit provides a \level shift." (The shape of the input signal$

Diode Circuits: Part 2

M. B. [email protected]

www.ee.iitb.ac.in/~sequel

Department of Electrical EngineeringIndian Institute of Technology Bombay

M. B. Patil, IIT Bombay

Peak detector (with Von =0V)

1050 15 20 25

time (msec)

4

2

0

−2

Vc(0) = 0 V

Von= 0 V

Ron→ 0Ω

Roff →∞Ω

C

DiD

Vc

VD

iD

VD

Vs

Vs < Vc

τ =RoffC

Vs > Vc

τ =RonC

Ron

Roff

C

Vs Vc

VcVs

SEQUEL file: ee101 peak detector 1.sqproj

Vo

1050 15 20 25

time (msec)

4

2

0

−2

Vc(0) = 0 V

Von= 0 V

Ron→ 0Ω

Roff →∞Ω

C

DiD

Vc

VD

iD

VD

Vs

Vs < Vc

τ =RoffC

Vs > Vc

τ =RonC

Ron

Roff

C

Vs Vc

VcVs

Vo

1050 15 20 25

time (msec)

4

2

0

−2

Vc(0) = 0 V

Von= 0 V

Ron→ 0Ω

Roff →∞Ω

C

DiD

Vc

VD

iD

VD

Vs

Vs < Vc

τ =RoffC

Vs > Vc

τ =RonC

Ron

Roff

C

Vs Vc

VcVs

Vo

1050 15 20 25

time (msec)

4

2

0

−2

Vc(0) = 0 V

Von= 0 V

Ron→ 0Ω

Roff →∞Ω

C

DiD

Vc

VD

iD

VD

Vs

Vs < Vc

τ =RoffC

Vs > Vc

τ =RonC

Ron

Roff

C

Vs Vc

VcVs

Vo

1050 15 20 25

time (msec)

4

2

0

−2

Vc(0) = 0 V

Von= 0 V

Ron→ 0Ω

Roff →∞Ω

C

DiD

Vc

VD

iD

VD

Vs

Vs < Vc

τ =RoffC

Vs > Vc

τ =RonC

Ron

Roff

C

Vs Vc

VcVs

Vo

1050 15 20 25

time (msec)

4

2

0

−2

Vc(0) = 0 V

Von= 0 V

Ron→ 0Ω

Roff →∞Ω

C

DiD

Vc

VD

iD

VD

Vs

Vs < Vc

τ =RoffC

Vs > Vc

τ =RonC

Ron

Roff

C

Vs Vc

VcVs

Vo

1050 15 20 25

time (msec)

4

2

0

−2

Vc(0) = 0 V

Von= 0 V

Ron→ 0Ω

Roff →∞Ω

C

DiD

Vc

VD

iD

VD

Vs

Vs < Vc

τ =RoffC

Vs > Vc

τ =RonC

Ron

Roff

C

Vs Vc

VcVs

Vo

Peak detector (with Von =0.7V)

1050 15 20 25

time (msec)

4

2

0

−2

Vc(0) = 0 V

Von= 0.7 V

Ron→ 0Ω

Roff →∞Ω

C

D

Vc

iD

VD

VDVon

Vs

iD

Vs

Vs < Vc + Von

τ =RoffC

Vs > Vc + Von

τ =RonC

RonVon

Roff

C

Vs Vc

Vo

1050 15 20 25

time (msec)

4

2

0

−2

Vc(0) = 0 V

Von= 0.7 V

Ron→ 0Ω

Roff →∞Ω

C

D

Vc

iD

VD

VDVon

Vs

iD

Vs

Vs < Vc + Von

τ =RoffC

Vs > Vc + Von

τ =RonC

RonVon

Roff

C

Vs Vc

Vo

1050 15 20 25

time (msec)

4

2

0

−2

Vc(0) = 0 V

Von= 0.7 V

Ron→ 0Ω

Roff →∞Ω

C

D

Vc

iD

VD

VDVon

Vs

iD

Vs

Vs < Vc + Von

τ =RoffC

Vs > Vc + Von

τ =RonC

RonVon

Roff

C

Vs Vc

Vo

1050 15 20 25

time (msec)

4

2

0

−2

Vc(0) = 0 V

Von= 0.7 V

Ron→ 0Ω

Roff →∞Ω

C

D

Vc

iD

VD

VDVon

Vs

iD

Vs

Vs < Vc + Von

τ =RoffC

Vs > Vc + Von

τ =RonC

RonVon

Roff

C

Vs Vc

Vo

1050 15 20 25

time (msec)

4

2

0

−2

Vc(0) = 0 V

Von= 0.7 V

Ron→ 0Ω

Roff →∞Ω

C

D

Vc

iD

VD

VDVon

Vs

iD

Vs

Vs < Vc + Von

τ =RoffC

Vs > Vc + Von

τ =RonC

RonVon

Roff

C

Vs Vc

Vo

1050 15 20 25

time (msec)

4

2

0

−2

Vc(0) = 0 V

Von= 0.7 V

Ron→ 0Ω

Roff →∞Ω

C

D

Vc

iD

VD

VDVon

Vs

iD

Vs

Vs < Vc + Von

τ =RoffC

Vs > Vc + Von

τ =RonC

RonVon

Roff

C

Vs Vc

Vo

1050 15 20 25

time (msec)

4

2

0

−2

Vc(0) = 0 V

Von= 0.7 V

Ron→ 0Ω

Roff →∞Ω

C

D

Vc

iD

VD

VDVon

Vs

iD

Vs

Vs < Vc + Von

τ =RoffC

Vs > Vc + Von

τ =RonC

RonVon

Roff

C

Vs Vc

Vo

Clamper circuits

Clamper VoVi0V 0V

* A clamper circuit provides a “level shift.” (The shape of the input signal is not altered.)

* The shift could be positive or negative.

Clamper circuits

Clamper VoVi0V 0V

Clamper circuits

Clamper VoVi0V 0V Clamper VoVi0V 0V

Clamper circuits

−5

0

5

10

time (msec)

1 2 3 4 0

iD

Vs(t)=Vm sinωt

Vc(0) = 0 V

Von = 0 V

Ron→ 0Ω

Roff →∞Ω

C

Vc

iD

VD

VD VoVs

Vs

Vo

Vc

* When D conducts, the capacitor charges instantaneously since Ron is small. In this phase,VD = 0→ Vc + Vs = 0→ Vc =−Vs .

* Vc can only increase since a decrease in Vc would require the diode to conduct in the reverse direction.

