University of Zagreb
Faculty of Electrical Engineering and Computing
Department of Electronics, Microelectronics,
Computer and Intelligent Systems
El t
i
Electronics
1
Ž. Butković, J. Divković Pukšec, A. Barić
8. Bipolar Junction Transistor
Circuits
Basic Circuit with Bipolar
Junction Transistor
input side
uI = RB iB + uBE
output side
uO = uCE = UCC − RLiC
8. Bipolar Junction Transistor Circuits
2
Output Characteristics
Q → bias point
example:
UCC = 15 V, RL = 1,5 kΩ
for uI ≈ uBE → IBQ = 40 μA,
ICQQ = 4,1 mA, UCEQQ = 8,8 V
between A and B → active
mode
point A → the transistor is in
the cutoff region
point B → BJT leaves the
active mode and enters
the saturation region
8. Bipolar Junction Transistor Circuits
3
Voltage Transfer Characteristic
voltage transfer characteristic –
uO = f(uI)
for uI < 0,5 V → transistor currents are
very small;
iC ≈ 0, uI = uBE , uO = uCE = UCC
‰ between A and B → active mode →
amplifier
⎛u ⎞
⎛u ⎞
iC = I S exp ⎜ BE ⎟ ≈ I S exp ⎜ I ⎟
⎝ UT ⎠
⎝ UT ⎠
⎛u ⎞
uO ≈ U CC − RL I S exp ⎜ I ⎟
⎝ UT ⎠
example: UIQ = 0,69 V, UOQ = 8,8 V
‰ left from A and right from B → switch
8. Bipolar Junction Transistor Circuits
4
Amplification
Graphical:
AV =
uo − U om sin ω t
U
1,88
=
= − om = −
= − 235
−
3
ui
U im sin ω t
U im
8 ⋅10
Analytical:
AV ≡
d uO
d uI
≈−
Q
⎛u ⎞
RL
I S exp ⎜ I ⎟
UT
⎝ UT ⎠
≈−
Q
RL I CQ
UT
= − g m RL
example:
l UT = 25 mV,
V ICQ = 4,1
4 1 mA
A → gm = 164 mA/V,
A/V RL = 1,5
1 5 kΩ → AV = − 246
AI ≡
d iO
d iI
= −
Q
d iC
d iB
= − h fe
Q
example: hfe ≈ 100 → AI = − 100
8. Bipolar Junction Transistor Circuits
5
Biasing with Fixed Base Current
input side
I BQ =
U CC − U BEQ
RB
U BEQ ≈ Uγ
output side
ICQ ≈ β I BQ
U CEQ = U CC − RC ICQ
active mode condition
U CEQ > U BEQ
8. Bipolar Junction Transistor Circuits
6
Example 8.1
Consider the circuit in the figure with UCC = 15 V and RC = 2 kΩ. Transistor
parameters are Uγ = 0,7 V and β = 100. Determine the value of resistor RB
that will ensure UCEQ = UCC/2 in the bias point
point.
8. Bipolar Junction Transistor Circuits
7
Example 8.2
If in the example 8.1 temperature is changed from 25ºC to 75ºC transistor
current gain increases from β1 = 100 to β2 = 150. Determine the new values
of transistor currents and voltages in the bias point at the 75
75ºC
C. Disregard
the current ICEO and the change of the voltage UBE with temperature.
8. Bipolar Junction Transistor Circuits
8
Bias Point Stabilization by
Emitter Resistor
8. Bipolar Junction Transistor Circuits
9
Application of
Thévenin's Theorem
U BB =
R2
UCC
R1 + R2
RB = R1 R2
8. Bipolar Junction Transistor Circuits
10
DC Analysis Schematic
U BB = RB I B + U BE + RE (I B + IC )
in bias point
U BB = RB I BQ + U BEQ + (1 + β ) RE I BQ
I BQ =
U BB − U BEQ
U BB − Uγ
=
RB + (1 + β ) RE RB + (1 + β ) RE
ICQ = β I BQ
UCEQ = UCC − RC ICQ − RE (I BQ + ICQ ) ≈ UCC − (RC + RE ) ICQ
8. Bipolar Junction Transistor Circuits
11
Example 8.3
Consider the circuit in the figure with UCC = 15 V,
R1 = 11 kΩ, R2 = 2 kΩ, RC = 2 kΩ and RE = 500 Ω.
