Exercise 4–1 Ex: 4.1 I 0, the (c)qli/ECE333/Chapter 04 ISM.pdf · 2016-02-27 · SEDRA-ISM: “E-CH04” — 2014/11/12 — 11:02 — PAGE1—#1 Exercise 4–1 Ex: 4.1 Refer to

SEDRA-ISM: “E-CH04” — 2014/11/12 — 11:02 — PAGE 1 — #1

Exercise 4–1

Ex: 4.1 Refer to Fig. 4.3(a). For v I ≥ 0, thediode conducts and presents a zero voltage drop.Thus vO = v I . For v I < 0, the diode is cut off,zero current flows through R, and vO = 0. Theresult is the transfer characteristic in Fig. E4.1.

Ex: 4.2 See Fig. 4.3a and 4.3b. During thepositive half of the sinusoid, the diode is forwardbiased, so it conducts resulting in vD = 0. Duringthe negative half cycle of the input signal v I , thediode is reverse biased. The diode does notconduct, resulting in no current flowing in thecircuit. So vO = 0 and vD = v I − vO = v I . Thisresults in the waveform shown in Fig. E4.2.

Ex: 4.3 iD = v I

R= 10 V

1 k�= 10 mA

dc component of vO = 1

πvO

= 1

πv I = 10

π

= 3.18 V

Ex: 4.4

(a)

�5 V

V � 0 V

� 2 mAI �2.5 k�5 � 0

2.5

�

�

(b)

�5 V

V � 5 V

I � 0 A2.5 k�

�

�

(c)

�5 V

V � 5 V

I � 0 A

�

�

2.5 k�

(d)

�5 V

V � 0 V

0 � 52.5

I �

�

�

2.5 k�� 2 mA

(e)

V � 3 V

3 V

2 V

1 V

31

I � 1 k� � 3 mA

0

0

I

(f)

�3 V

�2 V

�1 V

I � 1 k�

�5 V

0

0

I

V � 1 V

� 4 mA

5 � 11

�

�

Ex: 4.5 Vavg = 10

π

50+ R =10

π

1 mA= 10

πk�

∴ R = 3.133 k�


Exercise 4–2

Ex: 4.6 Equation (4.5)

V2 − V1 = 2.3 V T log

(I2

I1

)

At room temperature VT = 25 mV

V2 − V1 = 2.3× 25× 10−3 × log

(10

0.1

)

= 115 mV

Ex: 4.7 i = ISev/VT (1)

1 (mA) = ISe0.7/VT (2)

Dividing (1) by (2), we obtain

i (mA) = e(v−0.7)/VT

⇒ v = 0.7+ 0.025 ln(i)

where i is in mA. Thus,

for i = 0.1 mA,

v = 0.7+ 0.025 ln(0.1) = 0.64 V

and for i = 10 mA,

v = 0.7+ 0.025 ln(10) = 0.76 V

Ex: 4.8 �T = 125− 25 = 100◦C

IS = 10−14 × 1.15�T

= 1.17× 10−8A

Ex: 4.9 At 20◦C I = 1 V

1 M�= 1 μA

Since the reverse leakage current doubles forevery 10◦C increase, at 40◦C

I = 4× 1 μA = 4 μA

⇒ V = 4 μA × 1 M� = 4.0 V

@ 0◦C I = 1

4μA

⇒ V = 1

4× 1 = 0.25 V

Ex: 4.10 a. Use iteration:

Diode has 0.7 V drop at 1 mA current.

Assume VD = 0.7 V

ID = 5− 0.7

10 k�= 0.43 mA

Use Eq. (4.5) and note that

V1 = 0.7 V, I1 = 1 mA

ID

VCC � 5 V VD

R � 10 k�

�

�

V2 − V1 = 2.3× VT log

(I2

I1

)

V2 = V1 + 2.3× VT log

(I2

I1

)

First iteration

V2 = 0.7+ 2.3× 25× 10−3 log

(0.43

1

)= 0.679 V

Second iteration

I2 = 5− 0.679

10 k�= 0.432 mA

V2 = 0.7+ 2.3× 25.3× 10−3 log

(0.432

1

)

= 0.679 V � 0.68 V

we get almost the same voltage.

∴ The iteration yields

ID = 0.43 mA, VD = 0.68 V

b. Use constant voltage drop model:

VD = 0.7 V constant voltage drop

ID = 5− 0.7

10 k�= 0.43 mA

Ex: 4.11

RI

10 V

2.4 V

�

�

Diodes have 0.7 V drop at 1 mA

∴ 1 mA = ISe0.7/VT (1)

At a current I(mA),

I = ISeVD/VT (2)

Using (1) and (2), we obtain

I = e(VD−0.7)/VT


Exercise 4–3

For an output voltage of 2.4 V, the voltage drop

across each diode = 2.4

3= 0.8 V

Now I, the current through each diode, is

I = e(0.8−0.7)/0.025

= 54.6 mA

R = 10− 2.4

54.6× 10−3

= 139 �

Ex: 4.12

(a)

�5 V

V � 0.7 V

I � � 1.72 mA2.5 k�5 � 0.7

2.5

�

�

(b)

�5 V

V � 5 V

I � 0 A 2.5 k�

�

�

(c)

�5 V

V � 5 V

I � 0 A

�

�

2.5 k�

(d)

�5 V

�0.7 � 52.5

I �

�

�

2.5 k�

� 1.72 mA

V � 0.7 V

(e)

V � 3 � 0.7

3 V

2 V

1 V

2.31

I �

I

0

0

1 k�

� 2.3 mA

� 2.3 V

(f)

�3 V0

�2 V

�1 V

I � 1 k�

5 V

0

I

V � 1 � 0.7

� 1.7 V

� 3.3 mA

5 � 1.71

Ex: 4.13 rd = VT

ID

ID = 0.1 mA rd = 25× 10−3

0.1× 10−3 = 250 �

ID = 1 mA rd = 25× 10−3

1× 10−3 = 25 �

ID = 10 mA rd = 25× 10−3

10× 10−3 = 2.5 �

Ex: 4.14 For small signal model,

�iD = �vD/rd (1)

where rd = VT

ID

For exponential model,

iD = ISeV /V T


Exercise 4–4

iD2

iD1= e(V2−V1) /V T = e�vD/V T

�iD = iD2 − iD1 = iD1e�vD /V T − iD1

= iD1

(e�vD/V T − 1

)(2)

In this problem, iD1 = ID = 1 mA.

Using Eqs. (1) and (2) with VT = 25 mV, weobtain

�vD (mV) �iD (mA) �iD (mA)

small exponentialsignal model

a − 10 − 0.4 − 0.33

b − 5 − 0.2 − 0.18

c + 5 + 0.2 + 0.22

d + 10 + 0.4 + 0.49

Ex: 4.15

RIL

VO

�15 V

a. In this problem,�VO

�iL= 20 mV

1 mA= 20 �.

∴ Total small-signal resistance of the four diodes= 20 �

∴ For each diode, rd = 20

4= 5 �.

But rd = VT

ID⇒ 5 = 25 mV

ID.

∴ ID = 5 mA

and R = 15− 3

5 mA= 2.4 k�.

b. For VO = 3 V, voltage drop across each diode

= 3

4= 0.75 V

iD = ISeV /VT

IS = iD

eV /VT= 5× 10−3

e0.75/0.025= 4.7× 10−16 A

c. If iD = 5− i L = 5− 1 = 4 mA.

Across each diode the voltage drop is

VD = VT ln

(ID

IS

)

= 25× 10−3 × ln

(4× 10−3

4.7× 10−16

)

= 0.7443 V

Voltage drop across 4 diodes

= 4× 0.7443 = 2.977 V

so change in VO = 3− 2.977 = 23 mV.

Ex: 4.16 For a zener diode

VZ = VZ0 + IZ rZ

10 = VZ0 + 0.01× 50

VZ0 = 9.5 V

For IZ = 5 mA,

VZ = 9.5+ 0.005× 50 = 9.75 V

For IZ = 20 mA,

VZ = 9.5+ 0.02× 50 = 10.5 V

Ex: 4.17

IR

15 V

5.6 V

0 to 15 mA

The minimum zener current should be

5× IZk = 5× 1 = 5 mA

Since the load current can be as large as 15 mA,we should select R so that with IL = 15 mA, azener current of 5 mA is available. Thus thecurrent should be 20 mA, leading to

R = 15− 5.6

20 mA= 470 �

Maximum power dissipated in the diode occurswhen IL = 0 is

Pmax = 20× 10−3 × 5.6 = 112 mV


Exercise 4–5

Ex: 4.18

15 V

200 �

Thus, at no load VZ 5.1 V�

�

VZ

15 � 5.10.2 k�

I � � 49.5 mA 50 mA

For line regulation:

200 �

7 �vi

vo

��

Line regulation = vo

v i= 7

200+ 7= 33.8

mV

V

For load regulation:

200 �

VZ0

VO

rZ

�IL

�V O

�IL= −�IL(rZ ‖ 200 �)

�IL mA

= −6.8mV

mA

Ex: 4.19

�

vS

VD

2��t

Vs � 12 2

0

uu

a. The diode starts conduction at

vS = VD = 0.7 V

vS = Vs sin ωt, here Vs = 12√

2

At ωt = θ ,

vS = Vs sin θ = VD = 0.7 V

12√

2 sin θ = 0.7

θ = sin−1

(0.7

12√

2

)� 2.4◦

Conduction starts at θ and stops at 180− θ .

∴ Total conduction angle = 180− 2θ = 175.2◦

b. vO,avg = 1

2π

(π−θ)∫θ

(Vs sin φ − VD) dφ

= 1

2π[−Vs cos φ − VDφ]φ=π−θ

φ−θ

= 1

2π[Vs cos θ − Vs cos (π − θ)− VD (π − 2θ)]

But cos θ � 1, cos (π − θ) � − 1, and

π − 2θ � π

vO,avg = 2Vs

2π− VD

2

= Vs

π− VD

2

For Vs = 12√

2 and VD = 0.7 V

vO,avg = 12√

2

π− 0.7

2= 5.05 V

c. The peak diode current occurs at the peakdiode voltage.

∴ iD = Vs − VD

R= 12√

2− 0.7

100

= 163 mA

PIV = +V S = 12√

2

� 17 V

Ex: 4.20

( )

�

vS

Vs

�VS

�

� �VD

output

0

input

�t

a. As shown in the diagram, the output is zerobetween (π − θ) to (π + θ)

= 2θ


Exercise 4–6

Here θ is the angle at which the input signalreaches VD.

∴ Vs sin θ = VD

θ = sin−1

(VD

Vs

)

2θ = 2 sin−1

(VD

Vs

)

b. Average value of the output signal is given by

VO = 1

2π

⎡⎣2×

(π−θ)∫θ

(Vs sin φ − VD) dφ

⎤⎦

= 1

π[−Vs cos φ − VDφ]π−θ

φ=θ

� 2Vs

π− VD, for θ small.

c. Peak current occurs when φ = π

2.

Peak current

= Vs sin (π /2)− VD

R= Vs − VD

R

If vS is 12 V(rms),

then Vs =√

2× 12 = 12√

2

Peak current = 12√

2− 0.7

100� 163 mA

Nonzero output occurs for angle = 2 (π − 2θ)

The fraction of the cycle for which vO > 0 is

= 2 (π − 2θ)

2π× 100

=2

[π − 2 sin−1

(0.7

12√

2

)]2π

× 100

� 97.4 %

Average output voltage VO is

VO = 2Vs

π− VD = 2× 12

√2

π− 0.7 = 10.1 V

Peak diode current iD is

iD = Vs − VD

R= 12√

2− 0.7

100

= 163 mA

PIV = Vs − VD + VS

= 12√

2− 0.7+ 12√

2

= 33.2 V

Ex: 4.21

Vs

�Vs

� sin–1� �

2VD

output

02VD

VS

input

t

(a) VO,avg = 1

2π

∫(Vs sin φ − 2VD) dφ

= 2

2π[−Vs cos φ − 2VDφ]π−θ

φ=θ

= 1

π[Vs cos φ − Vs cos(π − θ)− 2VD(π − 2θ)]

But cos θ ≈ 1,

cos (π − θ) ≈ − 1

π − 2θ ≈ π . Thus

⇒ VO,avg � 2Vs

π− 2VD

= 2× 12√

2

π− 1.4 = 9.4 V

(b) Peak diode current = Peak voltage

R

= Vs − 2VD

R= 12√

2− 1.4

100

= 156 mA

PIV = Vs − VD = 12√

2− 0.7 = 16.3 V

Ex: 4.22 Full-wave peak rectifier:

R C�

�

vO

vS

�

�vS

D1

D2

Vp

Vr

assume ideal diodes

t

�t{

T2

The ripple voltage is the amount of voltagereduction during capacitor discharge that occurs


Exercise 4–7

when the diodes are not conducting. The outputvoltage is given by

vO = Vpe−t/RC

Vp − Vr = Vpe−T/2RC ← discharge is only half

the period. We also assumed �t T

2.

Vr = Vp

(1− e−

T/2RC

)

e−T/2RC � 1− T/2

RC, for CR� T/2

Thus Vr � Vp

(1− 1+ T/2

RC

)

Vr = Vp

2fRC(a) Q.E.D.

To find the average diode current, note that thecharge supplied to C during conduction is equalto the charge lost during discharge.