* After Vc reaches its maximum value (Vm), it cannot change any more. We then haveVo(t) =Vs(t) + Vc (t) =Vs(t) + Vm, i.e., a positive level shift.

* Note that we are generally interested only in the steady-state behaviour and not in the transient at thebeginning.

Clamper circuits

−5

0

5

10

time (msec)

1 2 3 4 0

iD

Vs(t)=Vm sinωt

Vc(0) = 0 V

Von = 0 V

Ron→ 0Ω

Roff →∞Ω

C

Vc

iD

VD

VD VoVs

Vs

Vo

Vc

Clamper circuits

−5

0

5

10

time (msec)

1 2 3 4 0

iD

Vs(t)=Vm sinωt

Vc(0) = 0 V

Von = 0 V

Ron→ 0Ω

Roff →∞Ω

C

Vc

iD

VD

VD VoVs

Vs

Vo

Vc

Clamper circuits

−5

0

5

10

time (msec)

1 2 3 4 0

iD

Vs(t)=Vm sinωt

Vc(0) = 0 V

Von = 0 V

Ron→ 0Ω

Roff →∞Ω

C

Vc

iD

VD

VD VoVs

Vs

Vo

Vc

Clamper circuits

−5

0

5

10

time (msec)

1 2 3 4 0

iD

Vs(t)=Vm sinωt

Vc(0) = 0 V

Von = 0 V

Ron→ 0Ω

Roff →∞Ω

C

Vc

iD

VD

VD VoVs

Vs

Vo

Vc

Clamper circuits

−5

0

5

10

time (msec)

1 2 3 4 0

iD

Vs(t)=Vm sinωt

Vc(0) = 0 V

Von = 0 V

Ron→ 0Ω

Roff →∞Ω

C

Vc

iD

VD

VD VoVs

Vs

Vo

Vc

Clamper circuits

−5

0

5

10

time (msec)

1 2 3 4 0

iD

Vs(t)=Vm sinωt

Vc(0) = 0 V

Von = 0 V

Ron→ 0Ω

Roff →∞Ω

C

Vc

iD

VD

VD VoVs

Vs

Vo

Vc

Clamper circuits

−5

0

5

10

time (msec)

1 2 3 4 0

iD

Vs(t)=Vm sinωt

Vc(0) = 0 V

Von = 0 V

Ron→ 0Ω

Roff →∞Ω

C

Vc

iD

VD

VD VoVs

Vs

Vo

Vc

Clamper circuits

−5

0

5

10

time (msec)

1 2 3 4 0

iD

Vs(t)=Vm sinωt

Vc(0) = 0 V

Von = 0 V

Ron→ 0Ω

Roff →∞Ω

C

Vc

iD

VD

VD VoVs

Vs

Vo

Vc

Clamper circuits

time (msec)

1 2 3 4 0

−5

0

5

10

iD

Vs(t)=Vm sinωt

Vc(0) = 0 V

Von= 0.7 V

Ron→ 0Ω

Roff →∞Ω

C

Vc

VD

iD

Von

VoVs

Vo

Vs

Vc

* When D conducts, the capacitor charges instantaneously since Ron is small (as in the last circuit).In this phase,Vc + Vs + Von = 0→ Vc =−Vs − Von.

* After Vc reaches its maximum value (Vm − Von), it cannot change any more. We then haveVo(t) =Vs(t) + Vc (t) =Vs(t) + Vm − Von. In this case, Vo gets clamped at −0.7 V.

Clamper circuits

time (msec)

1 2 3 4 0

−5

0

5

10

iD

Vs(t)=Vm sinωt

Vc(0) = 0 V

Von= 0.7 V

Ron→ 0Ω

Roff →∞Ω

C

Vc

VD

iD

Von

VoVs

Vo

Vs

Vc

Clamper circuits

time (msec)

1 2 3 4 0

−5

0

5

10

iD

Vs(t)=Vm sinωt

Vc(0) = 0 V

Von= 0.7 V

Ron→ 0Ω

Roff →∞Ω

C

Vc

VD

iD

Von

VoVs

Vo

Vs

Vc

Clamper circuits

time (msec)

1 2 3 4 0

−5

0

5

10

iD

Vs(t)=Vm sinωt

Vc(0) = 0 V

Von= 0.7 V

Ron→ 0Ω

Roff →∞Ω

C

Vc

VD

iD

Von

VoVs

Vo

Vs

Vc

Clamper circuits

5

0

−5

−10 0 1 2 3 4

time (msec)

Vs(t)=Vm sinωt

Vc(0) = 0 V

Von = 0 V

Ron→ 0Ω

Roff →∞Ω

CVo

iD iD

VD

Vc

Vs VD

Vs

Vo

Vc

* When D conducts, the capacitor charges instantaneously since Ron is small. In this phase,VD = 0→ Vc − Vs = 0→ Vc =Vs .

* After Vc reaches its maximum value (Vm), it cannot change any more. We then haveVo(t) =Vs(t)− Vc (t) =Vs(t)− Vm, i.e., a negative level shift.

Clamper circuits

5

0

−5

−10 0 1 2 3 4

time (msec)

Vs(t)=Vm sinωt

Vc(0) = 0 V

Von = 0 V

Ron→ 0Ω

Roff →∞Ω

CVo

iD iD

VD

Vc

Vs VD

Vs

Vo

Vc

Clamper circuits

5

0

−5

−10 0 1 2 3 4

time (msec)

Vs(t)=Vm sinωt

Vc(0) = 0 V

Von = 0 V

Ron→ 0Ω

Roff →∞Ω

CVo

iD iD

VD

Vc

Vs VD

Vs

Vo

Vc

Clamper circuits

5

0

−5

−10 0 1 2 3 4

time (msec)

Vs(t)=Vm sinωt

Vc(0) = 0 V

Von = 0 V

Ron→ 0Ω

Roff →∞Ω

CVo

iD iD

VD

Vc

Vs VD

Vs

Vo

Vc

Clamper circuits

5

0

−5

−10 0 1 2 3 4

time (msec)

Vs(t)=Vm sinωt

Vc(0) = 0 V

Von = 0 V

Ron→ 0Ω

Roff →∞Ω

CVo

iD iD

VD

Vc

Vs VD

Vs

Vo

Vc

Clamper circuits

5

0

−5

−10 0 1 2 3 4

time (msec)

Vs(t)=Vm sinωt

Vc(0) = 0 V

Von = 0 V

Ron→ 0Ω

Roff →∞Ω

CVo

iD iD

VD

Vc

Vs VD

Vs

Vo

Vc

Clamper circuits

5

0

−5

−10 0 1 2 3 4

time (msec)

Vs(t)=Vm sinωt

Vc(0) = 0 V

Von = 0 V

Ron→ 0Ω

Roff →∞Ω

CVo

iD iD

VD

Vc

Vs VD

Vs

Vo

Vc

Clamper circuits

0 1 2 3 4

time (msec)

5

0

−5

−10

Vs(t)=Vm sinωt

Vc(0) = 0 V

Von= 0.7 V

Ron→ 0Ω

Roff →∞Ω

CVo

iD

Vc

VD

iD

Von

Vs VD

Vo

Vs

Vc

* When D conducts, the capacitor charges instantaneously since Ron is small (as in the last circuit).In this phase,Vc + Von − Vs = 0→ Vc =Vs − Von.

