EE310
Solved Problems on BJT
Sedra/Smith 5th/6th ed.
By Turki Almadhi,
EE Dept., KSU,
Riyadh, Saudi Arabia
25/07/36
I E  3 mA; I SE 
I E  I SE e
VBE
VT
1

IS 
 1
I  1.01I S
 S
 VBE  VT  ln(
IE
)  0.747 V
I SE
VBE  VB  VE  0.747  0  VE  VE  0.747 V
IC   I E 

 1
I E  0.9901 3  2.9703 mA
VC  10  2.9703  2  4.0594 V (which verifies active mode)
I B  I E  I C  3  2.9703  29.7 103 mA=29.7  A
VEC  5, that means the pnp transistor is operating in the active mode.
Given that I E1  10 A  I B1 
I C1   I B1  15 
I C1  I S 1 e
VBE
VT
10
10

 0.625 A
  1 16
10
 9.375 A;
16
 I S 1  I C1 e
VBE
VT
0.85
 I S 1  9.375  e 2610  59.4 1015 A
3
AEBJ 1
59.4 1015 A 59.4 10 15


 29.3
VBE
15
AEBJ 2
2.03

10
I S 2  I C 2 e VT
The power BJT has an emitter-base junction area 29.3 times larger than the
small signal BJT.
a)
10.7  0.7
 1 mA
10
 is very large, we can assume I B is 0 
I1  I E 
I C  I E  1 mA
V2  10 1  (10.7)  0.7 V .
d)
Equating the collector and emitter currents:
IC  I E
10  V6 (V6  0.7)  (10)

 10  V6  3V6  27.9 
15
5
4V6  17.9  V6  4.475 V.
IC 
10  (4.475)
 0.965 mA = I E  I 5
15
To be able to find  , we must find two of the three currents:
I B , I C , and I E .
10  7
 3 mA.
1
The current following into the lower 1-k resistor is exactly
IE 
equal to I E ; why?
6.3  3
 0.033 mA
100
(I B is flowing out of the base for a pnp transistor.)
VC  3  1  3 V  I B 
I E  (   1) I B   
IE
 1  90.
IB
Assuming the transistor is in active mode:
0.8  (3) 2.2
VE  0.8  I E 

 1 mA
2.2
2.2
I
1
I B  E   19.61103 mA
  1 51
I C   I B  50  33.78 103  0.980 mA
VB  0
VC  3  2.2  0.980  0.844 V.
VBC  0  (0.844)  0.844  0.4  the CBJ is
reverse-biased  the transistor is in active mode as assumed!
(i) Note: the negative value of VB indicates that the base current
is going (into) the base which is the right direction for an npn BJT.
0  (1.5)
 0.15 mA
10
(current is in mA because the resistance is in k.)
VE  VB  0.7  1.5  0.7  2.2 V
IB 
VE  (9) 2.2  9 6.8


 0.68 mA
10
10
10
I C  I E  I B  0.68  0.15  0.53 mA
IE 
VC  9  0.53 10  3.7 V  VBC  1.5  3.7  5.2 V  0.4 V,
which means the transistor is operating in the active mode.

I C 0.53
I

3.5333

 3.5333   

 0.7794  C
I B 0.15
  1 4.5333
IE
(ii) I B  0  VB  0  VE  0.7 V  VC  0.7 V
(i) R B  100 k
Assuming the transistor is in active mode:
(   1) I B  I E
5  (0.7  VE ) VE
100

 4.3  VE 
VE
100
1
101
2.16
 VE  2.16 V  I E 
 2.16 mA
1
VB  VE  0.7  2.16  0.7  2.86 V
101
100
VC  5  1 I C  5  (
)  I E  2.86 V
101
VBC  VB  VC  0  0.4 V  the BJT is in active mode as assumed.
(ii) R B  10 k
Assuming the transistor is in active mode:
(   1) I B  I E
5  (0.7  VE ) VE
10

 4.3  VE 
VE
10
1
101
3.91
 VE  3.91 V  I E 
 3.91 mA
1
VB  VE  0.7  3.91  0.7  4.61 V
101
100
VC  5  1 I C  5  (
)  I E  1.13 V
101
VBC  VB  VC  3.48 V  0.4 V !  the BJT is saturated.
Restarting, and considering VCE ( sat )  0.2 V :
I E  IC  I B
VE 5  VC 5  VB


1
1
10
VE 5  (0.2  VE ) 5  (0.7  VE )


1
1
10
VE  4.8  VE  0.43  0.1VE  2.1VE  5.23  VE  2.49 V.
VC  0.2  VE  2.69 V;
VB  0.7  VE  3.19 V.
Useful relationships:
I B ( EdgeOfSaturation )  I B ( EOS ) 
Over Drive Factor (ODF)
I C ( sat )
 min
;  forced 
I C ( sat )
IB
;
IB
I B ( EOS )
5  0.2
 4.8 mA
1
I C ( sat ) 4.8


 0.24 mA  I B  I B ( EOS )  ODF  2.4 mA
 min
20
I C ( sat ) 
I B ( EOS )
IB 
5  0.7
4.3
 RB 
 1.792 k
RB
2.4
 forced 
I C ( sat )
IB

4.8 mA
 2;
2.4 mA
Note that  forced is only defined in saturation and it changes with I B and
always is less than  min .
We first draw the DC equivalent circuit (all caps are open-circuited), then
we find Thevenin's equivalent looking out of the base.
Rth  R1 R2  27 k 15 k 
Vth 
27 15
 9.64 k
27  15
R2
15
 VCC 
 9  3.21 V
R1  R2
27  15
Assuming the transistor is in active mode:
(   1) I B  I E
101
3.21  (0.7  VE ) VE

 2.51  VE  0.0796VE
9.64
1.2
DC equivalent circuit
 VE  2.325 V  I E 
2.325
 1.9375 mA;
1.2
100
IC  (
)  I E  1.918 mA
101
VB  VE  0.7  2.325  0.7  3.025 V
VC  9  2.2  I C ;  9  2.2  1.918  4.78 V
VBC  VB  VC  3.025  4.78  -1.755 V  0.4 V
 the BJT is in active mode as assumed.
I
1.918
gm  C 
 73.8 mS;
VT
26
r 
VT
V
 100
 T 

 103  1.355 k
IB
I C g m 73.8
Rin  R1 R2 r  Rth r  9.64 1.355  1.18 k
vbe
Rin
1.19


 0.106
vsig Rin  Rsig 1.19  10
vo   g m vbe  ( RC RL )  77.3vbe 
AV 
vo
 77.3
vbe
vo
v
v
 o  be  0.106  (77.3)  8.2 V/V
vsig vbe vsig
vo io  RL
i
R
v

 77.3  o  in  o  45.6 A/A
vbe ii  Rin
ii RL vbe
Small signal equivalent circuit