electronics ii
TRANSCRIPT
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ELECTRONICS II
Engr. Ryann AlimuinINSTRUCTOR
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BJT Modelings• 2 models
1. RE Model
2. Hybrid Model
Model – is the combination of the circuit elements properly chosen that best appropriate the actual behavior of the semiconductor devices under specific operating condition.
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Steps in getting the AC Equivalent Circuit
• Set all the DC source to zero and replace them by a short circuit equivalent.
• Replace all the capacitors by a circuit equivalent.• Remove all the elements bypassed by the short circuit
equivalent introduced by steps 1 and 2.• Redraw the networks In a more convenient and logical
form.
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1. Input Impedance, Zi
• for small signal analysis, once the input impedance has been determined the same numerical value can be used for changing levels of applied signals.
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• The input impedance of a BJT amplifier is purely resistive in nature and depending in the manner in which the transistor is employed, can vary form a few ohms to mse.
Note:
An ohmmeter can be used to measure the small signal ac input impedance since the ohmmeter operates in dc mode.
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2. Output impedance, Zo
• The output impedance is determined at the output terminals looking back into the system with the applied signal set to zero.
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3. Voltage Gain – Av
• One of the most important characteristics of an amplifier is the signal AC voltage gain as determined by
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• Condition is if Rl is approaching infinity.
• AVNL – No Load Voltage Gain
• The load has not been converted to the output terminals• • For Transistor amplifiers , The no load voltage gain is
greater than the loaded voltage gain
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Depending on the configuration, magnitude of the
voltage gain for a loaded single stage transitive amplifier
typically ranges from just less than 1 to a few hundred.
A multi-stage system, however can have a voltage gain in
thousands.
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4. Current Gain- Ai
• For BJT Amplifier, the current typically range from a level , just less than 1 to a level that may exceed 100
• Voltage gain
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Example:•Vi•Zi•Avnl•Avs (A.)Vi
Vs-IiRs-Vi=018mV-(10µA)(.65KΩ)-Vi
Vi= 11.5mV
(B.)ZiZi=Vi/Ii
=11mV/10mA=1.15KΩ
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(C.) Avnl
Avnl=Vo/Vi
=3.6V/11.5mV
=313V
(D.) Avs
= Vo/Vs
=3.6/18mV
=200V
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Systems Approach
• Effects of Rs and Rl
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• Using Voltage Divider
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Example
In the figure , a load impedance has been applied to the fixed bias transistor amplifier.Determine Av and Ai using two port systems. Determine Av and Ai using the RE Model and compare results.
• Rc= Ro• Ro=10.71• Avnl= -280.11• Zi= 1.071KΩ• Zo= 3KΩ
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• Av using current divider theorem
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2. For prefixed bias,
Determine:
A. Avnl,Zi and Zo
B. Sketch the two port model and parameters
C. Calculate Av.
D. Determine Ai
E. Calculate Av and Ai using AC analysis
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• DC Analysis
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• (B.)Sketch two port model and parameter• (C.)Calculate Av
• (D.)Determine Ai
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• (E.)Calculate Av and Ai using AC Analysis
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Effects of Source Impedance (Rs)• - the effect of an internal resistance on the gain of the
amplifier • - The parameter Zi and Avnl of a two port system are
unaffected by an internal resistance of the applied source.
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• In Fig. A source w/ an internal resistance has been applied to the fixed bias transistor
• Determine the voltage gain Avs= Vo/Vs.What % of the applied signal
• Determine the voltage gain Avs= Vo/Vs using the RE model
DC Analysis
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Combined Effects Of Rs and Rl
• A source of Rs and a load Rl have been applied in a two port systems for which the parameters Zi, Avnl, and Zo have been specified.
• Assume Zi and Zo are unaffected by Rl and Rs
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• Using Voltage Divider Bias
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• Example.
For a single stage amplifier with Rl= 4.7KΩ and Rs= 0.3KΩ.
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BJT And JFET Frequency Response
• Logarithms
• Example. B=10 , X= 2
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Decibels
• the relationship of logarithms to power and audio levels.
• the term (BEL) was derived from the surname of Alexander Grahambell.
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Power = 10• Voltage = 20• Current = 20
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Examples.
1.) Find the magnitude gain corresponding to a decibel gain of 100.
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2.) The input power to a device is 10000v at avoltage of 1000v. The output power is 500w, while the output impedance is 20Ω.