* After Vc reaches its maximum value (Vm − Von), it cannot change any more. We then haveVo(t) =Vs(t)− Vc (t) =Vs(t)− Vm + Von. In this case, Vo gets clamped at 0.7 V.

Clamper circuits

0 1 2 3 4

time (msec)

5

0

−5

−10

Vs(t)=Vm sinωt

Vc(0) = 0 V

Von= 0.7 V

Ron→ 0Ω

Roff →∞Ω

CVo

iD

Vc

VD

iD

Von

Vs VD

Vo

Vs

Vc

Clamper circuits

0 1 2 3 4

time (msec)

5

0

−5

−10

Vs(t)=Vm sinωt

Vc(0) = 0 V

Von= 0.7 V

Ron→ 0Ω

Roff →∞Ω

CVo

iD

Vc

VD

iD

Von

Vs VD

Vo

Vs

Vc

Clamper circuits

0 1 2 3 4

time (msec)

5

0

−5

−10

Vs(t)=Vm sinωt

Vc(0) = 0 V

Von= 0.7 V

Ron→ 0Ω

Roff →∞Ω

CVo

iD

Vc

VD

iD

Von

Vs VD

Vo

Vs

Vc

Voltage doubler (peak-to-peak detector)

Positive

clamper

Peak

Detector VoViVo

Vm

−Vm

2Vm

0V

2Vm

0VVi

V1

0V

* Input voltage: −Vm to Vm

* Output of positive clamper (V1): 0 to 2Vm

* The peak detector detects the peak of V1(t), i.e., 2Vm (dc).

Positive

clamper

Peak

Detector VoViVo

Vm

−Vm

2Vm

0V

2Vm

0VVi

V1

0V

Positive

clamper

Peak

Detector VoViVo

Vm

−Vm

2Vm

0V

2Vm

0VVi

V1

0V

Positive

clamper

Peak

Detector VoViVo

Vm

−Vm

2Vm

0V

2Vm

0VVi

V1

0V

Positive

clamper

Peak

Detector

Positive

clamper

Peak

Detector

VoV1Vi

V1VoVi

C1

D1 C2

D2

10

0

-10

20Von= 0V

0 1 2 3 4 5 6 7 8 9 10

10

0

-10

20Von= 0.7 V

time (msec)

SEQUEL file: ee101 voltage doubler.sqproj

Positive

clamper

Peak

Detector

Positive

clamper

Peak

Detector

VoV1Vi

V1VoVi

C1

D1 C2

D2

10

0

-10

20Von= 0V

0 1 2 3 4 5 6 7 8 9 10

10

0

-10

20Von= 0.7 V

time (msec)

Positive

clamper

Peak

Detector

Positive

clamper

Peak

Detector

VoV1Vi

V1VoVi

C1

D1 C2

D2

10

0

-10

20Von= 0V

0 1 2 3 4 5 6 7 8 9 10

10

0

-10

20Von= 0.7 V

time (msec)

Positive

clamper

Peak

Detector

Positive

clamper

Peak

Detector

VoV1Vi

V1VoVi

C1

D1 C2

D2

10

0

-10

20Von= 0V

0 1 2 3 4 5 6 7 8 9 10

10

0

-10

20Von= 0.7 V

time (msec)

Diode circuit example

R1

D1

R2

D2

Vo

C ID

VCiC

0.7 V VD

Vi

Vm

−Vm

T1 T2

Assuming R1C and R2C to be large compared to T , find Vo(t) insteady state.

* Charging time constant τ1 =R1C .

* Discharging time constant τ2 =R2C .

* Since τ1 T and τ2 T , we expect VC to be nearlyconstant in steady state,

i.e., VC (t) ≈ constant ≡ V 0C .

* Vo(t) = Vi (t)− VC (t) ≈ Vi (t)− V 0C .

Let us look at an example.

R1

D1

R2

D2

Vo

C ID

VCiC

0.7 V VD

Vi

Vm

−Vm

T1 T2

* Vo(t) = Vi (t)− VC (t) ≈ Vi (t)− V 0C .

R1

D1

R2

D2

Vo

C ID

VCiC

0.7 V VD

Vi

Vm

−Vm

T1 T2

* Vo(t) = Vi (t)− VC (t) ≈ Vi (t)− V 0C .

R1

D1

R2

D2

Vo

C ID

VCiC

0.7 V VD

Vi

Vm

−Vm

T1 T2

* Vo(t) = Vi (t)− VC (t) ≈ Vi (t)− V 0C .

R1

D1

R2

D2

Vo

C ID

VCiC

0.7 V VD

Vi

Vm

−Vm

T1 T2

* Vo(t) = Vi (t)− VC (t) ≈ Vi (t)− V 0C .

R1

D1

R2

D2

Vo

C ID

VCiC

0.7 V VD

Vi

Vm

−Vm

T1 T2

* Vo(t) = Vi (t)− VC (t) ≈ Vi (t)− V 0C .

15

10

5

0

−5

−10

R1

D1

R2

D2

Vo

C ID

0.7 V VD

VCiCVi

Vm

−Vm

T1 T2

R1= 5 kR2= 10 kVon = 0.7 VT1= 0.25msT2= 0.75msC= 10µF

Vo

Vi

VC

−1.82

−1.90

VC

0 1 2

Time (msec)

−1

0

1

2

3

ee101 diode circuit 13.sqproj

SEQUEL file:

iC (mA)

15

10

5

0

−5

−10

R1

D1

R2

D2

Vo

C ID

0.7 V VD

VCiCVi

Vm

−Vm

T1 T2

Vo

Vi

VC

−1.82

−1.90

VC

0 1 2

Time (msec)

−1

0

1

2

3

SEQUEL file:

iC (mA)

15

10

5

0

−5

−10

R1

D1

R2

D2

Vo

C ID

0.7 V VD

VCiCVi

Vm

−Vm

T1 T2

Vo

Vi

VC

−1.82

−1.90

VC

0 1 2

Time (msec)

−1

0

1

2

3

SEQUEL file:

iC (mA)

0 1 2

Time (msec)

−1

0

1

2

3

−1.82

−1.90

15

10

5

0

−5

−10

R1

D1

R2

D2

Vo

C ID

VCiC

0.7 V VD

Vi

R1= 5 kR2= 10 kVon= 0.7 VT1= 0.25msT2= 0.75msC= 10µF

iC (mA)

Vo

Vi

VC

Charge conservation:

∆Q =

∫ T

0iC dt =

∫ T1

0iC dt +

∫ T1+T2

T1

iC dt = 0.

