Microelectronic circuit design 3rd edition by r jaeger solutions mmzzhh

Thus silicon has an extra electron and will act as a donor impurity.. Thus germanium has one extra electron and will act as a donor impurity.. n-type silicon p-type silicon Si02Photoresi

Electronic door bell

Electronic gas pump

Pagers Personal computer Personal planner/organizer (PDA) Radar detector

Broadcast Radio (AM/FM/Shortwave) Razor

Satellite radio receiver Security systems Sewing machine Smoke detector Sprinkler system Stereo system Amplifier

Receiver

Stud sensor Talking toys Telephone Telescope controller Thermostats Toy robots Traffic light controller

TV receiver & remote control Variable speed appliances Blender Drill Mixer

Fan Vending machines Video game controllers Wireless headphones & speakers Wireless thermometer

Workstations Electromechanical Appliances*

Air conditioning and heating systems Clothes washer and dryer

0.1548= 6.46 years

1.6 F = 8.00x10−0.05806 2020−1970( )μm =10 nm

No, this distance corresponds to the diameter of only a few atoms Also, the wavelength of the

radiation needed to expose such patterns during fabrication is represents a serious problem

1.7

From Fig 1.4, there are approximately 600 million transistors on a complex Pentium IV

microprocessor in 2004 From Prob 1.4, the number of transistors/μP will be 8.85 x 1010

in

2020 Thus there will be the equivalent of 8.85x1010/6x108 = 148 Pentium IV processors

A 4 digit readout ranges from 0000 to 9999 and has a resolution of 1 part in 10,000 The

number of bits must satisfy 2B ≥ 10,000 where B is the number of bits Here B = 14 bits

1.25 The Thévenin equivalent resistance is found using the same approach as Problem 1.24,

1.26 (a)

R

βi v

R

βi v

6 kHz

1.41 Band-pass amplifier

f20

v O()t =10x5sin 2000( πt)+10x3cos 8000( πt)+ 0x3cos 15000( πt)

v O()t = 50sin 2000[ ( πt)+ 30cos 8000( πt) ] volts

1.44

v O()t = 20x0.5sin 2500( πt)+ 20x0.75cos 8000( πt)+ 0x0.6cos 12000( πt)

v O()t = 10.0sin 2500[ ( πt)+15.0cos 8000( πt) ] volts

1.45 The gain is zero at each frequency: vo(t) = 0

(a)

5V 1( −.05)≤ V ≤ 5V 1+ 05( ) or 5.75V≤ V ≤ 5.25V

V = 5.30 V exceeds the maximum range, so it is out of the specification limits

(b) If the meter is reading 1.5% high, then the actual voltage would be

V meter =1.015V act or Vact = 5.30

1.015= 5.22V which is within specifications limits

V2+

V

1.56 For one set of 200 cases using the equations in Prob 1.53

V =10 * 0.95 + 0.1* RAND()( ) R1= 22000 * 0.9 + 0.2 * RAND()( )

R1= 4700 * 0.9 + 0.2 * RAND()( ) R3 =180000 * 0.9 + 0.2 * RAND()( )

1.57 For one set of 200 cases using the Equations in Prob 1.54:

E BT

3

1062.8exp

For silicon, B = 1.08 x 1031 and EG = 1.12 eV:

cm A Q

j

2

10/

6 7

sec104

0

cm

MA cm

A x cm cm

C Qv

V cm

x

V E

K−3cm−6, k = 8.62x10-5eV/K and EG=1.12eV

This is a transcendental equation and must be solved numerically by iteration Using the HP

solver routine or a spread sheet yields T = 2701 K Note that this temperature is far above the

melting temperature of silicon

eV/K and EG=1.12eV

Using MATLAB as in Problem 2.5 yields T = 316.6 K

2.16

Si B

(a) Gallium is from column 3 and silicon is from column 4 Thus silicon has an extra electron

and will act as a donor impurity

(b) Arsenic is from column 5 and silicon is from column 4 Thus silicon is deficient in one

electron and will act as an acceptor impurity

2.18

Since Ge is from column IV, acceptors come from column III and donors come from column

V (a) Acceptors: B, Al, Ga, In, Tl (b) Donors: N, P, As, Sb, Bi

2.19

(a) Germanium is from column IV and indium is from column III Thus germanium has one

extra electron and will act as a donor impurity

(b) Germanium is from column IV and phosphorus is from column V Thus germanium has

one less electron and will act as an acceptor impurity

cm

A j

22 2

3 14

3 11 3

16 16

16

10502104

1010

4

102210

410105

/cm x

x

p

n | n /cm x

N

N

p

/cm x

n /cm

x x

N : N

N

N

i D

A

i D

No, the result is incorrect because of loss of significant digits

within the calculator It does not have enough digits

2.27

(a) Since boron is an acceptor, NA = 6 x 1018/cm3 Assume ND = 0, since it is not specified

