EXAMPLE 10.1OBJECTIVETo determine the excess minority-carrier electron concentration in the base of an npn bipolar transistor. Consider a uniformly doped silicon bipolar transistor at T = 300 K with impurity doping concentrations of NE = 1018 cm-3 and NB = 1016 cm-3. A forward-bias B-E voltage of VBE = 0.610 V is applied. Assume a neutral base width of xB = 1 m and a minority-carrier diffusion length of LB = 10 m. Determine the ratio of actual minority-carrier concentration at x = xB/2 [Equation (10.15a)] to the ideal case of a linear minority-carrier distribution [Equation (10.15b)]. SolutionWe find

For the actual distribution, we have

or

3-416

2102

0 cm1025.210

105.1

B

iB N

nn

10

5.0sinh

10

5.01sinh1

0259.0

610.0exp

sinh

1025.2

2 101

4B

B

xxn

314 cm109018.12

B

B

xxn

EXAMPLE 10.1 SolutionFor the linear approximation, we find

or

Taking the ratio of the actual to the linear approximation, we obtain

CommentWe see that for the case when xB = 1 m and LB = 10 m, the excess minority-carrier concentration is very nearly a linear function of distance through the base region..

9987.0109042.1

109018.1Ratio

14

14

314 cm109042.12

B

B

xxn

44

101

4

105.0105.010259.0

610.0exp

sinh

1025.2

2B

B

xxn

EXAMPLE 10.2OBJECTIVETo determine the excess minority-carrier concentration in the emitter compared to that in the base of a bipolar transistor. Consider a silicon bipolar transistor with the same parameters as given in Example 10.1. Determine the ratio pE (x = 0) / nB (x = 0). SolutionWe find from Equation (10.20a)

and we find from Equation (10.13a)

so

Then

CommentAs we continue our analysis of the bipolar transistor, we will see that this ratio needs to be fairly small for a “good” transistor.

1exp0 0 kT

eVpp BE

EE

1exp0 0 kT

eVnn BE

BB

18

16

2

2

0

0

10

10

0

0

B

E

Bi

Ei

B

E

B

E

N

N

Nn

Nn

n

p

n

p

01.0B

E

n

p

EXAMPLE 10.3OBJECTIVETo calculate a distance into the collector region. Consider the collector region of an npn bipolar transistor biased in the forward-active mode. At what value of x, compared to LC, does the magnitude of the minority-carrier concentration reach 95 percent of the thermal equilibrium value? SolutionCombining Equations (10.23) and (10.26), we find the minority-carrier concentration to be

or

For , we find

CommentIn order for the excess minority-carrier concentration in the collector to reach the steady-state value as assumed in the preceding analysis, the collector region must be fairly wide. This situation may not be valid in all cases.

CCCC L

xppxpxpc exp100

CC

C

L

x

p

xpexp1

0

3

CL

x 95.0

0

C

C

pxp

EXAMPLE 10.4OBJECTIVETo design the ratio of emitter doping to base doping to achieve an emitter injection efficiency factor equal to = 0.9967. Consider an npn bipolar transistor. Assume, for simplicity, that DE = DB, LE = LB, and xE = xB. SolutionEquation (10.35a) reduces to

so

Then

CommentThe emitter doping concentration must be much larger than the base doping concentration to achieve a high emitter injection efficiency.

Bi

Ei

B

E

NnNn

np

2

2

0

0 1

1

1

1

9967.01

1

E

B

NN

30200331.0 B

E

E

B

N

N

N

N

EXAMPLE 10.5OBJECTIVETo design the base width required to achieve a base transport factor equal to T = 0.9967. Consider a pnp bipolar transistor. Assume that DB = 10 cm2/s and B0 = 10-7 s. SolutionThe base transport factor applies to both pnp and npn transistors and is given by

Then

We have

so that the base width must then be

CommentIf the base width is less than approximately 0.8 m, then the required base transport factor will be achieved. In most cases, the base transport factor will not be the limiting factor in the bipolar transistor current gain.

9967.0cosh

1

BBT Lx

0814.0B

B

L

x

cm101010 370

BBB DL

m0.814cm10814.0 4 Bx

EXAMPLE 10.6OBJECTIVETo calculate the forward-bias B-E voltage required to achieve a recombination factor equal to = 0.9967. Consider an npn bipolar transistor at T = 300 K. Assume that Jr0 = 10-8 A/cm2 ad that Js0 = 10-11 A/cm2.

