CSE 477. VLSI Systems Design - GUC
Transcript of CSE 477. VLSI Systems Design - GUC
ELCT 503: Semiconductors German University in Cairo (GUC)
ELCT503
Semiconductors Fall 2014
Lecture 09: BJT Circuit Analysis
Dr. Hassan Mostafa
حسن مصطفى. د
ELCT 503: Semiconductors German University in Cairo (GUC)
Introduction
npn transistor
pnp transistor
ELCT 503: Semiconductors German University in Cairo (GUC)
DC Models
iC – vCB characteristics for common-base configuration
ELCT 503: Semiconductors German University in Cairo (GUC)
DC Models
iC – vCE characteristics for common-emitter configuration
(VA = Early voltage)
ELCT 503: Semiconductors German University in Cairo (GUC)
DC Models (npn)
Transistor OFF
(VBE < 0.7V) and (VBC < 0.4V)
IB = 0 IC = 0 IE = 0
Transistor in Active region
(VBE >= 0.7V) and (VBC < 0.4V)
VCE >=0.3V
IC = IS exp (VBE/VT)
IC = β IB
IE = IC + IB = (β+1) IB
IC = IE
β = / (1- ) and β >> 1
= β / (β+1) and <= 1
ELCT 503: Semiconductors German University in Cairo (GUC)
DC Models (npn)
Transistor in Saturation region
(VBE >= 0.7V) and (VBC >= 0.4V)
VCE =0.2V
ICsat = βforced IB
IE = IC + IB = (βforced+1) IB
ELCT 503: Semiconductors German University in Cairo (GUC)
DC Models (pnp)
Same equations as before but after changing letters order: VBE VEB
VBC VCB
VCE VEC
ELCT 503: Semiconductors German University in Cairo (GUC)
Example1
β = 100
VBE = 0.7V at IC = 1 mA
Design the circuit such that:
IC = 2 mA
VC = +5V
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Example1
Solution
Transistor works in the active region
kmA
RC 52
515
VVVVV
V
CBBC
BE
4.05
7.0
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Example1
Solution
kI
VR
mAmAII
V
VVVV
VmA
mAVV
E
EE
CE
E
EEBBE
TBE
07.7)15(
02.22*100
1100/
717.0
0717.0
717.0)1
2ln(*7.0
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Example2
β = 100
Find IC, IB, IE, VB, VC, and VE
Assume Active region VBE = 0.7V
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Example2
Check:
(VBE >= 0.7V)
(VBC = -1.3V < 0.4V)
Correct
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Example3
β = 100
Find IC, IB, IE,
VB, VC, and VE
Assume Active
region
Check:
(VBE >= 0.7V)
(VBC = 3.52V > 0.4V)
Incorrect
ELCT 503: Semiconductors German University in Cairo (GUC)
Example3
Assume
Saturation
region
Check:
(VBE >= 0.7V)
(VBC = 0.5V > 0.4V)
Correct
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BJT in Logic Gates RTL (Resistor Transistor Logic)
Increasing vBE decrease vo
C
T
BESCCCEo
T
BESC
BE
CCoC
BE
CCCCCEo
RV
vIVvv
V
vIi
ActiveAssume
ONVvWhen
Vvi
OFFVvWhen
RiVvv
*)exp(*
)exp(*
5.0
0
5.0
*
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When vBE increases such that vBC becomes larger than 0.4V
Transistor enters Saturation Region
BJT in Logic Gates
Vvv CEo 2.0
Which Logic gate this Circuit implements?
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Think…
Which Logic gate this Circuit implements?
BJT in Logic Gates
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Small Signal Models
Derivation : BJT must be in active region to use the small signal model
TV
CI
bev
ci
mg
V
vIi
iIi
VvV
vIi
V
vI
V
v
V
VIi
V
vVIi
V
vIiIi
T
beCc
cCC
Tbe
T
beCC
T
beC
T
be
T
BESC
T
beBESC
T
BEScCC
)1(*
)exp()exp()exp(
)exp(
)exp(
B
T
m
C
T
b
be
T
beCcb
cCCbBB
I
V
gr
I
V
i
vr
V
vIii
iIiiIi
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Small Signal Models
-model
C
Ao
mC
T
B
T
T
Cm
I
Vr
gI
V
I
Vr
V
Ig
||
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Small Signal Models
T-model
C
Ao
mC
T
E
Te
T
Cm
I
Vr
gI
V
I
Vr
V
Ig
||
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Common Emitter Amplifier
Why it is called common emitter?
Bypass capacitors
Open circuit at DC
Short circuit at small signal
How to draw the small signal equivalent circuit?
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Common Emitter Amplifier
Amplifier parameters:
Voltage gain (AV) = vo/vi <Units V/V>
Input resistance (Rin) = vi/ii <Units >
Output resistance (Rout) = vo/io|vs=0 <Units >
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Common Emitter Amplifier
Voltage gain (AV) = vo/vi = - gm*(ro//RC // RL) V/V
= - gm*(RC // RL) when ro >> RC and RL
= - gm*(RC ) when ro >> RC and RL =
ELCT 503: Semiconductors German University in Cairo (GUC)
Common Base Amplifier
VA = ro =
Voltage gain (AV) = vo/vi = (RC //RL)/ re = +gm (RC //RL)
= (RC )/ re = +gm (RC) when RL =