Lecture 20Lecture 20
Bipolar Junction Transistors (BJT): Part 4
Small Signal BJT Model
Reading:
Jaeger 13.5-13.6, Notes
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Further Model Simplifications(useful for circuit analysis)( y )
Ebers-MollForward Active
Mode
Neglect Small Terms
TEB
TEB
TCB
TEB
VV
SCRV
V
FFCV
V
RV
V
FFC eIIIeIIeIeII
0000 111
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Modeling the Early Effect (non-zero slopes in IV curves)
iB1< iB2< iB3IC Base width changes due to changes in the base-
iB3 (theory)
collector depletion width with changes in VCB.
This changes T, which
iB1 (theory)iB2 (theory)
VA VCE
g T,changes IC, DC and BF
A
Major BJT Circuit Relationships
TEB
TEB
TEB
VV
FO
S
F
CB
A
CEFOF
A
CEVV
SCV
V
SC eIi
iVv
Vv
eIieIi
11
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Small Signal Model of a BJT
Just as we did with a p-n diode, we can break the BJT up into a large signal analysis and a small signal analysis and linearize the non-linear behavior of the Ebers-Moll model.
S ll i l M d l l f l f F d i dSmall signal Models are only useful for Forward active mode and thus, are derived under this condition. (Saturation and cutoff are used for switches which involve very large voltage/current swings from the on to off states.)
Small signal models are used to determine amplifier characteristics (Example: Gain = Increase in the magnitude of a signal at the output of a circuit relative to its magnitude atof a signal at the output of a circuit relative to it s magnitude at the input of the circuit).
Warning: Just like when a diode voltage exceeds a certainWarning: Just like when a diode voltage exceeds a certain value, the non-linear behavior of the diode leads to distortion of the current/voltage curves (see previous lecture), if the i t / t t d t i li it th f ll Eb M ll d l
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inputs/outputs exceed certain limits, the full Ebers-Moll model must be used.
Consider the BJT as a two-port Network
i1 iTwo Port Network
+V1
-
i1 i2+V2
-
i1=y11v1 + y12v2 ib=y11vbe + y12vce
General y-parameter Network BJT y-parameter Network
i2=y21v1 + y22v2 ic=y21vbe + y22vce
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Consider the BJT as a two-port Network
iib=y11vbe + y12vceic=y21vbe + y22vce
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Consider the BJT as a two-port Network
o is most
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often taken as a constant, F
Alternative Representations
CT
Cm
V
IVI
yg 21
1
40
Transconductance
CEA
m
o
C
To
VVgI
Vy
r
11
1
1 Input Resistance
C
CEAo I
VVy
r
22
1Output Resistance
Y-parameter Model Hybrid-pi Model
v1
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Alternative Representations
bobmbem iirgvg
Voltage Controlled Current source version of Hybrid pi
Current Controlled Current source version of Hybrid pisource version of Hybrid-pi
Modelsource version of Hybrid-pi
Model
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Single Transistor Amplifier Analysis: Summary of Procedure Steps to Analyze a Transistor Amplifier
Important!
1.) Determine DC operating point and calculate small signal parameters (see next page)
Step
Step 1
2.) Convert to the AC only model.DC Voltage sources are shorts to groundDC Current sources are open circuitsLarge capacitors are short circuits
Step 2
Large capacitors are short circuitsLarge inductors are open circuits
3.) Use a Thevenin circuit (sometimes a Norton) where necessary. Ideally the
Step 3
Norton) where necessary. Ideally the base should be a single resistor + a single source. Do not confuse this with the DC Thevenin you did in step 1.
Step 4
4.) Replace transistor with small signal model5.) Simplify the circuit as much as necessary
Step 5
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necessary. 56.) Calculate the small signal parameters (r, gm, ro etc) and then gains etc
Single Transistor Amplifier AnalysisDC Bias Point
Important!Step 1 detail
Thevenin
Vbe Ie
Ib
RTH
3V=IERE+Vbe+IBRTH3V=IC((o+1)/o) Re+0.7V+IBRTH3V IC((o 1)/o) Re 0.7V IBRTH
3V=IB o((o+1)/o) Re+0.7V+IBRTH3V= IB(100+1)1300+0.7+ IB7500
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B B
IB=16.6 uA, IC= IB o=1.66 mA, IE=(o+1) c/ o=1.67 mA
Single Transistor Amplifier AnalysisCalculate small signal parameters
I CStep 6 detail Important!