QSUPPLIED = QLOST

iCav�t = CV r SUB (a)

(iD,av − IL

)�t = C

Vp

2fRC= Vp

2fR

= Vpπ

ωR

iD,av = Vpπ

ω�tR+ IL

where ω�t is the conduction angle.

Note that the conduction angle has the sameexpression as for the half-wave rectifier and isgiven by Eq. (4.30),

ω�t ∼=√

2Vr

Vp(b)

Substituting for ω�t, we get

⇒ iD,av = πVp√2Vr

Vp· R+ IL

Since the output is approximately held at Vp,Vp

R≈ IL · Thus

⇒ iD,av∼= π IL

√Vp

2Vr+ IL

= IL

[1+ π

√Vp

2Vr

]Q.E.D.

If t = 0 is at the peak, the maximum diode currentoccurs at the onset of conduction or at t = −ω�t.

During conduction, the diode current is given by

iD = iC + iL

iD,max = CdvS

dt

∣∣∣∣t=−ω�t

+ iL

assuming iL is const. iL � Vp

R= IL

= Cd

dt

(Vp cos ωt

)+ IL

= −C sin ω t × ωVp + IL

= −C sin(−ω�t)× ωVp + IL

For a small conduction angle

sin(−ω�t) ≈ − ω�t. Thus

⇒ iD,max = Cω�t × ωVp + IL

Sub (b) to get

iD,max = C

√2V r

VpωVp + IL

Substituting ω = 2π f and using (a) together withVp/R � IL results in

iDmax = IL

[1+ 2π

√Vp

2Vr

]Q.E.D.

Ex: 4.23

��

�

vS

vO

D2

D1

D3

D4

C

R

acline

voltage

The output voltage, vO, can be expressed as

vO =(Vp − 2VD

)e−t/RC

At the end of the discharge interval

vO = Vp − 2VD − Vr

The discharge occurs almost over half of the timeperiod � T/2.

For time constant RC � T

2


Exercise 4–8

e−t/RC � 1− T

2× 1

RC

∴ VP − 2VD − Vr =(Vp − 2VD

) (1− T

2× 1

RC

)

⇒ Vr =(Vp − 2VD

)× T

2RC

Here Vp = 12√

2 and Vr = 1 V

VD = 0.8 V

T = 1

f= 1

60s

1 = (12√

2− 2× 0.8)× 1

2× 60× 100× C

C = (12√

2− 1.6)

2× 60× 100= 1281 μF

Without considering the ripple voltage, the dcoutput voltage

= 12√

2− 2× 0.8 = 15.4 V

If ripple voltage is included, the output voltage is

= 12√

2− 2× 0.8− Vr

2= 14.9 V

IL = 14.9

100 �� 0.15 A

The conduction angle ω�t can be obtained usingEq. (4.30) but substituting Vp = 12

√2− 2× 0.8:

ω�t =√

2Vr

Vp=√

2× 1

12√

2− 2× 0.8

= 0.36 rad = 20.7◦

The average and peak diode currents can becalculated using Eqs. (4.34) and (4.35):

iDav = IL

(1+ π

√Vp

2Vr

), where IL = 14.9 V

100 �,

Vp = 12√

2− 2× 0.8, and Vr = 1 V; thus

iDav = 1.45 A

iDmax = I

(1+ 2π

√Vp

2Vr

)

= 2.76 A

PIV of the diodes

= VS − VDO = 12√

2− 0.8 = 16.2 V

To provide a safety margin, select a diode capableof a peak current of 3.5 to 4A and having a PIVrating of 20 V.

Ex: 4.24

�

�vA

iD

vI i�

i�

�vD

� vO

1 k�iR

The diode has 0.7 V drop at 1 mA current.

iD = ISevD/VT

iD

1 mA= e(vD−0.7)/V T

⇒ vD = VT ln

(iD

1 mA

)+ 0.7 V

For v I = 10 mV, vO = v I = 10 mV

It is an ideal op amp, so i+ = i− = 0.

∴ iD = iR = 10 mV

1 k�= 10 μA

vD = 25× 10−3 ln

(10 μA

1 mA

)+ 0.7 = 0.58 V

vA = vD + 10 mV

= 0.58+ 0.01

= 0.59 V

For v I = 1 V

vO = v I = 1 V

iD = vO

1 k�= 1

1 k�= 1 mA

vD = 0.7 V

VA = 0.7 V + 1 k�× 1 mA

= 1.7 V

For v I = −1 V, the diode is cut off.

∴ vO = 0 V

vA = −12 V

Ex: 4.25

�

�vO

R ILvI


Exercise 4–9

v I > 0 ∼ diode is cut off, loop is open, and theopamp is saturated:

vO = 0 V

v I < 0 ∼ diode conducts and closes the negativefeedback loop:

vO = v I

Ex: 4.26

vI vO

10 k�

10 k� 10 k�

�

�

�

�

�

�

�

�

5 V5 V

D1 D2

Both diodes are cut off

for −5V ≤ v I ≤ + 5V

and vO = v I

For v I ≤ −5 V, diode D1 conducts and

vO = −5+ 1

2(+v I + 5)

=(−2.5+ v I

2

)V

For v I ≥ 5 V, diode D2 conducts and

vO = +5+ 1

2(v I − 5)

=(

2.5+ v I

2

)V

Ex: 4.27 Reversing the diode results in the peakoutput voltage being clamped at 0 V:

t

vO

�10 V

Here the dc component of vO = VO = −5 V

SEDRA-ISM: “P-CH04-001-013” — 2014/11/12 — 15:18 — PAGE 1 — #1

Chapter 4–1

4.1

I

�

�

VD1.5 V

1 �

(a)

(b)

I�

�

VD1.5 V

1 �

(a) The diode is reverse biased, thus

I = 0 A

VD = −1.5 V

(b) The diode is forward biased, thus

VD = 0 V

I = 1.5 V

1 �= 1.5 A

4.2 Refer to Fig. P4.2.

(a) Diode is conducting, thus

V = −3 V

I = +3− (−3)

10 k�= 0.6 mA

(b) Diode is reverse biased, thus

I = 0

V = +3 V

(c) Diode is conducting, thus

V = +3 V

I = +3− (−3)

10 k�= 0.6 mA

(d) Diode is reverse biased, thus

I = 0

V = −3 V

4.3

(a)

2 k�2 k�

I �

Cutoff

Conducting

D1

D2

�3 V

�1 V

�2 V V � 2 V

� 2.5 mA

2 �(�3)

(b)

2 k�2

Cutoff

ConductingD1

D2

�3 V

�1 V

�2 V

V � 1 V

� 1 mA

3 �(�1)I �

4.4

(a)

5 V0 t

vO

V p+ = 5 V Vp− = 0 V f = 1 kHz

(b)

�5 V0 t

vO

Vp+ = 0 V Vp− = −5 V f = 1 kHz


Chapter 4–2

(c)

t0

vO

vO = 0 V

Neither D1 nor D2 conducts, so there is no output.

(d)

5 Vt

vO

Vp+ = 5 V, Vp− = 0 V, f = 1 kHz

Both D1 and D2 conduct when v I > 0

(e)

5 V

�5 Vt

vO

Vp+ = 5 V, Vp− = −5 V, f = 1 kHz

D1 conducts when v I > 0 and D2 conducts whenv I < 0. Thus the output follows the input.

(f)

5 Vt

vO

Vp+ = 5 V, Vp− = 0 V, f = 1 kHz

D1 is cut off when v I < 0

(g)

�5 Vt

vO

Vp+ = 0 V, Vp− = −5 V, f = 1 kHz

D1 shorts to ground when v I > 0 and is cut offwhen v I < 0 whereby the output follows v I .

(h)

t

vO � 0 V

vO = 0 V∼ The output is always shorted toground as D1 conducts when v I > 0 and D2

conducts when v I < 0.

(i)

5 V

t

vO

�2.5 V

Vp+ = 5 V, Vp− = −2.5 V, f = 1 kHz

When v I > 0, D1 is cut off and vO follows v I .

When v I < 0, D1 is conducting and the circuitbecomes a voltage divider where the negativepeak is

1 k�

1 k�+ 1 k�×−5 V = −2.5 V

(j)

�2.5 V

5 V

t

vO

Vp+ = 5 V, Vp− = −2.5 V, f = 1 kHz

When v I > 0, the output follows the input as D1

is conducting.

When v I < 0, D1 is cut off and the circuitbecomes a voltage divider.

(k)

5 V

�5 V

1 Vt

vO

�4 V

Vp+ = 1 V, Vp− = −4 V, f = 1 kH3

When v I > 0, D1 is cut off and D2 is conducting.The output becomes 1 V.

When v I < 0, D1 is conducting and D2 is cut off.The output becomes:

vO = v I + 1 V


Chapter 4–3

4.5

From Fig. P4.5 we see that when v I < VB; that is,v I < 3 V, D1 will be conducting the current I andiB will be zero. When v I exceeds the batteryvoltage (3 V), D1 cuts off and D2 conducts, thussteering I into the battery. Thus, iB will have thewaveform shown in the figure. Its peak value willbe 60 mA. To obtain the average value, we firstdetermine the conduction angle of D2, (π − 2θ),where

θ = sin−1

(3

6

)= 30◦

Thus

π − 2θ = 180◦ − 60 = 120◦

The average value of iB will be

iB|av = 60× 120◦

360◦= 20 mA

If the peak value of v I is reduced by 10%, i.e.from 6 V to 5.4 V, the peak value of iB does notchange. The conduction angle of D2, however,changes since θ now becomes

θ = sin−1

(3

5.4

)= 33.75◦

and thus

π − 2θ = 112.5◦

Thus the average value of iB becomes

iB|av = 60× 112.5◦

360◦= 18.75 mA

4.6

A B X Y

0 0 0 0

0 1 0 1

1 0 0 1

1 1 1 1

X = AB, Y = A+ B

X and Y are the same for

A = B

X and Y are opposite if A �= B

4.7 The case for the highest current in a singlediode is when only one input is high:

VY = 5 V

VY

R≤ 0.2 mA⇒ R ≥ 25 k�

4.8 The maximum input current occurs when oneinput is low and the other two are high.

5− 0

R≤ 0.2 mA

R ≥ 25 k�

4.9

(a)

�3 V

�3 V

12 k� 0.33 mA

0.33mA

I � 0 D2

ON

D1

OFF

�1 V

V � �1 V

6 k�

(a) If we assume that both D1 and D2 areconducting, then V = 0 V and the current in D2

will be [0− (−3)]/6 = 0.5 mA. The current inthe 12 k� will be (3− 0)/12 = 0.25 mA. A nodeequation at the common anodes node yields anegative current in D1. It follows that ourassumption is wrong and D1 must be off. Nowmaking the assumption that D1 is off and D2 is on,we obtain the results shown in Fig. (a):

I = 0

V = −1 V


Chapter 4–4

(b)

�3 V

�3 V

6 k�3 – 0

6

0.25 mA

0 V

D2D1

ONON

� 0.5 mA

V � 0

12 k�

0 – (–3)12

�0.25 mA

(b) In (b), the two resistors are interchanged.With some reasoning, we can see that the currentsupplied through the 6-k� resistor will exceedthat drawn through the 12-k� resistor, leavingsufficient current to keep D1 conducting.Assuming that D1 and D2 are both conductinggives the results shown in Fig. (b):

I = 0.25 mA

V = 0 V

4.10 The analysis is shown on the circuitdiagrams below.

4.11 R ≥ 120√

2

40≥ 4.2 k�

The largest reverse voltage appearing across thediode is equal to the peak input voltage:

120√

2 = 169.7 V

These figures belong to Problem 4.10.

V

� 2.5 V

I

10 � 10 � 5 k�

20 k�

(a)

10 10 � 10

2.5 5 � 20

5 �

V � 0.1 � 20 � 2 V

I � � 0.1 mA(b)

I � 0

V2.5 V 1.5 V

5 k�

reverse biased

5 k�

V � 1.5 � 2.5 � �1 V

� �

4.12

RR � RS

12

�

�

vO vI vI

RSD

�

R � RS

vIiD �

�vI ��

D starts to conduct when v I > 0

vO

vI0

4.13 For v I > 0 V: D is on, vO = v I , iD = v I/R

For v I < 0 V: D is off , vO = 0, iD = 0


Chapter 4–5

4.14

4.15

conduction occurs

R D

vI

vI

A12 V

0

2

��

��

�

�

��

��

12 V

v I = A sin θ = 12 ∼ conduction through Doccurs

For a conduction angle (π − 2θ) that is 25% of acycle

π − 2θ

2π= 1

4

θ = π

4

A = 12/sin θ = 17 V

∴ Peak-to-peak sine wave voltage

= 2A = 34 V

Given the average diode current to be

1

2π

2π∫0

A sin φ − 12

Rdφ = 100 mA

1

2π

(−17 cos φ − 12φ

R

)φ= 0.75π

φ= 0.25π

= 0.1

R = 8.3 �

Peak diode current = A− 12

R= 0.6 A

Peak reverse voltage = A+ 12 = 29 V

For resistors specified to only one significant digitand peak-to-peak voltage to the nearest volt,choose A = 17 so the peak-to-peak sine wavevoltage = 34 V and R = 8 �.

Conduction starts at v I = A sin θ = 12

17 sin θ = 12

θ =(π

4

)rad

Conduction stops at π − θ .

