Access full Solution Manual click on linkfiles.book4me.xyz/sample/Solution Manual for... · 11 The built‐in potential of an equally doped pn junction is 0.65 V at room temperature

https://www.book4me.xyz/solution-manual-microelectronics-international-student-version-razavi/

Access full Solution Manual click on link

2 The electrons in a piece of n‐type semiconductor take 10 ps to cross from one end to

another end when a potential of 1 V is applied across it. Find the length of the

semiconductor bar.

Solution

2

4

10 ps

But E and

10ps 10ps 1350 1

1.16 10 cm 1.16 m

n

n

L

VV

V EL

L V

L


3 A current of 0.05 μA flows through an n‐type silicon bar of length 0.2 μm and cross

section area of 0.01 μm 0.01 μm when a voltage of 1 V is applied across it. Find the doping

level at room temperature.

Solution

12 2

drift

2102

2 2

4

0.05 A

0.2 m

0.01 m 0.01 m 1 10 cm

----(1)

1.08 10

1350 / (V.s) 480 cm /(V.s)

15 10 V / cm

0.2 m

Substituting these values in Eq. (1) we ge

tot n p

i

n p

I

L

A

I I q n p AE

np

n n

cm

V VE

L

10 26 19 12 4

15 3

t

(1.08 10 ) 4800.05 10 1.6 10 1350 10 5 10

4.63 10 cm .

A nn

n


4. Repeat Problem 2.3 for Ge. Assume μn=3900cm2/(V·s) and μp =1900cm2/(V·s). In order to obtain the doping level proceed as follows. We use the formula

1( )n pq n p

For n-type semiconductor n>>p, and hence

1nnq

Given:-

2

2

2

0.05 A0.2 m

(0.01 m)1V

3900cm / (V s)

1900cm / (V s)n

p

Il

AV

Resistance of the Ge bar;

1V0.05 A20MΩ

VRI

Therefore resistivity,

2(0.01 m)20MΩ0.2 m

1 cm

ARl

Again, 1

nne

or, 1

n

ne


Substituting the values, we get

19 2

16 3

15 3

11 cm 1.6 10 3900cm / (V s)

16.24 10 cm

1.6 10 cm

DN n

Since at room temperature all the impurities are ionized and hence carrier concentration n is equal to the doping density.






9 A Si semiconductor cube with side equal to 1 μm is doped with 4 1017 cm–3

phosphorous impurities. Calculate the drift current when a voltage of 5 V is applied across it.

Solution

17 3

2 10 2

17

2 2

4

8 2

4 10 cm

(1.08 10 )291.6

4 10

1350cm /(V.s) and 480 cm /(V.s)

5V5 10 /cm

1 m

1 m 1 m 10 cm

Since , we can write

0.0432A

i

n p

n p

tot n

n

np

n

VE V

L

A

n

I qn AE


10 One side of pn junction is doped with pentavalent impurities which gives a net

doping of 5 1018 cm–3. Find the doping concentration to be added to the other side to get a

built‐in potential of 0.6 V at room temperature.

Solution

0 2

18 3 10 30

1811 3

10 2

ln

5 10 cm , 26 mV, 1.08 10 cm , 0.6 V

5 10 0.6 0.026 ln 2.45 10 cm

(1.08 10 )

A DT

i

D T i

AA

N NV V

n

N V n V

NN


11 The built‐in potential of an equally doped pn junction is 0.65 V at room temperature.

When the doping level in n‐side is doubled, keeping the doping of the p‐side unchanged,

calculate the new built‐in potential and the doping level in P and N region.

Solution

0 2

10 30

15 310 2

15 15

0 10 2

ln

, 26 mV, 1.08 10 cm , 0.65 V

0.65 0.026 ln 2.89 10 cm

(1.08 10 )

2 2.89 10 2.89 10 0.026 ln 0.668 V

(1.08 10 )

A DT

i

A D T i

A AA

N NV V

n

N N V n V

N NN

V


12 A silicon pn junction diode having ND = 1015 cm–3 and NA= 10

17 cm–3 gives a total

depletion capacitance of 0.41 pF/m2. Determine the voltage applied across the diode.

Solution

15 17

0 2 10 2

14 15 178 2

0 15 170

0

0

28

00

10 10ln 0.026ln 0.7144 V

(1.08 10 )

1 11.7 8.854 10 10 101.072 10 F/cm

2 2 (10 10 ) 0.7144

1

1.072 101 0.7144

0.41

A DT

i

si A Dj

A D

jj

R

jR

j

N NV V

n

q N NC

N N V

CC

V

V

CV V

C

8

4.167 V10




14 Two identical pn junction diodes are connected in series. Calculate the current

flowing through each diode when a forward bias of 1 V is applied across the series

combination. Assume the reverse saturation current is 1.44 10–17 A for each diode.

Solution

A 1023.311044.11 9026.0

5.017

eeII T

F

V

V

SD



16 Consider a pn junction in forward bias. Initially a current of 1 mA flows through it,

and the current increases by 10 times when the forward voltage is increased by 1.5 times.

Determine the initial bias applied and reverse saturation current.

Solution

1.5

7

1 mA (1)

10 mA (2)

ln(100) 0.239V0.5

From Eq. (1), we get 10 A

F

T

F

T

F

T

V

VD S

V

VS

V

VS

TF

S

I I e

I e

I e

VV

I

















Documents

Access full Solution Manual click on linkfiles.book4me.xyz/sample/Solution Manual for... · 11 The built‐in potential of an equally doped pn junction is 0.65 V at room temperature