• (a.)Find the power gain dB
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• b.)Find the voltage gain in dB
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CASCADED AMPLIFIER
AVT= (Av1)(Av2)
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V4.77V
15KΩ4.7KΩ
4.7KΩ20V
RR
R VV
21
2
ccTH
3.58KΩ15KΩKRTH //7.4
A(201)1KΩ3.58KΩ
0.7V4.77V
1)R(βR
VVI
ETH
BETH
B 89.19
3.94mAμA(201)19.89IE
6.50Ω3.94mA
26mVre
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953.69ΩΩ)(200)(6.50KBRZ reTHi //58.3//
102.3346.50Ω
953.69ΩK
re
ZiRcAv1
//2.2//
mVVAvVo i11 56.225334.102
338.466.50Ω
2.2KΩ
re
RcAv2
34,635338.46)102.334)((ΑvΤ
865.9mVV)5(34,635)(2ViAVo VT2 If a 10KΩ resistor is connected across the output. What is Vi?
709.75mV856.9mV2.2KΩ10KΩ
10KΩVo
RCRL
RLVi
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JFET(JUNCTION FIELD EFFECT TRANSISTOR)
A type of FET that operates with a reversed-biased junctio.n to control current in a channel.
GM= Forward transconductance change in drain current
for a given change in rate to source voltage
with the drain to source voltage constant. DI GSV
2
GS(OFF)
GS
DSS
GS
B
V
V1IIB
SIEMENSΔV
ΔIGM
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gm or Forward Transfer Admittance
DEPLETION MOSFET ( D-MOSFET)- The drain & source are diffused into the substrate material & then connected by a narrow channel adjacent to the insulated gate.
||V
2Igm
V
V1gmgm
GS(off)
DSS
o
GS(off)
GS
o
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n-channel-operates in the depletion mode when a negative gate-to-source voltage is applied & in enhancement mode when a positive gate to source voltage is applied
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FET AMPLIFIERS
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AC Equivalent
gmRs1
gmRdAv
2
GS(Off)
D
SDSD
V
RsI1II
RsIV DGS
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FET EQUIVALENT CIRCUIT
If no RL: Id
gmIdRd
gmId
IdRd
V
Vout
Vo
ViAv
Gs
gmRDAv
gmRdAv
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Internal Drain Resistance
Effects of the source resistance on gain
By KVL:
)gm(rds//RdAv
)//Rgm(rds//(RAv CD
IdRdVo
IdRsVgsVi
IdRsVgsVi
0
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( If there is Rs )
gmRs1
gmRdAv
IdRsgmId
IdRd
IdRsVgs
IdRd
Vi
VoAv
gmRs1
gmRdAv
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Example: Bypassed source Resistance
D
LO
gmRAv
)//Rgm(RAv
RdZo
RZi G
gmRs1
gmRdAv
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Example: The JFET has a gm=4mS w/ external ac drain resistance of 1.5K Ω , is the ideal voltage gain?
-6Av
4mS(1.5KAv
gmRDAv
)
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Example: An FET equivalent circuit is shown. Determine the volatage gain when the output is taken across Rd.
1.852Av
4mS(560Ωm1
4mS(1.5KΩm
gmRs1
gmRAv
D
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JFET SELF-BIAS CONFIGURATION
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AC Analysis
yosrd
1
yos
1rd
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Example: The fixed bias configuration has an operating point defined by VGSQ=-2V and
IDQ=5.625mA with IDSS=10mA & Vp=-8V, The network should be redrawn w/ an applied signal Vi. The Value of yos is provided as 40µS.
Calculate:
a. gm e. Av
b. rd f. Av if rd is IGNORED
c. Zi
d. Zo
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Circuit Diagram
d.)2KΩ
c.)1MΩ
25KΩ40μ0
1
yos
1b.)rd
1.875mS8
212.5mgm
2.5mS8
0.02
//V
2Ia.)gm
p
DSS
o
75.3)2)(875.18(.)
475.3
)25//2)(875.1()//(.)