∴ Fraction of cycle that current flows is

π − 2θ

2π× 100 = 25%

Average diode current =1

2π

(−17 cos φ − 12φ

8

)φ= 3π/4

φ=π/4

= 103 mA

Peak diode current

= 17− 12

8= 0.625 A

Peak reverse voltage =A+ 12 = 29 V

4.16

V RED GREEN

3 V ON OFF - D1 conducts

0 OFF OFF - No current flows

–3 V OFF ON - D2 conducts

4.17 VT = kT

q

wherek = 1.38× 10−23 J/K = 8.62× 10−5eV/K

T = 273+ x◦C

q = 1.60× 10−19C


Chapter 4–6

Thus

VT = 8.62× 10−5 × (273× x◦C), V

x [◦C] VT [mV]

−55 18.80 23.5

+55 28.3+125 34.3

for VT = 25 mV at 17◦C

4.18 i = I Sev/0.025

∴ 10,000IS = IS ev/0.025

v = 0.230 V

At v = 0.7 V,

i = IS e0.7/0.025 = 1.45× 1012IS

4.19 I1 = IS e0.7/VT = 10−3

i2 = ISe0.5/VT

i2

i1= i2

10−3 = e0.5− 0.7

0.025

i2 = 0.335 μA

4.20 I = ISeVD/VT

10−3 = ISe0.7/VT (1)

For VD = 0.71 V,

I = ISe0.71/VT (2)

Combining (1) and (2) gives

I = 10−3e(0.71− 0.7)/0.025

= 1.49 mA

For VD = 0.8 V,

I = ISe0.8/VT (3)

Combining (1) and (3) gives

I = 10−3 × e(0.8− 0.7)/0.025

= 54.6 mA

Similarly, for VD = 0.69 V we obtain

I = 10−3 × e(0.69− 0.7)/0.025

= 0.67 mA

and for VD = 0.6 V we have

I = 10−3e(0.6− 0.7)/0.025

= 18.3 μA

To increase the current by a factor of 10, VD mustbe increased by �VD,

10 = e�VD/0.025

⇒ �VD = 0.025 ln10 = 57.6 mV

4.21 IS can be found by using IS = ID · e−VD/VT .

Let a decrease by a factor of 10 in ID result in adecrease of VD by �V:

ID = ISeVD/VT

ID

10= ISe(VD−�V)/VT = ISeVD/VT · e−�V /V T

Taking the ratio of the above two equations, wehave

10 = e�V /V T ⇒ �V � 60 mV

Thus the result in each case is a decrease in thediode voltage by 60 mV.

(a) VD = 0.700 V, ID = 1 A⇒ IS = 6.91× 10−13 A;

10% of ID gives VD = 0.64 V

(b) VD = 0.650 V, ID = 1 mA⇒ IS = 5.11× 10−15 A;


(c) VD = 0.650 V, ID = 10 μA⇒ IS = 5.11× 10−17 A;


(d) VD = 0.700 V, ID = 100 mA⇒ IS = 6.91× 10−14 A;


4.22 IS can be found by using IS = ID · e−VD/VT .

Let an increase by a factor of 10 in ID result in anincrease of VD by �V:

ID = IS eVD/VT

10ID = IS e(VD+�V )/VT = IS eVD/VT · e�V /V T


10 = e�V /V T ⇒ �V � 60 mV

Thus the result is an increase in the diode voltageby 60 mV.

Similarly, at ID/10, VD is reduced by 60 mV.


Chapter 4–7

(a) VD = 0.70 V, ID = 10 mA⇒ IS = 6.91× 10−15 A;

ID × 10 gives V D = 0.76 V

ID/10 gives VD = 0.64 V

(b) VD = 0.70 V, ID = 1 mA⇒ IS = 6.91× 10−16 A;

ID × 10 gives V D = 0.76 V


(c) VD = 0.80 V, ID = 10 A⇒ IS = 1.27× 10−13 A;

ID × 10 gives V D = 0.86 V


(d) VD = 0.70 V, ID = 1 mA⇒ IS = 6.91× 10−16 A;

ID × 10 gives V D = 0.76 V


(e) VD = 0.6 V, ID = 10 μA⇒ IS = 3.78× 10−16 A

ID × 10 gives V D = 0.66 V


4.23 The voltage across three diodes in series is2.0 V; thus the voltage across each diode must be0.667 V. Using ID = ISeVD/VT , the requiredcurrent I is found to be 3.9 mA.

If 1 mA is drawn away from the circuit, ID will be2.9 mA, which would give VD = 0.794 V, givingan output voltage of 1.98 V. The change in outputvoltage is −22 mV.

4.24 Connecting an identical diode in parallelwould reduce the current in each diode by a factorof 2. Writing expressions for the currents, we have

ID = ISeVD/VT

ID

2= ISe(VD−�V)/VT = ISeVD/VT · e−�V /V T


2 = e�V /V T ⇒ �V = 17.3 mV

Thus the result is a decrease in the diode voltageby 17.3 mV.

4.25

VD

�

�

I

ID2ID1

D1 D2

ID1 = IS1eVD/VT (1)

ID2 = IS2eVD/VT (2)

Summing (1) and (2) gives

ID1 + ID2 = (IS1 + IS2)eVD/VT

But

ID1 + ID2 = I

Thus

I = (IS1 + IS2) eVD/VT (3)

From Eq. (3) we obtain

VD = VT ln

(I

IS1 + IS2

)

Also, Eq. (3) can be written as

I = IS1 eVD/VT

(1+ IS2

IS1

)(4)

Now using (1) and (4) gives

ID1 = I

1+ (IS2/IS1)= I

IS1

IS1 + IS2

We similarly obtain

ID2 = I

1+ (IS1/IS2)= I

IS2

IS1 + IS2

4.26

I

D2 D3

8I1

D4

4I12I1

I1 � 0.1 mA

I1

D1


Chapter 4–8

The junction areas of the four diodes must berelated by the same ratios as their currents, thus

A4 = 2A3 = 4A2 = 8 A1

With I1 = 0.1 mA,

I = 0.1+ 0.2+ 0.4+ 0.8 = 1.5 mA

4.27 We can write a node equation at the anodes:

ID2 = I1 − I2 = 7 mA

ID1 = I2 = 3 mA

We can write the following equation for the diodevoltages:

V = VD2 − VD1

If D2 has saturation current IS , then D1, which is10 times larger, has saturation current 10IS . Thuswe can write

ID2 = ISeVD2/VT

ID1 = 10I SeVD1/VT

Taking the ratio of the two equations above, wehave

ID2

ID1= 7

3= 1

10e(VD2−VD1)/VT = 1

10eV /V T

⇒ V = 0.025 ln

(70

3

)= 78.7 mV

To instead achieve V = 60 mV, we need

ID2

ID1= I1 − I2

I2= 1

10e0.06/0.025 = 1.1

Solving the above equation with I1 still at 10 mA,we find I2 = 4.76 mA.

4.28 We can write the following node equation atthe diode anodes:

ID2 = 10 mA − V /R

ID1 = V /R

We can write the following equation for the diodevoltages:

V = VD2 − VD1

We can write the following diode equations:

ID2 = ISeVD2/VT

ID1 = ISeVD1/VT

Taking the ratio of the two equations above, wehave

ID2

ID1= 10 mA − V /R

V /R= e(VD2−VD1)/VT = eV /V T

To achieve V = 50 mV, we need

ID2

ID1= 10 mA − 0.05/R

0.05/R= e0.05/0.025 = 7.39

Solving the above equation, we have

R = 42 �

4.29 For a diode conducting a constant current,the diode voltage decreases by approximately 2mV per increase of 1◦C.

T = −20◦C corresponds to a temperaturedecrease of 40◦C, which results in an increase ofthe diode voltage by 80 mV. Thus VD = 770 mV.

T = + 85◦C corresponds to a temperatureincrease of 65◦C, which results in a decrease ofthe diode voltage by 130 mV. Thus VD = 560 mV.

4.30

D2

D1

R1

I

�10 V

V1

�

�

V2

�

�

At 20◦C:

VR1 = V2 = 520 mV

R1 = 520 k�

I = 520 mV

520 k�= 1 μA

Since the reverse current doubles for every 10◦Crise in temperature, at 40◦C, I = 4 μA

V2

ID2

480

40°C

40 mV

20°C

520 mV

4 µA

1 µA

V2 = 480+ 2.3× 1× 25 log 4

= 514.6 mV

VR1 = 4 μA × 520 k� = 2.08 V

At 0◦C, I = 1

4μA

V2 = 560− 2.3× 1× 25 log 4

= 525.4 mV

VR1 = 1

4× 520 = 0.13 V


Chapter 4–9

4.31 For a diode conducting a constant current,the diode voltage decreases by approximately2 mV per increase of 1◦C.

A decrease in VD by 100 mV corresponds to ajunction temperature increase of 50◦C.

The power dissipation is given by

PD = (10 A) (0.6 V) = 6 W

The thermal resistance is given by

�T

PD= 50◦C

6 W= 8.33◦C/W

4.32 Given two different voltage/currentmeasurements for a diode, we can write

ID1 = ISeVD1/VT

ID2 = ISeVD2/VT


ID1

ID2= ISe(VD1−VD2)/VT ⇒ VD1 − VD2

= VT ln

(ID1

ID2

)

For ID = 1 mA, we have

�V = VT ln

(1× 10−3 A

10 A

)= −230 mV

⇒ VD = 570 mV

For ID = 3 mA, we have

�V = VT ln

(3× 10−3 A

10 A

)= −202 mV

⇒ VD = 598 mV

Assuming VD changes by –2 mV per 1◦C increasein temperature, we have, for ±20◦C changes:

For ID = 1 mA, 530 mV ≤ VD ≤ 610 mV

For ID = 3 mA, 558 mV ≤ VD ≤ 638 mV

Thus the overall range of VD is between 530 mVand 638 mV.

4.33 Given two different voltage/currentmeasurements for a diode, we have

ID1

ID2= IS e(VD1−VD2)/VT ⇒ VD1 − VD2

= VT ln

(ID1

ID2

)

For the first diode, with ID = 0.1 mA and

VD = 700 mV, we have

ID = 1 mA:

�V = VT ln

(1.0

0.1

)= 57.6 mV

⇒ VD = 757.6 mV

ID = 3 mA:

�V = VT ln

(3

0.1

)= 85 mV⇒ VD = 785 mV

For the second diode, with

ID = 1 A and VD = 700 mV, we have

ID = 1.0 mA:

�V = VT ln

(0.001

1

)= −173 mV

⇒ VD = 527 mV

ID = 3 mA:

�V = VT ln

(0.003

1

)= −145 mV

⇒ VD = 555 mV

For both ID = 1.0 mA and ID = 3 mA, thedifference between the two diode voltages isapproximately 230 mV. Since, for a fixed diodecurrent, the diode voltage changes withtemperature at a constant rate (–2 mV per ◦Ctemp. increase), this voltage difference will beindependent of temperature!

4.34

�

�VDD

R � 1 kΩ

1 Vi v

IS = 10−15A = 10−12 mA

Calculate some points

v = 0.6 V , i = ISev/V T

= 10−12e0.6/0.025

� 0.03 mA

v = 0.65 V, i � 0.2 mA

v = 0.7 V, i � 1.45 mA

Make a sketch showing these points and load lineand determine the operating point. The points forthe load line are obtained using

ID = VDD − VD

R


Chapter 4–10

Load line

0.40

0.20.40.60.81.0

2.0

i (mA)

0.6 0.8 1.0v (V)

Diode characteristic

From this sketch one can see that the operatingpoint must lie between v = 0.65 V to v = 0.7 V

For i = 0.3 mA, v = VT ln

(i

IS

)

= 0.025× ln

(3

10−12

)

= 0.661 V

For i = 0.4 mA, v = 0.668 V

Now we can refine the diagram to obtain a betterestimate

i (mA)

Load line

0.300.660 0.664 0.67

0.35

0.4

v (V)

From this graph we get the operating point

i = 0.338 mA, v = 0.6635 V

Now we compare graphical results with theexponential model.

At i = 0.338 mA

v = VT ln

(i

IS

)= 0.025× ln

(0.338

10−12

)

= 0.6637 V

The difference between the exponential modeland graphical results is = 0.6637− 0.6635

= 0.0002 V

= 0.2 mV

4.35R � 1 k�

1 V

�

�

�

�

VDID

IS = 10−15A = 10−12 mA

Use the iterative analysis procedure:

1. VD = 0.7 V, ID = 1− 0.7

1 K= 0.3 mA

2. VD = VT ln

(ID

IS

)= 0.025 ln

(0.3

10−12

)

|| = 0.6607 V

ID = 1− 0.6607

1 K= 0.3393 mA

3. VD = 0.025 ln

(0.3393

10−12

)= 0.6638 V

ID = 1− 0.6638

1 K= 0.3362 mA

4. VD = 0.025 ln

(0.3362

10−12

)= 0.6635 V

ID = 1− 0.6635

1 k�= 0.3365 mA

Stop here as we are getting almost same value ofID and VD

4.36500 �

1 V

�

�

�

�

vi

a) ID = 1− 0.7

0.5 k�= 0.6 mA

b) Diode has 0.7 V drop at 1 mA current. UseEq. (4.5):

v2 = v1 + 2.3VT log

(i2

i1

)

1. v = 0.7 V

1 i = 1− 0.7

0.5 k�= 0.6 mA

2. v = 0.7+ 2.3× 0.025 log

(0.6

1

)

= 0.6872 V

2 i = 1− 0.6872

0.5= 0.6255 mA


Chapter 4–11

3. v = 0.7+ 2.3× 0.025 log

(0.6255

1

)

= 0.6882 V

3 i = 1− 0.6882

0.5 k�= 0.6235 mA

4. v = 0.7+ 2.3× 0.025 log

(0.6235

1

)

= 0.6882 V

4 i = 1− 0.6882

0.5 k�= 0.6235 mA

Stop as we are getting the same result.

