Transcript 投影片 1

EXAMPLE 10.1
OBJECTIVE
To 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 minoritycarrier concentration at x = xB/2 [Equation (10.15a)] to the ideal case of a linear
minority-carrier distribution [Equation (10.15b)].
 Solution
We find
nB 0


10 2
n
1.5 10


NB
1016
2
i
 2.25104 cm-3
For the actual distribution, we have
4
  0.610  
x
2
.
25

10

 1  0.5 
 0.5 
B 
nB  x   
 exp
  1 sinh
  sinh 
1
2  sinh10    0.0259 

 10 
 10 
or


nB  x 
xB 
14
3
  1.901810 cm
2
EXAMPLE 10.1
 Solution
For the linear approximation, we find
xB  2.25104   0.610  

4
4 
nB  x   
 exp
  1 0.5 10  0.5 10 
1
2  sinh10    0.0259 



 
or


nB  x 
xB 
14
3
  1.904210 cm
2
Taking the ratio of the actual to the linear approximation, we obtain
1.90181014
Rat io 
 0.9987
14
1.904210
 Comment
We 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..

EXAMPLE 10.2
OBJECTIVE
To 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).
 Solution
We find from Equation (10.20a)


eVBE 
pE 0  pE 0 exp

1



kT




and we find from Equation (10.13a)


eVBE 
  1
 kT 

nB 0  nB 0 exp

so
Then
 Comment

pE 0
ni2 N E
pE 0
NE
1016

 2


nB 0
nB 0
ni N B
NB
1018
pE
 0.01
nB
As we continue our analysis of the bipolar transistor, we will see that this ratio needs to be
fairly small for a “good” transistor.
EXAMPLE 10.3
OBJECTIVE
To 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?
 Solution
Combining Equations (10.23) and (10.26), we find the minority-carrier concentration to
be

  x 
pc x  pC  x  pC 0  pC 0 1  exp
 L 

 C 

or
For
pC  x 
pC 0
  x 
pC x

 1  exp
pC 0
 LC 
 0.95
, we find
 Comment
x
3
LC
In order for the excess minority-carrier concentration in the collector to reach the steadystate value as assumed in the preceding analysis, the collector region must be fairly wide.
This situation may not be valid in all cases.
EXAMPLE 10.4
OBJECTIVE
To 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.
 Solution
Equation (10.35a) reduces to
1
1
 

2
pE 0
n
NE
1
1  i2
nB 0
ni N B
so
1
 
 0.9967
N
1 B
NE
Then
NB
NE
 0.00331
 302
NE
NB
 Comment
The emitter doping concentration must be much larger than the base doping
concentration to achieve a high emitter injection efficiency.
EXAMPLE 10.5
OBJECTIVE
To 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.
 Solution
The base transport factor applies to both pnp and npn transistors and is given by
1
T 
 0.9967
coshxB LB 
Then
We have
xB
 0.0814
LB
LB 
DB B 0 
10 10 7   10 3 cm
so that the base width must then be
xB  0.814104 cm  0.814m
 Comment
If 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.
EXAMPLE 10.6
OBJECTIVE
To 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.
 Solution
The recombination factor, from Equation (10.44), is
1
 
J
  eVBE 
1  r 0 exp

J s0
 2kT 
We then have
1
0.9967 
108
  eVBE 
1
exp


11
10
2
kT


We can rearrange this equation and write
3
  eVBE  0.996710
exp
 3.02105

1  0.9967
 2kT 
Then
VBE = 2(0.0259) ln (3.02  105) = 0.654 V
 Comment
This 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.
EXAMPLE 10.7
OBJECTIVE
To 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 V
The following parameters are calculated:
pE 0

1.5  10 

 2.25 102 cm3
nB 0

1.5 10 

 2.25 104 cm3
10 2
1 1018
10 2
11016
LE  DE E 0  103 cm
LB  DB B 0  3.54103 cm
 Solution
The emitter injection efficiency factor, from Equation (10.35a), is
 