gIV
yr
SIVI
yg
o
C
To
CT
Cm
15061
0664.040
11
21
Transconductance
Input Resistance
KIV
IVV
yr
gIy
C
A
C
CEAo
mC
2.451
22
11
Output Resistance
RTHVTH=0.88 VS
rRL= RC| R3| ro
Vbe gmVbe
Vout880
vvandR
rvvandRvgv SThThbeLbemout 88.0
rR
rRg
vv
vv
vv
vv
GainVoltageA
rRg
ThLm
S
th
th
be
be
out
S
outv
SThTh
ThbeLbemout
88.0
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VVA
A
v
v
ThSthbeS
/139
88.01506880
1506000,100||4300||200,450664.0
For Extra Examples see: Jaeger section 13.6, and
pages 627-630 (top of 630)
Supplement Added to describe more details of theImportant!
Supplement Added to describe more details of the Solution of this Problem
Bipolar Junction Transistors (BJT): Part 5
D t il f A lifi A l iDetails of Amplifier Analysis
Reading:
Jaeger 13.5-13.6, Notes
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Detailed Example: Single Transistor Amplifier Analysis
Important!
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Notes on slides 14-25 were prepared by a previous student.
Step 1: Determine DC Operating PointRemove the Capacitors
Important!
Remove the Capacitors
Because the impedance of a capacitor is Z = 1/(jC), capacitors have infinitehave infinite impedance or are open circuits in DC ( = 0).
Inductors (not present in this circuit) have an impedance Z = jLimpedance Z = jL, and are shorts in DC.
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Step 1: Determine DC Operating PointDetermine the DC Thevenin Equivalent
Important!Determine the DC Thevenin Equivalent
Replace all connections to the transistor with their Thevenin equivalents.
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Step 1: Determine DC Operating PointCalculate Small Signal Parameters
Important!
Calculate Small Signal Parameters
Identify the type of transistor (npn in this example) and draw the base, collector, and emitter currents in their
di i d h i di l+
ICproper direction and their corresponding voltage polarities.Applying KVL to the controlling loop (loop 1):
-
IB
( p )VTHB IBRTHB VBE IERE = 0
Applying KCL to the transistor:
+ -+
VBEB
IE
pp y gIE = IB + IC
Because IC = IB,
+
-BE
1IE = IB + IC = IB + IB = IB(1+)
Substituting for IE in the loop equation:
-
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VTHB IBRTHB VBE IB(1+)RE = 0
Step 1: Determine DC Operating PointPlug in the Numbers
Important!
Plug in the Numbers
VTHB IBRTHB VBE IB(1+)RE = 0VTHB VBE IB(RTHB + (1+ )RE) = 0
+
ICVTHB VBE IB(RTHB + (1+ )RE) 0VTHB = 12R1/(R1+R2) = 3 VRTHB = R1 || R2 = 7.5 kAssume VBE = 0.7 V
-
IB
BEAssume for this particular transistor is given to be 100. + -
+VBE3 0 7 I (7500 + (1+100)*1300) 0 B
IE
+
-BE3 0.7 IB(7500 + (1+100)*1300) = 0IB = 16.6 AIC = IB = 1.66 mAI = I + I = 1 676 mA
1 -IE = IB + IC = 1.676 mA
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Step 1: Determine DC Operating PointCheck Assumptions: Forward Active?
Important!
Check Assumptions: Forward Active?
VC = 12 ICRC = 12 - (1.66 mA)(4300) = 4.86 VV = I R = (1 67 mA)(1300) = 2 18 V
+
ICVE = IERE = (1.67 mA)(1300) = 2.18 VVB = VTHB IBRTHB = 3 (16.6A)(7500) = 2.88 V
-
Check: V
IB
+ -+
VBE
Check:For an npn transistor in forward active:
VC > VB
VC
VE
VB
B
IE
+
-BE4.86 V > 2.88 V
VB VE = VBE = 0.7 V
E
1 -2.88 V 2.18 V = 0.7 V
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Step 2: Convert to AC-Only ModelShort the Capacitors and DC Current Sources
Important!
Short the Capacitors and DC Current Sources
DC voltage sources are h t ( ltshorts (no voltage
drop/gain through a short circuit).
DC current sources areDC current sources are open (no current flow through an open circuit).
Large capacitors are shorts (if C is large, 1/jC is small).
Large inductors are Large inductors are open (if L is large, jL is large).
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Step 2: Convert to AC-Only Model