4.37 We first find the value of IS for the diode,given by IS = IDe−VD/VT with ID = 1 mA andVD = 0.75 V. This gives IS = 9.36× 10−17 A.

In order to have 3.3 V across the 4series-connected diodes, each diode drop must be0.825 V. Applying this voltage to the diode givescurrent ID = 20.1 mA. We can then find theresistor value using

R = 15 V − 3.3 V

20.1 mA= 582 �

4.38 Constant voltage drop model:

Using vD = 0.7 V,⇒ iD1 = V − 0.7

R

Using vD = 0.6 V,⇒ iD2 = V − 0.6

R

For the difference in currents to be only 1%,

⇒ iD2 = 1.01iD1

V − 0.6 = 1.01 (V − 0.7)

V = 10.7 V

For V = 3 V and R = 1 k�:

At VD = 0.7 V, iD1 = 3− 0.7

1= 2.3 mA

At VD = 0.6 V, iD2 = 3− 0.6

1= 2.4 mA

iD2

iD1= 2.4

2.3= 1.04

Thus the percentage difference is 4%.

4.39 Available diodes have 0.7 V drop at 2 mAcurrent since 2VD = 1.4 V is close to 1.3 V, use Nparallel pairs of diodes to split the 1 mA currentevenly, as shown in the figure next.

N

1 mA

�

�

V

The voltage drop across each pair of diodes is1.3 V. ∴ Voltage drop across each diode

= 1.3

2= 0.65 V. Using

I2 = I1e(V2−V1)/VT

= 2e(0.65−0.7)/0.025

= 0.2707 mA

Thus current through each branch is 0.2707 mA.

The 1 mA will split in= 1

0.2707= 3.69 branches.

Choose N = 4.

There are 4 pairs of diodes in parallel.

∴ We need 8 diodes.

Current through each pair of diodes

= 1 mA

4= 0.25 mA

∴ Voltage across each pair

= 2

[0.7+ 0.025 ln

(0.25

2

)]

= 1.296 V

SPECIAL NOTE: There is another possibledesign utilizing only 6 diodes and realizing avoltage of 1.313 V. It consists of the seriesconnection of 4 parallel diodes and 2 paralleldiodes.

4.40 Refer to Example 4.2.

(a)

10 k�

5 k�

�10 V

1.86�1 �0.86 mA

�10 V

V � 0 V

I �

� 1.86 mA

� 1 mA

��

�0.7 V�0.7 � 10

52

10 � 010

3

45

1


Chapter 4–12

(b)

Cutoff

5 k�

10 k�

ID2

�10 V

�10 V

0.7 V

V

I � 0

ID2

�

��

ID2 = 10− (−10)− 0.7

15= 1.29 mA

VD = −10+ 1.29 (10)+ 0.7 = 3.6 V

4.41

(a)

10 k�

�3 V

�3 V

V

I

V = −3+ 0.7 = −2.3 V

I = 3+ 2.3

10

= 0.53 mA

(b)

10 k�

Cutoff

�3 V

�3 V

V

I

I = 0 A

V = 3− I (10) = 3 V

(c)

10 k�

�3 V

I

�3 V

V

V = 3− 0.7 = 2.3 V

I = 2.3+ 3

10= 0.53 mA

(d)

�3 V

�3 V

Cutoff I

V

I = 0 A

V = −3 V

4.42

(a)

2 k�

Cutoff

�3 V

�1 V

�2 V

I

V

D1

D2

V = 2− 0.7

= 1.3 V

I = 1.3− (−3)

2

= 2.15 mA


Chapter 4–13

(b)

2 k�

Cutoff

�3 V

I

V�1 VD1

D2�2 V

V = 1+ 0.7 = 1.7 V

I = 3− 1.7

2= 0.65 mA

4.43

6 k�

+ 3 V

�

�

�3 V

0.7 V

12 k�

V

I � 0

ID2

ID2

D1 D2Cutoff

(a)

ON

12 k�

� 3 V

� 0.7 V

�3 V

6 k�

(b)

I � 0.383 � 0.25

V � 0 V

D1D2

ONON

6� 0.383 mA

� 0.133 mA

3�0.7

120.25 mA

0�(�3)�

(a) ID2 = 3− 0.7− (−3)

12+ 6= 0.294 mA

V = −3+ 0.294× 6 = −1.23 V

Check that D1 is off: Voltage at the anode ofD1 = V + VD2 = −1.23+ 0.7 = −0.53 V whichkeeps D1 off.

(b) See analysis on Fig. (b).

I = 0.133 mA

V = 0 V

4.44

I20 k�

10 � 10 � 5 k� 0.7 V

1010 � 105

(a)

V� �

� 2.5 V

1.5 V

� V �5 k� 5 k�

2.5 V0 mA

(b)

(a) I = 2.5− 0.7

5+ 20= 0.072 mA

V = 0.072× 20 = 1.44 V

(b) The diode will be cut off, thus

I = 0

V = 1.5− 2.5 = −1 V

4.45

vIiD R

��

iD,peak = v I ,peak − 0.7

R≤ 40 mA

R ≥ 120√

2− 0.7

40= 4.23 k�

Reverse voltage = 120√

2 = 169.7 V.

The design is essentially the same since thesupply voltage� 0.7 V

4.46 Use the exponential diode model to find thepercentage change in the current.

iD = ISev/VT

iD2

iD1= e(V2−V1)/VT = e�v/VT

For +5 mV change we obtain


Chapter 4–14

iD2

iD1= e5/25 = 1.221

% change

= iD2 − iD1

iD1× 100 = 1.221− 1

1× 100

= 22.1%

For –5 mV change we obtain

iD2

iD1= e−5/25 = 0.818

% change = iD2 − iD1

iD1× 100 = 0.818− 1

1× 100

= −18.1%

Maximum allowable voltage signal change whenthe current change is limited to ±10% is found asfollows:

The current varies from 0.9 to 1.1

iD2

iD1= e�V /VT

For 0.9, �V = 25 ln (0.9) = −2.63 mV

For 1.1, �V = 25 ln (1.1) = +2.38 mV

For ±10% current change the voltage signalchange is from –2.63 mV to +2.38 mV

4.47 0.1 A

Ten diode connected in parallel and fed with atotal current of 0.1 A. So the current through each

diode = 0.1

10= 0.01 A

Small signal resistance of each diode

= VT

iD= 25 mV

0.01 A= 2.5 �

Equivalent resistance, Req, of 10 diodes connectedin parallel is given by

Req = 2.5

10= 0.25 �

If there is one diode conducting 0.1 A current,then the small signal resistance of this diode

= 25 mV

0.1 A= 0.25 �

This value is the same as of 10 diodes connectedin parallel.

If 0.2 � is the resistance for making connection,the resistance in each branch= rd + 0.2 = 2.5+ 0.2 = 2.7 �

For a parallel combination of 10 diodes,equivalent resistance, Req, is

Req = 2.7

10= 0.27 �

If there is a single diode conducting all the 0.1 Acurrent, the connection resistance needed for thesingle diode will be = 0.27− 0.25 = 0.02 �.

4.48 The dc current I flows through the diode

giving rise to the diode resistance rd = VT

Iand

the small-signal equivalent circuit is representedby

vovs rd

�

�

Rs

vo = vs × rd

rd + RS= vs

VT /IVT

I+ RS

= vsVT

VT + IRS

Now, vo = 10 mV × 25 mV

25 mV + 103I

I vo

1 mA 0.24 mV

0.1 mA 2.0 mV

1 μA 9.6 mV

For vo = 1

2v s = vs × 0.025

0.025+ 103I

⇒ I = 25 μA

4.49 As shown in Problem 4.48,

vo

v i= VT

VT + RSI= 0.025

0.025+ 104I(1)

Here RS = 10 k�

The current changes are limited ±10%. Usingexponential model, we get

iD2

iD1= e�v/VT = 0.9 to 1.1

�v = 25× 10−3 ln

(iD2

iD1

)and here

For 0.9, �v = −2.63 mV

For 1.1, �v = 2.38 mV

The variation is –2.63 mV to 2.38 mV for ±10%current variation. Thus the largest symmetricaloutput signal allowed is 2.38 mV in amplitude. To


Chapter 4–15

obtain the corresponding input signal, we dividethis by (vo/v i):

vs = 2.38 mV

vo/v i(2)

Now for the given values of(vo/v i

)calculate I

and vS using Equations (1) and (2)

vo

v iI in mA vs in mV

0.5 0.0025 4.76

0.1 0.0225 23.8

0.01 0.2475 238

0.001 2.4975 2380

4.50

vo

vi

C21 mA

D1 D2

I

C1

When both D1 and D2 are conducting, thesmall-signal model is

vovi

rd2

rd1

where we have replaced the large capacitors C1

and C2 by short circuits:

vo

v i= rd2

rd1 + rd2=

VT

1 m − IVT

I+ VT

1 m − I

= I

1 m

Thusvo

v i= I , where I is in mA

Now I = 0 μA,vo

v i= 0

I = 1 μA,vo

v i= 1× 10−3 = 0.001 V/V

I = 10 μA,vo

v i= 10× 10−3 = 0.01 V/V

I = 100 μA,vo

v i= 100× 10−3 = 0.1 V/V

I = 500 μA,vo

v i= 500× 10−3 = 0.5 V/V

I = 600 μA,vo

v i= 600× 10−3 = 0.6 V/V

I = 900 μA,vo

v i= 900× 10−3 = 0.9 V/V

I = 990 μA,vo

v i= 990× 10−3 = 0.99 V/V

I = 1 mA

= 1000 μA,vo

v i= 1000× 10−3 = 1 V/V

4.51

I

D1 D3

D2

vo

vi

D4

I

R 10 k�

a. The current through each diode isI

2:

rd = VT

I

2

= 2VT

I= 0.05

I

From the equivalent circuit

vo

v i= R

R+ (2rd ‖ 2rd )= R

R+ rd

I rdvo

v i

0 μA ∞ 0

1 μA 50 k� 0.167

10 μA 5 k� 0.667

100 μA 500 � 0.952

1 mA 50 � 0.995

10 mA 5 � 0.9995


Chapter 4–16

Equivalent Circuit

rd rd

rd rdvI

vo

R 10 k��

b. For signal current to be limited to ±10% of I (Iis the biasing current), the change in diodevoltage can be obtained from the equation

iD

I= e�VD/V T = 0.9 to 1.1

�vD = −2.63 mV to +2.32 mV

� ±2.5 mV

so the signal voltage across each diode is limitedto 2.5 mV when the diode current remains within10% of the dc bias current.

∴ vo = 10− 2.5− 2.5 = 5 mV

and i = 5 mV

10 k�= 0.5 μA

vo

R 10 k��

�2.5�mV

�2.5�mV

�2.5�mV

�2.5�mV

10 mV

i

The current through each diode

= 0.5

2μA = 0.25 μA

The signal current i is 0.5 μA, and this is 10% ofthe dc biasing current.

∴ DC biasing current I = 0.5× 10 = 5 μA

c. Now I = 1 mA.

∴ ID = 0.5 mA

Maximum current derivation 10%.

∴ id = 0.5

10= 0.05 mA

and i = 2id = 0.1 mA.

∴ Maximum vo = i × 10 k�

= 0.1× 10

= 1 V

From the results of (a) above, for I = 1 mA,vo/v i = 0.995; thus the maximum input signalwill be

v i = vo/0.995 = 1/0.995

= 1.005 V

The same result can be obtained from the figureabove where the signal across the two seriesdiodes is 5 mV, thusv i = vo + 5 mV = 1 V + 5 mV = 1.005 V. Seealso the figure below.

4.52

I

v1

D1

i1 i3

i2 i4

D3

D2

iO vO

D4

R 10 k�v2 v4

v3

�

� �

�

�

� �

�

I

vI

I = 1 mA

Each diode exhibits 0.7 V drop at 1 mA current.Using diode exponential model we have

v2 − v1 = VT ln

(i2

i1

)

and v1 = 0.7 V, i1 = 1 mA

⇒ v = 0.7+ VT ln

(i

1

)

= 700+ 25 ln(i)


Chapter 4–17

Calculation for different values of vO:

vO = 0, iO = 0, and the current I = 1 mA divideequally between the D3, D4 side and the D1, D2

side.

i1 = i2 = i3 = i4 = I

2= 0.5 mA

v = 700+ 25 ln(0.5) � 683 mV

v1 = v2 = v3 = v4 = 683 mV

From the circuit, we have

v I = −v1 + v3 + vO = −683+ 683+ 0 = 0 V

For vO = 1 V, iO = 1

10 k�= 0.1 mA

Because of symmetry of the circuit, we obtain

i3 = i2 = I

2+ iO

2= 0.5+ 0.05 = 0.55 mA and

i4 = i1 = 0.45 mA

v3 = v2 = 700+ 25 ln

(i2

1

)= 685 mV

v4 = v1 = 700+ 25 ln

(i4

1

)= 680 mV

vO iO i3 = i2 i4 = i1 v3 = v2 v4 = v1 v I = −v1+(V) (mA) (mA) (mA) (mV) (mV) v3 + vO (V)

0 0 0.5 0.5 683 683 0

+ 1 0.1 0.55 0.45 685 680 1.005

+ 2 0.2 0.6 0.4 ∼ 687 677 2.010

+ 5 0.5 0.75 0.25 ∼ 693 665 5.028

+ 9 0.9 0.95 0.05 ∼ 699 ∼ 625 9.074

+ 9.9 0.99 0.995 0.005 ∼ 700 568 10.09

9.99 0.999 0.9995 0.0005 ∼ 700 510 10.18

10 1 1 0 700 0 10.7

v I = −v1 + v2 + vO = −0.680

+0.685+ 1 = 1.005 V

Similarly, other values are calculated as shown inthe table. The largest values of vO on positive andnegative side are +10 V and −10 V, respectively.This restriction is imposed by the currentI = 1 mA

A similar table can be generated for the negativevalues. It is symmetrical.