KmSAvf
KKmSRdRDgmAve
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CASCODE CONNECTION (BJT)
- Cascode Connection has a transistor on top or in series with another.-A Common Emitter ( CE ) stage feeding a Common Base
(CB ) stage
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Ex.Calculate the Av for the cascode amplifier
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6.74Ω3.86mA
26mV
Ie
26mvre
3.86mA1.1KΩ
0.74.95
R
VVI
0RIVVusing:
10.89V6.8KΩ4.7KΩ5.6KΩ
4.7KΩ.18v(5.6KΩ
RRR
)RVcc(RVB2
4.95V18V6.8KΩ5.6KΩ4.7KΩ
4.7KΩVcc
RRR
RV
E
BEB1
E
EEBEB1
B3B2B1
B3B2
B3B2B1
B3
B1
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sdsdsdsds
267.29267.29(1)))(A(AA
267.296.74KΩ
1.8KΩ
re
RcA
1re
re
re
RcA
V2V1VT
V2
V1
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Darlington Connection- A super beta transistor-The composite transistor acts as a single unit with a current gain that the product of the current gain of the individual transistor.
if
21 D
2
21
D
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What current gain is provided by a darlington connection of two identical transistor each having a current gain of
200
000,40
20022
D
D
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EEE
BDE
BEEB
EBB
BECCB
EBBBEcc
EBBeBBcc
EEBeBBcc
RIV
II
VVV
RR
VVI
RRIVV
RIVRIV
RIVRIV
1
0)1((
0)1(
0
DC Bias of a darlington circuit
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Calculate the dc bias voltage and current ECCB VVII ,,,
VV
IV
RIVV
VV
mARIV
mAI
AIII
AI
MRBR
VVI
CC
CCC
CCCCC
E
EEE
C
BDEC
B
EDB
BECCB
18
0
957.7
390403.20
403.20
55.2180001
55.2
39080003.3
6.118
VVBE
D
6.1
8000
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AC equivalent circuit
- For a darlington emitter-follower the ac input signal is applied to the base of the darlington transistor through capacitor C1. with the ac output Vo obtained from the emitter through capacitor Cr
- The darlington transistor is replaced by an ac equivalent circuit composed of an input resistance, ri and an output current source BDIB
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BDI
EDBO
DRBO
EDEBO
EEDEBO
OiBC
RIV
RIV
RRIV
RIRIV
VrIV
1
0
0
![Page 78: ELECTRONICS II](https://reader031.vdocuments.us/reader031/viewer/2022012308/551ec188497959335b8b4826/html5/thumbnails/78.jpg)
BEDi
BDi
DEiB
B
i
B
i
oD
BDBo
i
B
B
o
i
oi
DEiBi
DEiB
i
DEiBi
DEBiBi
oiBi
RRr
RA
RrR
R
I
I
I
I
III
I
I
I
I
I
IA
RrRZ
RrI
V
RrIV
RIrIV
VrIV
)//(
0
If ri is so small if not given = 0
BED
BDi RR
RA
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VVBE
D
6.1
8000
4112
3.3)390(8
)3.7)(8(
61.1Z
)390)(8(5//3.3Z
5r if impedanceinput theCalculate
i
i
i
i
i
A
Mk
MkA
M
kkM
k
![Page 80: ELECTRONICS II](https://reader031.vdocuments.us/reader031/viewer/2022012308/551ec188497959335b8b4826/html5/thumbnails/80.jpg)
1
0given not is r i
v
EDi
EDv
EDiB
EDB
i
ov
A
if
Rr
RA
RrI
RI
V
VA
998.0A
)390)(8000(5
)390(8000A
Av Calculate
v
v
k
![Page 81: ELECTRONICS II](https://reader031.vdocuments.us/reader031/viewer/2022012308/551ec188497959335b8b4826/html5/thumbnails/81.jpg)
Feedback Pair
2121
2222
1111
CCCEC
EBC
EBC
IIIII
III
III
180
140
2
1
CB
EBB
CBBEB
BBCBEB
BBEBCB
BBEBCC
RR
VI