a)) D
Determine
t
i th
the ttransistor
i t currents
t and
d voltages
lt
in the bias point at 25ºC. Transistor parameters
are Uγ = 0,7 V and β1 = 100.
b) Redo the a) part at temperature of 75ºC where
β2 = 150. Disregard the change of the voltage
Uγ = 0,7
0 7 V with
ith temperature.
temperat re
8. Bipolar Junction Transistor Circuits
12
Condition for Good
Bias Point Stabilization
β (U BB − U BEQ )
ICQ = β I BQ =
RB + (1 + β ) RE
Current ICQ is independent of β if RB << (1 + β) RE
ICQ ≈
β (U BB − U BEQ ) U BB − U BEQ
≈
(1 + β ) RE
RE
For RB << (1 + β) RE base is at fixed potential UBB
U BB ≈ U BE + RE IC
8. Bipolar Junction Transistor Circuits
13
Biasing Using
Two Supply Voltages
input side
U EE = RB I B + U BE + RE (I B + IC )
with IC = β IB
IB =
U EE − U BE
RB + (1 + β ) RE
output side
U CE = U CC + U EE − (RC + RE ) IC
8. Bipolar Junction Transistor Circuits
14
Common-Emitter Amplifier
8. Bipolar Junction Transistor Circuits
15
Common-Emitter Amplifier –
Voltage Gain
RB = R1 R2
uo = − g m ube
b ( rce RC RL )
AV =
uo
= − g m ( rce RC RL )
ui
ui = ube
b
AV =
uo
≈ − g m ( RC RL )
ui
With dependent source hfe ib and gm = hfe/rbe → AV =
8. Bipolar Junction Transistor Circuits
h (R R )
uo
≈ − fe C L
ui
rbe
16
Common-Emitter Amplifier –
Current Gain
io = − g m ube
AI =
rce RC
rce RC + RL
ube = ( RB rbe ) ii
rce RC
io
= − gm
RB rbe )
(
ii
rce RC + RL
AI =
io
RC
≈ − gm
RB rbe )
(
ii
RC + RL
With dependent source hfe ib and gm = hfe/rbe → AI =
8. Bipolar Junction Transistor Circuits
io
RC
RB
≈ − h fe
ii
RC + RL RB + rbe
17
Common-Emitter Amplifier –
Input and Output Resistance
Ri =
ui
= RB rbe
ii
current gain → AI =
io uo / RL
R
=
= AV i
ii ui / Ri
RL
Equivalent circuit for determining the output resistance
Ro =
u
= rce RC
i
8. Bipolar Junction Transistor Circuits
18
Example 8.4
Consider the amplifier in the figure: UCC = 15 V, Rs = 500 Ω, R1 = 30 kΩ,
R2 = 11 kΩ, RC = 2 kΩ, RL = 1,2 kΩ and RE = 1 kΩ. Parameters of npn
transistor are β ≈ hfe = 100, Uγ = 0,7
0 7 V and UA = 200 V. Thermal voltage is
UT = 25 mV. Determine the gains AV = uo/ui, AI = io/ii, AVs = uo/us, and input
and output resistance.
8. Bipolar Junction Transistor Circuits
19
Example 8.5
Consider the amplifier from the example 8.4. Draw the dc and ac load line at
the output characteristics graph. Calculate the largest output voltage and
current swing so that operating point does not leave transistor active
mode area.
8. Bipolar Junction Transistor Circuits
20
Setting the Bias Point for the
Largest Output Signal Swing (1)
dc load line:
ac load line:
UCEQ = UCC − (RC + RE ) ICQ
uce = − (RC RT )ic
8. Bipolar Junction Transistor Circuits
21
Setting the Bias Point for the
Largest Output Signal Swing (2)
uce = − ( RC RL ) ic
uCE − U CEQ = − ( RC RL ) ( iC − I CQ )
In point B → for uCE = 0, iC = 2 ICQ
0 − U CEQ = − ( RC RL ) ( 2 I CQ − I CQ )
U CEQ = ( RC RL ) I CQ
I CQ =
U CC
RC + RE + RC RL
8. Bipolar Junction Transistor Circuits
U CEQ =
RC RT
U CC
RC + RE + RC RL
22
Example 8.6
Consider the amplifier in the figure. Determine the bias point for largest output
signal swing. Calculate the largest allowed voltage swing of uo and largest
allowed current swing of io. Determine the values of resistors R1 and R2 to
achieve the bias point. Circuit parameters are; UCC = 15 V, RC = 2 kΩ,
RL = 1,2 kΩ and RE = 1 kΩ. Parameters of npn transistor are β = 100 and
Uγ = 0,7 V.
