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8/12/2019 Chapter 13 Linear Amplifiers BJT-1-1
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A :
A
A / C
A (C )
, .
, :
,
, .
.
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(.) A
A B
, ,
B ( )
, ,
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The BJT is biased in the active region by dc voltage
source VBE. Q-point is set at (IC, VCE) = (1.5 mA, 5 V)
withIB = 15 A (F = 100)
Total base-emitter voltage is: vBE= VBE+ vbe
Collector-emitter voltage is: vCE= 10 iCRC This
is the load line equation.
Assuming:
VBE=0.7 V,IB=15A
(peak) = 8 mV Causes 5A change in vBE
varies between [0.70.008 and 0.7 + 0.008] = [0.692 and 0.708]iB varies between [15A 5A and 15A + 5A] = [10A and 20A]
iC= iB iC varies between [1 mA and 2mA]
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If changes in operating currents and voltages
are small enough, then iCand vCEwaveforms
are undistorted replicas of the input signal.
A small voltage change at the base causes a
large voltage change at collector.
is 180o out of phase of
Assuming:
VBE=0.7 V,IB=15A
(peak) = 8 mV Causes 5A change in vBE
varies between [0.70.008 and 0.7 + 0.008] = [0.692 and 0.708]iB varies between [15A 5A and 15A + 5A] = [10A and 20A]
iC= iB iC varies between [1 mA and 2mA]
vCE varies between [3.4 V and 6.7V] (peak) =
= 1.65V
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Av =Vce
Vbe=
1.65180o
0.0080o= 206180o = 206
180
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C
.
1 2
.
3
,
(
)
.
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A
, , , 3 .
C
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==
==
k100k22
k300k100
3
21
RRL
R
RRR
C
B
Capacitors are replaced by short circuits
C
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:
rd=1
gd
:
dDDi
dDD
iI
vVv
+=
+=
:
8/12/2019 Chapter 13 Linear Amplifiers BJT-1-1
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= 0
. A ,
:
gd = IS
VT and rd =
VT
IS
8/12/2019 Chapter 13 Linear Amplifiers BJT-1-1
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iD =IS exp vDVT
1
ID+id=IS exp
VD+vdVT
1
ID+i
d=I
Sexp
VDVT
1
+ISexp
vD
VT
vd
VT
+12
vd
VT
2
+16
vd
VT
3
+...
SubtractingID from both sides of the equation,
id= (I
D+I
S)
vd
VT
+12
vd
VT
2
+16
vd
VT
3
+...
For idto be a linear function of signal voltage vd, vd
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( )
The bipolar transistor is assumed to be operating in the Forward-Active Region:
Using a two-porty-parameter network:
The port variables can represent either time-varying part of total voltages and
currents or small changes in them away from Q-point values.
ic = gmvbe + govce
ib = gvbe + grvce
iC IS exp vBE
VT
1+
vCE
VA
and iB
iC
F VCE VBE( )
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&
o= g
mr= F
1IC
1
F
F
iC
Qpo int
IM is the collector current at which is maximum.
o >Ffor iC
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B
Transconductance:
Input resistance:
Output resistance:
gm=IC
VT
40IC
r=oVT
IC=ogm or
o=
gmr
ro=VA+
VCEIC
VAIC
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,
B B
.
vbe = ibr
gmvbe =gmibr =oib
ic =oib +vce
rooib
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For linearity, ic should be proportional to vbe with vbe
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.
A .
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DefineRiB as the input resistance looking into
the base of the transistor:
The input resistance presented to vi is:
The signal source voltage gain is:
(o +1)RE
RiB
=v
b
ib= r
Rin=R
I+R
B R
iB=R
I+R
B r
Av
CE=v
o
vi
=v
o
vb
vb
vi
=Avt
CE R
B r
RI+RB r
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AvCE=Avt
CE RB r
RI +RB r
AvtCE Avt
CE= gmRL RL = ro RC R3
Typically: ro >>RC and R3 >>RC AvCE gmRC = 40ICRC
ICRCrepresents the voltage dropped across collector resistor RC
A typical design point is ICRC =VCC
3
AvCE 40
VCC
3
=13.3VCC
To help account for all the approximations and have a number that is easy to
remember, our "rule-of-thumb" estimate for the voltage gain becomes
AvCE
10VCC
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C ,
= 100, = 75 , (0.245 A, 3.39 )
,
=. . .
gm = 40IC = 40 0.245mA( ) = 9.80 mS r =ogm =
100
9.8mS=10.2 k
ro =VA +VCE
IC=
75V+3.39V
0.245mA= 320 k RB =R1 R2 =160k 300k=104 k
RL = ro RC RL = 320k 22k 100k=17.1 k RB r =104k 10.2k= 9.29 k
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(.)
Av = gmRLRB r
RI +RB r
= 9.8mS 17.1k( )
9.29k
1k+ 9.29k
= 168 0.903( ) = 151
Rin =RI +RB r =10.3 k
vbe = viRB r
RI +RB r
vbe 0.005V vi 5mV
10.3k
9.29k
= 5.54 mV
Check the rule-of-thumb estimate: AvCE
10 12( ) = 120 (ballpark estimate)What is the maximum amplitude of the output signal: vo 5.54mV 151( ) = 0.837 V
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(.)
5, 10 .
,
.
.
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,
= 65, = 50
A , = 0.7 , .
Analysis: To find the Q-point, the
dc equivalent circuit is constructed.
IB=3.71 A
IC=65I
B=241 A
IE=66I
B=245 A
105IB+VBE+(F+1)IB(1.6104)=5
5104ICV
CE(1.6104)I
E(5)=0
VCE
=3.67 V
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.
Rin=R
Br=6.31 k
Rout =RC ro = 9.57 k
AvCE=
vo
v i= gm Rout R3( )
Rin
RI +Rin
= 84.0
Gain Estimate: AvCE
10 VCC+VEE( ) = 100
6.74 k
()
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.
(a) Total power dissipated in the C-B and E-B
junctions is:
PD = VCEIC+ VBEIB where VCE = VCB + VBE
Total power supplied is:
PS = VCCIC + VEEIE
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vCE =VCE Vm sin t where Vm is the
output signal. Active region operationrequires vCE vBE So: Vm VCE VBE
Also: vRc t( ) =ICRCVm sint 0
Vm
min ICRC, VCE
VBE( )
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