For v I > +10.7, vO will be saturated at +10 Vand it is because I = 1 mA. Similarly, forv I < −10.7 V, vO will be saturated at −10 V.

For I = 0.5 mA, the output will saturate at0.5 mA×10 k� = 5 V.

vo (V )

v1 (V )

10

5

�5

�5.68 5.68�10.7 10.7

I � 1 mA

I � 0.5 mA

4.53 Representing the diode by the small-signalresistances, the circuit is

vi

VT

vo

ID

C

�

�

��

rd

rd

��

Vo

Vi=

1

sC

rd + 1

sC

= 1

1+ sCrd

Vo

Vi= 1

1+ jωCrd

Phase shift = −tan−1

(ωCrd

1

)

= −tan−1

(ωC

VT

I

)

For phase shift of −45◦, we obtain

−45 = −tan−1

(2π × 100× 103 × 10

× 10−9 × 0.025

I

)

⇒ I = 157 μA

Now I varies from157

10μA to 157× 10 μA

Range is 15.7 μA to 1570 μA

Range of phase shift is −84.3◦ to −5.71◦


Chapter 4–18

4.54V�

VO

R

�

�

(a)�VO

�V+= rd

R+ rd= VT /I

R+ VT

I

= VT

IR+ VT

For no load, I = V+ − VO

R= V+ − 0.7

R.

∴ �VO

�V+= VT

VT + (V+ − 0.7)

Small-signal model

�

�

R

�V rd �VO��

(b) If m diodes are in series, we obtain

�VO

�V+= mrd

mrd + R= mVT

mVT + IR

= mVT

mVT + (V+ − 0.7m)

(c) For m = 1

�VO

�V+= VT

VT + V+ − 0.7= 1.75 mV/V

For m = 4

�VO

�V+= mVT

mVT + 15− m× 0.7= 8.13 mV/V

4.55

R

Small-signal model

�

�

IL

IL

rdR

�VO

V�

VO

(a) From the small-signal model

�VO = −IL (rd ‖ R)

�VO

IL= − (rd ‖ R)

(b) At no load, ID = V+ − 0.7

R

rd = VT

ID

�VO

IL= − (rd ‖ R) = − 1

1

rd+ 1

R

= − 1ID

V+ − 0.7+ ID

VT

= −VT

ID× 1

VT

V+ − 0.7+ 1

= −VT

ID× V+ − 0.7

VT + V+ − 0.7

For�VO

IL≤ 5

mV

mA

i.e.,VT

ID× V+ − 0.7

VT + V+ − 0.7≤ 5 mV

mA[25

ID× 15− 0.7

0.025+ 15− 0.7

]≤ 5

mV

mA

ID ≥ 4.99 mA

ID � 5 mA

R = V+ − 0.7

ID= 15− 0.7

5 mA

R = 2.86 k�


Chapter 4–19

Diode should be a 5-mA unit; that is, it conducts5 mA at VD = 0.7 V, thus 5× 10−3 = ISe0.7/0.025.

⇒ IS = 3.46× 10−15 A

(c) For m diodes connected in series we have

ID = V+ − 0.7m

R

and rd = VT

ID

So now�VO

IL= −(R ‖ mrd ) = − 1

1

R+ 1

mrd

= − 1ID

V+ − 0.7m+ ID

mVT

= −mV T

ID mV T

V+ − 0.7m+ 1

= −mVT

ID

V+ − 0.7m

V+ − 0.7m+ mVT

4.56

IL

ID

Vo

I

�5 V

RL

R

Diode has 0.7 V drop at 1 mA current

VO = 1.5 V when RL = 1.5 k�

ID = ISeV /V T

1× 10−3 = ISe0.7/0.025

⇒ IS = 6.91× 10−16 A

Voltage drop across each diode =1.5

2= 0.75 V.

∴ ID = ISeV /V T = 6.91× 10−16 × e0.75/0.025

= 7.38 mA

IL = 1.5/1.5 = 1 mA

I = ID + IL = 7.39 mA + 1 mA

= 8.39 mA

∴ R = 5− 1.5

8.39 mA= 417 �

Use a small-signal model to find voltage �VO

when the value of the load resistor, RL, changes:

rd = VT

ID= 0.025

7.39= 3.4 �

When load is disconnected, all the current I flowsthrough the diode. Thus

�ID = 1 mA

�VO = �ID × 2rd

= 1× 2× 3.4

= 6.8 mV

With RL = 1 k�,

IL � 1.5 V

1= 1.5 mA

�IL = 0.5 mA

�ID = −0.5 mA

�VO = −0.5× 2× 3.4

= −3.4 mV

With RL = 750 �,

IL � 1.5

0.75= 2 mA

�IL = 1 mA

�ID = −1 mA

�VO = −1× 2× 3.4

= −6.8 mV

With RL = 500 �,

IL � 1.5

0.5= 3 mA

�IL = 2.0 mA

�ID = −2.0 mA

�VO = −2× 2× 3.4

= −13.6 mV


Chapter 4–20

4.57

IL

IL varies from 2 to 7 mA

ID

I

� 1.5 V � �VOvO

To supply a load current of 2 to 7 mA, the currentI must be greater than 7 mA. So I can be only10 mA or 15 mA.

Now let us evaluate �VO for both 10-mA and15-mA sources. For the 10-mA source:

Since IL varies from 2 to 7 mA, the current ID willvarry from 8 to 3 mA.

Correspondingly, the voltage across each diodechanges by �VD where

3

8= e�VD/VT

⇒ �VD = 25 ln

(3

8

)= −24.5 mV

and the output voltage changes by

�VO = 2×�VD = −49 mV

With I = 15 mA, the diodes current changes from13 to 8 mA. Correspondingly, the voltage acrosseach diode changes by �VD where

8

13= e�VD/VT

⇒ �VD = 25 ln

(8

13

)= −12.1 mV

and the output voltage changes by

�VO = 2×�VD = −24.2 mV

which is less than half that obtained with the10-mA supply. Thus, from the point of view ofreducing the change in VO as IL changes, wechoose the 15-mA supply. Note, however, that theprice paid is increased power dissipation.

4.58

IL

ID

D1

D2

I

�5 V

�

�

RL � 150 �

R � 200 �

VO

(a) Iteration #1:

VD = 0.7 V

VO = 2VD = 1.4 V

IL = VO

RL= 1.4

0.15= 9.33 mA

I = 5− VO

R= 5− 1.4

0.2= 18 mA

ID = I − IL = 18− 9.33 = 8.67 mA

Iteration #2:

VD = 0.7+ 0.025 ln

(8.67

10

)= 0.696 V

VO = 1.393 V

IL = 9.287 mA

I = 5− 1.393

0.2= 18.04 mA

ID = 18.04− 9.287 = 8.753 mA

Iteration #3:

VD = 0.7+ 0.025 ln

(8.753

10

)= 0.697

VO = 1.393 V

IL = 9.287

I = 18.04 mA

ID = 8.753

No further iterations are necessary and

VO = 1.39 V

(b) With no load:

Iteration #1:

VD = 0.7 V

VO = 1.4 V


Chapter 4–21

I = 5− 1.4

0.2= 18 mA

ID = I = 18 mA

Iteration #2:

VD = 0.7+ 0.025 ln

(18

10

)= 0.715 V

VO = 1.429 V

I = 17.85 mA

ID = 17.85 mA

Iteration #3:

VD = 0.7+ 0.025 ln

(17.85

10

)= 0.714 V

VO = 1.43 V

I = 17.86 mA

ID = 17.86 mA

No further iterations are warranted and

VO = 1.43 V

(c) VO = 1.39− 0.1 = 1.29 V

IL = 1.29

0.15= 8.6 mA

VD = 1.29

2= 0.645 V

ID = 10× e(0.645−0.7)/0.025

= 1.11 mA

I = IL + ID = 8.6+ 1.11 = 9.71 mA

VSupply = VO + IR = 1.29+ 9.71× 0.2

= 3.232 V

which is a reduction of 1.768 V or −35.4%.

(d) For VSupply = 5+ 1.786 = 6.786 V,

Iteration #1:

VD = 0.7 V

VO = 1.4 V

IL = 9.33 mA

I = 6.768− 1.4

0.2= 26.84

ID = I − IL = 26.84− 9.33 = 17.51 mA

Iteration #2:

VD = 0.7+ 0.025 ln

(17.51

10

)= 0.714 V

VO = 1.428 V

IL = 9.52 mA

I = 26.70 mA

ID = 17.18 mA

Iteration #3:

VD = 0.7+ 0.025 ln

(17.18

10

)= 0.714 V

VO = 1.428 V

No further iterations are needed and

VO = 1.43 V

(e) From the above we see that as VSupply changesfrom 5 V to 3.232 V (a change of −35.4%) theoutput voltage changes from 1.39 V to 1.29 V (achange of −7.19%).

As VSupply changes from 5 V to 6.786 V (a changeof +35.4%) the output voltage changes from 1.39V to 1.43 V (a change of +2.88%).

Thus the worst-case situation occurs when VSupply

is reduced, and

Percentage change in VO per 1% change in

VSupply = 7.19

35.4= 0.2%

4.59 VZ = VZ0 + IZT rz

(a) 10 = 9.6+ 0.05× rz

⇒ rz = 8 �

For IZ = 2IZT = 100 mA,

VZ = 9.6+ 0.1× 8 = 10.4 V

P = 10.4× 0.1 = 1.04 W

(b) 9.1 = VZ0 + 0.01× 30

⇒ VZ0 = 8.8 V

At IZ = 2IZT = 20 mA,

VZ = 8.8+ 0.02× 30 = 9.4 V

P = 9.4× 20 = 188 mW

(c) 6.8 = 6.6+ IZT × 2

⇒ IZT = 0.1 A

At IZ = 2IZT = 0.2 A,

VZ = 6.6+ 0.2× 2 = 7 V

P = 7× 0.2 = 1.4 W

(d) 18 = 17.6+ 0.005× rz

⇒ rz = 80 �


Chapter 4–22

At IZ = 2IZT = 0.01 A,

VZ = 17.6+ 0.01× 80 = 18.4 V

P = 18.4× 0.01 = 0.184 W = 184 mW

(e) 7.5 = VZ0 + 0.2× 1.5

⇒ VZ0 = 7.2 V

At IZ = 2IZT = 0.4 A,

VZ = 7.2+ 0.4× 1.5 = 7.8 V

P = 7.8× 0.4 = 3.12 W

4.60 (a) Three 6.8-V zeners provide 3× 6.8 =20.4 V with 3 ×10 = 30-� resistance. NeglectingR, we have

Load regulation = −30 mV/mA.

(b) For 5.1-V zeners we use 4 diodes to provide20.4 V with 4 ×30 = 120-� resistance.

Load regulation = −120 mV/mA

4.61

Small-signal model

�

�

82 �

8 � �vO�vS ��

From the small-signal model we obtain

�vO

�vS= 8

8+ 82= 8

90

Now �vS = 1.0 V.

∴ �vO = 8

90�vS = 8

90× 1.0

= 88.9 mV

4.62 VZ = VZ0 + IZT rZ

9.1 = VZ0 + 0.02× 10

⇒ VZ0 = 8.9 V

At IZ = 10 mA,

VZ = 8.9+ 0.01× 10 = 9.0 V

At IZ = 50 mA,

VZ = 8.9+ 0.05× 10 = 9.4 V

4.63

IZ

I

�10 V

VO � VZ

(a)

RL

R

IL

I

�11 V

VZ0

rZ

(b)

R

VO � VZ

I

�9 V

VO � VZ

(c)

RL

R

IL0.5 mA


Chapter 4–23

To obtain VO = 7.5 V, we must arrange forIZ = 10 mA (the current at which the zener isspecified).

Now,

IL = VO

RL= 7.5

1.5= 5 mA

Thus

I = IZ + IL = 10+ 5 = 15 mA

and

R = 10− VO

I= 10− 7.5

15= 167 �

When the supply undergoes a change �VS , thechange in the output voltage, �VO, can bedetermined from

�VO

�VS= (RL ‖ rz)

(RL ‖ rz)+ R

= 1.5 ‖ 0.03

(1.5 ‖ 0.03)+ 0.167= 0.15

For �VS = +1 V (10% high), �VO = +0.15 Vand VO = 7.65 V.

For �VS = −1 V (10% low), �VO = −0.15 Vand VO = 7.35 V.