8. Bipolar Junction Transistor Circuits
23
Common-Emitter Amplifier
with a Emitter Degeneration
8. Bipolar Junction Transistor Circuits
24
Amplifier with Emitter Degeneration –
Voltage Gain
ui = ib rbe + (1 + h ffe ) ib RE
AV =
uo = − h fef ib ( RC RL )
RC RL
uo
= − h fe
ui
rbe + (1 + h fe ) RE
With: gm RE >> 1 → AV ≈ −
8. Bipolar Junction Transistor Circuits
AV ≈
− g m ( RC RL )
1 + g m RE
RC RL
RE
25
Amplifier with Emitter Degeneration –
Current Gain
Ri′ =
ui
= rbe + (1 + h fe ) RE
ib
io = − h fe ib
AI =
RC
RC + RL
ib = ii
RB
RB
= ii
RB + Rul′l
RB + rbe
b + (1 + h fe
f ) RE
io
RC
RB
= − h fe
ii
RC + RL RB + rbe + (1 + h fe ) RE
8. Bipolar Junction Transistor Circuits
26
Amplifier with emitter degeneration –
Input and Output Resistance
Ri =
i=
ui
= RB Rul′ = RB ⎡⎣ rbe + (1 + h fe ) RE ⎤⎦
ii
u
+ h fe ib
RC
ue = (1 + h fe )ib RE = − ib (Rg RB + rbe ) → ib = 0
Ro =
u
= RC
i
8. Bipolar Junction Transistor Circuits
27
Example 8.7
Consider the amplifier in the figure with the following parameters: UCC = 15 V,
Rs = 500 Ω, R1 = 25 kΩ, R2 = 2,2 kΩ, RC = 3 kΩ, RL = 2 kΩ and RE = 200 Ω.
Parameter of npn transistor are β ≈ hfe = 100 and Uγ = 0,7
0 7 V. Disregard the
increase of the collector current in the active mode. Thermal voltage is
equal to UT = 25 mV. Determine the gains AV = uo/ui, AI = io/ii, AVs = uo/us,
and input and output resistance.
8. Bipolar Junction Transistor Circuits
28
Common-Base Amplifier
with one power supply
8. Bipolar Junction Transistor Circuits
with two power supplies
29
Common-Base Amplifier –
Voltage Gain
ui = − ib rbe
AV =
uo = − h fe ib ( RC RL )
R R
uo
= h fe C L = g m ( RC RL )
ui
rbe
8. Bipolar Junction Transistor Circuits
30
Common-Base Amplifier –
Current Gain
Ri′ =
ui
− ib rbe
r
r
1
=
= be ≈ be =
ie − (1 + h fe ) ib 1 + h fe h fe g m
io = − h fe ib
AI =
RC
RC + RL
ie = − (1 + h fe ) ib = iul
RE
RE + Ri′
h fe
io
RC
RE
=
ii 1 + h fef RC + RL RE + Ri′
8. Bipolar Junction Transistor Circuits
31
Common-Base Amplifier –
Input and Output Resistance
Ri =
ui
rbe
1
= RE Ri′ = RE
≈ RE
1 + h fe
ii
gm
ue = − ib rbe = (1 + h fe ) ib ( Rs RE ) → ib = 0
Ro =
u
= RC
i
8. Bipolar Junction Transistor Circuits
32
Example 8.8
Consider the amplifier in the figure with the
following parameters: UCC = UEE = 15 V,
Rs = 500 Ω, RC = 2 kΩ and RL = 1,2
1 2 kΩ.
