Transistors
Lesson #8
Chapter 4
BME 372 Electronics I –
J.Schesser
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Configurations of the BJT
• Common Emitter and Emitter Follower
Collector
C
o
B
iB
o
Base
+
vCE
-
+
vBE -
o
iC
iE
E
Emitter
npn
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J.Schesser
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Configurations of the BJT
• Common Collector
Emitter
E
o
+
B
iE
iB vBE vCE
o
Base
+
o
ic
C
Collector
• Common Base
npn
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Characteristics of the BJT
npn
Common Emitter Configuration
iC
iB
100μA
25
iB
10mA
20
50μA
5mA
15
vBE
.6
.8
20V
1
Base-emitter junction looks
like a forward biased diode
vCE
Collector-emitter is a family of
curves which are a function of
base current
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BJT Equations
iE  iC  iB
Collector
C
iC

iE
o
iB
B
o
Base
iB  (1   )iE

iC 
iB   iB
1


1
+
vCE
-
+
vBE -
o
Ic=β iB =α iE
iE
E
Emitter
Since α is less than unity then β
will be greater than unity and
there is current gain from base
to collector.
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Example
• Calculate the values of β and α from the
transistor shown in the previous graphs.
iC
10mA
β = ic / ib
= 5m/50µ = 100 5mA
α = β /(β+1) = 100/101
=.99
BME 372 Electronics I –
J.Schesser
100μA
iB
50μA
20V
vCE
230
BJT Analysis
RC
2k
+
RB
VCC
10V
Vin
+
50k
-
-
1.6V
+
VBB
Here is a common emitter BJT amplifier:
What are the steps?
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BJT Analysis – Inputs and Outputs
RC
Vin
ib
50k
-
1.6V
+
iC
+
+
RB
+
2k
+
VBE
-
VCE
-
VCC
10V
-
VBB
We would want to know the collector current (ic), collector-emitter voltage
(VCE), and the voltage across RC.
To get this we need to fine the base current (ib) and the base-emitter voltage
(VBE).
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BJT Analysis – Input Equation
RC
Vin
iB
50k
-
1.6V
+
iC
+
+
RB
+
2k
+
VBE
-
VCE
VCC
10V
-
-
VBB
To start, let’s write KVL around the base circuit.
Vin(t) + VBB = iB(t)RB + VBE(t)
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BJT Analysis – Output Equations
RC
Vin
iB
50k
-
1.6V
+
iC
+
+
RB
+
2k
+
VBE
-
VCE
VCC
10V
-
-
VBB
Likewise, we can write KVL around the collector circuit.
VCC= iC(t)RC + VCE(t)
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BJT Analysis
Use Superposition: DC & AC sources
• Note that both equations are written so as to calculate the
transistor parameters (i.e., base current, base-emitter
voltage, collector current, and the collector-emitter voltage)
for both the DC signal and the AC signal sources.
• Let’s use superposition, calculate the parameters for each
separately, and add up the results
– First, the DC analysis to calculate the DC Q-point
• Short Circuit any AC voltage sources
• Open Circuit any AC current sources
– Next, the AC analysis to calculate gains of the amplifier.
• Depends on how we perform AC analysis
– Graphical Method
– Equivalent circuit method for small AC signals
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BJT DC Analysis
•
Using KVL for the input and output circuits and the
transistor characteristics, the following steps apply:
1. Draw the load lines on the transistor characteristics
2. For the input characteristics determine the Q point for the
input circuit from the intersection of the load line and the
characteristic curve (Note that some transistor do not need
an input characteristic curve.)
3. From the output characteristics, find the intersection of
the load line and characteristic curve determined from the
Q point found in step 2, determine the Q point for the
output circuit.
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BJT DC Analysis
Base-Emitter Circuit Q point
39
34
VBE = 0;
iB = VBB / RB
= 1.6 /50k = 32A
iB
 amps
29
24
19
14
9
4
Q POINT
IBQ =
20A
iB = 0; VBE = VBB = 1.6 V
VBEQ = 0.6 V
-1
-0. -0. -0. 0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1 1.1 1.2 1.3 1.4 1.5 1.6 1.7 1.8
3 2 1
v BE volts
First let’s set Vin(t) =0 to get the Q-point for the BJT.