When the load is removed and VS = 11 V, we canuse the zener model to determine VO. Refer toFig. (b). To determine VZ0, we use

VZ = VZ0 + IZT rz

7.5 = VZ0 + 0.01× 30

⇒ VZ0 = 7.2 V

From Fig. (b) we have

I = 11− 7.2

0.167+ 0.03= 19.3 mA

Thus

VO = VZ0 + Irz

= 7.2+ 0.0193× 30 = 7.78 V

To determine the smallest allowable value of RL

while VS = 9 V, refer to Fig. (c). Note thatIZ = 0.5 mA, thus

VZ = VZK � VZ0 = 7.2 V

I = 9− 7.2

0.167= 10.69 mA

IL = I − IZ = 10.69− 0.5 = 10.19 mA

RL = VO

IL= 7.2

10.19= 707 �

VO = 7.2 V

4.649 V ± 1 V

VO

IZ

R

GIVEN PARAMETERS

VZ = 6.8V, rz = 5 �

IZ = 20 mA

At knee,

IZK = 0.25 mA

rz = 750 �

FIRST DESIGN: 9-V supply can easily supplycurrent

Let IZ = 20 mA, well above knee.

∴ R = 9− 6.8

20= 110 �

Line regulation = �VO

�VS= rZ

rZ + R

= 5

5+ 110

= 43.5mV

V

SECOND DESIGN: limited current from 9-Vsupply

IZ = 0.25 mA

VZ = VZK � VZO − calculate VZ0 from

VZ = VZ0 + rZ IZT

6.8 = VZ0 + 5× 0.02

VZ0 = 6.7 V

∴ R = 8− 6.7

0.25= 5.2 k�

LINE REGULATION = �VO

�VS= 750

750+ 5200

= 126mV

V


Chapter 4–24

4.65

IZ

I

VS � �15 V � 10%

VO � VZ

RL

R

IL

VZ = VZ0 + IZT rz

9.1 = VZ0 + 0.009× 40

⇒ VZ0 = 8.74 V

For IZ = 10 mA,

VZ = 8.74+ 0.01× 40 = 9.14 V

IL = 9.14

1 k�= 9.14 mA

I = IZ + IL = 10+ 9.14 = 19.14 mA

R = 15− 9.14

19.4= 306 �

Select R = 300 �

Denoting the resulting output voltage VO, weobtain

I = 15− VO

0.3(1)

IL = VO

1(2)

IZ = VO − VZ0

rz= VO − 8.74

0.04(3)

Since I = IZ + IL, we can use (1)–(3) to obtainVO:

15− VO

0.3= VO − 8.74

0.04+ VO

⇒ VO = 9.15 V

�VO = �VSrz ‖ RL

(rz ‖ RL)+ R

= ±1.5× (0.04 ‖ 1)

(0.04 ‖ 1)+ 0.3

= ±0.17 V

If IL is reduced by 50%, then

IL = 1

2× 9.15

1= 4.6 mA

I = 15− VO

0.3

IZ = VO − 8.74

0.04

15− VO

0.3= VO − 8.74

0.04+ 4.6

⇒ VO = 9.31 V

which is an increase of 0.16 V. When the supplyvoltage is low,

VS = 13.5 V

and RL is at its lowest value, to maintainregulation, the zener current must be at least equalto IZK , thus

IZ = 0.5 mA

VZ = VZK � VZ0 � 8.74

I = 13.5− 8.74

0.3= 15.87 mA

IL = I − IZ = 15.87− 0.5 = 15.37 mA

RL = VZ

IL= 8.74

15.37= 589 �

The lowest value of output voltage = 8.74 V

Line regulation = 170 mV

1.5 V

= 113 mV/V

Load regulation = −(rz ‖ R)

= −(40 ‖ 300) = −35 mV/mA

Or using the results obtained in this problem:

For a reduction in IL of 4.6 mA, �VO = +0.16 V,thus

Load regulation = −160

4.6= −35 mV/mA

4.66 (a) VZT = VZ0 + rzIZT

10 = VZ0 + 7 (0.025)

⇒ VZ0 = 9.825 V


Chapter 4–25

(b) The minimum zener current of 5 mA occurswhen IL = 20 mA and VS is at its minimum of20(1− 0.25) = 15 V. See the circuit below:

20 mA

RL

VS � 15 V

VO

IZIZmin � 5 mA

R

R ≤ 15− VZ0

20+ 5

where we have used the minimum value of VS , themaximum value of load current, and the minimumrequired value of zener diode current, and weassumed that at this current VZ � VZ0. Thus,

R ≤ 15− 9.825+ 7

25

≤ 207 �.

∴ use R = 207 �

(c) Line regulation = 7

207+ 7= 33

mV

V

±25% change in vS ≡ ± 5 V

VO changes by ±5× 33 = ±0.165 mV

corresponding to±0.165

10× 100 = ±1.65%

(d) Load regulation = − (rZ ‖ R)

= −(7 ‖ 207) = −6.77 �

or –6.77 V/A

�VO = −6.77× 20 mA = −135.4 mV

corresponding to −0.1354

10× 100 = −1.35%

(e) The maximum zener current occurs at no loadIL = 0 and the supply at its largest value of

20+ 1

4(20) = 25 V.

VZ = VZ0 + rZ IZ

25 V

207 �

VZ

IL � 0

= 9.825+ 7× 25− VZ

207

207VZ = 207 (9.825)+ 7 (25)− 7VZ

⇒ VZ = 10.32 V

IZmax = 25− 10.32

0.207= 70.9 mA

PZ = 10.32× 70.9

= 732 mW

4.67

�

�

vS R vO��

Using the constant voltage drop model:

1 k�

ideal

R

�

�

vO��

VD

� 0.7 V

v S

(a) vO = vS + 0.7 V, For vS ≤ − 0.7 V

vO = 0, for vS ≥ −0.7 V


Chapter 4–26

slope � 1

�0.7 V

0

vO

vS

(b)

0.7 V

�0.7 V

�10 V

� 10 V

t

vO

vS

u

(c) The diode conducts at an angle

θ = sin−1

(0.7

10

)= 4◦ and stops

at π − θ = 176◦

Thus the conduction angle is π − 2θ

= 172◦ or 3 rad.

vO,avg = −1

2π

π−θ∫θ

(10 sin φ − 0.7) dφ

= −1

2π[−10 cos φ − 0.7φ]π−θ

θ

= −2.85 V

(d) Peak current in diode is

10− 0.7

1= 9.3 mA

(e) PIV occurs when vS is at its the peak andvO = 0.

PIV= 10 V

4.68

R

��

�

vS vO

vD

��

D

iD = ISevD/V T

iD

iD (1 mA)= e[vD−vD(at 1 mA)]/V T

vD − vD(at 1 mA) = VT ln

[iD

1 mA

]

vD = vD(at 1 mA)+ VT ln

[vO/R

1

]vO = vS − vD

= vS − vD (at 1 mA)− VT ln(vO

R

)where R is in k�.

4.69

R � 1 k�

��

�

vS vO��

0.7 V

�2.5

2.51.8

0.70 t

t1 t2 TT4

T2

vS (V)

First find t1 and t2

2.5T

4

= 0.7

t1

⇒ t1 = 0.07 T

t2 = T

2− t1

= T

2− 0.07 T

t2 = 0.43 T

vO(ave.) = 1

T× area of shaded triangle

= 1

T× (2.5− 0.7)×

(T

4− t1

)

= 1

T× 1.8× T

(1

4− 0.07

)

= 0.324 V


Chapter 4–27

4.70

1 k�

ideal

120 Vrms

60 Hz

0.7 V12 : 1

�

�

�

�

�

�

vS vO10 Vrms

vO = 10√

2− 0.7 = 13.44 V

Conduction begins at

10√

2 sin θ = 0.7

θ = sin−1

(0.7

10√

2

)= 2.84◦

= 0.0495 rad

Conduction ends at π − θ .

∴ Conduction angle = π − 2θ = 3.04 rad

The diode conducts for

3.04

2π× 100 = 48.4% of the cycle

vO,avg = 1

2π

π−θ∫θ

(10√

2sinφ − 0.7) dφ

= 4.15 V

iD,avg = vO,avg

R= 4.15 mA

4.71

1 k�10 Vrms

6 : 1

120 Vrms

60 Hz10 Vrms

�

�

�

�D2

D1

0 �

0.7 V

10 V

t� ��

vs

vo

vS , vO (V)

vo

vO = 10√

2− VD = 13.44 V

Conduction starts at θ = sin−1 0.7

10√

2=

2.84◦ = 0.05 rad

and ends at π − θ . Conduction angle =π − 2θ = 3.04 rad in each half cycle. Thus thefraction of a cycle for which one of the two

diodes conduct = 2(3.04)

2π× 100 = 96.8%

Note that during 96.8% of the cycle there will beconduction. However, each of the two diodesconducts for only half the time, i.e., for 48.4% ofthe cycle.

vO,avg = 1

π

π−θ∫θ

(10√

2sinφ − 0.7)dφ

= 8.3 V

iL,avg = 8.3

1 k�= 8.3 mA

4.72

1 k�

VD � 0.7 V

120 Vrms

12 : 1

10 Vrms

��

� � D2

D1

D3

D4

R

vs

Peak voltage across R = 10√

2− 2VD

= 10√

2− 1.4

= 12.74 V

1.4 V

10 2 V

t�

vS

θ = sin−1 1.4

10√

2= 5.68◦ = 0.1 rad

Fraction of cycle that D1 & D2 conduct is

π − 2θ

2π× 100 = 46.8%

Note that D3 & D4 conduct in the other half cycleso that there is 2 (46.8) = 93.6% conductioninterval.

vO,avg = 2

2π

π−θ∫θ

(10√

2sinφ − 2VD) dφ


Chapter 4–28

= 1

π

[−12√

2 cos φ − 1.4φ]π−θ

θ

= 2(12√

2 cos θ)

π− 1.4 (π − 2θ)

π

= 7.65 V

iR,avg = vO,avg

R= 7.65

1= 7.65 mA

4.73

R��

�

�

�

�

�

vS

vS vO120 Vrms

Refer to Fig. 4.24.

For VD � Vs, conduction angle � π , and

vO,avg = 2

πVs − VD = 2

πVs − 0.7

(a) For vO,avg = 10 V

Vs = π

2× 10.7 = 16.8 V

Turns ratio = 120√

2

16.8= 10.1 to 1

(b) For vO,avg = 100 V

Vs = π

2× 100.7 = 158.2 V


2

158.2= 1.072 to 1

4.74 Refer to Fig. 4.25

For 2VD � Vs

VO,avg = 2

πVs − 2VD = 2

πVs − 1.4

(a) For VO,avg = 10 V

10 V = 2

π· Vs − 1.4

∴ Vs = 11.4(π

2

)= 17.9 V


2

17.9= 9.5 to 1

(b) For VO,avg = 100 V

100 V = 2

π· Vs − 1.4

⇒ Vs = 101.4(π

2

)= 159 V


2

159= 1.07 to 1

4.75 120√

2± 10%: 20√

2± 10%

⇒ Turns ratio = 6:1

vS = 20√

2

2± 10%

PIV= 2Vs − VD

= 2× 20√

2

2× 1.1− 0.7

= 30.4 V

Using a factor of 1.5 for safety, we select a diodehaving a PIV rating of approximately 45 V.

4.76 The circuit is a full-wave rectifier withcenter tapped secondary winding. The circuit canbe analyzed by looking at v+O and v−O separately.

D1

D3

�

�

�

�

vS

vS

D4 R

D2

�

�

�

�

vS

vS

vO�

vO, avg = 1

2π

∫(VSsinφ − 0.7) dφ = 12

= 2Vs

π− 0.7 = 12

where we have assumed Vs � 0.7 V and thus theconduction angle (in each half cycle) is almost π .

Vs = 12+ 0.7

2π = 19.95 V


Chapter 4–29

Thus voltage across secondary winding

= 2VS � 40 V

Looking at D4,

PIV= VS − V−O

= VS + (VS − 0.7)

= 2VS − 0.7

= 39.2 V

If choosing a diode, allow a safety margin bymoving a factor of 1.5, thus

PIV� 60 V

4.77

1 k�120 Vrms

12 : 1

10 Vrms � VS

R

RC

�

�

�

�

�

�

vO

Vp�VD

VpVr

(i) Vr∼= (Vp − VD

) T

CR[Eq. (4.28)]

0.1(Vp − VD

) = (Vp − VD

) T

CR

C = 1

0.1× 60× 103 = 166.7 μF

(ii) For

Vr = 0.01(Vp − VD

) =(Vp − VD

)T

CR

C = 1667 μF

(a) (i) vO, avg = Vp − VD − 1

2VΓ

= 10√

2− 0.7− 1

2

(10√

2− 0.7)

0.1

=(

10√

2− 0.7)(

1− 0.1

2

)

= 12.77 V

(ii) vO, avg =(

10√

2− 0.7)(

1− 0.01

2

)

= 13.37 V

(b) (i) Using Eq (4.30), we have the conductionangle =ω�t ∼=

√2Vr/

(Vp − VD

)

=√

2× 0.1(Vp − 0.7

)(Vp − 0.7

)= √0.2

= 0.447 rad

∴ Fraction of cycle for

conduction = 0.447

2π× 100

= 7.1%

(ii) ω�t �√

2× 0.01

(Vp − 0.7

)Vp − 0.7

= 0.141 rad

Fraction of cycle = 0.141

2π× 100 = 2.24%

(c) (i) Use Eq (4.31):

iD,avg = IL

⎛⎝1+ π

√2(Vp − VD

)Vr

⎞⎠

= vO,avg

R

(1+ π

√2(Vp − VD

)0.1

(Vp − VD

))

= 12.77

103

(1+ π

√2

0.1

)

= 192 mA

(ii) iD,avg = 13.37

103

(1+ π

√200

)= 607 mA

(d) Adapting Eq. (4.32), we obtain

(i) iD,peak = IL

⎛⎝1+ 2π

√2(Vp − VD

)Vr

⎞⎠

= 12.77

103

(1+ 2π

√2

0.1

)

= 371 mA

(ii) iD,peak = 13.37

103

(1+ 2π

√2

0.01

)

= 1201 mA � 1.2 A

4.78 (i) Vr = 0.1(Vp − VD

) =(Vp − VD

)2fCR

The factor of 2 accounts for discharge occurring

only during half of the period, T/2 = 1

2f.