T1

(Vm − VC − Von

R1

)− T2

(0− (−Vm − VC )− Von

R2

)= 0.(

T1

R1−

T2

R2

)(Vm − Von) = VC

(T1

R1+

T2

R2

).

→ VC =

(T1

R1−

T2

R2

)(T1

R1+

T2

R2

) (Vm − Von) = −1.86 V.

0 1 2

Time (msec)

−1

0

1

2

3

−1.82

−1.90

15

10

5

0

−5

−10

R1

D1

R2

D2

Vo

C ID

VCiC

0.7 V VD

Vi

iC (mA)

Vo

Vi

VC

∆Q =

∫ T

0iC dt =

∫ T1

0iC dt +

∫ T1+T2

T1

iC dt = 0.

T1

(Vm − VC − Von

R1

)− T2

(0− (−Vm − VC )− Von

R2

)= 0.(

T1

R1−

T2

R2

)(Vm − Von) = VC

(T1

R1+

T2

R2

).

→ VC =

(T1

R1−

T2

R2

)(T1

R1+

T2

R2

) (Vm − Von) = −1.86 V.

0 1 2

Time (msec)

−1

0

1

2

3

−1.82

−1.90

15

10

5

0

−5

−10

R1

D1

R2

D2

Vo

C ID

VCiC

0.7 V VD

Vi

iC (mA)

Vo

Vi

VC

∆Q =

∫ T

0iC dt =

∫ T1

0iC dt +

∫ T1+T2

T1

iC dt = 0.

T1

(Vm − VC − Von

R1

)− T2

(0− (−Vm − VC )− Von

R2

)= 0.

(T1

R1−

T2

R2

)(Vm − Von) = VC

(T1

R1+

T2

R2

).

→ VC =

(T1

R1−

T2

R2

)(T1

R1+

T2

R2

) (Vm − Von) = −1.86 V.

0 1 2

Time (msec)

−1

0

1

2

3

−1.82

−1.90

15

10

5

0

−5

−10

R1

D1

R2

D2

Vo

C ID

VCiC

0.7 V VD

Vi

iC (mA)

Vo

Vi

VC

∆Q =

∫ T

0iC dt =

∫ T1

0iC dt +

∫ T1+T2

T1

iC dt = 0.

T1

(Vm − VC − Von

R1

)− T2

(0− (−Vm − VC )− Von

R2

)= 0.(

T1

R1−

T2

R2

)(Vm − Von) = VC

(T1

R1+

T2

R2

).

→ VC =

(T1

R1−

T2

R2

)(T1

R1+

T2

R2

) (Vm − Von) = −1.86 V.

0 1 2

Time (msec)

−1

0

1

2

3

−1.82

−1.90

15

10

5

0

−5

−10

R1

D1

R2

D2

Vo

C ID

VCiC

0.7 V VD

Vi

iC (mA)

Vo

Vi

VC

∆Q =

∫ T

0iC dt =

∫ T1

0iC dt +

∫ T1+T2

T1

iC dt = 0.

T1

(Vm − VC − Von

R1

)− T2

(0− (−Vm − VC )− Von

R2

)= 0.(

T1

R1−

T2

R2

)(Vm − Von) = VC

(T1

R1+

T2

R2

).

→ VC =

(T1

R1−

T2

R2

)(T1

R1+

T2

R2

) (Vm − Von) = −1.86 V.

Rectifiers

ACmains

step-downtransformer

Vo

Rectifierwithoutfilter

half-wave rectifier full-wave rectifier

ACmains

Vo

Rectifierwithfilter

* A rectifier is used to convert an AC voltage to a DC voltage (typically 5 to 20 V), e.g., a mobile phonecharger.

* AC mains → step-down transformer → DC voltage ORAC mains → DC voltage → lower DC voltage

* A voltage regulator would be typically used to remove the ripple riding on the DC output.

Rectifiers

ACmains

Vo

ACmains

Vo

Rectifierwithfilter

Rectifiers

ACmains

Vo

ACmains

Vo

Rectifierwithfilter

Rectifiers

ACmains

Vo

half-wave rectifier

full-wave rectifier

ACmains

Vo

Rectifierwithfilter

Rectifiers

ACmains

Vo

ACmains

Vo

Rectifierwithfilter

Rectifiers

ACmains

Vo

ACmains

Vo

Rectifierwithfilter

Rectifiers

ACmains

Vo

ACmains

Vo

Rectifierwithfilter

half-wave rectifier

full-wave rectifier

Rectifiers

ACmains

Vo

ACmains

Vo

Rectifierwithfilter

Rectifiers

ACmains

Vo

ACmains

Vo

Rectifierwithfilter

Half-wave rectifier without filter

slope=1

Vi Vo

Von

Vi

Vo

t

D on D off

Vi

Vo

D

R

* D conducts only if Vi > Von, and in that case Vo =Vi − Von, a straight line with slope = 1.

* If Vi < Von, D does not conduct → Vo = 0.

slope=1

Vi Vo

Von

Vi

Vo

t

D on D off

Vi

Vo

D

R

slope=1

Vi Vo

Von

Vi

Vo

t

D on D off

Vi

Vo

D

R

Full-wave (bridge) rectifier without filter

Vs

D4

D3D2

D1

R

Vo

A

B

Q P

Vs

D4

D3

D2

D1

B

P

Q

A

Vo Vs

D4

D3

D2

D1

B

P

Q

A

Vo

D1, D2 on D3, D4 onVm

0

−Vs

Vs

t

Vo

Vs

D4

D3D2

D1

R

Vo

A

B

Q P Vs

D4

D3

D2

D1

B

P

Q

A

Vo

Vs

D4

D3

D2

D1

B

P

Q

A

Vo

0

−Vs

Vs

t

Vo

Vs

D4

D3D2

D1

R

Vo

A

B

Q P Vs

D4

D3

D2

D1

B

P

Q

A

Vo

Vs

D4

D3

D2

D1

B

P

Q

A

Vo

0

−Vs

Vs

t

Vo

Vs

D4

D3D2

D1

R

Vo

A

B

Q P Vs

D4

D3

D2

D1

B

P

Q

A

Vo Vs

D4

D3

D2

D1

B

P

Q

A

Vo

0

−Vs

Vs

t

Vo

Vs

D4

D3D2

D1

R

Vo

A

B

Q P Vs

D4

D3

D2

D1

B

P

Q

A

Vo Vs

D4

D3

D2

D1

B

P

Q

A

Vo

0

−Vs

Vs

t

Vo

Half-wave rectifier with capacitor filter

RC

iRiC

DiD

VD

Vs Vo

* Similar to the peak detector except that the loadresistance provides a discharge path for thecapacitor in this case.

