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Lecture 20Lecture 20
Bipolar Junction Transistors (BJT): Part 4
Small Signal BJT Model
Reading:
Jaeger 13.5-13.6, Notes
ECE 3040 - Dr. Alan DoolittleGeorgia Tech
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
ECE 3040 - Dr. Alan DoolittleGeorgia Tech
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)
VAVCE
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
ECE 3040 - Dr. Alan DoolittleGeorgia Tech
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 d•Small 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 it’s 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
ECE 3040 - Dr. Alan DoolittleGeorgia Tech
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
ECE 3040 - Dr. Alan DoolittleGeorgia Tech
Consider the BJT as a two-port Network
iib=y11vbe + y12vce
ic=y21vbe + y22vce
ECE 3040 - Dr. Alan DoolittleGeorgia Tech
Consider the BJT as a two-port Network
o is most
ECE 3040 - Dr. Alan DoolittleGeorgia Tech
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
ECE 3040 - Dr. Alan DoolittleGeorgia Tech
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
ECE 3040 - Dr. Alan DoolittleGeorgia Tech
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 ground•DC Current sources are open circuits•Large capacitors are short circuits
Step 2
•Large capacitors are short circuits•Large 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
ECE 3040 - Dr. Alan DoolittleGeorgia Tech
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+IBRTH
3V=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+IBRTH
3V= IB(100+1)1300+0.7+ IB7500
ECE 3040 - Dr. Alan DoolittleGeorgia Tech
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 C
Step 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
VbegmVbe
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
ECE 3040 - Dr. Alan DoolittleGeorgia Tech
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
ECE 3040 - Dr. Alan DoolittleGeorgia Tech
Detailed Example: Single Transistor Amplifier Analysis
Important!
ECE 3040 - Dr. Alan DoolittleGeorgia Tech
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/(jωC), capacitors have infinitehave infinite impedance or are open circuits in DC (ω = 0).
Inductors (not present in this circuit) have an impedance Z = jωLimpedance Z = jωL, 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 = βIB,
+
-BE
1IE = IB + IC = IB + βIB = IB(1+β)
Substituting for IE in the loop equation:
-
ECE 3040 - Dr. Alan DoolittleGeorgia Tech
VTHB – IBRTHB – VBE – IB(1+β)RE = 0
Step 1: Determine DC Operating PointPlug in the Numbers
Important!
Plug in the Numbers
VTHB – IBRTHB – VBE – IB(1+β)RE = 0VTHB – VBE – IB(RTHB + (1+ β)RE) = 0
+
ICVTHB VBE IB(RTHB + (1+ β)RE) 0VTHB = 12R1/(R1+R2) = 3 VRTHB = R1 || R2 = 7.5 kΩAssume VBE = 0.7 V
-
IB
BEAssume β 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 μAIC = βIB = 1.66 mAI = I + I = 1 676 mA
1 -IE = IB + IC = 1.676 mA
ECE 3040 - Dr. Alan DoolittleGeorgia Tech
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 – (16.6μA)(7500) = 2.88 V
-
Check: V
IB
+ -+
VBE
Check:For an npn transistor in forward active:
VC > VB
VC
VE
VB
B
IE
+
-BE
4.86 V > 2.88 V
VB – VE = VBE = 0.7 V
E
1 -2.88 V – 2.18 V = 0.7 V
ECE 3040 - Dr. Alan DoolittleGeorgia Tech
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/jωC is small).
• Large inductors are• Large inductors are open (if L is large, jωL is large).
ECE 3040 - Dr. Alan DoolittleGeorgia Tech
Step 2: Convert to AC-Only Model(Optional) Simplify Before Thevenizing
Important!