RRIV
RIRIV
RIVRI
RIVRI
21
CC1
21CC
21CC
21CC
CC
V
V
V
0V
0V
:KVLby
![Page 82: ELECTRONICS II](https://reader031.vdocuments.us/reader031/viewer/2022012308/551ec188497959335b8b4826/html5/thumbnails/82.jpg)
Calculate the dc bias current and voltages
AI
MRR
VVI
B
CDB
EBCCB
465.4
)75)(180)(140(2
7.18
1
1
mAI
mAAIII
mAI
AI
AI
III
C
CCC
C
C
C
BBC
143.113
518.1121.625
518.112
1.625180
1.625
)465.4(140
21
2
2
1
1121
![Page 83: ELECTRONICS II](https://reader031.vdocuments.us/reader031/viewer/2022012308/551ec188497959335b8b4826/html5/thumbnails/83.jpg)
VV
VVV
VVV
VV
mA
RIVV
dci
EBodc
i
oEBdc
i
dco
CCCCdc
o
81.8
7.051.9
0
51.9
)75)(143.113(18
)(
)(
)(
)(
\)(
![Page 84: ELECTRONICS II](https://reader031.vdocuments.us/reader031/viewer/2022012308/551ec188497959335b8b4826/html5/thumbnails/84.jpg)
AC Operation
11 BI 22 BI
BCi
CB
o
CBo
CB
CB
CBBo
CBBCCo
RRZ
RI
V
RIV
RI
RI
RIIV
RIIRIV
//
1
21
211
22
211
211
11121
1122
![Page 85: ELECTRONICS II](https://reader031.vdocuments.us/reader031/viewer/2022012308/551ec188497959335b8b4826/html5/thumbnails/85.jpg)
211
211
211
211
1121
111112
11122
1
11
1
B
o
Bo
B
B
B
BBB
BBBo
I
I
II
I
I
I
III
IIII
![Page 86: ELECTRONICS II](https://reader031.vdocuments.us/reader031/viewer/2022012308/551ec188497959335b8b4826/html5/thumbnails/86.jpg)
iC
CV
iBCB
CBoV
iBiiBoiB
Bi
i
B
B
oi
iB
B
i
B
rR
RA
rIRI
RIVA
rIVrIVZR
RA
I
I
I
IA
ZR
R
I
I
21
21
21
21
i
oi21
V
V , V
![Page 87: ELECTRONICS II](https://reader031.vdocuments.us/reader031/viewer/2022012308/551ec188497959335b8b4826/html5/thumbnails/87.jpg)
Calculate the ac circuit values of Zi , Zo and Av and Ai assume that ri = 3kΩ
KZ
MZ
i
i
72.971
2//)75)(180)(140(180
140
2
1
048.119
)180)(140(
3
)180)(140(
o
io
Z
KrZ
88.1695929.9712
)2)(140(180
KM
MAi
9998.03)750)(180)(140(
)750)(180)(140(
KAV
![Page 88: ELECTRONICS II](https://reader031.vdocuments.us/reader031/viewer/2022012308/551ec188497959335b8b4826/html5/thumbnails/88.jpg)
Differential Amplifier Circuit
-If an output signal is applied to either input with the other input connected to ground, the operator is referred to as “single ended”.
-If two opposite polarity input signals are applied, the operation is referred to as “double-ended”.
-If the same input is applied to both inputs, the operator is called “common mode”.
-In double ended operation two input signals are applied the difference of the inputs resulting in outputs from both collectors due to the difference of the signals applied is both input.
![Page 89: ELECTRONICS II](https://reader031.vdocuments.us/reader031/viewer/2022012308/551ec188497959335b8b4826/html5/thumbnails/89.jpg)
- In common-mode operation, the common-input signal results in opposite signals to each collector, these signals canceling so that the resulting output signal is zero.
- The main future of the differential amplifier is the very large gain when opposite signals are applied to the inputs as compared to the very small gain resulting common inputs.
![Page 90: ELECTRONICS II](https://reader031.vdocuments.us/reader031/viewer/2022012308/551ec188497959335b8b4826/html5/thumbnails/90.jpg)
Common-mode rejection ratio- ratio of the difference gain to common gain DC bias.
