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BME 372 Electronics I –J.Schesser
270
Transistors
Lesson #10Chapter 4
BME 372 Electronics I –J.Schesser
271
Small Signal Equivalent Circuits and Parameters for the BJT
• If the variable portion of the input signal is small (in amplitude), it is possible to approximate the (non-linear) transistor as a linear device by representing it by an equivalent circuit
• Here’s how we do it:– Recall
– Since the base-emitter junction is a forward-biased diode and the current that crosses this junction is IE, then we can use the Shockley equation
1 where1 and
)1( then ; :Define
BBC
EBE
C
BCE
iii
iiii
iii
)1( T
BEVv
ESE eIi
BME 372 Electronics I –J.Schesser
272
Small Signal Equivalent Circuits and Parameters for the BJT
• Then:• Before we continue let’s define the following notation:
– A lower case signal (voltage or current) with upper case subscripts represents the total of the DC portion of the signal and the AC portion of the signal
– An upper case signal with upper case subscripts represents the DC portion (the Q-point)
– A lower case signal with lower case subscripts represents the AC portion
iB(t) = IBQ + ib(t)vBE(t) = VBEQ + vbe(t)
)1()1( T
BEVv
ESB eIi
BME 372 Electronics I –J.Schesser
273
Small Signal Equivalent Circuits and Parameters for the BJT
T
beBQb
T
beBQBQ
T
beBQbBQ
x
Vtv
BQbBQ
VV
ESBQ
Vtv
VV
ESbBQ
VtvV
ESbBQ
Vv
ESB
VtvIti
VtvIIV
tvItiI
xe
eItiI
eII
eeItiI
eItiI
eIi
T
be
T
BE
T
be
T
BE
T
beBE
T
BE
)()(
)())(1()(
1 :function lexponentia theofexpansion Taylor theof termsfirst two theUsing
)(
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equation,Shockley esatisfy th alsomust espoint valu-Q that theNoting
)()1()(
termlexponentia the tocompared neglible isit since term1- theIgnoring
)1()1()(
)1()1(
)(
)(
)(
BME 372 Electronics I –J.Schesser
274
Small Signal Equivalent Circuits and Parameters for the BJT
ip.relationsh thisrepresent tocollector in the sourcecurrent dependent a have weAnd)()(
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Q
IVr
rtiIVtitvV
tvIti
bc
bBcC
BC
BQ
T
bBQ
Tbbe
T
beBQb
BME 372 Electronics I –J.Schesser
275
Small Signal Equivalent Circuits and Parameters for the BJT
• Here is our equivalent circuit:
• Here is a second version equivalent circuit:
B
E
Crπ β ib
o
o
o
rπ β ib = β / rπ vbe= gm vbe
B
E
Co
o
o
BME 372 Electronics I –J.Schesser
276
Small Signal Equivalent CircuitsExample
• We see that we have capacitors and extra resistors.
• The extra resistors represent – the load to the next stage (input
impedance of the next stage)– the source resistance (output
impedance of the previous stage)• The extra capacitors are to:
– protect the DC design of our amplifier
– allow the passing the AC portion of the input signal from the source to our amplifier and on to the load (coupling capacitors)
– eliminate the loading of the emitter resistor for the AC signal (bypass capacitors)
VCC = 15V
RC 1kR1 10k
β = 300
RE 1kR2
500 RL
2k
RS
vin = 0.001sinωt
+
-5k
+vo
--
+vi
--
BME 372 Electronics I –J.Schesser
277
Small Signal Equivalent Circuits - ExampleDC Analysis
VCC = 15V
RC 1kR1 10k
β = 300
RE 1kR2
500 RL 2k
RS
vin+
-5k
• For DC the capacitors are open-circuits so for DC analysis the circuit becomes the same circuit as we have just analyzed
• The components not affected by the DC voltage are “grayed out”.
• The next step is to perform the AC analysis.
BME 372 Electronics I –J.Schesser
278
Small Signal Equivalent Circuits-ExampleAC Analysis
• The next step is to perform the AC analysis by
– Shorting any DC voltage sources– Opening any DC current sources – Shorting the capacitors– Eliminating any resistors which
are shorted by the capacitors (e.g., the emitter resistor
– Connecting the source, source resistance, and load resistance
– Replace the transistor with its small signal equivalent circuit.
• Short circuits are drawn in thick lines, the removed capacitor is “greyed-out”, and the equivalent circuit is in red.