SolutionThe recombination factor, from Equation (10.44), is

We then have

We can rearrange this equation and write

Then VBE = 2(0.0259) ln (3.02 105) = 0.654 V

CommentThis example demonstrates that the recombination factor may be an important limiting factor in the bipolar current fain. In this example, if VBE is smaller than 0.654 V, then the recombination factor will fall below the desired 0.9967 value.

kTeV

JJ BE

s

r

2exp1

1

0

0

kTeVBE

2exp

1010

1

19967.0

11

8

53

1002.39967.01

109967.0

2exp

kT

eVBE

EXAMPLE 10.7OBJECTIVETo calculate the common-emitter current gain of a silicon npn bipolar transistor at T = 300 K given a set of parameters. Assume the following parameters: DE = 10 cm2/s xB = 0.70 m DB = 25 cm2/s xE = 0.50 m E0 = 1 10-7 s NE = 1 1018 cm-3

B0 = 5 10-7 s NB = 1 1016 cm-3

Jr0 = 5 10-8 A/cm2 VBE = 0.65 VThe following parameters are calculated:

cm1054.3

cm10

cm1025.2101

105.1

cm1025.2101

105.1

30

30

3416

210

0

3218

210

0

BBB

EEE

B

E

DL

DL

n

p

SolutionThe emitter injection efficiency factor, from Equation (10.35a), is

The base transport factor, from Equation (10.39a) is

The recombination factor, from Equation (10.44), is

where

9944.0

050.0tanh0198.0tanh

10251025.21054.3101025.2

1

1

34

32

9998.0

1054.31070.0

cosh

1

3

4

T

0259.02065

exp105

1

1

0

8

sJ

2923

4190

0 A/cm1029.110977.1tanh1054.3

1025.225106.1

tanh

B

BB

BBs

Lx

L

neDJ

SolutionWe can now calculate = 0.99986. The common-base current gain is then

= T = (0.9944)(0.9998)(0.99986) = 0.99406which gives a common-emitter current gain of

CommentIn this example, the emitter injection efficiency is the liming factor in the current gain.

16799406.01

99406.0

1

15102

EXAMPLE 10.8OBJECTIVETo calculate the change in the neutral base width with a change in C-B voltage. Consider a uniformly doped silicon bipolar transistor at T = 300 K with a base doping of NB = 5 1016 cm-3 and a collector doping of NC = 2 1015 cm-3. Assume the metallurgical base width is 0.70 m. Calculate the change in the neutral base width as the C-B voltage changes from 2 to 10 V. SolutionThe space chare width extending into the base region can be written as

or

which becomesxdB = [(9.96 10-12)(Vbi + VCB)]1/2

The built-in potential is

21

12

CBB

CCBbisdB NNN

N

e

VVx

21

151616

15

19

14

102105

1

105

102

106.1

1085.87.112

CBbi

dB

VVx

V718.0ln2

i

CBbi n

NN

e

kTV

SolutionFor VCB = 2 V, we find xdB = 0.052 m, and for VCB = 10 V, we find xdB = 0.103 m. If we neglect the B-E space charge region, which will be small because of the forward-biased junction, then we can calculate the neutral base width. For VCB = 2 V.

xB = 0.70 0.052 = 0.648 mand for VCB = 10 V,

xB = 0.70 0.103 = 0.597 m CommentThis example shows that the neutral base width can easily change by approximately 8 percent as the C-B voltage changes from 2 to 10 V.

EXAMPLE 10.9OBJECTIVETo calculate change in collector current with a change in neutral base width, and to estimate the Early voltage. Consider a uniformly doped silicon npn bipolar transistor with parameters described in Example 10.8. Assume DB = 25 cm2/s, and VBE = 0.60 V, and also assume that xB << L