2E
C
II
0BVE
EEEE
EEEEE
R
VVI
VRIV
0
7.0
E
BEE
BEBE
V
VV
VVV
CE
CCC
CCCCC
RI
VV
RIVV
2
![Page 91: ELECTRONICS II](https://reader031.vdocuments.us/reader031/viewer/2022012308/551ec188497959335b8b4826/html5/thumbnails/91.jpg)
Solve for Ie and Vc
mVIK
I
E
E
515.23.3
)7.0(9
VV
Km
V
C
C
096.4
)3.3(2
515.29
![Page 92: ELECTRONICS II](https://reader031.vdocuments.us/reader031/viewer/2022012308/551ec188497959335b8b4826/html5/thumbnails/92.jpg)
Single Ended
11 BI 22 BI
eii
iii
BBB
rZr
rrr
III
21
21
![Page 93: ELECTRONICS II](https://reader031.vdocuments.us/reader031/viewer/2022012308/551ec188497959335b8b4826/html5/thumbnails/93.jpg)
i
iB
iBi
iBiBi
r
VI
rIV
rIrIV
2
02
0
e
CV
i
C
i
OV
Ci
iO
CBO
CCO
BC
r
RA
r
R
V
VA
Rr
VV
RIV
RIV
II
2
2
2
![Page 94: ELECTRONICS II](https://reader031.vdocuments.us/reader031/viewer/2022012308/551ec188497959335b8b4826/html5/thumbnails/94.jpg)
Calculate the single ended output voltage = Vo
AK
I
R
VVI
E
E
EEEE
02.19343
7.09
mAI
II
C
EC
51.962
02.193
2
39.26951.96
26
mA
mVre
mVV
mVV
VAV
KA
o
o
iVo
V
56.174
)2(23.87
23.87)39.269(2
47
![Page 95: ELECTRONICS II](https://reader031.vdocuments.us/reader031/viewer/2022012308/551ec188497959335b8b4826/html5/thumbnails/95.jpg)
Low Frequency Reponse- BJT Amplifier
![Page 96: ELECTRONICS II](https://reader031.vdocuments.us/reader031/viewer/2022012308/551ec188497959335b8b4826/html5/thumbnails/96.jpg)
• Effect of Cs on low frequency response
![Page 97: ELECTRONICS II](https://reader031.vdocuments.us/reader031/viewer/2022012308/551ec188497959335b8b4826/html5/thumbnails/97.jpg)
• Getting AC Equivalent
![Page 98: ELECTRONICS II](https://reader031.vdocuments.us/reader031/viewer/2022012308/551ec188497959335b8b4826/html5/thumbnails/98.jpg)
• Effect of Cc on the Low Frequency Response
![Page 99: ELECTRONICS II](https://reader031.vdocuments.us/reader031/viewer/2022012308/551ec188497959335b8b4826/html5/thumbnails/99.jpg)
Low Frequency Response
• Can be establish for each capacitive element and the frequency at which the output voltage drops to 0.707 of its maximum value
![Page 100: ELECTRONICS II](https://reader031.vdocuments.us/reader031/viewer/2022012308/551ec188497959335b8b4826/html5/thumbnails/100.jpg)
![Page 101: ELECTRONICS II](https://reader031.vdocuments.us/reader031/viewer/2022012308/551ec188497959335b8b4826/html5/thumbnails/101.jpg)
![Page 102: ELECTRONICS II](https://reader031.vdocuments.us/reader031/viewer/2022012308/551ec188497959335b8b4826/html5/thumbnails/102.jpg)
• -3db drop in gain from the midband level w hen f=f an RC network will determine the low-frequency cut-off frequency for a BJT transistor, f will be
![Page 103: ELECTRONICS II](https://reader031.vdocuments.us/reader031/viewer/2022012308/551ec188497959335b8b4826/html5/thumbnails/103.jpg)
![Page 104: ELECTRONICS II](https://reader031.vdocuments.us/reader031/viewer/2022012308/551ec188497959335b8b4826/html5/thumbnails/104.jpg)
![Page 105: ELECTRONICS II](https://reader031.vdocuments.us/reader031/viewer/2022012308/551ec188497959335b8b4826/html5/thumbnails/105.jpg)
![Page 106: ELECTRONICS II](https://reader031.vdocuments.us/reader031/viewer/2022012308/551ec188497959335b8b4826/html5/thumbnails/106.jpg)
• Example.– Determine the lower out off frequency using:
![Page 107: ELECTRONICS II](https://reader031.vdocuments.us/reader031/viewer/2022012308/551ec188497959335b8b4826/html5/thumbnails/107.jpg)
– Solution:
![Page 108: ELECTRONICS II](https://reader031.vdocuments.us/reader031/viewer/2022012308/551ec188497959335b8b4826/html5/thumbnails/108.jpg)
![Page 109: ELECTRONICS II](https://reader031.vdocuments.us/reader031/viewer/2022012308/551ec188497959335b8b4826/html5/thumbnails/109.jpg)
![Page 110: ELECTRONICS II](https://reader031.vdocuments.us/reader031/viewer/2022012308/551ec188497959335b8b4826/html5/thumbnails/110.jpg)
![Page 111: ELECTRONICS II](https://reader031.vdocuments.us/reader031/viewer/2022012308/551ec188497959335b8b4826/html5/thumbnails/111.jpg)
![Page 112: ELECTRONICS II](https://reader031.vdocuments.us/reader031/viewer/2022012308/551ec188497959335b8b4826/html5/thumbnails/112.jpg)
OP-AMP BASICS
• A very high gain differential amplifier with very high input ompedance and low output impedance.
