BJT v-i characteristics

File type ppt

1. Lecture 2626 - 1I-V Characteristics of BJT

Common-Emitter Output Characteristics
i
B
B
C
E
C
i
CE
v
B
C
E
i
B
C
i
EC
v

2. Lecture 2626 - 2To illustrate the I
C
-V
CE
characteristics, we use an enlarged β
R

0510 -5
-1
0
1
2
V
CE
(V)
Reverse-Active
Region
Saturation
Region
Cutoff
I
B
= 100µA
I
B
=80µA
I
B
=60µA
I
B
=40µA
I
B
=20µA
I
B
=0µA
Forward Active
Region
Saturation
Region
β
F
= 25;β
R
=5
Collector Current (mA)
v
CE
v
BE
≥
i
C
β
F
i
B
=
v
CE
v
BE
≤
v
CE
v
BE
≤
i
C
β
R
1+ ()–i
B
=
v
CE
v
BE
0 ≤≤

3. Lecture 2626 - 3Common Base Output Characteristics
i
E
B
C E
v
CB
Ci
B
C E
v
BC
C
i
i
E

4. Lecture 2626 - 4
Forward-Active
Region
I
E
=0mA
I
E
=0.2mA
I
E
=0.4mA
I
E
=0.6mA
I
E
=0.8mA
I
E
=1.0mA
β
F
= 25;β
R
=5
v
CB
or v
BC
(V)
-20246810
0
0.5
1.0
Collector Current (mA)

5. Lecture 2626 - 5Common-Emitter Transfer Characteristic i
C
- v
BE
. BE voltage changes as -1.8 mV/
o
C - this is its temperature coef-
ficient (recall from diodes).
v
BC
=0
Collector Current I
C(mA)
-2
0
4
6
8
10
2
60 mV/decade
Base-Emitter Voltage(V)
0.00.20.40.60.81.0
10
-11
10
-9
10
-8
10
-6
10
-4
10
-2
Log(I
C)

6. Lecture 2626 - 6Common-Emitter Transfer Characteristic i
C
- v
BE
(p. 180)
. BE voltagechangesas-1.8 mV/
o
C - thisisits temperature coef-
ficient (recall from diodes).
I
C
I
S
v
BE
V
T
---------



1– exp



=
v
BC
=0
Collector Current I
C(mA)
-2
0
4
6
8
10
2
60 mV/decade
Base-Emitter Voltage(V)
0.00.20.40.60.81.0
10
-11
10
-9
10
-8
10
-6
10
-4
10
-2
Log(I
C)

7. Lecture 2626 - 7Junction Breakdown - BJT has two diodes back-to-back. Each diode has a
breakdown. The diode (BE) with higher doping concentrations has the lower
breakdown voltage (5 to 10 V).
In forward active region, BC junction is reverse biased.
In cut-off region, BE and BC are both reverse biased.
The transistor must withstand these reverse bias voltages.

8. Lecture 2626 - 8Junction Breakdown - BJT has two diodes back-to-back. Each diode has a
breakdown. The diode (BE) with higher doping concentrations has the lower
breakdown voltage (5 to 10 V).
In forward active region, BC junction is reverse biased.
In cut-off region, BE and BC are both reverse biased.
The transistor must withstand these reverse bias voltages.

9. Lecture 2626 - 9Junction Breakdown - BJT has two diodes back-to-back. Each diode has a
breakdown. The diode (BE) with higher doping concentrations has the lower
breakdown voltage (5 to 10 V).
In forward active region, BC junction is reverse biased.
In cut-off region, BE and BC are both reverse biased.
The transistor must withstand these reverse bias voltages.

10. Lecture 2626 - 10Minority Carrier Transport in Base Region

Inj.
Elec.
recombined electrons
Coll.
Elec.
i
T
I
F/
β
F
I
R/
β
R
I
REC
NP
N
EmitterBase Collector
Space Charge regions
(p
no
,n
po
)
+-
+-
i
E
i
B
i
C
v
BE
v
BC
n(x)
x
n(W
B
)
W
B
i
T
qAD
n
dn
dx
------ =
n(0)
0
(p
no
,n
po
)
Electron conc.
in base (neglects
recombination)
Inj.
Holes
I
REC
n0()n
bo
v
BE
V
T
-------- -



exp =

11. Lecture 2626 - 11Transport current i
T
results from diffusion of minority carriers (holes in npn)
across base region.
Base current i
B
is composed of holes injected back into E and C and I
REC
needed to replenish holes lost to recombination with electrons in B.
The minority carrier concentrations at two ends of base are
and where is the equilib-
rium electron density in the base region.
The junction voltages establish a minority carrier concentration gradient at
ends of base region. For a narrow base, we get
is the B width; is the cross-sectional area of B region.
The saturation current is
n
bo
W
B
A