Parameters of npn transistor are β ≈ hfe = 100
and Uγ = 0,7 V. Disregard the increase of
collector current in the active mode. Thermal
voltage is equal to UT = 25 mV. Determine
the resistance of resistor RE that will ensure
ICQ = 3 mA. Calculate the gains AV = uo/ui,
AI = io/ii, AVg = uiz/ug, and input and output
resistance.
i t
8. Bipolar Junction Transistor Circuits
33
Common-Collector Amplifier –
Emitter Follower
For dc: UCE ≈ UCC − RE IC
8. Bipolar Junction Transistor Circuits
34
Common-Collector Amplifier –
Voltage Gain
uo = (1 + h fe ) ib ( RE RL )
ui = ib rbe + (1 + h fe ) ib ( RE RL )
AV =
(1 + h fe ) ( RE RL )
uo
=
ui rbe + (1 + h fe ) ( RE RL )
AV ≈
h fe ( RE RL )
g m ( RE RL )
=
rbe + h fe ( RE RL ) 1 + g m ( RE RL )
8. Bipolar Junction Transistor Circuits
35
Common-Collector Amplifier –
Current Gain
Ri′ =
ui
= rbe + (1 + h fe ) ( RE RL )
ib
io = (1 + h fe ) ib
AI =
RE
RE + RL
ib = ii
RB
RB + Ri′
io
RE
RB
= (1 + h fe )
ii
RE + RL RB + Ri′
8. Bipolar Junction Transistor Circuits
36
Common-Collector Amplifier –
Input and Output Resistance
Ri =
ui
= RB Ri′ = RB ⎡⎣ rbe + (1 + h fe ) ( RE RL ) ⎤⎦
ii
u = − ib ( Rs RB + rbe )
Ro =
1 + h fe
i
1 (1 + h fe ) ib 1
=
−
=
+
u RE
u
RE Rs RB + rbe
R R +r
u
= RE s B be
1 + h fe
i
8. Bipolar Junction Transistor Circuits
37
Example 8.9
Consider the amplifier in the figure with the following parameters: UCC = 15 V,
Rs = 500 Ω, R1 = 70 kΩ, R2 = 100 kΩ, RE = 4 kΩ and RL = 1 kΩ. Parameters
of the npn transistor are β ≈ hfe = 100 and Uγ = 0,7
0 7 V. Disregard the increase
of the collector current in the active mode. Thermal voltage is equal to
UT = 25 mV. Determine the gains AV = uo/ui, AI = io/ii, AVs = uo/us, and input
and output resistance.
8. Bipolar Junction Transistor Circuits
38
Example 8.10
Consider the amplifier in the figure with the
following parameters: UCC = UEE = 15 V,
Rs = 500 Ω, RE = 4 kΩ and RL = 1 kΩ.
Parameters of the npn transistor are
β ≈ hfe = 100 i Uγ = 0,7 V. Disregard the
increase of the collector current in the active
mode. Thermal voltage is equal to
UT = 25 mV. Determine the gains AV = uo/ui,
AI = io/ii, and input and output resistance.
8. Bipolar Junction Transistor Circuits
39
Comparison of Single-Stage Bipolar
Junction Transistor Amplifiers
Common-emitter
AV
AI
(
− g m RC RL
− h fe
)
RC
RB
RC + RL RB + rbe
Rul
RB rbe
Riz
RC
Common-base
(
g m RC RL
8. Bipolar Junction Transistor Circuits
rbe
b
1 + h fe
RC
(
)
h fe
RC
RE
1 + h fe RC + RL R + rbe
E
1 + h fe
RE
Common-collector
g m RE RL
(
)
1 + g m RE RL
(1 + h fe )
)
RE
RB
RE + RL RB + rbe + (1 + h fe ) RE RL
(
(
)
)
RB ⎡ rbe + (1 + h fe ) RE RL ⎤
⎣
⎦
RE
Rs RB + rbe
1 + h ffe
40
Differential Amplifier
Differential amplifier
‰ one of the most widely used amplifiers
‰ first (input) stage in operational amplifiers,
amplifiers comparators
comparators, regulators etc.
etc
‰ applications in measurement circuits
2 inputs → ui1 and ui2
2 outputs → uo1 and uo2
uses:
‰ only uo11 or uo22 →
single-ended output
‰ difference uo = uo2 − uo1
→ differential output
8. Bipolar Junction Transistor Circuits
41
DC Analysis
For dc analysis → us1 = us2 = 0
Symmetrical configuration :
T1 = T2, Rs1 = Rs2, RC1 = RC2 →
IB1 = IB2, IC1 = IC2
for the input
p circuit of transistor T1:
U EE = I BQ1 Rs1 + U BEQ1 + 2 (1 + β ) I BQ1 RE
I BQ1 = I BQ 2 =
U EE − U BEQ1
Rs1 + 2 (1 + β ) RE
I CQQ1 = I CQQ 2 = β I BQQ1
U CEQ1 = U CEQ 2 = U CC + U EE − [β RC1 + 2 (1 + β ) RE ] I BQ1 ≈
≈ U CC + U EE − ( RC1 + 2 RE ) I CQ1.