Chapter 4–30

C = 1

(2fR) 0.1= 1

2 (60) 103 × 0.1= 83.3 μF

(ii) C = 1

2 (60)× 103 × 0.01= 833 μF

(a) (i) VO = Vp − VD − 1

2Vr

= (Vp − VD

) (1− 0.1

2

)

= (13.44)

(1− 0.1

2

)

= 12.77 V

(ii) VO = (13.44)

(1− 0.01

2

)= 13.37 V

(b) (i) Fraction of cycle = 2ω�t

2π× 100

=√

2Vr/(Vp − VD

)π

× 100

= 1

π

√2 (0.1)× 100 = 14.2%

(ii) Fraction of cycle = 2√

2 (0.01)

2π× 100

= 4.5%

(c) Use Eq. (4.34):

(i) iD, avg = IL

(1+ π

√Vp − VD

2Vr

)

= 12.77

1

(1+ π

√1

2 (0.1)

)= 102.5 mA

(ii) iD, avg = 13.37

1

(1+ π

1√2 (0.01)

)= 310 mA

(d) Use Eq. (4.35):

(i) iD = IL

(1+ 2π

1√2 (0.1)

)= 192 mA

(ii) iD = IL

(1+ 2π

1√0.02

)= 607 mA

4.79 (i) Vr = 0.1(Vp − VD × 2

) = Vp − 2VD

2fCR

C =(Vp − 2VD

)(Vp − 2VD

) 1

2 (0.1) fR= 83.3 μF

(ii) C = 1

2 (0.01) fR= 833 μF

(a) VO = Vp − 2VD − 1

2Vr

(i) VO = Vp − 2VD − 1

2

(Vp − 2VD

)× 0.1

= (Vp − 2VD

)× 0.95

= (10√

2− 2× 0.7)× 0.95 = 12.1 V

(ii) VO = (10√

2− 2× 0.7)× 0.995 = 12.68 V

(b) (i) Fraction of cycle = 2ω�t

2π× 100

=√

2 (0.1)

π× 100 = 14.2%

(ii) Fraction of cycle =√

2 (0.01)

π× 100 = 4.5%

(c) (i) iD, avg = 12.1

1

(1+ π

√1

0.2

)= 97 mA

(ii) iD, avg = 12.68

1

(1+ π/

√0.02

)= 249 mA

(d) (i) iD = 12.1

1

(1+ 2π

√1

0.2

)= 182 mA

(ii) iD = 12.68

1

(1+ 2π

√1

0.02

)= 576 mA

4.80

200 �120 Vrms

0.7 V

vS

RC

�

�

�

�

�

�

60 Hz vO

VO = 12 V± 1 V (ripple)

RL = 200 �

(a) VO = Vp − VD − 1

⇒ Vp = 13+ 0.7 = 13.7 V

Vrms = 13.7√2= 9.7 V

(b) Vr = Vp − VD

fCR

2 = 13.7− 0.7

60× C × 200

⇒ C = 13

2× 60× 200= 542 μF


Chapter 4–31

Vp�VD

Vp

�VpPIV

t

(c) When the diode is cut off, the maximumreverse voltage across it will occur whenvS = −Vp. At this time, vO = VO and themaximum reverse voltage will be

Maximum reverse voltage = VO + Vp

= 12+ 13.7 = 25.7 V

Using a factor of safety of 1.5 we obtain

PIV = 1.5× 25.7

= 38.5 V

(d) iDav = IL

[1+ π

√2(Vp − VD)

Vr

]

= VO

RL

[1+ π

√2(Vp − VD)

Vr

]

= 12

0.2

[1+ π

√2(13.7− 0.7)

2

]

= 739 mA

(e) iDmax = IL

[1+ 2π

√2(Vp − VD)

Vr

]

= 12

0.2

[1+ 2π

√2(13.7− 0.7)

2

]

= 1.42 A

4.81

0.7 V

0.7 V

120 Vrms

60 Hz

RC�

�Vs

�

�Vs

�

�

D1

D2

vO

(a) VO = Vp − VD − 1

⇒ Vp = VO + VD + 1 = 13+ 0.7 = 13.7 V

Vrms = 13.7√2= 9.7 V

This voltage appears across each half of thetransformer secondary. Across the entiresecondary we have 2× 9.7 = 19.4 V (rms).

PIV

(b) Vr = Vp − VD

2fCR

2 = 13.7− 0.7

2× 60× 200× C

⇒ C = 12

2× 2× 60× 200= 271 μF

(c) Maximum reverse voltage across D1 occurswhen vS = −Vp. At this point vO = VO. Thusmaximum reverse voltage = VO + Vp =12+ 13.7 = 25.7. The same applies to D2.

In specifying the PIV for the diodes, one usuallyuses a factor of safety of about 1.5,

PIV = 1.5× 25.7 = 38.5 V

(d) iDav = IL

[1+ π

√Vp − VD

2 Vr

]

= 12

0.2

[1+ π

√13.7− 0.7

2× 2

]

= 399 mA

(e) iDmax = IL

[1+ 2π

√Vp − VD

2 Vr

]

= 12

0.2

[1+ 2π

√13.7− 0.7

2× 2

]

= 739 mA

4.82

120 Vrms

��

�

��

�

vS vO

D2

D1

D3

D4

R

C

60 Hz

(a) VO = Vp − 2VD − 1

⇒ Vp = VO+2VD+1 = 12+2×0.7+1 = 14.4 V


Chapter 4–32

Vrms = 14.4√2= 10.2 V

(b) Vr = Vp − 2 VD

2fCR

⇒ C = 14.4− 1.4

2× 2× 60× 200= 271 μF

(c) The maximum reverse voltage across D1

occurs when Vs = −Vp = −14.4 V. At this timeD3 is conducting, thus

Maximum reverse voltage = −Vp + VD3

= −14.4+ 0.7 = −13.7 V

The same applies to the other three diodes. Inspecifying the PIV rating for the diode we use afactor of safety of 1.5 to obtain

PIV = 1.5× 13.7 = 20.5 V

(d) iDav = IL

[1+ π

√Vp − 2 VD

2 Vr

]

= 12

0.2

[1+ π

√14.4− 1.4

2× 2

]

= 400 mA

(e) iDmax = IL

[1+ 2π

√Vp − 2 VD

2 Vr

]

= 12

0.2

[1+ 2π

√14.4− 0.7

2× 2

]

= 740 mA

4.83

(a)

C R

�

�

�

�

100 F 100 �vI vO

vI

Vr vO

�12 V

12 V11.3 V

t

vI

��

(b)

T 1 ms

VrVr

12 V

11.311.3 �

� 10.2 V

� 1.13 V

0

�t

}

T4

(c)

During the diode’s off interval (which is almostequal to T ) the capacitor discharges and itsvoltage is given by

vO(t) = 11.3 e−t/CR

where C = 100 μF and R = 100 �, thus

CR = 100× 10−6 × 100 = 0.01 s

At the end of the discharge interval, t � T and

vO = 11.3 e−T/CR

Since T = 0.001 s is much smaller than CR,

vO � 11.3

(1− T

CR

)

The ripple voltage Vr can be found as

Vr = 11.3− 11.3

(1− T

CR

)

= 11.3T

CR= 11.3× 0.001

0.01= 1.13 V

The average dc output voltage is

vO = 11.3− Vr

2= 11.3− 1.13

2= 10.74 V

To obtain the interval during which the diodeconducts, t, refer to Fig. (c).

12

T /4= Vr

t

⇒ t = Vr × (T/4)

12= 1.13× 1

12× 4

= 23.5 μs

Now, using the fact that the charge gained by thecapacitor when the diode is conducting is equal tothe charge lost by the capacitor during itsdischarge interval, we can write

iCav ×t = C Vr

⇒ iCav = C Vr

t= 100× 10−6 × 1.13

23.5× 10−6= 4.8 A

iDav = iCav + iLav

where iLav is the average current through R duringthe short interval t. This is approximately11.3

R= 11.3

100= 0.113 A. Thus


Chapter 4–33

iDav = 4.8+ 0.113 = 4.913 A

Finally, to obtain the peak diode current, we use

iDmax = iCmax + iLmax

= Cdv I

dt+ 11.3

R

= C × 12

T/4+ 11.3

R

= 100× 10−6 × 12× 4

1× 10−3+ 11.3

100

= 4.8+ 0.113 = 4.913 A

which is equal to the average value. This is aresult of the linear v I which gives rise to aconstant capacitor current during the diodeconduction interval. Thus iCmax = iCav = 4.8 A.Also, the maximum value of iL is approximatelyequal to its average value during the shortinterval t.

4.84 Refer to Fig. P4.76 and let a capacitor C beconnected across each of the load resistors R. Thetwo supplies v+O and v−O are identical. Each is afull-wave rectifier similar to that of thetapped-transformer circuit. For each supply,

VO = 12 V

Vr = 1 V (peak to peak)

Thus

vO = 12± 0.5 V

It follows that the peak value of vS must be12.5+ 0.7 = 13.2 V and the total rms voltageacross the secondary will be

= 2× 13.2√2= 18.7 V (rms)

Transformer turns ratio = 120

18.7= 6.43:1

To deliver 100-mA dc current to each load,

R = 12

0.1= 120 �

Now, the value of C can be found from

Vr = Vp − 0.7

2fCR

1 = 12.5

2× 60× C × 120

⇒ C = 868 μF

To specify the diodes, we determine iDav and iDmax,

iDav = IL(1+ π√

(Vp − 0.7)/2 Vr )

= 0.1(1+ π√

12.5/2 )

= 785 mA

iDmax = IL(1+ 2π√

(Vp − 0.7)/2 Vr )

= 0.1(1+ 2π√

12.5/2 )

= 1.671 A

To determine the required PIV rating of eachdiode, we determine the maximum reversevoltage that appears across one of the diodes, sayD1. This occurs when vS is at its maximumnegative value −Vp. Since the cathode of D1 willbe at +12.5 V, the maximum reverse voltageacross D1 will be 12.5+ 13.2 = 25.7 V. Using afactor of safety of 1.5, then each of the fourdiodes must have

PIV = 1.5× 25.7 = 38.6 V

4.85 Refer to Fig. P4.85. When v I is positive, vA

goes positive, turning on the diode and closing thenegative feedback loop around the op amp. Theresult is that v− = v I , vO = 2v− = 2v I , andvA = vO + 0.7. Thus

(a) v I = +1 V, v− = +1 V, vO = +2 V, andvA = +2.7 V.

(b) v I = +3 V, v− = +3 V, vO = +6 V, andvA = +6.7 V.

When v I goes negative, vA follows, the diodeturns off, and the feedback loop is opened. The opamp saturates with vA = −13 V, v− = 0 V andvO = 0 V. Thus

(c) v I = −1 V, v− = 0 V, vO = 0 V, andvA = −13 V.

(d) v I = −3 V, v− = 0 V, vO = 0 V, andvA = −13 V.

Finally, if v I is a symmetrical square wave of1-kHz frequency, 5-V amplitude, and zeroaverage, the output will be zero during thenegative half cycles of the input and will equaltwice the input during the positive half cycles.See figure.

vO

vI

�10 V

�5 V

�5 V1 ms

t0

Thus, vO is a square wave with 0-V and +10-Vlevels, i.e. 5-V average and, of course, the samefrequency (1 kHz) as the input.


Chapter 4–34

4.86 v I > 0: D1 conducts and D2 cutoff

v I < 0: D1 cutoff,

D2 conducts ∼ vO

v I= −1

slope � �1

vO

vI

(a) v I = +1 V

vO = 0 V

vA = −0.7 V

Keeps D2 off so no current flows through R

⇒ v− = 0 V

Virtual ground as feedback loop is closedthrough D1

(b) v I = +3 V

vO = 0 V

vA = −0.7 V

v− = 0 V

(c) v I = −1 V

vO = +1 V

vA = 1.7 V

v− = 0 V

∼ Virtual ground as negative feedback loop isclosed through D2 and R.

(d) v I = −3 V⇒ vO = +3 V

vA = +3.7 V

v− = 0 V

4.87 (a) See figure (a) on next page. Forv I ≤ 3.5 V, i = 0 and vO = v I . At v I = 3.5 V, thediode begins to conduct. At vO = 3.7 V, the diodeis conducting i = 1 mA and thus

v I = vO + i × 1 k� = 4.7 V

For v I > 4.7 V the diode current increases but thediode voltage remains constant at 0.7 V, thus vO

flattens and vO vs. v I becomes a horizontal line.

