Chapter 9 - BJT Bipolar Junction Transistor
Chapter 9
Bipolar Junction Transistor
•
•
•
•
BJT Characteristics
NPN, PNP BJT
DC Biasing
Collector Characteristic and Load Line
Bipolar Junction Transistor (BJT)
BJT is a three-terminal device, which consists a collector (C), an emitter (E)
and a base (B). There are two types of BJTs: NPN type and PNP type. Fig. 1
shows the symbols of both NPN and PNP types BJTs.
C
iB
C
iC
B
B
E
C
N
iB
P
B
C
E
P
B
N
iE
iC
E
iE
N
P
E
PNP
NPN
The terminal currents are illustrated in the above figure. It should be noted
that iE is always greater than iC and iB.
1. Characteristics of BJT
• BJT is a current-controlled device; the values of iC and iE are determined
primarily by iB.
• iC = β iB, where β is the ratio between the collector and base currents.
• iE = iC +iB = (1 + β) iB . For some cases, β can be written as hFE. In practice, β
is much greater than 1. So, iE ≈ iC.
• VCE, VBE and VCB are the voltages across collector-to-emitter, base-to-emitter,
and collector-to-base, respectively.
-1-
Chapter 9 - BJT Bipolar Junction Transistor
• Emitter is heavily doped and collector is relatively lightly doped with the
same material. The base is with the smallest size and lightest doped in a BJT.
2. Operation of a NPN BJT
Fig. 1 shows a NPN BJT, which is connected to two external dc source and
they are VBE and VCE. This circuit configuration is so called common emitter
configuration.
(1) VC > VB
(2) VBE is kept
constant
RB
IB
RC
VRC
By using KVL, two equations
can be obtained
IC
VB = VBE + I B RB
VC
VCE
VB
VBE = 0.7V
VC = VCE + I C RC
Increasing VB causes
decreasing VCE
IE
Fig. 1 Common Emitter Configuration
•
When IB increases, IC and VRC also increase. So, VCE decreases.
•
Conversely, decreasing IC causes increasing of VCE.
•
So, if VB is an ac signal, e.g. sinusoid, VCE will be an inverted sinusoid.
•
Such arrangement is rarely used for actual application.
3. DC load line and Collector Characteristic Curve
DC load line – a graph that represents all the possible combinations of IC and
VCE. DC load line is constructed from the equation
VCC = VCE + ICRC
RC is a fixed value
When IC = 0, VCE (off)= VCC. When VCE = 0, IC (sat)= VCC/RC.
-2-
Chapter 9 - BJT Bipolar Junction Transistor
Collector Characteristic Curve
DC load line
Collector Characteristic Curve illustrates the relationship among IC, IB and VCE.
When IB is fixed, a constant amount of IC is drawn from the supply source. The
characteristic curve shown in the last page illustrates that different values of RC
give different VCE.
Fig. 2 Combination of dc load line and collector characteristic curve
The graph shown in Fig. 2 is used to determine the operating point (Q point)
of a BJT. Once RC is determined, the dc load is kept unchanged. The intersection
between the dc load line and the collector characteristic curve is the operating
point (Q-point). That means, when the BJT is subject to an input current IB,
-3-
Chapter 9 - BJT Bipolar Junction Transistor
respective VCE and IC can be obtained at the collector terminal of BJT. With the
use of this graph, we can determine the desired operating condition for a BJT. In
practice, Q-point is selected at the mid-point of VCC and IC.
4. DC Biasing
The ac operation of an amplifier depends on its initial values of IB, IC, and
VCE. The function of dc biasing is to set the initial value of IB, IC, and VCE. Two
dc biasing methods are introduced: base bias and voltage divider bias.
Fig. 3(a) and 3(b) show the circuit configurations of base bias and voltage
divider bias, respectively. The primary goal of circuit analysis is to determine
the Q-point values of IC and VCE for a given IB.
Fig. 3(a) Base bias
Fig. 3(b) Voltage divider bias
Base bias
Fig. 3(a) shows a base-biased BJT circuit. After applying KVL, two
equations can be formulated:
VCC = I B RB + VBE ,
VCC = I C RC + VCE
The base current IB can be obtained as, I B =
VCC − VBE
RB
So, VCE can be obtained as, VCE = VCC − β I B RC where I C = β I B .
-4-
Chapter 9 - BJT Bipolar Junction Transistor
We can say that the obtained IC and VCE equal to ICQ and VCEQ. The subscript
Q means the Q-point. Appropriate choosing the value of RB can adjust the
location of Q-point at the midpoints of VCE and IC.
Disadvantage: The value of β is temperature dependent. IC could be changed
under different operating temperature such that the Q-point could shift along the
dc load line. Such problem is called Q-point shift. When an ac signal is injected
to the base terminal, the resultant VCE will be either saturated or cutoff.
Voltage-divider bias
Voltage-divider bias circuit is by far the most commonly used. Fig. 3(b)
shows the circuit configuration. The analysis procedures are slightly different for
different ratios between R2 and RE.