8. Bipolar Junction Transistor Circuits
42
Example 8.11
Consider the differential amplifier in the figure with the following parameters :
UCC = UEE = 15 V, Rs1 = Rs2 = 500 Ω, RC1 = RC2 = 1,5 kΩ and RE = 4,5 kΩ.
Parameters of the bipolar junction transistors are β = 100 and Uγ = 0,7
0 7 V.
Determine currents and voltages of the transistors in bias point.
8. Bipolar Junction Transistor Circuits
43
AC Analysis –
Small-Signal Equivalent Circuit
8. Bipolar Junction Transistor Circuits
44
Common-Mode and
Differential Signals
Voltages us1 and us2 can be separated as:
‰ common-mode signal ucm and
‰ diff
differential-mode
ti l
d signal
i
l ud
u + us 2
ucm = s1
2
ud = us 2 − us1
Individual input voltages us1 and us2:
us1 = ucm − ud / 2
us 2 = ucm + ud / 2
Analysis
A
l i iis b
based
d on superposition
iti principle
i i l – separate
t analysis
l i ffor
common-mode and differential signal
8. Bipolar Junction Transistor Circuits
45
Common-Mode Gain
Common-mode signal (connected to both inputs) → us1 = us2 = ucm
Symmetry → ibcm1 = ibcm2.; for uo = uo2
uo = − h fe ibcm2 RC 2
ucm = ibcm2 ⎡⎣ Rs 2 + rbe2 + 2 RE (1 + h ffe ) ⎤⎦
AVcm =
8. Bipolar Junction Transistor Circuits
−h fe RC 2
uo
=
ucm Rs 2 + rbe 2 + 2 RE (1 + h fe )
46
Differential Gain
Differential signal ud → us2 = − us1 = ud / 2
Symmetry
y
y → ibd2 = − ibd1 → no voltage
g drop
p on RE .; for uo = uo2
uo = − h fe ibd 2 RC 2
ud / 2 = ibd 2 ( Rs 2 + rbe2 )
AVd =
8. Bipolar Junction Transistor Circuits
− h fe RC 2
u o 1 uo
=
=
ud 2 ud / 2 2 ( Rs 2 + rbe 2 )
47
Common-Mode
Rejection Ratio (CMRR)
Output voltage → superposition of voltages with common-mode and differential
signals
uo = AVd ud + AVcm ucm
Common-mode rejection ratio: CMMR ≡
CMMR =
AVd
AVcm
Rs 2 + rbe 2 + 2 RE (1 + h fe ) 1 RE (1 + h fe )
= +
2 ( Rs 2 + rbe 2 )
2
Rs 2 + rbe 2
With Rs1 = Rs2 = 0
1 RE (1 + h fe ) 1
1 I
CMMR = +
≈ + g m2 RE = + CQ 2 RE
rbe2
2
2
2 UT
8. Bipolar Junction Transistor Circuits
48
Example 8.12
Consider the differential amplifier from the example 8.11 with single-ended output
uo = uo2. Calculate
C l l t common-mode
d and
d diff
differential
ti l voltage
lt
gain
i AVcm and
d AVd,
and commo-mode rejection ratio CMMR. Circuit parameters are hfe = 100 and
UT = 25 mV. Disregard the increase of the collector current in the active mode.
8. Bipolar Junction Transistor Circuits
49
Example 8.13
Consider the differential amplifier from the example 8.12 with sinusoidal inputs
us1 = Us1m sin
i ωt and
d us2 = Us2m sin
i ωt. Calculate
C l l t th
the output
t t voltage
lt
uo = uo2 for
f
a)
Us1m = − 5 mV and Us2m = 5 mV,
b)
Us1m
1 = 20 mV and Us2m
2 = 30 mV.
8. Bipolar Junction Transistor Circuits
50
Single-Ended and Differential
Output Gains
Differential output → uo = uo2 − uo1 → uo2 and uo1 are individual output voltages.