* Because of the load current iR , there is a drop inthe output voltage → “ripple”

* The peak diode current is much larger than theaverage load current.

* VD(t) =Vs(t)− Vo(t)≈Vs(t)− Vm

→ The maximum reverse bias (“Peak InverseVoltage” or PIV) across the diode is 2Vm.

* With Von = 0.7 V, the plots are slightly different.

iD

−Vm

−2Vm

0

VD

VoVs

VRVm

0

TcT

T2 t

iC

0VVon:

0.7 V

RC

iRiC

DiD

VD

Vs Vo

iD

−Vm

−2Vm

0

VD

VoVs

VRVm

0

TcT

T2 t

iC

0VVon:

0.7 V

RC

iRiC

DiD

VD

Vs Vo

iD

−Vm

−2Vm

0

VD

VoVs

VRVm

0

TcT

T2 t

iC

0VVon:

0.7 V

RC

iRiC

DiD

VD

Vs Vo

iD

−Vm

−2Vm

0

VD

VoVs

VRVm

0

TcT

T2 t

iC

0VVon:

0.7 V

RC

iRiC

DiD

VD

Vs Vo

iD

−Vm

−2Vm

0

VD

VoVs

VRVm

0

TcT

T2 t

iC

0VVon:

0.7 V

RC

iRiC

DiD

VD

Vs Vo

iD

−Vm

−2Vm

0

VD

VoVs

VRVm

0

TcT

T2 t

iC

0VVon:

0.7 V

RC

iRiC

DiD

VD

Vs Vo

iD

−Vm

−2Vm

0

VD

VoVs

VRVm

0

TcT

T2 t

iC

0VVon:

0.7 V

RC

iRiC

DiD

VD

Vs Vo

iD

−Vm

−2Vm

0

VD

VoVs

VRVm

0

TcT

T2 t

iC

0VVon:

0.7 V

RC

iRiC

DiD

VD

Vs Vo

Vm = 16 V, f = 50 Hz, R = 100 Ω. For a ripple voltage VR = 2 V,find (a) the filter capacitance C , (b) average and peak diodecurrents, (c) maximum reverse voltage across the diode.(Let Von = 0 V.)

(a) filter capacitance

1. In the discharge phase,

Vo(t) = Vme−t/τ ≈ Vm

(1−

t

τ

).

The drop in Vo(t) is given by the second term.

Using T2≈T ,

VR = VmT

τ= Vm

T

RC.

t= 0 t=T

iD

−Vm

−2Vm

0

VD

Vo

Vs

VRVm

0

TcT

T2 t

iC

RC

iRiC

DiD

VD

Vs Vo

(1−

t

τ

).

Using T2≈T ,

VR = VmT

τ= Vm

T

RC.

t= 0 t=T

iD

−Vm

−2Vm

0

VD

Vo

Vs

VRVm

0

TcT

T2 t

iC

RC

iRiC

DiD

VD

Vs Vo

(1−

t

τ

).

Using T2≈T ,

VR = VmT

τ= Vm

T

RC.

t= 0 t=T

iD

−Vm

−2Vm

0

VD

Vo

Vs

VRVm

0

TcT

T2 t

iC

RC

iRiC

DiD

VD

Vs Vo

(a) Ripple voltage VR

2. Assuming iC = iR =Vo

R≈

Vm

Rin the discharge phase,

we get

iC =Vm

R= C

∆Vo

∆t≈ C

VR

T→ VR = Vm

T

RC.

→ C =Vm

VR

T

R=

16 V

2 V

20 ms

100 Ω= 1600µF.

t= 0 t=T

iD

−Vm

−2Vm

0

VD

Vo

Vs

VRVm

0

TcT

T2 t

iC

RC

iRiC

DiD

VD

Vs Vo

R≈

Vm

we get

iC =Vm

R= C

∆Vo

∆t≈ C

VR

T→ VR = Vm

T

RC.

→ C =Vm

VR

T

R=

16 V

2 V

20 ms

100 Ω= 1600µF.

t= 0 t=T

iD

−Vm

−2Vm

0

VD

Vo

Vs

VRVm

0

TcT

T2 t

iC

RC

iRiC

DiD

VD

Vs Vo

R≈

Vm

we get

iC =Vm

R= C

∆Vo

∆t≈ C

VR

T→ VR = Vm

T

RC.

→ C =Vm

VR

T

R=

16 V

2 V

20 ms

100 Ω= 1600µF.

t= 0 t=T

iD

−Vm

−2Vm

0

VD

Vo

Vs

VRVm

0

TcT

T2 t

iC

RC

iRiC

DiD

VD

Vs Vo

(b) Average diode current

Using charge balance,∫ T

T−Tc

(iD − iR) dt =

∫ T−Tc

0iR dt

→∫ T

T−Tc

iD dt =

∫ T

0iR dt.

iavD =

1

T

∫ T

0iD dt =

1

T

∫ T

T−Tc

iD dt

=1

T

∫ T

0iR dt ≈

Vm

R.

iavD ≈

16 V

100 Ω= 160 mA.

t= 0 t=T

iD

−Vm

−2Vm

0

VD

Vo

Vs

VRVm

0

TcT

T2 t

iC

RC

iRiC

DiD

VD

Vs Vo

T−Tc

(iD − iR) dt =

∫ T−Tc

0iR dt

→∫ T

T−Tc

iD dt =

∫ T

0iR dt.

iavD =

1

T

∫ T

0iD dt =

1

T

∫ T

T−Tc

iD dt

=1

T

∫ T

0iR dt ≈

Vm

R.

iavD ≈

16 V

100 Ω= 160 mA.

t= 0 t=T

iD

−Vm

−2Vm

0

VD

Vo

Vs

VRVm

0

TcT

T2 t

iC

RC

iRiC

DiD

VD

Vs Vo

T−Tc

(iD − iR) dt =

∫ T−Tc

0iR dt

→∫ T

T−Tc

iD dt =

∫ T

0iR dt.