1. Provide voltage amplitude amplitude changes
2. Oscillators
3. Filter circuits
4. Instrumentation circuit
![Page 113: ELECTRONICS II](https://reader031.vdocuments.us/reader031/viewer/2022012308/551ec188497959335b8b4826/html5/thumbnails/113.jpg)
• Single-ended input (mode)– Results when the input signal is connected to one
input with the order input connected to the ground.
![Page 114: ELECTRONICS II](https://reader031.vdocuments.us/reader031/viewer/2022012308/551ec188497959335b8b4826/html5/thumbnails/114.jpg)
• Differential mode, two opposite-polarity(out of phase signals are appliedto the inputs. Refered as double-ended).
![Page 115: ELECTRONICS II](https://reader031.vdocuments.us/reader031/viewer/2022012308/551ec188497959335b8b4826/html5/thumbnails/115.jpg)
• Common mode input voltage range– Range of input voltages which, when applied to both
inputs will not cause clipping or other output distortion.
• Input offset voltage– Differential dc voltage required between the inputs to
force the output to zero volts.
• Input offset voltage drift– Specifies how much change occurs in the input offset
voltage for each degree change in temperature.
![Page 116: ELECTRONICS II](https://reader031.vdocuments.us/reader031/viewer/2022012308/551ec188497959335b8b4826/html5/thumbnails/116.jpg)
• Input bias current– DC current required by the inputs of the amplifier to
properly operate the first stage.
• Input impedance ( Differential input impedance )– Total resistance between the inverting and non-
inverting inputs.
![Page 117: ELECTRONICS II](https://reader031.vdocuments.us/reader031/viewer/2022012308/551ec188497959335b8b4826/html5/thumbnails/117.jpg)
• Input offset current– Difference of the input bias currents.
![Page 118: ELECTRONICS II](https://reader031.vdocuments.us/reader031/viewer/2022012308/551ec188497959335b8b4826/html5/thumbnails/118.jpg)
• Output impedance– Resistance viewed from the output terminal.
![Page 119: ELECTRONICS II](https://reader031.vdocuments.us/reader031/viewer/2022012308/551ec188497959335b8b4826/html5/thumbnails/119.jpg)
• Double Ended Differential Input
![Page 120: ELECTRONICS II](https://reader031.vdocuments.us/reader031/viewer/2022012308/551ec188497959335b8b4826/html5/thumbnails/120.jpg)
• Double Ended output
![Page 121: ELECTRONICS II](https://reader031.vdocuments.us/reader031/viewer/2022012308/551ec188497959335b8b4826/html5/thumbnails/121.jpg)
• Common Mode Operation– Two inputs one equally amplified and since they result
in opposite polarity signals at the output, these signals cancel and results in 0V output.
![Page 122: ELECTRONICS II](https://reader031.vdocuments.us/reader031/viewer/2022012308/551ec188497959335b8b4826/html5/thumbnails/122.jpg)
• CMR ( Common Mode Rejection )
– Amplifier the difference signal while rejecting the common signal at the two inputs.
![Page 123: ELECTRONICS II](https://reader031.vdocuments.us/reader031/viewer/2022012308/551ec188497959335b8b4826/html5/thumbnails/123.jpg)
Common and Differential Mode Operation
• Differential Inputs– difference of the two signals.
• Common Input– Average of the sum of the two signals.
![Page 124: ELECTRONICS II](https://reader031.vdocuments.us/reader031/viewer/2022012308/551ec188497959335b8b4826/html5/thumbnails/124.jpg)
• Output Voltage
![Page 125: ELECTRONICS II](https://reader031.vdocuments.us/reader031/viewer/2022012308/551ec188497959335b8b4826/html5/thumbnails/125.jpg)
• CMMR – Common Mode Rejection Ratio
![Page 126: ELECTRONICS II](https://reader031.vdocuments.us/reader031/viewer/2022012308/551ec188497959335b8b4826/html5/thumbnails/126.jpg)
• The output for Vo
![Page 127: ELECTRONICS II](https://reader031.vdocuments.us/reader031/viewer/2022012308/551ec188497959335b8b4826/html5/thumbnails/127.jpg)
• Example.– Determine the output voltage of an op-amp for input
voltage of Vi=150μV, Vi2=140μV. The amplifier has a differential gain of Ad= 4000 and CMRR is
(a)100
(b)10^5
Solution:
![Page 128: ELECTRONICS II](https://reader031.vdocuments.us/reader031/viewer/2022012308/551ec188497959335b8b4826/html5/thumbnails/128.jpg)
![Page 129: ELECTRONICS II](https://reader031.vdocuments.us/reader031/viewer/2022012308/551ec188497959335b8b4826/html5/thumbnails/129.jpg)
![Page 130: ELECTRONICS II](https://reader031.vdocuments.us/reader031/viewer/2022012308/551ec188497959335b8b4826/html5/thumbnails/130.jpg)
• Slew Rate– Maximum rate of change of the output voltage in
response to a step input voltage.