For common-mode signal us1 = us2 = ucm
AVcm1 =
−h fe RC1
uo1
=
ucm Rs1 + rbe1 + 2 RE (1 + h fe )
AVcm =
uo 2 − uo1
= AVcm 2 − AVcm1 = 0
ucm
AVcm 2 =
−h fe RC 2
uo 2
=
ucm Rs 2 + rbe 2 + 2 RE (1 + h fe )
For differential signal
g
us2 = − us1 = ud / 2
AVd1 =
+ h fe RC1
uo1
=
ud 2 ( Rs1 + rbe1 )
AVd =
− h fe RC 2
uo 2 − uo1
= AVd 2 − AVd1 =
ud
Rs 2 + rbe 2
8. Bipolar Junction Transistor Circuits
AVd 2 =
− h fe RC 2
uo 2
=
ud 2 ( Rs 2 + rbe 2 )
51
Transfer Characteristic (1)
I0 =
u E + U EE U EE
≈
RE
RE
I 0 ≈ iC1 + iC 2
⎛u ⎞
iC1 = I S exp
e p ⎜⎜ BE1 ⎟⎟
⎝ UT ⎠
⎛u
⎞
iC 2 = I S exp ⎜⎜ BE 2 ⎟⎟
⎝ UT ⎠
⎛u
⎞
⎛ u
⎛ u
− u BE1 ⎞
− u BE1 ⎞
⎟⎟ = iC 2 exp ⎜⎜ − BE 2
⎟⎟
iC1 = I S exp ⎜⎜ BE 2 ⎟⎟ exp ⎜⎜ − BE 2
U
U
U
⎝ T ⎠
⎝
⎠
⎝
⎠
T
T
u D = u B 2 − u B1 = u BE 2 − u BE1
8. Bipolar Junction Transistor Circuits
52
Transfer Characteristic (2)
iC 2 ≈
I0
⎛ u ⎞
1 + exp ⎜ − D ⎟
⎝ UT ⎠
iC1 ≈
I0
⎛u ⎞
1 + exp ⎜ D ⎟
⎝ UT ⎠
For uD = 0
Gm ,max =
d iC1 d iC 2
I
=
= 0
d u D d u D 4U T
AVd ,max = − Gm, max RC 2 =
=−
8. Bipolar Junction Transistor Circuits
I0
RC 2
4U T
53
Bipolar Junction Transistor
as a Switch
u I − u BE
RB
uO = uCE = U CC − RC iC
iB =
Point A: iC ≈ 0, uO = uCE = UCC → switch is off
Point B: uO = uCE = UCEsat, iC = ICsat = (UCC − UCEsat)/RC → switch is on
Saturation condition: IBsat ≥ ICsat/β
8. Bipolar Junction Transistor Circuits
54
Example 8.14
BJT switch in the figure has the following parameters: UCC = 5 V and RC = 1 kΩ.
Determine the maximum value of resistance RB for which the input voltage
UI = UCC will ensure that T1 is in saturation
saturation. The value of current gain β can
be between 50 and 150. Assume that UCEsat = 0,2 V and UBEsat = 0,8 V.
8. Bipolar Junction Transistor Circuits
55
Voltage Transfer Characteristic
Bipolar junction transistor switch is an inverter
Voltage transfer characteristic → uI = f(uO)
‰ for uI < UIL → iC ≈ 0, uO = UCC = UOH= U1
‰ for UIL < uI < UIH → iC = β iB
uO = uCE = U CC − RC iC = U CC − β RC
u I − u BE
RB
for uI = UIL = UBE → iC ≈ 0 → UO = UCC
for uI = UIH → UO = UCEsat
IB =
U IH
U IH − U BE I Csat U CC − U CEsat
=
=
β
β RC
RB
R
= B (U CC − U CEsat ) + U BE
β RC
‰ for uI > UIH → IC = ICsat = (UCC − UCEsat)/RC,
IBsat ≥ ICsat/β , uO = UCEsat = UOL = U0
8. Bipolar Junction Transistor Circuits
56
Example 8.15
Determine the voltage transfer characteristic breakpoint voltages of the bipolar
junction transistor inverter shown on the figure. Voltage supply is UCC = 5 V,
and the resistance values are RB = 10 kΩ and RC = 1 kΩ. Assume β = 100,
and UCEsat = 0,2 V. Assume that in active mode UBE = 0,7 V. Calculate the
voltage noise margins.