B. SolutionThe excess minority-carrier electron concentration in the base is given by w”quation (10.15) as

If xB << LB, then (xB x) << LB so we can write the approximations

The expression for nB(x) can then be approximated as

B

B

BB

BBEB

B

Lx

Lx

Lxx

kTeV

n

xn

sinh

sinhsinhexp0

B

B

B

B

B

B

B

B

L

xx

L

xx

L

x

L

xsinhandsinh

xxx

kT

eV

x

nxn B

BE

B

BB 1exp0

The collector current is now

The value of nB0 is calculated as

If we let xB = 0.648 m when VCB = 2 V (VCE = 2.6 V), then

Now let xB = 0.597 m when VCB = 10 V (VCE = 10.6 V). In this case we have |JC| = 3.47 A/cm2. From Equation (10.45a), we can write

Using the calculated values of current and voltage, we have

The Early voltage is then determined to be VA 92 V CommentThis example indicates how much the collector current can change as the neutral base width changes with a change in the B-C space charge width, and it also indicates the magnitude of the Early voltage.

kT

eV

x

neD

dx

xndeDJ BE

B

BBBBC exp0

3316

2102

0 cm105.4105

105.1

B

iB N

nn

24

319

A/cm20.30259.0

60.0exp

10648.0

105.425106.1

CJ

AACE

C

CE

C

VVV

J

V

J

6.2

20.3

6.26.10

20.347.3

CE

C

ACE

C

CE

C

V

J

VV

J

dV

dJ

EXAMPLE 10.10OBJECTIVETo determine the increase in pE0 in the emitter due to bandgap narrowing. Consider a silicon emitter at T = 300 K. Assume the emitter doping increases from 1018 cm-3 to 1019 cm-3. Calculate the change in the pE0 value. SolutionFor emitter doping of NE = 1018 cm-3 and 1019 cm-3, we have, neglecting bandgap narrowing.

and

Taking into account the bandgap narrowing, we obtain, respectively, for NE = 1018 cm-3 and NE = 1019 cm-3

and

CommentIf the emitter doping increases from 1018 to 1019 cm-3, the thermal equilibrium minority carrier concentration decreases by approximately a factor of 2 rather than a factor of 10. This effect is due to bandgap narrowing.

3218

2102

0 cm1025.210

105.1

E

iE N

np

3119

210

0 cm1025.210

105.1

Ep

3218

210

0 cm1016.70259.0

030.0exp

10

105.1

Ep

3219

210

0 cm1094.40259.0

08.0exp

10

105.1

Ep

EXAMPLE 10.11OBJECTIVETo determine the effect of emitter current crowding. Consider the geometry shown in Figure 10.33. The base doping concentration is NB = 1016 cm-3, the neutral base width is xB = 0.80 m, the emitter width is S = 10 m, and the emitter length is L = 10 m. (a) Determine the resistance of the base between x = 0 and x = S/2. Assume a hole mobility of p = 400 cm2/V-s. (b) If the base current in this region is uniform and given by IB/2 = 5 A, determine the potential difference between x = 0 and x = S/2. (c) Using the results of part (b), what is the ratio of emitter current density at x = 0 and x = S/2? Solution(a) The resistance is found from

or

R = 9.77 103 = 9.77 k

44

4

1619 1010108.0

105

10400106.1

1

21

Lx

S

NeA

lR

BBp

(b) The potential difference is

or

V = 4.885 10-2 V = 48.85 mV

36 1077.91052

R

IV B

n+ emitter

L

S

xB

x = S/2x = 0

p base

IB/2

n collectorFigure 10.33 Geometry used for Example 10.11 and Exercise Problem EX 10.11.

(c) The ratio of emitter current at x = 0 to that at x = S/2 is found to be

or

CommentBecause the B-E voltage at the emitter edge (x = 0) is larger than that in the center of the emitter (x = S/2), the current at the edge is larger than that in the center of the emitter.

0259.0

04885.0expexp

2

0

tE

E

V

V

SxI

xI

59.6

2

0

SxI

xI

E

E

EXAMPLE 10.12OBJECTIVETo design the collector doping and collector width to meet a punch-through voltage specification. Consider a uniformly doped silicon bipolar transistor with a metallurgical base width of 5 m and a base doping of NB = 1016 cm-3. The punch-through voltage is to be Vpt = 25 V. SolutionThe maximum collector doping concentration can be determined from Equation (10.54) as

or

which yields NC = 8.38 1014 cm-3

This n-type doping concentration in the collector must extend at least as far as the depletion width extends into the collector to avoid breakdown in the collector region. We have, using results from Chapter 5.