The material is p-type

3 3

18

6 20 2

3 18

i 3

18 3

10

7.1610

6

10

and 10

6

So

2n

>>

/106

and10

re, temperatu

room

At

/cm /cm

x

/cm p

n n /cm

x

p

cm x

N N /cm

n

i

D A i

17

6 20 2

3 17

i 3

17 3

10 i

33310

3

10

and 10

3

So

2n

>>

/103

and/

10nre, temperatu

room

At

/cm /cm

x

/cm n

n p /cm

x

n

cm x

N N cm

i

A D

(a) Arsenic is a donor, and boron is an acceptor ND = 2 x 1018/cm3, and NA = 8 x 1018/cm3

Since NA > ND, the material is p-type

3 3

18

6 20 2

3 18

i 3

18 3

10 i

71610

6

10

and 10

6

So

2n

>>

/106

and/

10nre, temperatu

room

At

(b)

/cm /cm x

/cm p

n n /cm

x

p

cm x

N N cm

i

D A

(a) Phosphorus is a donor, and boron is an acceptor ND = 2 x 1017/cm3, and NA = 5 x 1017/cm3

Since NA > ND, the material is p-type

3 3

17

6 20 2

3 17

i 3

17 3

10

33310

3

10

and 10

3

So

2n

>>

/103

and10

re, temperatu

room

At

(b)

/cm /cm

x

/cm p

n n /cm

x

p

cm x

N N /cm

n

i

D A i

ND = 4 x 1016/cm3 Assume NA = 0, since it is not specified

Based upon the value of its resistivity, the material is an insulator However, it is not intrinsic

because it contains impurities Addition of the impurities has increased the resistivity

2.3 x 10 15 1330 3.06 x 10 18

Yes, by adding equal amounts of donor and acceptor impurities the mobilities are reduced, but

the hole and electron concentrations remain unchanged See Problem 2.37 for example

However, it is physically impossible to add exactly equal amounts of the two impurities

To change the resistivity to 0.25 ohm-cm:

Additional acceptor concentration = 1.1x1017- 1.7x1016 = 9.3 x 1016/cm3

(b) If donors are added:

So ND = 4.1 x 1016/cm3 must be added to change achieve a resistivity of 0.25 ohm-cm The

silicon is converted to n-type material

An n-type ion implantation step could be used to form the n+ region following step (f) in Fig

2.17 A mask would be used to cover up the opening over the p-type region and leave the

opening over the n-type silicon The masking layer for the implantation could just be

photoresist

n-type silicon

p-type silicon

Si02Photoresist

Structure after exposure and

development of photoresist layer

Mask

n-type silicon p-type silicon

Structure following ion implantation of n-type impurity

n +

Ion implantation

Side view

Top View Mask for ion implantation

06 -1.003E+04 9.426E-04 9.992E-

16 -1.003E+04

0.8 0.6

0.4 0.2

which is the equation of a straight line with slope 1/nVT and x-axis intercept at -ln (IS) The

values of n and IS can be found from any two points on the line in the figure: e g iD = 10-4 A

for vD = 0.60 V and iD = 10-9 A for vD = 0.20 V Then there are two equations in two

1.039 7.606E-15

V D I D -Measured I D -Calculated Squared Error

Total Squared Error 1.1622E-06

Using trial and error with a spreadsheet yields T = 4.27 K, 14.6 K, and 30.7 K to increase the

saturation current by 2X, 10X, and 100X respectively

5 anode contacts and 14 cathode contacts

Resistance of anode contacts =10Ω

-1 -2 -3

-4

-5

-6

-1 mA -2 mA (b) Q-point

-1 -2 -3

-4

-5

-6

-1 mA (b) Q-point

iD

(c) Q-point

-1 -2 -3 -4

The load line equation: V = iD R + vD We need two points to plot the load line

(a) V = 6 V and R = 4kΩ: For vD = 0, iD = 6V/4 kΩ = 1.5 mA and for iD = 0, vD = 6V

Plotting this line on the graph yields the Q-pt: (0.5 V, 1.4 mA)

(b) V = -6 V and R = 3kΩ: For vD = 0, iD = -6V/3 kΩ = -2 mA and for iD = 0, vD = -6V