8. Bipolar Junction Transistor Circuits
57
Influence of Load on
the Logic Voltage Levels
Operation points of inverter with transistor T1
‰ T0 in cutoff → T1 in saturation →
T2 in cutoff
U CC − U BEsat
RC + RB
I
U − U CEsat
≥ C1sat = CC
β
β RC
I B1sat =
I B1sat
uO1 = uCE1 = U CEsat = U 0
‰ T0 in saturation → T1 in cutoff →
T2 in saturation
u I 1 = uCE 0 = U CEsat
uO1 = U CC − RC I B 2 sat =
= U CC
U − U BEsat
− RC CC
= U1
RC + RB
8. Bipolar Junction Transistor Circuits
If T1 is loaded with N inverters
U1 = U CC − N RC
U CC − U BEsat
N RC + RB
58
Example 8.16
In the inverter chain shown in the figure the following values are set: UCC = 5 V,
RB = 20 kΩ and RC = 1 kΩ. The parameters of all bipolar junction tranistors are
the same β = 100,
100 UCEsat = 0,2
0 2 V and UBEsat = 0,8
0 8 V. Check whether the inverters
work correctly. Determine the transistor T1 voltages of the logic levels 0 and 1.
8. Bipolar Junction Transistor Circuits
59
Impulse Response
Transient response to square pulse
8. Bipolar Junction Transistor Circuits
60
Bipolar Junction Transistor
Turning On
‰ For t < 0 → voltage uI = − UI2 sets the reverse biasing of emitter-base
junction → iB ≈ 0, uBE = − UI2; voltage UCC sets the reverse biasing of
collector-base junction → iC ≈ 0 → transistor is in cutoff region
‰ For t = 0 → instantaneous change of voltage uI from − UI2 to UI1 causes
instant change of current iB → emitter-base junction is gradually being
forward bised → iC is g
gradually
y increased → transistor p
passes through
g
active region and with
U − U BEsat I Csat U CC − U CEsat
iB1 = UL1
>
=
β
β RC
RB
goes to saturation region
‰ Switching times
¾ delay time td → from t = 0 to iC = 0,1
0 1 Icsat
¾ rise time tr → from iC = 0,1 ICzas to iC = 0,9 Iczas
¾ turn-on time ton = td + tr
8. Bipolar Junction Transistor Circuits
61
Bipolar Junction Transistor
Turning Off
‰ At t = t1 → instantaneous change of voltage uI from UI1 to − UI2 → causes
instant change of current iB to
− I B2 =
− UUL 2 − U BEsat
RB
emitter-base junction stays at the beginning forward biased due to the
injected minority carriers → base current − IB2 removes excess charge in
saturation region → currents iB and iC start to reduce when transistor is
entering the active region → after passes through active region the
t
transistor
i t ends
d in
i cutoff
t ff regiontimes
i ti
‰ Switching times
¾ storage time ts → from t = t1 to iC = 0,9 ICsat
¾ fall time tf → from iC = 0,9 ICsat to iC = 0,1 Icsat
¾ turn-off time toff = ts + tf
8. Bipolar Junction Transistor Circuits
62
Switching Times
‰ The ratio between turn-on and turn-off time is set by the ratio of currents
IB2 / IB1 , i.e. voltages UI2 / UI1 → higher amount of current IB1 →
shorter turn
turn-on
on time and longer turn-off
turn off time →
higher amount of current IB2 → shorter turn-off time and longer turn-on time
‰ Switch can be controlled by only a positive impulse uI – with high voltage
level uI = UI1 and low voltage level uI = 0 → lesser amounts of current IB2
and longer turn-off time
‰ Longest time is the storage time ts → due to the removal of excess charge
of minority carriers in the saturation mode
8. Bipolar Junction Transistor Circuits
63
Reducing the Storage Time –
Schottky Transistor
Schottky diode between collector and base
When the collector-base junction is forward biased, the Schottky diode limits the
voltage uBC to the diode knee voltage – blocking the transistor to enter deep
into the saturation mode
Voltage uCE of the transistor in saturation uCE = uBE − uD ≈ 0,7 − 0,4 = 0,3 V
Application: Schottky TTL
8. Bipolar Junction Transistor Circuits
64
Digital Integrated
Bipolar Logic Families
‰ First digital integrated circuits – bipolar junction transistor circuits