![Page 131: ELECTRONICS II](https://reader031.vdocuments.us/reader031/viewer/2022012308/551ec188497959335b8b4826/html5/thumbnails/131.jpg)
• Negative Feedback– The inverting (-) input effectively makes the
feedback signal 180 degrees out of phase with the input signal.
![Page 132: ELECTRONICS II](https://reader031.vdocuments.us/reader031/viewer/2022012308/551ec188497959335b8b4826/html5/thumbnails/132.jpg)
• Closed-Loop Voltage Gain, Acl– Voltage gain of an op-amp with external feedback.
• Non-Inverting Amplifier– Op-amp connected in a closed-loop configuration.
![Page 133: ELECTRONICS II](https://reader031.vdocuments.us/reader031/viewer/2022012308/551ec188497959335b8b4826/html5/thumbnails/133.jpg)
• Example.– Determine the gain of the amplifier. The open-loop
voltage gain of the op-amp is 100,000.
![Page 134: ELECTRONICS II](https://reader031.vdocuments.us/reader031/viewer/2022012308/551ec188497959335b8b4826/html5/thumbnails/134.jpg)
– Solution:
![Page 135: ELECTRONICS II](https://reader031.vdocuments.us/reader031/viewer/2022012308/551ec188497959335b8b4826/html5/thumbnails/135.jpg)
• Voltage Follower– A special case of a non-inverting amplifier where all of
the output voltage is fed back to the inverting (-) input by a straight connection.
![Page 136: ELECTRONICS II](https://reader031.vdocuments.us/reader031/viewer/2022012308/551ec188497959335b8b4826/html5/thumbnails/136.jpg)
• Inverting Amplifier– Configuration where there is a controlled amount of
voltage gain.
![Page 137: ELECTRONICS II](https://reader031.vdocuments.us/reader031/viewer/2022012308/551ec188497959335b8b4826/html5/thumbnails/137.jpg)
• Example.– Given the op-amp configuration determine the value
of Rf required to produce a closed-loop voltage gain of -100.
![Page 138: ELECTRONICS II](https://reader031.vdocuments.us/reader031/viewer/2022012308/551ec188497959335b8b4826/html5/thumbnails/138.jpg)
– Solution:
![Page 139: ELECTRONICS II](https://reader031.vdocuments.us/reader031/viewer/2022012308/551ec188497959335b8b4826/html5/thumbnails/139.jpg)
• Impedance of Non-Inverting Amplifier– Input impedance
![Page 140: ELECTRONICS II](https://reader031.vdocuments.us/reader031/viewer/2022012308/551ec188497959335b8b4826/html5/thumbnails/140.jpg)
– Output impedance
![Page 141: ELECTRONICS II](https://reader031.vdocuments.us/reader031/viewer/2022012308/551ec188497959335b8b4826/html5/thumbnails/141.jpg)
• Example.– Determine the input and output impedance of the
amplifier . The op-amp datasheet gives Zin=2MΩ, Zout=75Ω and Acl= 200,000.
![Page 142: ELECTRONICS II](https://reader031.vdocuments.us/reader031/viewer/2022012308/551ec188497959335b8b4826/html5/thumbnails/142.jpg)
– Solution:
![Page 143: ELECTRONICS II](https://reader031.vdocuments.us/reader031/viewer/2022012308/551ec188497959335b8b4826/html5/thumbnails/143.jpg)
BASIC OP-AMP CIRCUITS
• Comparator– To determine when an input voltage exceeds a
certain level the (-) inverting input is grounded to produce a zero level and that the input signal voltage is applied to the non-inverting (+) input.
![Page 144: ELECTRONICS II](https://reader031.vdocuments.us/reader031/viewer/2022012308/551ec188497959335b8b4826/html5/thumbnails/144.jpg)
• Non-Zero Level Detection
– The zero level detector can be modified to detect voltages other than zero by connecting a fixed reference voltage source to the (-) inverting input.
![Page 145: ELECTRONICS II](https://reader031.vdocuments.us/reader031/viewer/2022012308/551ec188497959335b8b4826/html5/thumbnails/145.jpg)
• Example. The input signal is applied to the comparator circuit make a sketch of the output showing its proper relationship to the input signal. Assume the maximum output levels of the op-amp are 12V.