Plotting this line on the graph yields the Q-pt: (-4 V, -0.67 mA)

(c) V = -3 V and R = 3kΩ: Two points: (0V, -1mA), (-3V, 0mA); Q-pt: (-3 V, 0 mA)

(d) V = +12 V and R = 8kΩ: Two points: (0V, 1.5mA), (4V, 1mA); Q-pt: (0.5 V, 1.4 mA)

(e) V = -25 V and R = 10kΩ: Two points: (0V, -2.5mA), (-5V, -2mA); Q-pt: (-4 V, -2.1 mA)

1 2 3 4 5 6 7 -1

-2 -3 -4 -5

Load line for (b) Q-Point

(-4V,-2.1 mA)

(e)

(c)

Q-Point (-3V,0 mA)

(d)

Q-Point (0.5V,1.45 mA)

Using the equations from Table 3.1, (f = 10-10-9 exp , etc.)

VD = 0.7 V requires 12 iterations, VD = 0.5 V requires 22 iterations,

VD = 0.2 V requires 384 iterations - very poor convergence because the second iteration (VD =

1.0000E+00 -9.991E+03 -1.000E+04

9.2766E-04 1.496E-01 -1.003E+04

9.4258E-04 3.199E-06 -1.003E+04

9.4258E-04 9.992E-16 -1.003E+04

9.4258E-04 9.992E-16 -1.003E+04

Ideal diode model: ID = 1V/10kΩ = 100μA; (100μA, 0 V)

Constant voltage drop model: ID = (1-0.6)V/10kΩ = 40.0μA; (40.0μA, 0.6 V)

+

-V

I 1 kΩ 1.2 k Ω

1.5 V 1.2 V

0.3 V

+ -

+

-V

I 2.2 k Ω

(a) Ideal diode model: The 0.3 V source appears to be forward biasing the diode, so we will

assume it is "on" Substituting the ideal diode model for the forward region yields

I =

2.2kΩ 0.3V = 0.136 mA This current is greater than zero, which is consistent with the diode

being "on" Thus the Q-pt is (0 V, +0.136 mA)

Ideal Diode:

0.3 V

+ -

+

-V I

2.2 k Ω

CVD:

0.3 V +

-0.6 V

I 2.2 k Ω

+ -

on

V

(b) CVD model: The 0.3 V source appears to be forward biasing the diode so we will assume it

is "on" Substituting the CVD model with Von = 0.6 V yields

I =

2.2kΩ

0.3V − 0.6V = −136 μA

This current is negative which is not consistent with the assumption that the diode is "on"

Thus the diode must be off The resulting Q-pt is: (0 mA, -0.3 V)

0.3 V +

-I=0 2.2 k Ω

- V +

(c) The second estimate is more realistic 0.3 V is not sufficient to forward bias the diode into

significant conduction For example, let us assume that IS = 10-15 A, and assume that the full

0.3 V appears across the diode Then

For maximum current, we make the Thévenin equivalent voltage at the diode anode as large as

possible and that at the cathode as small as possible

For minimum current, we make the Thévenin equivalent voltage at the diode anode as small as

possible and that at the cathode as large as possible

16kΩ = 0.625 mA (c) Diode is reverse biased : I = 0 | V = −5 +16kΩ I( )= −5 V | VD= −10 V

(d ) Diode is reverse biased : I = 0 | V = 7 −16kΩ I( )= 7 V | VD= −10 V

(c) Diode is reverse biased : I = 0 | V = −5 +16kΩ I( )= −5 V | VD= −10 V

(d ) Diode is reverse biased : I = 0 | V = 7 −16kΩ I( )= 7 V | VD= −10 V

(c) Diode is reverse biased : I = 0 A | V = −5 +100kΩ I( )= −5 V | V D = −10 V

(d ) Diode is reverse biased : I = 0 A | V = 7 −100kΩ I( )= 7 V | V D = −10 V

(b)