iavD =

1

T

∫ T

0iD dt =

1

T

∫ T

T−Tc

iD dt

=1

T

∫ T

0iR dt ≈

Vm

R.

iavD ≈

16 V

100 Ω= 160 mA.

t= 0 t=T

iD

−Vm

−2Vm

0

VD

Vo

Vs

VRVm

0

TcT

T2 t

iC

RC

iRiC

DiD

VD

Vs Vo

T−Tc

(iD − iR) dt =

∫ T−Tc

0iR dt

→∫ T

T−Tc

iD dt =

∫ T

0iR dt.

iavD =

1

T

∫ T

0iD dt =

1

T

∫ T

T−Tc

iD dt

=1

T

∫ T

0iR dt ≈

Vm

R.

iavD ≈

16 V

100 Ω= 160 mA.

t= 0 t=T

iD

−Vm

−2Vm

0

VD

Vo

Vs

VRVm

0

TcT

T2 t

iC

RC

iRiC

DiD

VD

Vs Vo

T−Tc

(iD − iR) dt =

∫ T−Tc

0iR dt

→∫ T

T−Tc

iD dt =

∫ T

0iR dt.

iavD =

1

T

∫ T

0iD dt =

1

T

∫ T

T−Tc

iD dt

=1

T

∫ T

0iR dt ≈

Vm

R.

iavD ≈

16 V

100 Ω= 160 mA.

t= 0 t=T

iD

−Vm

−2Vm

0

VD

Vo

Vs

VRVm

0

TcT

T2 t

iC

RC

iRiC

DiD

VD

Vs Vo

(b) Peak diode current

ipeakD = C

d

dt(Vm cosωt)

∣∣∣∣t=−Tc

+Vm

R

= −ωC Vm sin(−ωTc ) +16 V

100 Ω= ωC Vm sinωTc + 0.16

Vm cos(−ωTc ) = Vm − VR , giving

ωTc = cos−1

(1−

VR

Vm

)= cos−1

(1−

2

16

)= 29.

ipeakD = 2π × 50× 1600× 10−6 × 16× sin 29 + 0.16

= 3.89 + 0.16 = 4.05 A.

t= 0 t=T

iD

−Vm

−2Vm

0

VD

Vo

Vs

VRVm

0

TcT

T2 t

iC

RC

iRiC

DiD

VD

Vs Vo

ipeakD = C

d

dt(Vm cosωt)

∣∣∣∣t=−Tc

+Vm

R

ωTc = cos−1

(1−

VR

Vm

)= cos−1

(1−

2

16

)= 29.

ipeakD = 2π × 50× 1600× 10−6 × 16× sin 29 + 0.16

= 3.89 + 0.16 = 4.05 A.

t= 0 t=T

iD

−Vm

−2Vm

0

VD

Vo

Vs

VRVm

0

TcT

T2 t

iC

RC

iRiC

DiD

VD

Vs Vo

ipeakD = C

d

dt(Vm cosωt)

∣∣∣∣t=−Tc

+Vm

R

ωTc = cos−1

(1−

VR

Vm

)= cos−1

(1−

2

16

)= 29.

ipeakD = 2π × 50× 1600× 10−6 × 16× sin 29 + 0.16

= 3.89 + 0.16 = 4.05 A.

t= 0 t=T

iD

−Vm

−2Vm

0

VD

Vo

Vs

VRVm

0

TcT

T2 t

iC

RC

iRiC

DiD

VD

Vs Vo

ipeakD = C

d

dt(Vm cosωt)

∣∣∣∣t=−Tc

+Vm

R

ωTc = cos−1

(1−

VR

Vm

)= cos−1

(1−

2

16

)= 29.

ipeakD = 2π × 50× 1600× 10−6 × 16× sin 29 + 0.16

= 3.89 + 0.16 = 4.05 A.

t= 0 t=T

iD

−Vm

−2Vm

0

VD

Vo

Vs

VRVm

0

TcT

T2 t

iC

RC

iRiC

DiD

VD

Vs Vo

ipeakD = C

d

dt(Vm cosωt)

∣∣∣∣t=−Tc

+Vm

R

ωTc = cos−1

(1−

VR

Vm

)= cos−1

(1−

2

16

)= 29.

ipeakD = 2π × 50× 1600× 10−6 × 16× sin 29 + 0.16

= 3.89 + 0.16 = 4.05 A.

t= 0 t=T

iD

−Vm

−2Vm

0

VD

Vo

Vs

VRVm

0

TcT

T2 t

iC

RC

iRiC

DiD

VD

Vs Vo

(b) Peak diode current: analytic expression

Vm cos(−ωTc ) = Vm − VR → cosωTc = 1−VR

Vm≡ 1− x

sinωTc =√

1− cos2 ωTc =√

1− (1− x)2

=√

1− (1− 2x + x2) ≈√

2x =

√2VR

Vm

ipeakD = iR + C

d

dt(Vm cosωt)

∣∣∣∣t=−Tc

= iR + ωC Vm sinωTc

= iR + ωC Vm

√2VR

Vm

(c) Maximum reverse bias≈ 2Vm = 32 V.

SEQUEL file: ee101 half rectifier.sqproj

t= 0 t=T

iD

−Vm

−2Vm

0

VD

Vo

Vs

VRVm

0

TcT

T2 t

iC

RC

iRiC

DiD

VD

Vs Vo

Vm≡ 1− x

sinωTc =√

1− cos2 ωTc =√

1− (1− x)2

=√

1− (1− 2x + x2) ≈√

2x =

√2VR

Vm

ipeakD = iR + C

d

dt(Vm cosωt)

∣∣∣∣t=−Tc

= iR + ωC Vm

√2VR

Vm

t= 0 t=T

iD

−Vm

−2Vm

0

VD

Vo

Vs

VRVm

0

TcT

T2 t

iC

RC

iRiC

DiD

VD

Vs Vo

Vm≡ 1− x

sinωTc =√

1− cos2 ωTc =√

1− (1− x)2

=√

1− (1− 2x + x2) ≈√

2x =

√2VR

Vm

ipeakD = iR + C

d

dt(Vm cosωt)

∣∣∣∣t=−Tc

= iR + ωC Vm

√2VR

Vm

t= 0 t=T

iD

−Vm

−2Vm

0

VD

Vo

Vs

VRVm

0

TcT

T2 t

iC

RC

iRiC

DiD

VD

Vs Vo

Vm≡ 1− x

sinωTc =√

1− cos2 ωTc =√

1− (1− x)2

=√

1− (1− 2x + x2) ≈√

2x =

√2VR

Vm

ipeakD = iR + C

d

dt(Vm cosωt)

