Session 15
The transistor as a singlestage amplifier (BJT)
Electronic Components and Circuits
José A. Garcia-Souto
www.uc3m.es/portal/page/portal/dpto_tecnologia_electronica/Personal/JoseAntonioGarcia
The transistor as a single-stage
amplifier (BJT)
OBJECTIVES
•
•
•
•
To understand the principle of amplification by BJT.
To know and use small-signal equivalent circuits of BJT.
To know the basic parameters of small signal equivalent
circuit: hfe, β0, gm, rπ, r0. To calculate them from the data
of the bias point.
To analyze small-signal circuits for single-stage BJT
amplifiers: common emitter.
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Concept of amplification with BJT
SATURATES
Vo ≈ 0 V
IC (mA)
60
50
40
AMPLIFIES
Vo = G·Vi
30
20
CUT-OFF
Vo = Vcc
10
0
0,2
0,4
0,6
0,8 VBE (V)
Small changes of Vi result in greater variation of
Vo, thus gain Vo/Vi is provided
It should be around a bias point VBE-Q, VCE-Q
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The transistor as an amplifier
IC
IB (µA)
60
IB
50
VCE
VBE
VBB is a continuous source that
with RB provides a bias point or
vg = 0
polarization:
40
IB
30
IE
20
10
VBE
VBB
VBE (V)
A variable signal is coupled to
VBB
VBB + v
g
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Dynamic load line
IB (µA)
60
Variations in input voltage
result in displacement of the
load line:
50
IBQ
∆IB
Q
40
(VBB + vi ) − vBE
iB =
RB
30
20
10
∆VBE
0
0,2
0,4
0,6
0,8 VBE (V)
VBEQ
Small changes in base-emitter voltage and in base current are
produced around the bias point of the device .
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Load line (II)
Static load line
(Bias point)
Dynamic load line
(Variations in the output)
IC (mA)
6
60 µ A
5
50 µ A
IC 4
IB =40 µ A
3
30 µ A
IC (mA)
6
60µA
5
50µA
ICQ
Q
∆IC
4
∆IB
3
IBQ=40µA
30µA
RECTA DE CARGA ALT.
2
2
20 µ A
10 µ A
1
20µA
∆VCE
10µA
1
0 µA
0
2
4
6
8
10
VCE
VCC − vCE
iC =
RC
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14
16
V CE (V)
0
2
VCC
4
6
8
10 12 14
0µA
16 18
VCE (V)
VCEQ
The collector current varies proportionally to the base
current (and base-emitter voltage).
There are variations of the collector-emitter voltage (output)
amplified related to the input voltage signal.
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Example: common-emitter amplifier
IB (µA)
60
∆Vi ≈ ±200mV
50
∆IB
IBQ
VBEQ ≈ 0,6V
Q
40
I BQ ≈ 40 µA
30
20
∆VBE ≈ ±50mV
10
∆VBE
0
0,2
0,4
0,6
0,8 VBE (V)
∆I B ≈ ±10µA
VBEQ
IC (mA)
6
60µA
5
50µA
EXAMPLE
RC = 3 kΩ
RB = 15 kΩ
VCC = 18 V
VBB = 1,2 V
ICQ
Q
∆IC
4
∆IB
3
VCEQ ≈ 6V
IBQ=40µA
∆I C ≈ ±1mA
30µA
RECTA DE CARGA ALT.
2
10µA
1
0
2
4
6
∆VCE ≈ ±3V
20µA
∆VCE
β = 100
vi = 0,2 V (peak)
I CQ ≈ 4mA
8
10 12 14
0µA
16 18
VCE (V)
VCEQ
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Small-signal variations
IC (mA)
Relationship between variations
of the collector current and
changes in the base-emitter
voltage (transfer curve).