12. Lecture 2626 - 12Transport current i
T
results from diffusion of minority carriers (electrons in
npn) across base region.
Base current i
B
is composed of holes injected back into E and C and I
REC
needed to replenish holes lost to recombination with electrons in B.
The minority carrier concentrations at two ends of base are
and where is the equilib-
rium electron density in the base region.
The junction voltages establish a minority carrier concentration gradient at
ends of base region. For a narrow base, we get
is the B width; is the cross-sectional area of B region.
The saturation current is
n0()n
bo
v
BE
V
T
---------



exp =
nW
B
()n
bo
v
BC
V
T
---------



exp =
n
bo
W
B
A

13. Lecture 2626 - 13Transport current i
T
results from diffusion of minority carriers (holes in npn)
across base region.
Base current i
B
is composed of holes injected back into E and C and I
REC
needed to replenish holes lost to recombination with electrons in B.
The minority carrier concentrations at two ends of base are
and where is the equilib-
rium electron density in the base region.
The junction voltages establish a minority carrier concentration gradient at
ends of base region. For a narrow base, we get
.
is the B width; is the cross-sectional area of B region.
The saturation current is
n0()n
bo
v
BE
V
T
---------



exp =
nW
B
()n
bo
v
BC
V
T
---------



exp =
n
bo
i
T
qAD
n
dn
dx
------qAD
n
n
bo
W
B
--------
v
BE
V
T
---------


v
BC
V
T
---------



exp – exp



– ==
W
B
A

14. Lecture 2626 - 14Transport current i
T
results from diffusion of minority carriers (holes in npn)
across base region.
Base current i
B
is composed of holes injected back into E and C and I
REC
needed to replenish holes lost to recombination with electrons in B.
The minority carrier concentrations at two ends of base are
and where is the equilib-
rium electron density in the base region.
The junction voltages establish a minority carrier concentration gradient at
ends of base region. For a narrow base, we get
.
is the B width; is the cross-sectional area of B region.
The saturation current is .
n0()n
bo
v
BE
V
T
---------



exp =
nW
B
()n
bo
v
BC
V
T
---------



exp =
n
bo
i
T
qAD
n
dn
dx
------qAD
n
n
bo
W
B
--------
v
BE
V
T
---------


v
BC
V
T
---------



exp – exp



– ==
W
B
A
I
S
qAD
n
n
bo
W
B
--------qAD
n
n
i
2
N
AB
W
B
-------------------- ==

15. Lecture 2626 - 15Base Transit Time

Forward transit time is time associated with storing charge Q in Base region
and it is
with .
Using we get

τ
F
Q
i
T
---- =
QqAn0()n
bo
– []
W
B
2
-------- =
QqAn
bo
v
BE
V
T
---------



1– exp


W
B
2
-------- =

16. Lecture 2626 - 16
Using we get
and .
Q
n(x)
x
n
bo
n(0)
n(
W
B
)=n
bo
0
W
B
Q = excess minority
charge in Base
QqAn
bo
v
BE
V
T
---------



1– exp


W
B
2
-------- =
i
T
qAD
n
W
B
-------------- -n
bo
v
BE
V
T
---------



1– exp



=
τ
F
W
B
2
2D
n
----------
W
B
2
2V
T
µ
n
---------------- - ==

17. Lecture 2626 - 17This defines an upper limit on frequency . f
1
2πτ
F
------------ - ≤
Q
n(x)
x
n
bo
n(0)
n(
W
B
)=n
bo
0
W
B
Q = excess minority
charge in Base

18. Lecture 2626 - 18PSPICE EXAMPLE *Libraries:
* Local Libraries :
.LIB ".\example10.lib"
* From [PSPICE NETLIST] section of C:\Program Files\OrcadLite\PSpice\PSpice.ini file:
.lib "nom.lib"
*Analysis directives:
.DC LIN V_V1 0 5 0.05
+ LIN I_I1 10u 100u 10u
.PROBE V(*) I(*) W(*) D(*) NOISE(*)
.INC ".\example10-SCHEMATIC1.net"
**** INCLUDING example10-SCHEMATIC1.net ****
* source EXAMPLE10
Q1
V1
0Vdc
BF=100
I1
0Adc
IS=1.0e-15
VAF=80

19. Lecture 2626 - 19PSPICE EXAMPLE (Cont’d) Q_Q1 N00060 N00159 0 Qbreakn
V_V1 N00060 0 0Vdc
I_I1 0 N00159 DC 0Adc
**** RESUMING example10-SCHEMATIC1-Example10Profile.sim.cir ****
.END
**** BJT MODEL PARAMETERS
******************************************************************************
Qbreakn
NPN
IS 1.000000E-15
BF 100
NF 1
VAF 80
BR 3
NR 1
VAR 30
CN 2.42
D .87
JOB CONCLUDED
TOTAL JOB TIME .21

20. Lecture 2626 - 20PSPICE EXAMPLE (Cont’d)
V_V1
0V
0.5V
1.0V
1.5V
2.0V
2.5V
3.0V
3.5V
4.0V
4.5V
5.0V
-I(V1)
-4mA
0A
4mA8mA
12mA