∣∣∣∣t=−Tc

= iR + ωC Vm

√2VR

Vm

t= 0 t=T

iD

−Vm

−2Vm

0

VD

Vo

Vs

VRVm

0

TcT

T2 t

iC

RC

iRiC

DiD

VD

Vs Vo

Vm≡ 1− x

sinωTc =√

1− cos2 ωTc =√

1− (1− x)2

=√

1− (1− 2x + x2) ≈√

2x =

√2VR

Vm

ipeakD = iR + C

d

dt(Vm cosωt)

∣∣∣∣t=−Tc

= iR + ωC Vm

√2VR

Vm

t= 0 t=T

iD

−Vm

−2Vm

0

VD

Vo

Vs

VRVm

0

TcT

T2 t

iC

RC

iRiC

DiD

VD

Vs Vo

Vm≡ 1− x

sinωTc =√

1− cos2 ωTc =√

1− (1− x)2

=√

1− (1− 2x + x2) ≈√

2x =

√2VR

Vm

ipeakD = iR + C

d

dt(Vm cosωt)

∣∣∣∣t=−Tc

= iR + ωC Vm

√2VR

Vm

t= 0 t=T

iD

−Vm

−2Vm

0

VD

Vo

Vs

VRVm

0

TcT

T2 t

iC

Full-wave (bridge) rectifier with capacitor filter

C C

R RVsVs

D4 D4

D3 D3

D2 D2

D1 D1

T/2 T/2

Vo Vo

TC2TC1

0

t

iD3, iD4

iD1, iD2

−Vm

Vm

VD3VD1

−Vs

Vs

B B

VRVo

A A

Q Q

P P

t= 0

* As in the half-wave rectifier case, we have charging and dischargingintervals, and Vo ≈Vm is maintained.

* Charging through D1, D2 takes place when Vo(t) falls below Vs(t).

* Charging through D3, D4 takes place when Vo(t) falls below −Vs(t).

* The discharging interval is typically much longer than the chargingintervals (TC1 and TC2).

* The maximum reverse bias across any of the diodes is Vm.

C C

R RVsVs

D4 D4

D3 D3

D2 D2

D1 D1

T/2 T/2

Vo Vo

TC2TC1

0

t

iD3, iD4

iD1, iD2

−Vm

Vm

VD3VD1

−Vs

Vs

B B

VRVo

A A

Q Q

P P

t= 0

C C

R RVsVs

D4 D4

D3 D3

D2 D2

D1 D1

T/2 T/2

Vo Vo

TC2TC1

0

t

iD3, iD4

iD1, iD2

−Vm

Vm

VD3VD1

−Vs

Vs

B B

VRVo

A A

Q Q

P P

t= 0

C C

R RVsVs

D4 D4

D3 D3

D2 D2

D1 D1

T/2 T/2

Vo Vo

TC2TC1

0

t

iD3, iD4

iD1, iD2

−Vm

Vm

VD3VD1

−Vs

Vs

B B

VRVo

A A

Q Q

P P

t= 0

C C

R RVsVs

D4 D4

D3 D3

D2 D2

D1 D1

T/2 T/2

Vo Vo

TC2TC1

0

t

iD3, iD4

iD1, iD2

−Vm

Vm

VD3VD1

−Vs

Vs

B B

VRVo

A A

Q Q

P P

t= 0

C C

R RVsVs

D4 D4

D3 D3

D2 D2

D1 D1

T/2 T/2

Vo Vo

TC2TC1

0

t

iD3, iD4

iD1, iD2

−Vm

Vm

VD3VD1

−Vs

Vs

B B

VRVo

A A

Q Q

P P

t= 0

Full-wave rectifier with capacitor filter

C C

R RVsVs

D4 D4

D3 D3

D2 D2

D1 D1

T/2 T/2

Vo Vo

TC2TC1

0

t

iD3, iD4

iD1, iD2

−Vm

Vm

VD3VD1

−Vs

Vs

B B

VRVo

A A

Q Q

P P

t= 0

Vm = 16 V, f = 50 Hz, R = 100 Ω. For a ripple voltage VR = 2 V, find (a) thefilter capacitance C , (b) average and peak diode currents, (c) maximumreverse voltage across the diode. (Let Von = 0 V.)

(a) filter capacitance:

Assuming iC = iR =Vo

R≈

Vm

Rin the discharge phase, we get

iC =Vm

R= C

∆Vo

∆t≈ C

VR

T/2→ VR = Vm

T

2RC.

→ C =1

2

Vm

VR

T

R=

1

2

16 V

2 V

20 ms

100 Ω= 800µF.

C C

R RVsVs

D4 D4

D3 D3

D2 D2

D1 D1

T/2 T/2

Vo Vo

TC2TC1

0

t

iD3, iD4

iD1, iD2

−Vm

Vm

VD3VD1

−Vs

Vs

B B

VRVo

A A

Q Q

P P

t= 0

R≈

Vm

iC =Vm

R= C

∆Vo

∆t≈ C

VR

T/2→ VR = Vm

T

2RC.

→ C =1

2

Vm

VR

T

R=

1

2

16 V

2 V

20 ms

100 Ω= 800µF.

C C

R RVsVs

D4 D4

D3 D3

D2 D2

D1 D1

T/2 T/2

Vo Vo

TC2TC1

0

t

iD3, iD4

iD1, iD2

−Vm

Vm

VD3VD1

−Vs

Vs

B B

VRVo

A A

Q Q

P P

t= 0

R≈

Vm

iC =Vm

R= C

∆Vo

∆t≈ C

VR

T/2→ VR = Vm

T

2RC.

→ C =1

2

Vm

VR

T

R=

1

2

16 V

2 V

20 ms

100 Ω= 800µF.

C C

R RVsVs

D4 D4

D3 D3

D2 D2

D1 D1

T/2 T/2

Vo Vo

TC2TC1

0

t

iD3, iD4

iD1, iD2

−Vm

Vm

VD3VD1

−Vs

Vs

B B

VRVo

A A

Q Q

P P

t= 0

R≈

Vm

iC =Vm

R= C

∆Vo

∆t≈ C

VR

T/2→ VR = Vm

T

2RC.

→ C =1

2

Vm

VR

T

R=

1

2

16 V

2 V

20 ms

100 Ω= 800µF.