Case 1: R2 ≤ 0.1 β RE
From Fig. 3(b), we can formulate
VB = VCC ×
R2
where VB is the voltage drop across R2.
R1 + R2
(1)
The voltage of VE can be found as
VE = VB – VBE = VB – 0.7V
(2)
Using ohm’s law, the emitter current can be obtained as
I E = VE RE
(3)
As mentioned before, ICQ ≅ IC ≅ IE, VCEQ can be found as
VCEQ = VCC − I CQ ( RC + RE )
(4)
It can be shown that the β term is not involved. That means, Q-point shift is
no longer presence.
-5-
Chapter 9 - BJT Bipolar Junction Transistor
Case 2: R2 ≥ 0.1 β RE
Calculate the parallel equivalent resistance from the base of the BJT to
ground. This resistance is found as
Req = R2 β RE
(5)
Solve for the base voltage VB as follows: VB = VCC
Req
R1 + Req
(6)
After solving for VB, substituting (6) into (2) to continue the sequence of
calculations to find the values of ICQ and VCEQ.
Example 1: Consider the circuit of Fig. 3(a), determine the Q-point values of IC
and VCE when RC = 2kΩ, RB = 360kΩ, VCC = 8V and β = 100.
Solution: IB can be found as
IB =
VCC − VBE (8 − 0.7)V
=
= 20.28 µA
360kΩ
RB
Next, IC is found as
IC = β IB = 100 × 20.28µA = 2.028mA
Finally, VCE is found as
VCE = VCC – ICRC = 8V – (2.028mA)(2kΩ) = 3.94V
The dc load line can be obtained as
VCC = I C RC + VCE ⇒ 8 = 2000 I C + VCE
When IC = 0, VCE(off) = 8V. When VCE = 0, IC(sat) = 8/2000 = 4mA.
IC
IC(sat) = 4mA
ICQ = 2.028mA
IB = 20µA
VCE
VCEQ = 3.94V
-6-
VCE (off) = 8V
Chapter 9 - BJT Bipolar Junction Transistor
Example 2: Consider the circuit of Fig. 3(b), determine the Q-point of ICQ and
VCEQ when R1 = 18kΩ, R2 = 4.7kΩ, RC = 3kΩ, RE = 1.1kΩ, VCC = 10V, and β =
50.
Solution: Check the value R2 and (0.1 β RE),
0.1 × 50 × 1.1kΩ = 5.5kΩ > R2
So, the base voltage VB can be found as
VB = 10V ×
4.7 kΩ
= 2.07 V
18kΩ + 4.7kΩ
VE is found as
VE = VB – VBE = 2.07V – 0.7V = 1.37V
ICQ is then found as
I CQ = VE RE = 1.37 V 1.1kΩ = 1.25mA
∴VCEQ = VCC − I CQ ( RC + RE ) = 10V − (1.25mA)(4.1kΩ) = 4.87 V
The dc load line can be obtained as
VCC = I C ( RC + RE ) + VCE ⇒ 10 = 4100 I C + VCE
When IC = 0, VCE(off) = 10V. When VCE = 0, IC(sat) = 10/4100 = 2.44mA
IC
IC(sat) =2.44 mA
IB = 25µA
ICQ = 1.25 mA
VCE
VCEQ = 4.87V
VCE (off) = 10V
Example 3: Consider the circuit of Fig. 3(b), determine the Q-point of ICQ and
VCEQ when R1 = 68kΩ, R2 = 10kΩ, RC = 6.2kΩ, RE = 1.1kΩ, VCC = 20V, and β =
50.
-7-
Chapter 9 - BJT Bipolar Junction Transistor
Solution: Check the value R2 and (0.1 β RE),
0.1 × 50 × 1.1kΩ = 5.5kΩ < R2
So, the equivalent resistance Req can be found as
Req = R2 β RE = (10kΩ) (55kΩ) = 8.46kΩ
Then, VB can be found as
VB = VCC ×
Req
Req + R1
= 20V ×
8.46kΩ
= 2.21V
68kΩ + 8.46kΩ
VE is found as
VE = VB – VBE = 2.21V – 0.7V = 1.51V
ICQ is then found as
I CQ = VE RE = 1.51V 1.1kΩ = 1.37mA
∴VCEQ = VCC − I CQ ( RC + RE ) = 20V − (1.37mA)(7.3kΩ) = 9.99V
The equation of dc load line is
VCC = VCE + I C ( RC + RE )
⇒ 20 = VCE + 7300 I C
When VCE = 0, IC(sat) = 2.74mA.
When IC = 0, VCE(off) = 20V.
5. Interfacing
Apart from the application of amplification, BJT can be used as a ON/OFF
switch. With the aid of dc load line, two phenomena can be observed:
(1) If IB = 0, IC will be zero and VCE will be equaled to VCC.
(2) If IB is sufficient large, IC will reach to its maximum value and VCE will
become zero.
That means, if a BJT is subject to a “square” pulse train with sufficient large
in magnitude, the measurement of VCE will be an “inverted” square pulse train.
-8-