(Optional) Simplify Before Thevenizing
rrc
4 3 kΩ
rs
r2
rL
30 kΩ
100 kΩ2 kΩ
4.3 kΩ
2 kΩ 4.12 kΩ
rs
r ||r
rc||rL
vsr1
kΩ2 kΩ10 kΩ 7.5 kΩ
r1||r2vs
ECE 3040 - Dr. Alan DoolittleGeorgia Tech
Step 3: Thevenize the AC-Only Model Important!
|| ||
rthC = rc||rL
4 12 kΩ
2 kΩ 4.12 kΩ
rs
r ||r
rc||rL
rthB = rs||r1||r2
1.58 kΩ
4.12 kΩ
vthC = 0 V
7.5 kΩr1||r2
vsrthE = 0 ΩvthE = 0 V
vthB = vs * (r1||r2)/(rs+[r1||r2])
ECE 3040 - Dr. Alan DoolittleGeorgia Tech
Step 4: Replace Transistor With Small Signal ModelImportant!
r hB = r ||r1||r2
rthC = rc||rL
4.12 kΩ v hC = 0 V
Important!
rthB rs||r1||r2
1.58 kΩ
r = 0 Ω*
vthC 0 V
After replacing the transistor, apply Ohm’s Law: V = IR to find vout.
rthE = 0 ΩvthE = 0 V
vthB = vs * [(r1||r2)/(rs+r1||r2)]
ro and rthC are in parallel, so that Ohm’s Law becomes: vout = -IR = -(gmvBE)(ro||rthC)
Because rthC = rc||rL
rthB B C
+ +
vout = -(gmvBE)(ro||rc||rL)
vout/vBE = -gm(ro||rc||rL) is the gain from transitor input (vBE)
vthB rthC
E E
-
vBE rπgmvBE
ro
-
voutto transistor/circuit output vout)
ECE 3040 - Dr. Alan DoolittleGeorgia Tech
TRANSISTOREXTERNAL
Step 5: Calculate Gain and Small Signal ParametersImportant!
rthB B C
+ +
As previously determined: vthB/vs = (r1||r2)/([r1||r2] + rs)
vthB rthC
E E
-
vBE rπgmvBE
ro
-
vout
G i / ( / )( / )( / )
Applying a voltage divider: vBE/vthB = rπ/(rπ+rthB)
G i f tTRANSISTOREXTERNAL
Gain = vout/vs = (vthB/vs)(vBE/vthB)(vout/vBE) Gain factor:vout/vBE = -gm(ro||rc||rL)
Because calculating the DC operating point was done first, we have equations for gm,Because calculating the DC operating point was done first, we have equations for gm, rπ, and ro in terms of previously calculated DC currents and voltages.
Plugging in the numbers:
Gain = vout/vs = -139 V/V
ECE 3040 - Dr. Alan DoolittleGeorgia Tech
Interpretation/Analysis of Results Important!
Gain = vout/vs = (vthB/vs)(vBE/vthB)(vout/vBE) = -139 V/V vthB/vs = (r1||r2)/([r1||r2] + rs)
v /v = r /(r +r )vBE/vthB = rπ/(rπ+rthB)
vout/vBE = -gm(ro||rc||rL)Both terms are loss factors, i.e. they can never be greater than 1
This term is the gain factorand is responsible for amplifying the signalnever be greater than 1
in magnitude and thus cause the gain to decrease.
amplifying the signal.The AC input signal has been amplified 139 times in magnitude. The negative i i di h h bsign indicates there has been
a phase shift of 180°.
ECE 3040 - Dr. Alan DoolittleGeorgia Tech
Completing the Small Signal Model of the BJTBase Charging Capacitance (Diffusion Capacitance)
In active mode when the emitter-base is forward biased, the capacitance of the emitter-base junction is dominated by the diffusion capacitance (not
Recall for a diode we started out by saying:
y p (depletion capacitance).
'dvdQC
D
DDiffusion
Neglect charge injected from the base into the
Sum up all the minority carrier h ith
'
''
dvdt
dtdQ
DD xvxv
D
D
D base to t eemitter due to p+ emitter in pnp
charges on either side of the junction
0011 dxeenqAdxeepqAQ nT
DpT
DL
xV
v
poL
xV
v
noD
h d i d l i l h
ECE 3040 - Dr. Alan DoolittleGeorgia Tech
Excess charge stored is due almost entirely to the charge injected from the emitter.