C C

R RVsVs

D4 D4

D3 D3

D2 D2

D1 D1

T/2 T/2

Vo Vo

TC2TC1

0

t

iD3, iD4

iD1, iD2

−Vm

Vm

VD3VD1

−Vs

Vs

B B

VRVo

A A

Q Q

P P

t= 0

Half of the charge lost by the capacitor is supplied by iD1 (= iD2), andthe other half by iD3 (= iD4).

iavD =

1

T×

1

2× (Charge lost in one cycle)

≈1

T×

1

2×(Vm

R× T

)=

Vm

2R=

16 V

2× 100 Ω= 80 mA.

C C

R RVsVs

D4 D4

D3 D3

D2 D2

D1 D1

T/2 T/2

Vo Vo

TC2TC1

0

t

iD3, iD4

iD1, iD2

−Vm

Vm

VD3VD1

−Vs

Vs

B B

VRVo

A A

Q Q

P P

t= 0

iavD =

1

T×

1

≈1

T×

1

2×(Vm

R× T

)=

Vm

2R=

16 V

2× 100 Ω= 80 mA.

C C

R RVsVs

D4 D4

D3 D3

D2 D2

D1 D1

T/2 T/2

Vo Vo

TC2TC1

0

t

iD3, iD4

iD1, iD2

−Vm

Vm

VD3VD1

−Vs

Vs

B B

VRVo

A A

Q Q

P P

t= 0

iavD =

1

T×

1

≈1

T×

1

2×(Vm

R× T

)=

Vm

2R=

16 V

2× 100 Ω= 80 mA.

C C

R RVsVs

D4 D4

D3 D3

D2 D2

D1 D1

T/2 T/2

Vo Vo

TC2TC1

0

t

iD3, iD4

iD1, iD2

−Vm

Vm

VD3VD1

−Vs

Vs

B B

VRVo

A A

Q Q

P P

t= 0

iavD =

1

T×

1

≈1

T×

1

2×(Vm

R× T

)=

Vm

2R=

16 V

2× 100 Ω= 80 mA.

C C

R RVsVs

D4 D4

D3 D3

D2 D2

D1 D1

T/2 T/2

Vo Vo

TC2TC1

0

t

iD3, iD4

iD1, iD2

−Vm

Vm

VD3VD1

−Vs

Vs

B B

VRVo

A A

Q Q

P P

t= 0

ipeakD1 = C

d

dt(Vm cosωt)

∣∣∣∣t=−TC1

+Vm

R

= −ωC Vm sin(−ωTC1) +16 V

100 Ω= ωC Vm sinωTC1 + 0.16

ωTC1 = cos−1

(1−

VR

Vm

)= cos−1

(1−

2

16

)= 29.

ipeakD1 = 2π × 50× 800× 10−6 × 16× sin 29 + 0.16

= 1.95 + 0.16 = 2.1 A.

C C

R RVsVs

D4 D4

D3 D3

D2 D2

D1 D1

T/2 T/2

Vo Vo

TC2TC1

0

t

iD3, iD4

iD1, iD2

−Vm

Vm

VD3VD1

−Vs

Vs

B B

VRVo

A A

Q Q

P P

t= 0

ipeakD1 = C

d

dt(Vm cosωt)

∣∣∣∣t=−TC1

+Vm

R

ωTC1 = cos−1

(1−

VR

Vm

)= cos−1

(1−

2

16

)= 29.

ipeakD1 = 2π × 50× 800× 10−6 × 16× sin 29 + 0.16

= 1.95 + 0.16 = 2.1 A.

C C

R RVsVs

D4 D4

D3 D3

D2 D2

D1 D1

T/2 T/2

Vo Vo

TC2TC1

0

t

iD3, iD4

iD1, iD2

−Vm

Vm

VD3VD1

−Vs

Vs

B B

VRVo

A A

Q Q

P P

t= 0

ipeakD1 = C

d

dt(Vm cosωt)

∣∣∣∣t=−TC1

+Vm

R

ωTC1 = cos−1

(1−

VR

Vm

)= cos−1

(1−

2

16

)= 29.

ipeakD1 = 2π × 50× 800× 10−6 × 16× sin 29 + 0.16

= 1.95 + 0.16 = 2.1 A.

C C

R RVsVs

D4 D4

D3 D3

D2 D2

D1 D1

T/2 T/2

Vo Vo

TC2TC1

0

t

iD3, iD4

iD1, iD2

−Vm

Vm

VD3VD1

−Vs

Vs

B B

VRVo

A A

Q Q

P P

t= 0

ipeakD1 = C

d

dt(Vm cosωt)

∣∣∣∣t=−TC1

+Vm

R

ωTC1 = cos−1

(1−

VR

Vm

)= cos−1

(1−

2

16

)= 29.

ipeakD1 = 2π × 50× 800× 10−6 × 16× sin 29 + 0.16

= 1.95 + 0.16 = 2.1 A.M. B. Patil, IIT Bombay

C C

R RVsVs

D4 D4

D3 D3

D2 D2

D1 D1

T/2 T/2

Vo Vo

TC2TC1

0

t

iD3, iD4

iD1, iD2

−Vm

Vm

VD3VD1

−Vs

Vs

B B

VRVo

A A

Q Q

P P

t= 0

(c) Maximum reverse bias = Vm = 16 V.

SEQUEL file: diode rectifier 4.sqproj

C C

R RVsVs

D4 D4

D3 D3

D2 D2

D1 D1

T/2 T/2

Vo Vo

TC2TC1

0

t

iD3, iD4

iD1, iD2

−Vm

Vm

VD3VD1

−Vs

Vs

B B

VRVo

A A

Q Q

P P

t= 0

(c) Maximum reverse bias = Vm = 16 V.

SEQUEL file: diode rectifier 4.sqproj

Comparison of half-wave and full-wave (bridge) rectifiers with capacitive filter

For the same source voltage (Vm sinωt), load (R), and ripple voltage (VR),compare the half-wave and full-wave rectifiers.

Parameter Half-wave Full-wave

Number of diodes 1 4

Filter capacitance C C/2

Average diode current iavD iav

D /2

Peak diode current ipeakD ipeak

D /2

Maximum reverse voltage 2Vm Vm

Comparison of half-wave and full-wave (bridge) rectifiers with capacitive filter

For the same source voltage (Vm sinωt), load (R), and ripple voltage (VR),compare the half-wave and full-wave rectifiers.

Parameter Half-wave Full-wave

Number of diodes 1 4

Filter capacitance C C/2

Average diode current iavD iav

D /2

Peak diode current ipeakD ipeak

D /2

Maximum reverse voltage 2Vm Vm

diode circuits: part 2sequel/ee101/mc_diodes_2.pdfclamper circuits 0v vi clamper vo 0v 0v vi clamper...

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