‰ Bipolar junction transistor logic families
¾ TTL – transistor-transistor logic circuits
– higher speed of operation due to the application of Schottky transistors
¾ ECL –emitter-coupled logic circuits
– fastest bipolar junction transistor digital circuits → transistors are
operating in active region and saturation region
8. Bipolar Junction Transistor Circuits
65
Current Switch
The same topology as differential amplifier
uI → input voltage, UR → reference voltage
iC1
⎛ u − u BE 2 ⎞
⎛ uI − U R ⎞
exp
= exp ⎜ BE1
=
⎟
⎜ U
⎟
iC 2
UT
⎝
⎠
⎝
⎠
T
for iC1 + iC2 ≈ I0
iC1
≈
I0
1
⎛ U − uI ⎞
1 + exp ⎜ R
⎟
⎝ UT ⎠
iC 2
1
≈
⎛ u − UR ⎞
I0
1 + exp ⎜ I
⎟
⎝ UT ⎠
8. Bipolar Junction Transistor Circuits
66
Current Switch –
Logic Level Voltages
for UR – uI > 4 UT →
iC1 ≈ 0 i iC2 ≈ I0
uO1 = U1 ≈ UCC
uO2 = U0 ≈ UCC − I0 RC2
for uI – UR > 4 UT →
iC1 ≈ I0 i iC2 ≈ 0
uO1 = U0 ≈ UCC − I0 RC1
uO2 = U1 ≈ UCC
Transistors are operating in active and in saturation region
U R − u BE 2 + U EE
RE
R
U 0 ≈ U CC − C (U R − u BE 2 + U EE )
RE
I0 =
8. Bipolar Junction Transistor Circuits
ΔU = U 1 − U 0 ≈
RC
(U R − uBE 2 + U EE )
RE
67
ECL 10K Logic-Gate –
Schematic
8. Bipolar Junction Transistor Circuits
68
ECL 10K Logic-Gate –
Description (1)
‰ Based on the current switch
‰ Logic function → transistor T1 replaced by parallel transistors T1i
‰ Outputs Y and Y connected to the collectors of transistors T1i and T2 through
emitter followers (transistors T3 and T4)
‰ For
F correctt operation
ti U0 > UR > U1
‰ For at least one of the inputs at logic level 1 → current flows through RC1
and not flow through RC2 → U Y = U 0 → U Y = U1
‰ For all inputs at logic level 0 → the current does not flow through RC1
and flows through RC2 → U Y = U 1 → U Y = U 0
‰ Logic functions OR and NOR Y = A + B, Y = A + B
8. Bipolar Junction Transistor Circuits
69
ECL 10K Logic-Gate –
Description (2)
‰ The role of the emitter followers T3 and T4 →
¾ voltages at the emitters of transistors T3 and T4 are more negative than
the voltages at the collectors of transistors T1i and T2 → shift of voltage
level ensures the same voltages for logic 0 and 1 at the input and output
of the gate
¾ D
Due to the
h llarge current gain
i off the
h ffollower
ll
→ higher
hi h output currents
affect less the voltages at the collectors of transistors T1i and T2 →
enables higher fan-out.
‰ The termination emitter resistances at the of transistors T3 and T4 are the
resistors RA and RB of the input of the next gate → for higher speed of
operation the resistors are decreased by using the parallel additional
resistors → for transmission lines – resistance of 50 Ω is connecting to the
negative power supply voltage of − 2 V.
8. Bipolar Junction Transistor Circuits
70
ECL 10K Logic-Gate –
Description (3)
‰ Transistors do not enter saturation region → if at the base of one of the
input transistors → UB = U1 → UY = U 0 → UC = U0 + UBE
U CB = U 0 + U BE − U1 = U BE − ΔU
UCE = UCB + U BE = 2U BE − ΔU = 2Uγ − ΔU
for active region → UCE ≥ UCEsat
example: for UCEmin = Uγ / 2 → ΔUmax = 1,5 Uγ
‰ Negative power supply voltages are used due to the less noise immunity at
this node.
node
‰ The power supply voltages for current switches and referent voltage sources
(that do not generate current changes) are separated from the power supply
f emitter
for
itt followers
f ll
(which
( hi h generates
t llarge currentt changes).
h
)
8. Bipolar Junction Transistor Circuits
71
ECL 10K Logic-Gate –
Characteristics
Basic characteristics
‰ Reference voltage→ UR = − 1,32 V
‰ Logic voltage levels → U1 = − 0,88 V and U0 = − 1,77 V, ΔU = U1 − U2 = 0,89 V
‰ Fan-out → N = 10
‰ Propagation delay → td = 2 ns
‰ Power dissipation→ P = 25 mW
‰ Power
Power-delay
delay product → P
P·ttd = 50 pJ
8. Bipolar Junction Transistor Circuits
72