• This circuit looks too messy, let’s redraw
RC 1kR1 10k
β = 300
RE 1kR2
500 RL 2k
RS
+
-5k
vin = 0.001sinωt
rπβ ib
BME 372 Electronics I –J.Schesser
279
Small Signal Equivalent Circuits-ExampleAC Analysis
RC
1k
R110k
R2500
RL
2k
RS
+
-5k
vin = 0.001sinωt
rπ β ib+vi
--
+vo
--
RL’RB
500
RS
+
-
vin = 0.001sinωt
rπ β ib+vi
--
+vo
--
1 21 2
1 2
1 2' || 6671 2
5 10|| 3.335 10
26 184414.1
'
' 109
|| 1.19
' '
c LL c L
c L
B
T
BQ
o b L
i b
o Lv
i
Bin B
B
o b L i iLo v
L L L L
ii
in
i
R R kR R RR RR R kR R R k
R RV mrI
v i Rv i r
v RAv r
R rZ R r kR r
v i R v vRi AR R R r RviZ
iA
64.4
7000
i
o inLv v
ii L
in
i v
vZRA Avi R
ZG A A
ii ib
ioii
ib
Zin
BME 372 Electronics I –J.Schesser
280
Small Signal Equivalent Circuits-ExampleAC Analysis Alternative Method for Ai
1 21 2
1 2
1 2' || 6671 2
5 10|| 3.335 10
26 184414.1
By current division
( )
( )
1 3.33001 2
c LL c L
c L
B
T
BQ
Co b
C L
Bb i
B
C Bo i
C L B
o C Bi
i C L B
R R kR R RR RR R kR R R k
R RV mrI
Ri iR R
Ri iR r
R Ri iR R R ri R RAi R R R r
kk k
3 64.363.33 1.844
kk k
RLRB
500
RS
+
-
vin = 0.001sinωt
rπ β ib+vi
--
+vo
--
ii ib
Zin
io
RC
Note they are the same:'in BL
i vL L B
C L
C L B
L B
C B
C L B
Z R rRA AR r R R rR R
R R R rr R R r
R RR R R r
BME 372 Electronics I –J.Schesser
281
Small Signal Equivalent Circuits-ExampleAC Analysis
RL’RB
500
RS
+
-
vin = 0.001sinωt
rπ β ib+vi
--
+vo
--
mVtvvRZ
ZAvvA
vv
vv
vvA
RZZ
vv
vRZ
Zv
vv
vv
vvA
krR
rRrRZ
rR
vvA
io
sin
inv
in
iv
in
i
i
o
in
ovs
sin
in
in
i
in
sin
ini
in
i
i
o
in
ovs
B
BBin
L
i
ov
sin4.764.76
4.76
7.
19.1||
109'
ii ib
Zin
BME 372 Electronics I –J.Schesser
282
Small Signal Equivalent Circuits-ExampleAC Analysis
kRivZ
Rvii
iRvi
ivZ
c
x
xo
c
xxb
b
c
xx
x
xo
1
;0
RC
RB
500
RS
rπ β ib+vi
--
Ib=0
Z0+vx
--
ix
BME 372 Electronics I –J.Schesser
283
Small Signal Equivalent CircuitsEmitter Follower Example
• We see that the load is across the emitter resistor
• The extra resistors and capacitors are represent – the load to the next stage (input
impedance of the next stage)– the source resistance (output
impedance of the previous stage)
– Coupling capacitors• Our analysis plan
– First, DC analysis to determine Q-point and equivalent circuit parameters
– AC analysis to calculate the gains
vin
VCC = 20V
R1 100k
β = 200
RE 2kR2
10k
RL
1k
RS
+
-100k +
vo
--
+vi
--
VBEQ = 0.7 V
BME 372 Electronics I –J.Schesser
284
Small Signal Equivalent CircuitsEmitter Follower Example DC Analysis
• Remove the circuit elements which are not affects by the DC voltages
• Opening the coupling capacitors
VVRR
RV
kRRR
CCTH
TH
10
50||
12
2
21
VCC = 20V
R1 100k
β = 200
RE 2kR2
10k
RL
1k
RS
vin
+
-100k +
vo
--
+vi
--
VBEQ = 0.7 V
100k
VCC = 20V
RE 2k
RTH
VTH
β IB
VCC = 20V
RE 2k
RTHVTHVBE
12606.20
26 ;7.11
14.4)1( ;12.4
6.202)2001(50
7.010)1(
)1(
m
IVrVRIVV
mAIImAII
AkkRR
VVI
RIVRIRIVRIV
BEQ
TEEQCCCEQ
BQEQBQCQ
ET
BEQTHBQ
EBQBEQTHBQEEQBEQTHBQTH
• Next, use the DC equivalent circuit for the active region and then write KVL for the base circuit