We start with the base circuit.
VBB = iBRB + VBE
And the intercepts occur at iB = 0; VBE = VBB = 1.6 V
and at VBE = 0; iB = VBB / RB = 1.6 /50k = 32A
The Load Line intersects the Base-emitter characteristics at VBEQ = 0.6 V and IBQ = 20A
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BJT DC Analysis
Collector-Emitter Circuit Q point
7.E-03
VCC  5 ma
RC
i C amps
6.E-03
iB=50 μa
40 μa
5.E-03
30 μa
4.E-03
3.E-03
20 μa
Q POINT
I CQ  2.5 ma
2.E-03
10 μa
1.E-03
VCEQ = 5.9 V
0 μa
0.E+00
0
2
4
6
v CE volts
8
10
VCC  10 v
Now that we have the Q-point for the base circuit, let’s proceed to the collector circuit.
VCC= iCRC + VCE
The intercepts occur at iC = 0; VCE = VCC = 10 V; and at VCE = 0; iC = VCC / RC = 10 /2k = 5mA
The Load Line intersects the Collector-emitter characteristic, iB=20mA at VCEQ = 5.9 V and ICQ = 2.5mA
β = 2.5m/20m = 125
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BJT DC Analysis
Summary
• Calculating the Q-point for BJT is the first step in
analyzing the circuit
• To summarize:
– We ignored the AC (variable) source
• Short circuit the voltage sources
• Open Circuit the current sources
– We applied KVL to the base-emitter circuit and using load line
analysis on the base-emitter characteristics, we obtained the base
current Q-point
– We then applied KVL to the collector-emitter circuit and using
load line analysis on the collector-emitter characteristics, we
obtained the collector current and voltage Q-point
• This process is also called DC Analysis
• We now proceed to perform AC Analysis
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BJT AC Analysis
• How do we handle the variable source Vin(t) ?
• When the variations of Vin(t) are large we will use
the base-emitter and collector-emitter characteristics
using a similar graphical technique as we did for
obtaining the Q-point
• When the variations of Vin(t) are small we will
shortly use a linear approach using the BJT small
signal equivalent circuit.
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BJT AC Analysis
• Let’s assume that Vin(t) = 0.2 sin(ωt).
• Then the voltage sources at the base vary from a
maximum of 1.6 + .2 = 1.8 V to a minimum of
1.6 -.2 = 1.4 V
• We can then draw two “load lines” corresponding
the maximum and minimum values of the input
sources
• The current intercepts then become for the:
– Maximum value: 1.8 / 50k = 36 µa
– Minimum value: 1.4 / 50k = 28 µa
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BJT AC Analysis Base-Emitter Circuit
39
34
iB
 amps
“load line” for minimum input voltage
29
24
19
14
9
4
“load line” for maximum input voltage
IBQ =
20μA
VBEQ = 0.6 V
-1
-0. -0. -0. 0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1 1.1 1.2 1.3 1.4 1.5 1.6 1.7 1.8
3 2 1
v BE volts
From this graph, we find:
At Maximum Input Voltage: VBE = 0.63 V, iB = 24μa
At Minimum Input Voltage: VBE = 0.59 V, iB = 15μa
Recall: At Q-point: VBE = 0.6 V, iB = 20μa
Note the asymmetry around the Q-point of the Max and Min Values for the base current and
voltage which is due to the non-linearity of the base-emitter characteristics
ΔiBmax=24-20=4μa; ΔiBmin=20-15=5μa
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BJT AC Analysis Base-Emitter Circuit
39
34
iB
 amps
“load line” for minimum input voltage
29
24
19
14
9
“load line” for maximum input voltage
IBQ =
20A
VBEQ = 0.6 V
4
-1
-0. -0. -0. 0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1 1.1 1.2 1.3 1.4 1.5 1.6 1.7 1.8
3 2 1
v BE volts
2
1.8
1.6
1.4
1.2
1
0.8