1
 0.9944
2
3
2.2510 10 3.5410
t anh0.0198
1

2.25104 25 103
t anh0.050

 
 



The base transport factor, from Equation (10.39a) is
1
T 
 0.7010 4
cosh
3
3
.
54

10




 0.9998
The recombination factor, from Equation (10.44), is

where
J s0
eDB nB 0

 xB
LB tanh
 LB
1
  065 
5 108
1
exp

J s0
 20.0259 

1.6 10 252.2510 

 1.2910
 3.5410 tanh1.97710 

19
3


4
2
9
A/cm2
 Solution
We can now calculate  = 0.99986. The common-base current gain is then
 = T = (0.9944)(0.9998)(0.99986) = 0.99406
which gives a common-emitter current gain of


0.99406

 167
1   1  0.99406
 Comment
In this example, the emitter injection efficiency is the liming factor in the current gain.
EXAMPLE 10.8
OBJECTIVE
To 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.
 Solution
The space chare width extending into the base region can be written as
 2 s Vbi  VCB
 NC


1
210
x dB  


e
 N B N B  NC 

12
15
or


 211.7  8.8510 Vbi  VCB   2 10
1
x dB  


19
16
16
15
1
.
6

10
5

10
5

10

2

10


which becomes
The built-in potential is
14
15

xdB = [(9.96  10-12)(Vbi + VCB)]1/2
kT  N B N C
V bi
ln
2
e
n
i


  0.718V

12




 Solution
For 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 forwardbiased junction, then we can calculate the neutral base width. For VCB = 2 V.
xB = 0.70  0.052 = 0.648 m
and for VCB = 10 V,
xB = 0.70  0.103 = 0.597 m
 Comment
This 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 10.9
OBJECTIVE
To 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 <<
LB.
 Solution
The excess minority-carrier electron concentration in the base is given by w”quation
(10.15) as






 eV
nB 0 exp BE
 kT


nB  x  
xB  x
x




sinh

sinh

 L

L

B


 B
x 
sinh B 
 LB 
If xB << LB, then (xB  x) << LB so we can write the approximations
 xB
sinh
L
 B
  xB 


L 

  B 
and
 xB  x   xB  x 


sinh
 L

 L

B
B

 

The expression for nB(x) can then be approximated as
nB 0
nB x  
xB




 eVBE  
  1 xB  x   x 
exp
 kT  


The collector current is now
d nB x  eDB nB 0
 eV 

exp BE 
dx
xB
 kT 
The value of nB0 is calculated as
2
J C  eDB
nB 0

ni2
1.5 1010


NB
5 1016

 4.5 103 cm3
If we let xB = 0.648 m when VCB = 2 V (VCE = 2.6 V), then
1.6 10 254.5 10  exp

19
JC
3
0.648104
0.60 
2
  3.20A/cm
 0.0259
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
dJC
JC
J C


dVCE
VCE  VA
VCE
Using the calculated values of current and voltage, we have
J C
JC
3.47  3.20
3.20



VCE
10.6  2.6
VCE  VA
2.6  VA
The Early voltage is then determined to be VA  92 V
 Comment
This 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.
EXAMPLE 10.10
OBJECTIVE
To 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.
 Solution
For emitter doping of NE = 1018 cm-3 and 1019 cm-3, we have, neglecting bandgap narrowing.
2
ni2

1.5 1010 
2
3
pE 0 


2
.
25

10
cm
NE
1018
and
1.5 10 

10 2
pE 0
19
10
 2.25101 cm3
Taking into account the bandgap narrowing, we obtain, respectively, for NE = 1018 cm-3 and
2
NE = 1019 cm-3