• Redraw the circuit and replace the base circuit with its Thevinen’s equivalent
BME 372 Electronics I –J.Schesser
285
Small Signal Equivalent CircuitsEmitter Follower Example AC Analysis
• The next step is to perform the AC analysis by– Shorting any DC voltage
sources– Opening any DC current
sources – Shorting the capacitors– Connecting the source, source
resistance, and load resistance– Replace the transistor with its
small signal equivalent circuit.• And Redraw to simplify
R1 100k
β = 200
RE 2kR2
10k
RL
1k
RS
vin
+
-100k
+vo
--
+vi
--
VBEQ = 0.7 V
rπβ ib
BME 372 Electronics I –J.Schesser
286
Small Signal Equivalent CircuitsEmitter Follower Example AC Analysis
R1
100kRE2k
R210k
RL
1k
RS
vin
+
-100k
+vo
--
+vi
--
rπ
β ib
ie= (1+β)ibib
RB10k
RL’
667
RS
vin
+
-
50k
+vo
--
+vi
--
rπ
β ib
ie= (1+β)ibib
1 2|| 50' || 667
(1 ) '
(1 ) '(1 ) '
(1 ) '(1 200) 667 .991
(1 200) 667 1260
(1 ) ' 134
|| 36.5
;
B
L E L
o b L
oi b o o
L
o Lv
i L
v
iit L
b
in B it
o v i io i
L L in
oi
R R R kR R Rv i R
vv i r v r vR
v RAv R r
A
vZ r R ki
Z R Z kv A v vi iR R ZiAi
36.2
35.8
inv
i L
v i
ZAR
G A A
Zin
ii iO
Zit
ii
BME 372 Electronics I –J.Schesser
287
Small Signal Equivalent CircuitsEmitter Follower Example AC Analysis
(1 200) 667 .991(1 200) 667 1260
0.991
36.5 0.7836.5 10
0.991 0.78 0.78
ov
i
o o i ivs
s i s s
i in
s in s
o o ivs
s i s
vAvv v v vAv v v v
v Zv Z R
v v vAv v v
RB10k
RL’
667
RS
vin
+
-
50k
+vo
--
+vi
--
rπ
β ib
ie= (1+β)ibib
Zin Zit
10k
RS
vin
+
-+vi
--
Zin
BME 372 Electronics I –J.Schesser
288
Small Signal Equivalent CircuitsEmitter Follower Example AC Analysis
6.461
'||)'()1(1
1
rRRrRR
Z SE
SE
o
RB10k
RE
2k
RS
50k
rπ
β ib
ie= (1+β)ibib
ZO +vx
--
ix
RS’
))'()1(1
1(
')1(
)'(33.8||'
)1(
rRRiv
rRv
Rvi
rRivkRRR
iRvi
Rvi
ivZ
SE
xx
S
x
E
xx
Sbx
BSS
b
E
xe
E
xx
x
xo
BME 372 Electronics I –J.Schesser
289
The BJT as a Digital Switch
VCC
RC
RB
LoadIo
Vo
+
-
0.E+00
1.E-03
2.E-03
3.E-03
4.E-03
5.E-03
6.E-03
7.E-03
0 1 2 3 4 5 6 7 8 9 10
v CE volts
i C amps
Vin
+
-
in
In Saturation, if
.2
Then, .2
BEQo CE CC C C CC C B CC C
B
o CE
V VV V V R i V R i V R V
RV V
CCo VVVV
Then,7.
if Cutoff,In
in
• Operating between cutoff and saturation (i.e., bypassing the active region), the BJT acts like an inverter.
• From this behavior, logic circuits such as NOR gates can be developed
0 μa
iB=50 μa
40 μa
30 μa
10 μa
20 μa
BME 372 Electronics I –J.Schesser
290
The BJT as a Digital Switch
CCo VVVV
Then,7.
if Cutoff,In
in
VCC= 3
RC
RB
LoadIo
Vo
+
-Vin
+
-
7.77.2
2.35
2.5
7.23
2.
if ,SaturationIn
in
in
in
V
VV
VR
VVRVV
o
B
BEQCCCo
2k
5k
β =10
Vo
Vin1.4
VCC=3
VCC=30.7
0.2
=10
=100
BME 372 Electronics I –J.Schesser
291
Switching and Timing
VOL
VOL
VOH
VOH
(VOH + VOL)/2
(VOH + VOL)/2
50%
10%
90%
100%
0%
trtf
td tPLH
vi
vo
tr = rise time
tf = fall time
td = delay time (propagation delay from High to Low)
ts = storage time (propagation delay from Low to High)
BME 372 Electronics I –J.Schesser
292
Homework
• Probs. 4.40, 4.42, 4.43, 4.45, 4.46, 4.51, 4.53, 4.54, 4.56,