0.6
0.4
0.2
0
0.00004
0.000035
0.00003
0.000025
0.00002
0.000015
vin
ib
0.00001
0.000005
0
0
10
20
30
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BJT Characteristics-Collector Circuit
7.E-03
i C amps
iB=50 μa
6.E-03
40 μa
5.E-03
30 μa
4.E-03
3.E-03
24 μa
20 μa
15 μa
10 μa
Q POINT
I CQ  2.5 ma
2.E-03
1.E-03
VCEQ = 5.9 V
0 μa
0.E+00
0
2
4
6
8
10
v CE volts
Using these maximum and minimum values for the base current on the collect circuit load line, we find:
At Maximum Input Voltage: VCE = 5 V, iC = 2.7ma
At Minimum Input Voltage: VCE = 7 V, iC = 1.9ma
Recall: At Q-point: VCE = 5.9 V, iC = 2.5ma
Note that in addition to the asymmetry around the Q-point there is an inversion between the input voltage and
the collector to emitter voltage
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BJT Characteristics-Collector Circuit
7.E-03
i C amps
iB=50 μa
6.E-03
40 μa
5.E-03
30 μa
4.E-03
3.E-03
24 μa
20 μa
15 μa
10 μa
Q POINT
I CQ  2.5 ma
2.E-03
1.E-03
VCEQ = 5.9 V
0 μa
0.E+00
0
2
4
6
8
10
v CE volts
0.025
0.00004
0.000035
0.02
4
8
3.5
7
0.00003
0.000025
0.015
0.00002
0.01
0.000015
3
6
ic
2.5
5
ib
2
4
1.5
3
1
2
0.5
1
0
0
0.00001
0.005
0.000005
0
0
0
10
20
30
0
BME 372 Electronics I –
J.Schesser
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20
30
245
vin
vce
BJT AC Analysis
Amplifier Gains
• From the values calculated from the base and collector
circuits we can calculate the amplifier gains:
– β = 125
– Current gain = Δic / Δib = (2.7 – 1.9) m / (24 -15) μ
= 0.8/9 ✕ 103 = 88.9
– Voltage gain = Vo / Vi = ΔVCE / ΔVBE
= (5 – 7) / (.63 - .59) = -2/0.04 = - 50
– Voltage gain = Vo / Vs = ΔVCE / ΔVin
= (5 – 7) / .4 = -2 / .4 = - 5
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BJT AC Analysis Summary
• Once we complete DC analysis, we analyze the
circuit from an AC point of view.
• AC analysis can be performed via a graphical
processes
– Find the maximum and minimum values of the input
parameters (e.g., base current for a BJT)
– Use the transistor characteristics to calculate the output
parameters (e.g., collector current for a BJT).
• Calculate the gains for the amplifier
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BJT Characteristics - Distortion
7.E-03
i C amps
iB=50 μa
6.E-03
40 μa
5.E-03
30 μa
4.E-03
I CQ  2.5 ma
24 μa
Q POINT
3.E-03
20 μa
2.E-03
15 μa
10 μa
VCEQ = 5.9 V
1.E-03
0 μa
0.E+00
0
2
4
6
8
10
v CE volts
Note that even though the input
(base current) is symmetrical
about its Q-point, the output
voltage is uneven.
BME 372 Electronics I –
J.Schesser
This is due to the nonlinear spacing of the
characteristic curves
and leads to signal
distortion
248
The pnp Transistor
• Basically, the pnp transistor is similar to the npn
except the parameters have the opposite sign.
– The collector and base currents flows out of the
transistor; while the emitter current flows into the
transistor
– The base-emitter and collector-emitter voltages are
negative
• Otherwise the analysis is identical to the npn
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PNP Bipolar Junction Transistors
•
Two junctions
– Collector-Base and Emitter-Base
•
Collector
Biasing
– vBE Forward Biased
– vCB Reverse Biased
Collector
C
o
iB
p-type
B
o
Base
n-type
Base
vCE
+
vBE +
o
iC
RB
iE
Vin
+
10V
50k
-
E
p-type
RC 2k
VCC
+
1.6V
-
+
Emitter
pnp
Emitter
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Homework
• Probs. 4.4, 4.5, 4.8, 4.10, 4.14, 4.15, 4.19,
4.20, 4.21, 4.22,
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