In practice, the diode voltage increases slowlyand the line will have a small nonzero slope.

(b) See figure (b) on next page. Here vO = v I forv I ≥ 2.5 V. At v I = 2.5 V, vO = 2.5 V and thediode begins to conduct. The diode will beconducting 1 mA and exhibiting a drop of 0.7 atvO = 2.3 V. The corresponding value of v I

v I = vO − iR = 2.3− 1× 1 = +1.3 V

As v I decreases below 1.3 V, the diode currentincreases, but the diode voltage remains constantat 0.7 V. Thus vO flattens at about 2.3 V.

(c) See figure (c) on next page. For v I ≤ −2.5 V,the diode is off, and vO = v I . At v I = −2.5 V thediode begins to conduct and its current reaches 1mA at v I = −1.3 V (corresponding to vO = −2.3V). As v I further increases, the diode currentincreases but its voltage remains constant at 0.7 V.Thus vO flattens, as shown.

(d) See figure (d) on next page.

4.88

vI vo

R = 0.5 k�

D2

(a)

D1

�3 V

�3 V

i

From Fig. (a) we see that for−3.5 V ≤ v I ≤ +3.5 V, diodes D1 and D2 will becut off and i = 0. Thus, vO = v I . For v I ≥ +3.5V, diode D1 begins to conduct and its voltagereaches 0.7 V (and thus vO = +3.7 V) ati = 1 mA. The corresponding value of v I is

v I = vO − iR

v I = 3.7+ 1× 0.5 = +4.2 V

For v I ≥ 4.2 V, the voltage of diode D1 remains0.7 V and vO saturates at +3.7 V.

A similar description applies for v I ≤ −3.5 V.Here D2 conducts at v I = −3.5 V and its voltagebecomes 0.7 V, and hence vO = −3.7 V,at i = 1 mA (in the direction into v I )at v I = −4.2 V. For v I ≤ −4.2 V, vO = −3.7 V.


Chapter 4–35

These figures belong to Problem 4.87.

3.5

3.5

3.7

4.7slope � 1

(a)

vO (V)

vI (V)vI vO

D

R � 1 k�

�3 V

i

2.5

2.3

1.3 2.5

(b)

slope � 1

vO (V)

vI (V)

vI vO

D

R � 1 k�

�3 V

i

(c)

slope � 1

vO (V)

vI (V)

vI vO

D

R � 1 k�

�3 V

i

�2.5

�2.5

�2.3

�1.3

(d)

slope � 1

vO (V)

vI (V)vI vO

D

R � 1 k�

�3 V

�4.7 �3.5

�3.5

�3.7

i


Chapter 4–36

This figure belongs to Problem 4.88, part b.

�3.7

�3.50�3.5

�3.5 (Not to scale)

slope � �1

�3.7

�4.2�4.2

�3.5

vO (V)

vI (V)

D1 OFFD2 ON

D1 and D2 OFF

(b)

D1 OND2 OFF

Figure (b) shows a sketch of the transfercharacteristic of this double limiter.

4.89 See figure.

vI vO

�3 V

D1 D2

(a)R � 0.5 k�

vI (V)

vO (V)

�3.7�3.5

�2.5 (Not to scale)

(b)

slope � �1

�2.3

�1.8

D1 OND2 OFF

D1 &D2 OFF

D1 OFFD2 ON

�2.5 �3.5 �4.2

4.90

D4 D1

D2 D3

1 k�

Z

(a)

vI vO

The limiter thresholds and the output saturationlevels are found as 2× 0.7+ 6.8 = 8.2 V. Thetransfer characteristic is given in Fig. (b). Seefigure on next page.

4.91vI vO

D2

1 k�

D1


Chapter 4–37

This figure belong to Problem 4.90, part b.

ZZ

All diodes and zener are OFF

Diodes have 0.7 V drop at 1 mA current

∴ For diode D1

iD

1 mA= e(vO−0.7)/VT

iD = 1× 10−3 e(vO−0.7)/VT

vO = 0.7+ VT ln

(iD

1 mA

)

v I = vO + iD × 1 k�

Using these equations, calculate v I for thedifferent values of vO. For D2,v I = vO − iD × 1 k�

vO (V)

vI (V)1�1�2 2

0.8 to 55.4

�0.8

It is a soft limiter with a gain K � 1 andL+ � 0.7 V, L− � −0.7 V

4.92

(a)

�vO

�

� �

10 k�

vI


Chapter 4–38

(b)

�

vOvI

�

� �

10 k�

(c)

�

vOvI

�

� �

10 k�

4.93

vI

�

�

�1.5 V 1.5 V vO

�

�

R

1 k�

In the nonlimiting region

vO

v I= 1000

1000+ R≥ 0.94

R ≤ 63.8 �

4.94

VA

I

VB

VC

D3

5 k�

1 k�

D1

I1 D4D2

I2

Figure 1

When VA > 0, D1 and D2 are cut off and D3 andD4 conduct a current I2. Since the diodes are0.1-mA devices, the current I2 is related to thediode voltage VD as follows:

I2 = 0.1× e(VD − 0.7)/0.025, mA (1)

The voltage VC is given by

VC = I2 × 1 k� = I2, V (2)

where I2 is in mA, and the voltage VB is given by

VB = 2VD + VC (3)

and the voltage VA is given by

VA = VB + I × 5k�

VA = VB + 5I2 (4)

Equations (1), (2), (3), and (4) can be used to findVB and VC versus VA. We start with a value forVD, use (1) to determine I2, use (2) to determineVC , use (3) to determine VB, and finally use (4) todetermine VA. The results are given in Table 1.

Table 1

VD3, VD4 I2 (mA) VC (V) VB = VC VA (V)(V) +VD3+

VD4

0.4 � 0 0 0.8 0.8

0.5 0.00003 � 0 1.0 1.0

0.6 0.002 0.002 1.202 1.212

0.7 0.1 0.1 1.5 2.0

0.73 0.332 0.332 1.792 3.452

0.735 0.406 0.406 1.876 3.91

0.74 0.495 0.495 1.975 4.45

0.745 0.605 0.605 2.095 5.12

For VA < 0, D3 and D4 are cutoff, I2 = 0, VC = 0,and D1 and D2 are conducting a current I1,

I1 = 0.1 e(VD−0.7)/0.025, mA (5)

The voltage VB is given by

VB = −2VD (6)

and the voltage VA is

VA = VB − 5I1 (7)

Equations (5)–(7) can be used to obtain VB versusVA for negative values of VA. The results aregiven in Table 2.

Table 2

VD1, VD2 I 1 VB (V) VA

(V) (mA) (V)

0.4 � 0 – 0.8 – 0.8

0.5 � 0 –1.0 –1.0

0.6 0.002 –1.2 –1.21

0.7 0.1 –1.4 –1.9

0.73 0.332 –1.46 –3.12

0.74 0.495 –1.48 –3.955

0.75 0.739 –1.5 –5.20


Chapter 4–39

Figure 2 shows plots for VB and VC versus VA

using the data in Tables 1 and 2. Finally, Figure 3shows the waveforms obtained at B and C when a5-V peak, 100-Hz sinusoid is applied at A.

2

1

�1

�1 10 2 3 4 5VA (V)

VC

VB

VB, VC (V)

VC

�2�3�4�5

�2

Figure 2

5

4

32

1

�1t

VB

VA

VA, VB, VC(V)

VB

VC

�2

�3

�4

�5

Figure 3

4.95 Refer to the circuit in Fig. P4.95. Forv I > 0, D2 and D3 are cut off, and the circuitreduces to that in Fig. 1.

vI vO

1 k�

3 k�

�1 V

D1i1i1

Figure 1

Now, for v I < 1.5 V, diode D1 will be off, i1 � 0,and

vO = v I

As v I exceeds 1.5 V, diode D1 will turn on and

vD1 = 0.7+ 0.025 ln

(i1

1

)(1)

vO = 1+ vD1 + i1 × 1 = 1+ vD1 + i1 (2)

v I = vO + 3i1 (3)

where i1 is in mA and vD1, vO and v I are in volts.Using these relationships we obtain:

i1 (mA) v D1 (V) v O (V) v I (V)

0.01 0.584 1.594 1.625

0.1 0.642 1.742 2.042

0.2 0.660 1.860 2.460

0.5 0.682 2.182 3.682

1.0 0.700 2.700 5.700

1.5 0.710 3.210 7.710

1.8 0.715 3.515 8.915

2.0 0.717 3.717 9.717

For v I < 0, D1 will be cut off and the circuitreduces to that in Fig. 2.

vI vO

1 k�

3 k�

D2

D3

i2

i3

i4

i2

�2 V

Figure 2

Here, for v I < −2.5, D2 will be cut off, i2 � 0and D3 also will be cut off, and

vO = v I

As v I reduces below about −2.5 V, D2 begins toconduct and eventually D3 also conducts. Thedetails of this segment of the vO − v I

characteristic can be obtained using the followingrelationships:

vD3 = 0.7+ 0.025 ln

(i3

1

)

i4 = vD3

1 k�= vD3

i2 = i3 + i4

vD2 = 0.7+ 0.025 ln

(i2

1

)

vO = −2− vD3 − vD2

v I = vO − 3i2


Chapter 4–40

This figure belongs to Problem 4.95, part c.

10987654321

4

3

2

1 slope � 1

All diodes are

cutoff

slope 14

D1 has an almost constantvoltage and the 1 k� andthe 3 k� form a voltage

divider

D2 and D3 are conducting,

limiting vO �2�1.4 � �3.4 V

vt (V)

vO (V)

�1

�2

�3

�3.6

�1�2�3�4�5�6�7�8�9�10

slope 0.013

0

Figure 3

In all these equations, currents are in mA andvoltages are in volts. The numerical resultsobtained are as follows:

i3 v D3, i4 i2 v D2 v O v I

� 0 0.01 0.01 0.585 −2.60 −2.63

� 0 0.05 0.05 0.625 −2.68 −2.85

� 0 0.1 0.1 0.642 −2.74 −3.04

� 0 0.2 0.2 0.660 −2.86 −3.46

� 0 0.3 0.3 0.670 −2.97 −3.87

� 0 0.4 0.4 0.677 −3.08 −4.28

� 0 0.5 0.5 0.683 −3.18 −4.68

0.01 0.585 0.595 0.687 −3.28 −5.07

0.1 0.642 0.742 0.693 −3.35 −5.56

0.2 0.660 0.860 0.696 −3.36 −5.94

0.5 0.682 1.182 0.704 −3.38 −6.93

1.0 0.700 1.700 0.713 −3.41 −8.51

1.5 0.710 2.210 0.720 −3.43 −10.06

The complete transfer characteristic vO versus v I

can be plotted using the data in the tables above.The result is displayed in Fig. 3.

Slopes: For v I near +10 V the slope isapproximately determined by the voltage dividercomposed of the 1 k� and the 3 k�,

Slope = 1

3+ 1= 0.25 V/V

For v I near −10 V, the slope is approximatelygiven by

Slope � − rd2 + rd3

rd2 + rd3 + 3 k�

From the table above,

i3 � 1.5 mA ⇒ rd3 = 25

1.5= 16.7 �

i2 � 2.21 mA ⇒ rd2 = 25

2.21= 11.3 �

Thus

Slope = 11.3+ 16.7

11.3+ 16.7+ 3000= 0.009 V/V

which is reasonably close to the value found fromthe graph.


Chapter 4–41

4.96

5 2

0 tt

vI

C

�

�

DvI

�

�

vO

�5 2

0

vO

From the figure we see that

vOav = −5√

2 = −7.07 V

4.97

(a)�10 V

�10 V

(b)�20 V

0 V

(c)0 V

�20 V

(d)0 V

�20 V

(e)�10 V

�10 V

(f) Here there are two different time constantsinvolved. To calculate the output levels, we shallconsider the discharge and charge wave forms.

During T1, vO = V1e−t/RC

At = T1 = T , vO = V′1

= V1e−T/RC

tV 1

V 2

V1

V20

T1 T2

where for T CR

V′1 � V1(1− T/CR) = V1(1− α)

where α 1

During the interval T2, we have

|vO| = |V2| e−t/(CR/2)

At the end of T2, t = T , and vO = |V ′2|where

|V ′2| = |V2| e−T/(CR/2)

� |V2|(

1− T

RC/2

)= |V2| (1− 2α)

Now

V′1 + |V2| = 20⇒ V1 + |V2| − αV 1 = 20 (1)

and∣∣∣V ′2∣∣∣+ V1 = 20⇒ V1 + |V2| − 2α |V2| = 20 (2)

From (1) and (2) we find that

V1 = 2|V2|Then using (1) and neglecting αV1 yields

3 |V2| = 20⇒ |V2| = 6.67 V

V1 = 13.33 V

The result is

�13.33 V

�6.67 V

(g)

�18 V

�2 V

(h) Using a method similar to that employed forcase (f) above, we obtain

�13.33 V

�6.67 V

Documents

Exercise 4–1 Ex: 4.1 I 0, the (c)qli/ECE333/Chapter 04 ISM.pdf · 2016-02-27 · SEDRA-ISM: “E-CH04” — 2014/11/12 — 11:02 — PAGE1—#1 Exercise 4–1 Ex: 4.1 Refer to