unit 1 amplifiers
TRANSCRIPT
-
8/2/2019 Unit 1 Amplifiers
1/46
Unit 1:
Amplifiers
-
8/2/2019 Unit 1 Amplifiers
2/46
Differential Amplifier
is a type of electronic amplifier that multiplies the difference between twoinputs by some constant factor (the differential gain).
is the basic building block of operational amplifiers.
A Schematic Symbol of aDifferential Amplifier
where:Vs+ and Vs_= supply voltagesA= amplifier gainV
+= non-inverting input
V_ = inverting input
-
8/2/2019 Unit 1 Amplifiers
3/46
Basic Differential Amplifier Circuit
DC Bias
-
8/2/2019 Unit 1 Amplifiers
4/46
Example page 651
Calculate the dc voltages and currents in the circuit below.
-
8/2/2019 Unit 1 Amplifiers
5/46
Single-Ended AC Voltage Gain
AC Equivalent
-
8/2/2019 Unit 1 Amplifiers
6/46
-
8/2/2019 Unit 1 Amplifiers
7/46
Double-Ended AC Voltage Gain
AC Equivalent
21
2
iidid
od
VVV
r
Rc
V
VA
where:Ad= Differential Mode Gain
-
8/2/2019 Unit 1 Amplifiers
8/46
Common-Mode Operation
Common-Mode ConnectionAC Circuit in Common-Mode Connection
Ei
C
i
oC
Rr
R
V
VA
12
-
8/2/2019 Unit 1 Amplifiers
9/46
Example page 656
Calculate the common-mode gain for the amplifier circuit shown below.
1= 2=75ri1
= ri2= ri
=20k
-
8/2/2019 Unit 1 Amplifiers
10/46
Use of ConstantCurrent Source
Differential Amplifier with constant-current source AC Equivalent
EiC
i
oC
Rr
R
V
VA
12
-
8/2/2019 Unit 1 Amplifiers
11/46
Example page 657
Calculate the common-mode gain for the differential amplifier in the circuit below.
1= 2= =75ri1
= ri2= ri
=11k
Q3
ro=200k3=75
-
8/2/2019 Unit 1 Amplifiers
12/46
Infinite input impedance
Infinite open-loop gain
High Common Mode Rejection Ratio (CMMR)
Differential input voltage is zero
Zero output impedance
Infinite Bandwidth
Characteristics of an Ideal op-amp
-
8/2/2019 Unit 1 Amplifiers
13/46
Inverting Amplifier-is a constant gain amplifier circuit.
Example Page 685
If the circuit above has R1=100k and Rf=500k, what output voltage resultsfor an input of V1=2V?
VV
k
kV
R
RVo
f10)2(
100
5001
1
-
8/2/2019 Unit 1 Amplifiers
14/46
-
8/2/2019 Unit 1 Amplifiers
15/46
Noninverting Amplifier-is a constant gain amplifier circuit.
Example Page 686
If the circuit above has R1=100k and Rf=500k, what output voltage resultsfor an input of V1=2V?
-
8/2/2019 Unit 1 Amplifiers
16/46
Example page 716
Calculate the output voltage from the circuit below for an input of 120V.
240k
2.4k +16V
-16V
-
8/2/2019 Unit 1 Amplifiers
17/46
-
8/2/2019 Unit 1 Amplifiers
18/46
-
8/2/2019 Unit 1 Amplifiers
19/46
Example page 717
Show the connection of an LM124 quad op-amp as a three-stageamplifier with gains of +10, -18 and -27. Use 270-k feedback resistor for
all three circuits. What output voltage will result for an input of 150V?
-
8/2/2019 Unit 1 Amplifiers
20/46
Example page 718
Show the connection of three op-amp stages using an LM348 IC toprovide outputs that are 10, 20 and 50 times larger than the input. Use a
feedback resistor of Rf=500k
in all stages.