Completing the Small Signal Model of the BJTBase Charging Capacitance (Diffusion Capacitance)
Th BJT t lik ffi i t “ i h ” A j it i•The BJT acts like a very efficient “siphon”: As majority carriers from the emitter are injected into the base and become “excess minority carriers”, the Collector “siphons them” out of the base.
•We can view the collector current as the amount of excess charge in the base collected by the collector per unit time.
•Thus, we can express the charge due to the excess hole us, we c e p ess e c ge due o e e cess o econcentration in the base as:
FCB iQ or the excess charge in the base depends on the magnitude ofor the excess charge in the base depends on the magnitude of current flowing and the “forward” base transport time, F, the average time the carriers spend in the base.
•It can be shown (see Pierret section 12 2 2) that:•It can be shown (see Pierret section 12.2.2) that:
whereD
W
BF ,
2
2
ECE 3040 - Dr. Alan DoolittleGeorgia TechtcoefficiendiffusioncarrierMinorityD
widthregionneutralQuasiBaseW
B
Completing the Small Signal Model of the BJTBase Charging Capacitance (Diffusion Capacitance)
Thus, the diffusion capacitance is,
CB iWQC
2
poQBE
C
BpoQ
BE
BB
IvDv
QC
intint 2
mFT
CFB g
VI
C
The upper operational frequency of the transistor is limited by the forward base transport time: f
21
Note the similarity to the Diode Diffusion capacitance we found previously:
F2
ECE 3040 - Dr. Alan DoolittleGeorgia Tech
timetransittheis
IqALnLp
wheregCS
npopnottdDiffusion
Completing the Small Signal Model of the BJTBase Charging Capacitance (Total Capacitance)
In active mode for small forward biases the depletion capacitance of the base-emitter junction can contribute to the total capacitance
V
CC
EB
jEojE
1
total capacitance
where
V baseemitterforbi
,
1
junctionBEtheforvoltageinbuiltV
cecapacidepletionbiaszeroC
baseemitterforbi
jEo
tan
CCC
Thus, the total emitter-base capacitance is:
ECE 3040 - Dr. Alan DoolittleGeorgia Tech
jEB CCC
Completing the Small Signal Model of the BJTBase Charging Capacitance (Depletion Capacitance)
In active mode when the collector-base is reverse biased, the capacitance of the collector-base junction is dominated by the depletion capacitance (not diffusion capacitance)depletion capacitance (not diffusion capacitance).
CC o
h
VV
basecollectorforbi
CB
1
junctionCBtheforvoltageinbuiltV
cecapacidepletionbiaszeroCwhere
o
tan,
junctionCBtheforvoltageinbuiltV basecollectorforbi
ECE 3040 - Dr. Alan DoolittleGeorgia Tech
Completing the Small Signal Model of the BJTCollector to Substrate Capacitance (Depletion Capacitance)
In some integrated circuit BJTs (lateral BJTs in particular) the device has a capacitance to the substrate wafer it is fabricated in This results from a “buried” reverse biased junction Thusin. This results from a buried reverse biased junction. Thus, the collector-substrate junction is reverse biased and the capacitance of the collector-substrate junction is dominated by the depletion capacitance (not diffusion capacitance). Emitterp p ( p )
VC
CCS
CSCS
1
p
p-collectorn-base
where
V substratecollectorforbi
,
1 p-collector
n-substrate
junctionsubstrateCtheforvoltageinbuiltVcecapacidepletionbiaszeroC
substratecollectorforbi
CS
tan
ECE 3040 - Dr. Alan DoolittleGeorgia Tech
Completing the Small Signal Model of the BJTParasitic Resistances
•rb = base resistance between metal interconnect and B- E junction•r = parasitic collector resistancerc parasitic collector resistance •rex = emitter resistance due to polysilicon contact•These resistance's can be included in SPICE sim lations b t are s all ignored in handsimulations, but are usually ignored in hand calculations.
ECE 3040 - Dr. Alan DoolittleGeorgia Tech
Completing the Small Signal Model of the BJTComplete Small Signal Model
ECE 3040 - Dr. Alan DoolittleGeorgia Tech
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