60
50
ICQ
Q
40
30
∆ IC
20
10
∆VBE
VBEQ
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If there are small variations
around the bias point (small
signal) a linear approximation of
transconductance gm can be
established
Small-signal changes in the
transfer function is generalizable
to other transistor devices (FET)
VBE (V)
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Small-signal equivalent circuit
GENERIC: BJT, JFET, MOSFET, Others.
rin
gm
IB (µA)
IC (mA)
60
50
IBQ
Q
40
∆IB
ICQ
30
60
IC (mA) 6
50
5
Q
40
30
20
10
0
ro
∆VBE
0,2
0,4
0,6
VBEQ
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VBE (V)
IC
∆ IC
10
1
∆VBE
CCE - Session 15
20 µ A
10 µ A
3
2
VBE (V)
30 µ A
4
20
VBEQ
IB =40 µA
0 µA
-V A
0
2
4
6
8
10 12 14
V CE (V)
9
Transconductance model of BJT:
π - Hybrid Model
B
C
E
Cπ
n+
Cs
n+
rb
rc
p
Cµ
n
n+
p
BODY
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BJT full model: With parasitic capacitances and
parasitic resistances (Common-Emitter)
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BJT simplified model: Without negligible parasitic
elements and for low frequency (Common-Emitter)
re, rc → 0, ZCs→ ∞
Low frequency: Real(ZCπ), Real(ZCµ ) → ∞
Still simplifiable: rb → 0 , ro → ∞
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π - Hybrid Model
(Ebers-Moll Equations)
 vVBE

iC = I S  e T − 1




Equations:
∂ic
βo =
∂ib
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vce =0,VCEQ
VT =
KT
q
vπ = ib· rπ
vbe = ib·(rπ+rb)
ic = gm·vπ = gm· rπ· ib
∂i
gm = c
∂vbe
vce = 0 ,VCEQ
CCE - Session 15
∂vce
r0 =
∂ic
ib = 0 , I BQ
13
Small Signal Parameters
– gm
– β0
– rπ
– r0
gm =
rπ =
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VT
βo
gm
(Ω −1 )
(Ω )
VT
To = 300 K
≈ 25.6mV ≈ 25mV
βo = gm · rπ
VA
r0 =
( Ω)
I CQ
• Datasheet
– hfe, hie
– Cob, Cib
I CQ
BC547
BD335
2N222
Search BC547, BD335, 2N222 in
http://www.fairchildsemi.com/
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Summary : Small-signal equivalent
gm =
rπ =
I CQ
VT
βo
gm
(Ω −1 )
=
VT
β o ·VT
I CQ
(Ω )
To = 300 K
≈ 25mV
β o = g m ·rπ
VA
(Ω)
r0 =
I CQ
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Example: small signal analysis
RC = 3 kΩ
VBE-on = 0,6 V
RB = 15 kΩ
βF = 100
VCC = 18 V
VBB = 1,2 V
Vg = 0,2 V (pico)
VBEQ ≈ 0,6V
I BQ ≈ 40 µA
(1) BIAS ANALYSIS
Vg = 0V
I CQ ≈ 4mA
VCEQ ≈ 6V
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Example: small signal analysis
βo = 100
VA = 100 V
(2) SMALL SIGNAL
PARAMETERS
4mA
gm =
= 160mA / V
25mV
100·25mV
rπ =
= 625Ω
4mA
r0 =
100V
= 25kΩ
4mA
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(3) SMALL SIGNAL
ANALYSIS
V0 = − g mVπ (r0 // RC )
Vπ = Vg
rπ
RB + rπ
CCE - Session 15
VBB = 0V
VCC = 0V
V0
rπ
= − g m (r0 // RC )
≈ −17V
V
Vg
RB + rπ
V0
= − g m (r0 // RC ) ≈ −428V
V
Vπ
ic
= β o = 100 A
A
ib
17
Small signal analysis of amplifier
circuits
METHODOLOGY
1.
2.
3.
4.
Analyze the bias circuit (DC) with the signal sources
removed (superposition) and the coupling and
decoupling capacitors as open circuits. Obtain the bias
(quiescent) point.
Calculate the small signal parameters of the transistor
with the data of the DC analysis.
Represent the small signal equivalent circuit of the
devices with external signal sources. Cancel DC
sources (superposition). Coupling and decoupling
capacitors as open circuits at mid-frequency.
Obtain the characteristics of the amplifier.
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Exercise: Gm amplifier
DATA:
BJT Transistor
VEB-ON = 0,7 V
VEC-SAT = 0,2 V
VA = 100 V
C→∞
• Calculate RE that makes the current Io through the load RL to be 1 mA.
•What is the operating region and plot the output characteristic and load line.
• Draw the small-signal equivalent circuit. Obtain the small-signal transconductance
gain io/vg (io is the current signal through RL).
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