-
8/2/2019 Unit 1 Amplifiers
21/46
-
8/2/2019 Unit 1 Amplifiers
22/46
Example
5V
10
3mA
1M
100k
20k
1k+ v
x-
100
1M
100
-
8/2/2019 Unit 1 Amplifiers
23/46
Voltage Summing Amplifier
33
f2
2
f1
1
fo V
R
RV
R
RV
R
RV
The output is the sum of individual signals times the gain:
The summing amplifier is used to add the voltages.Since the input resistance is very large V1=V2=0, therefore
-
8/2/2019 Unit 1 Amplifiers
24/46
-
8/2/2019 Unit 1 Amplifiers
25/46
Example Page 720
Calculate the output voltage for the circuit of the figure below. The inputsare V1= 50mV sin(1000t) and V2=10mV sin(3000t).
330k
33k
10k
+9V
-9V
741
4
56
11
10
-
8/2/2019 Unit 1 Amplifiers
26/46
-
8/2/2019 Unit 1 Amplifiers
27/46
Subtractor Amplifier
21
22
41
2
42
31
3
VVVo
RRandRRif
VR
RV
R
RR
RR
RVo
4231
-is used to subtract two voltages.
Vo
V2
V1
113
22
22
113
VRR
RfRV
R
RVo
VR
RV
R
R
R
RVo
ff
fff
-
8/2/2019 Unit 1 Amplifiers
28/46
Example page 721Determine the output for the circuit below with the components Rf=1 M, R1=100k,R2=50k and R3=500k.
-
8/2/2019 Unit 1 Amplifiers
29/46
Example page 721Determine the output voltage for the circuit shown below.
Vo
100k100k
20k
20k Vo741
-
8/2/2019 Unit 1 Amplifiers
30/46
-
8/2/2019 Unit 1 Amplifiers
31/46
Example 7-15 page 279
In the circuit of Figure 1, R1CF= 1 second and the input is a step (dc) voltageas shown in Figure 2. Determine the output voltage and sketch it. Assume thatthe op-amp is initially nulled.
Figure 1
Figure 2
-
8/2/2019 Unit 1 Amplifiers
32/46
Differentiator Amplifier
frequencylimitinggainCRCR
f
0dBisgainthewhichatfrequencyCR
f
dt
tdvRCtv
FFb
Fa
o
,2
1
2
1
,2
1
)()(
11
1
1
-is used to produce a voltage output proportional to the input voltage's rate of change-is used in waveshaping circuits to detect high frequency components in an input signaland also as a rate-of-change detector in FM modulators.
Input-output waveforms
-
8/2/2019 Unit 1 Amplifiers
33/46
-
8/2/2019 Unit 1 Amplifiers
34/46
-
8/2/2019 Unit 1 Amplifiers
35/46
-
8/2/2019 Unit 1 Amplifiers
36/46
-
8/2/2019 Unit 1 Amplifiers
37/46
Precision Rectifier
Why?
Signals of few millivolts (peak)
High open loop gain Rectifier output same as input
-
8/2/2019 Unit 1 Amplifiers
38/46
Precision Rectifier
Precision Half-wave rectifier
-
8/2/2019 Unit 1 Amplifiers
39/46
Precision Rectifier
Precision Full-wave rectifier
Vi
-
8/2/2019 Unit 1 Amplifiers
40/46
Example page 728
-
8/2/2019 Unit 1 Amplifiers
41/46
Example page 728
Calculate the output voltage expression for the instrumentation amplifier circuitgiven below. Assume that all resistors are 5k.
-
8/2/2019 Unit 1 Amplifiers
42/46
Log AmplifierA logarithmic amplifier has an output voltage that is proportional to the logarithmof the input.
Logarithmic Amplifier circuit
ampopofcurrentbiasinput
VR
I
VR
i
c
i
_____
ma x
ma x
ma x
1
1
The capacitor across the npn transistor is used to reduce the ac gain.
The diode protects the transistor against excessive reverse base-to-emitter voltage.
Resistor R1 is determined by the inequality pair
VBE = A log (Ic)
-
8/2/2019 Unit 1 Amplifiers
43/46
-
8/2/2019 Unit 1 Amplifiers
44/46
-
8/2/2019 Unit 1 Amplifiers
45/46
Multiplication of two input signals using log and antilog amplifier
log (AB) = log A + log B
log (A/B) =log A - log B
B i M lti li Ci it
-
8/2/2019 Unit 1 Amplifiers
46/46
Basic Multiplier Circuitry