(c) Diode is reverse biased : I = 0 | V = −5 +100kΩ I( )= −5 V | V D = −10 V

(d ) Diode is reverse biased : I = 0 | V = 7 −100kΩ I( )= 7 V | V D = −10 V

MODEL DIODE DIODE DIODE

ID 9.90E-04 -1.92E-12 7.98E-04

VD 7.14E-01 -1.02E+00 7.09E-01

NAME D1 D2 D3

MODEL DIODE DIODE DIODE

ID 4.74E-04 -4.22E-13 2.67E-11

VD 6.95E-01 -4.21E-01 2.63E-01

NAME D1 D2 D3

MODEL DIODE DIODE DIODE

ID 8.79E-03 1.05E-03 7.96E-04

VD 7.11E-01 7.16E-01 7.09E-01

NAME D1 D2 D3

MODEL DIODE DIODE DIODE

ID -4.28E-13 -8.55E-13 1.15E-03

VD -4.27E-01 -8.54E-01 7.18E-01

For all cases, the results are very similar to the hand analysis

MODEL DIODE DIODE DIODE

ID 1.47E-03 -4.02E-12 9.35E-04

VD 6.65E-01 -4.01E+00 6.53E-01

The simulation results are very close to those given in Ex 3.8

i D -6 -5 -4 -3 -2 -1

16.7ms 1ms

(a) V dc = − V( P −V on)= − 6.3 2 −1( )= −7.91V (b) C = I T

V r =7.910.55

10.5

260

+0.000e+000 +10.000m +20.000m +30.000m +40.000m +50.000m +60.000m +70.000m

-10.000 -5.000 +0.000e+000 +5.000 +10.000

SPICE Graph Results: VDC = 9.29 V, Vr = 1.05 V, IP = 811 A, ISC = 1860 A

260

1

1.3ms = 923A

V(1) V(2) *REAL(Rectifier)*

Time (s) Circuit3_93b-Transient-11

SPICE Graph Results: VDC = -6.55 V, Vr = 0.58 V, IP = 150 A, ISC = 370 A

Note that a significant difference is caused by the diode series resistance

3.94

(a) V dc = − V( P −V on)= − 6.3 2 −1( )= −7.91V (b) C = I T

V r =7.910.25

10.5

1

400= 0.158F (c) PIV ≥ 2V P = 2 ⋅ 6.3 2 =17.8V (d ) I surge =ωCV P = 2π( )400 (0.158) ( )6.3 2 = 3540 A

2400

194.3μs = 839A

3.95

(a) V dc = − V( P −V on)= − 6.3 2 −1( )= −7.91V (b) C = I T

V r =7.910.25

10.5

2

105

10.377μs = 839A

3.96

160

The doubler circuit is effectively two half-wave rectifiers connected in series Each capacitor is

discharged by I = 3000V/3000 = 1 A for 1/60 second The ripple voltage on each capacitor is

33.3 V With two capacitors in series, the output ripple should be 66.6 V, which is close to the

Simulation Results: VDC = -12.9 V, Vr = 0.20 V, IP = 33.3 A, ISC = 362 A RS results in a

significant reduction in the values of IP and ISC

The circuit is behaving like a half-wave rectifier The capacitor should charge during the first

1/2 cycle, but it is not Therefore, diode D1 is not functioning properly It behaves as an open

10.25

3.112

3.3-V, 15-A power supply with Vr ≤ 10 mV Assume Von = 1 V

Filter Capacitor

(i) The large value of C suggests we avoid the half-wave rectifier This will reduce the cost

and size of the circuit

(ii) The PIV ratings are all low and do not indicate a preference for one circuit over another

(iii) The peak current values are lower for the full-wave and full-wave bridge rectifiers and

also indicate an advantage for these circuits

(iv) We must choose between use of a center-tapped transformer (full-wave) or two extra

diodes (bridge) At a current of 15 A, the diodes are not expensive and a four-diode bridge

should be easily found The final choice would be made based upon cost of available

components

3.113

200-V, 3-A power supply with Vr ≤ 4 V Assume Von = 1 V

Filter Capacitor

(i) The the half-wave rectifier requires a larger value of C which may lead to more cost

(ii) The PIV ratings are all low enough that they do not indicate a preference for one circuit

over another

(iii) The peak current values are lower for the full-wave and full-wave bridge rectifiers and

also indicate an advantage for these circuits

(iv) We must choose between use of a center-tapped transformer (full-wave) or two extra

diodes (bridge) At a current of 3 A, the diodes are not expensive and a four-diode bridge

should be easily found The final choice would be made based upon cost of available

components

3.114

3000-V, 1-A power supply with Vr ≤ 120 V Assume Von = 1 V

Filter Capacitor

(i) A series string of multiple capacitors will normally be required to achieve the voltage

rating

(ii) The PIV ratings are high, and the bridge circuit offers an advantage here

(iii) The peak current values are lower for the full-wave and full-wave bridge rectifiers but

neither is prohibitively large

(iv) We must choose between use of a center-tapped transformer (full wave) or extra diodes

(bridge) With a PIV of 3000 or 6000 volts, multiple diodes may be required to achieve the

require PIV rating