![Page 146: ELECTRONICS II](https://reader031.vdocuments.us/reader031/viewer/2022012308/551ec188497959335b8b4826/html5/thumbnails/146.jpg)
• Summing Amplifier– Has two or more inputs, its output voltage is
proportional to the negative of the algebraic sum of its input voltages.
• Summing Amplifier with Unity Gain
![Page 147: ELECTRONICS II](https://reader031.vdocuments.us/reader031/viewer/2022012308/551ec188497959335b8b4826/html5/thumbnails/147.jpg)
![Page 148: ELECTRONICS II](https://reader031.vdocuments.us/reader031/viewer/2022012308/551ec188497959335b8b4826/html5/thumbnails/148.jpg)
![Page 149: ELECTRONICS II](https://reader031.vdocuments.us/reader031/viewer/2022012308/551ec188497959335b8b4826/html5/thumbnails/149.jpg)
• Example.– Determine the output voltage.
![Page 150: ELECTRONICS II](https://reader031.vdocuments.us/reader031/viewer/2022012308/551ec188497959335b8b4826/html5/thumbnails/150.jpg)
• Summing Amplifier with Gain Greater than Unity.
Since
![Page 151: ELECTRONICS II](https://reader031.vdocuments.us/reader031/viewer/2022012308/551ec188497959335b8b4826/html5/thumbnails/151.jpg)
• Example.– Determine the output voltage for the summing
amplifier.
![Page 152: ELECTRONICS II](https://reader031.vdocuments.us/reader031/viewer/2022012308/551ec188497959335b8b4826/html5/thumbnails/152.jpg)
– Solution :
![Page 153: ELECTRONICS II](https://reader031.vdocuments.us/reader031/viewer/2022012308/551ec188497959335b8b4826/html5/thumbnails/153.jpg)
• Averaging Amplifier– A summing amplifier can be make to produce the
mathematical average of the input voltages. This is done by setting the ratio equal to the reciprocal of the number of inputs(n).
![Page 154: ELECTRONICS II](https://reader031.vdocuments.us/reader031/viewer/2022012308/551ec188497959335b8b4826/html5/thumbnails/154.jpg)
• Example.– Show that the amplifier produces an output whose
magnitude is the mathematical average of the input voltages.
![Page 155: ELECTRONICS II](https://reader031.vdocuments.us/reader031/viewer/2022012308/551ec188497959335b8b4826/html5/thumbnails/155.jpg)
– Solution :
![Page 156: ELECTRONICS II](https://reader031.vdocuments.us/reader031/viewer/2022012308/551ec188497959335b8b4826/html5/thumbnails/156.jpg)
• Scaling Adder
– Example. Determine the weight of each input and the output voltage.
![Page 157: ELECTRONICS II](https://reader031.vdocuments.us/reader031/viewer/2022012308/551ec188497959335b8b4826/html5/thumbnails/157.jpg)
• Solution:
![Page 158: ELECTRONICS II](https://reader031.vdocuments.us/reader031/viewer/2022012308/551ec188497959335b8b4826/html5/thumbnails/158.jpg)
• Multiple-Stage Gain
![Page 159: ELECTRONICS II](https://reader031.vdocuments.us/reader031/viewer/2022012308/551ec188497959335b8b4826/html5/thumbnails/159.jpg)
• Example– Determine the output voltage using the circuit for
resistor components of value Rf = 470kΩ, R1= 4.3kΩ, R2 = 33kΩ, and R3 = 38kΩ for an input of 80μΩ
![Page 160: ELECTRONICS II](https://reader031.vdocuments.us/reader031/viewer/2022012308/551ec188497959335b8b4826/html5/thumbnails/160.jpg)
• Voltage Subtraction
![Page 161: ELECTRONICS II](https://reader031.vdocuments.us/reader031/viewer/2022012308/551ec188497959335b8b4826/html5/thumbnails/161.jpg)
• Example.– Determine the output for the circuit with components
Rf= 1MΩ, R1= 100kΩ, and R3= 500kΩ.
solution:
![Page 162: ELECTRONICS II](https://reader031.vdocuments.us/reader031/viewer/2022012308/551ec188497959335b8b4826/html5/thumbnails/162.jpg)
Submitted by:
Octavo, Antonio
Pasquile, Ronald
Pineda, John
Ricarde, Julius Clarrence Ricarde B.
Rogador, Mark
Trinanes, Nomeer
EC32FB1