1.5 1010 
 0.030 
2
3
pE 0 
exp

  7.1610 cm
18
10
 0.0259
and
2

1.5 1010 
 0.08 
2
3
pE 0 
exp

4
.
94

10
cm


1019
 0.0259
 Comment
If 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.
EXAMPLE 10.11
OBJECTIVE
To 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
 S 2

 x L 
 B
1
5 10 4

1.6 1019 400 1016 0.8 10 4 1010 4
 1
R

A  e p N B
l
or




R = 9.77  103  = 9.77 k


(b) The potential difference is


I 
V   B  R  5 106 9.77 103
 2
or

V = 4.885  10-2 V = 48.85 mV
IB/2
L
S
n+ emitter
p base
xB
x=0
x = S/2
n collector
Figure 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
 V
I E x  0
 exp
I E x  S 2
 Vt
or

0.04885
  exp

 0.0259 

I E x  0
 6.59
I E x  S 2
 Comment
Because 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.
EXAMPLE 10.12
OBJECTIVE
To 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.
 Solution
The maximum collector doping concentration can be determined from Equation (10.54)
as
1.6 1019 0.5 104 1016 N  1016
25 
or
which yields


 
211.78.8510 N
C

14
C
1016
12.94  1 
NC
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.
12
 2 s Vbi  VR   N B

1


xn  

e
 N C N B  N C 

Neglecting Vbi compared to VR = Vpt, we obtain

 211.7 8.8510
xn  
19
1
.
6

10

or
14
25 

10
1

 8.381014  1016  8.381014 



16
1/ 2
xn = 5.97 m
 Comment
From 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.
EXAMPLE 10.13
OBJECTIVE
To 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.
 Solution
From Equation (10.63), the minimum open-emitter junction breakdown voltage must be
BVCB0  n  BVCE 0
Assuming the empirical constant n is 3, we find
BVCB 0  3 10015  69.6V
From Figure 9.30, the maximum collector doping concentration should be approximately 7
 1015 cm-3 to achieve this breakdown voltage.
 Comment
In 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.
EXAMPLE 10.14
OBJECTIVE
To determine, to a first approximation, the frequency at which the small-signal current
gain decreases to 1/ 2 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.
B
Ib
+
Vbe

Ic
C
r
C
gmVbe
E
Figure 10.42 Simplified hybrid-pi equivalent circuit.
 Solution
At very low frequency, we may neglect C so that
Vbe = Ibr
and
Ic = gmVbe = gmrIb
 Solution
We can then write
h fe0
Ic
  g m r
Ib
where hfe0 is the low-frequency, small-signal common-emitter current gain.
Taking into account C, we have


r

Vbe  I b 
 1  jr C 
Then
h fe0


I c  g mVbe  I b 
 1  jr C
or the small-signal current gain can be written as
h fe0

Ic 


Ai   
I b  1  jr C 



The magnitude of the current gain drops to 1/ 2 of its low-frequency value at f =
1/2rC.
If, for example, r = 2.6 k and C = 4 pF, then
f = 15.3 MHz
 Comment
the frequency calculated in this example is called the beta cutoff frequency. Highfrequency transistors must have small diffusion capacitances, implying the use of small
devices.
EXAMPLE 10.15
OBJECTIVE
To 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
 Solution
We will initially calculate the various time-delay factors. If we neglect the parasitic
capacitance, the emitter-base junction charging time is
e = reCje
where
Then
The base transit time is
kT 1
0.0259
re 


 25.9
3
e I E 110
e = (25.9)(1012) = 25.9 ps

xB2
0.5 104
b 

2 Dn
225

2
 50ps
The collector depletion region transit time is
2.4 104
b 

 24ps
7
s
10
xdc
The collector capacitance charging time is
c = rc(C + Cs) = (20)(0.2  1012) = 4 ps
The total emitter-to-collector time delay is then
ec = 25.9 + 50 + 24 + 4 = 103.9 ps
so that the cutoff frequency is calculated as
fT 
1
2 ec
1

 1.53GHz
12
2 103.9 10


If we assume a low-frequency common-emitter current gain of  = 100, then the beta
cutoff frequency is
9
1.5310
f 

 15.3MHz
0
100
fT
 Comment
The design of high-frequency transistors requires small device geometries to reduce
capacitances and narrow base widths to reduce the base transit time.