C

C

N

N14

1616419

1085.87.112

1010105.0106.125

CN

1610194.12

21

12

CBC

BRbisn NNN

N

e

VVx

Neglecting Vbi compared to VR = Vpt, we obtain

orxn = 5.97 m

CommentFrom Figure 9.30, the expected avalanche breakdown voltage for this junction is greater than 300 V. Obviously punch-through will occur before the normal breakdown voltage in this case. For a larger punch-through voltage, a larger metallurgical base width will be required, since a lower collector doping concentration is becoming impractical. A larger punch-through voltage will also require a larger collector width in order to avoid premature breakdown in this region.

2/1

141614

16

19

14

1038.810

1

1038.8

10

106.1

251085.87.112

nx

EXAMPLE 10.13OBJECTIVETo design a bipolar transistor to meet a breakdown voltage specification. Consider a silicon bipolar transistor with a common-emitter current gain of = 100 and a base doping concentration of NB = 1017 cm-3. The minimum open-base breakdown voltage is to be 15 V. SolutionFrom Equation (10.63), the minimum open-emitter junction breakdown voltage must be

Assuming the empirical constant n is 3, we find

From Figure 9.30, the maximum collector doping concentration should be approximately 7 1015 cm-3 to achieve this breakdown voltage. CommentIn a transistor circuit, the transistor must be designed to operate under a worst-case situation. In this example, the transistor must be able to operate in an open-base configuration without going into breakdown. As we determined previously, an increase in breakdown voltage can be achieved by decreasing the collector doping concentration.

00 CEn

CB BVBV

V6.691510030 CBBV

2

EXAMPLE 10.14OBJECTIVETo determine, to a first approximation, the frequency at which the small-signal current gain decreases to 1/ of its low frequency value. Consider the simplified hybrid-pi circuit shown in Figure 10.42. We are ignoring C, Cs, r, Cje, r0, and the series resistances. We must emphasize that this is a first-order calculation and that C normally cannot be neglected.

SolutionAt very low frequency, we may neglect C so that

Vbe = Ibr and Ic = gmVbe = gmrIb

C rgmVbeVbe

+

B C

E

Ib Ic

Figure 10.42 Simplified hybrid-pi equivalent circuit.

SolutionWe can then write

where hfe0 is the low-frequency, small-signal common-emitter current gain. Taking into account C, we have

Then

or the small-signal current gain can be written as

The magnitude of the current gain drops to 1/ of its low-frequency value at f = 1/2rC. If, for example, r = 2.6 k and C = 4 pF, then

f = 15.3 MHz Commentthe frequency calculated in this example is called the beta cutoff frequency. High-frequency transistors must have small diffusion capacitances, implying the use of small devices.

rgI

Ih m

b

cfe 0

Crj

rIV bbe 1

Crj

hIVgI fe

bbemc 10

Crj

h

I

IA fe

b

ci 1

0

2

EXAMPLE 10.15OBJECTIVETo calculate the emitter-to-collector transit time and the cutoff frequency of a bipolar transistor, given the transistor parameters. Consider a silicon npn transistor at T = 300 K. Assume the following parameters: IE = 1 mA Cje = 1 pF xB = 0.5 m Dn = 25 cm2/s xdc = 2.4 m rc = 25 C = 0.1 pF Cs = 0.1 pF SolutionWe will initially calculate the various time-delay factors. If we neglect the parasitic capacitance, the emitter-base junction charging time is

e = reCje

where

Thene = (25.9)(1012) = 25.9 ps

The base transit time is

9.25101

0259.013

Ee Ie

kTr

ps50252

105.0

2

242

n

Bb D

x

The collector depletion region transit time is

The collector capacitance charging time isc = rc(C + Cs) = (20)(0.2 1012) = 4 ps

The total emitter-to-collector time delay is thenec = 25.9 + 50 + 24 + 4 = 103.9 ps

so that the cutoff frequency is calculated as

If we assume a low-frequency common-emitter current gain of = 100, then the beta cutoff frequency is

CommentThe design of high-frequency transistors requires small device geometries to reduce capacitances and narrow base widths to reduce the base transit time.

GHz53.1109.1032

1

2

112

ecTf

ps2410

104.27

4

s

dcb

x

MHz3.15100

1053.1 9

0

Tff