op-amps (v.1a) 1
CENG4480_A2 Op Amps and Analog interfacing
Analog interfacing techniques
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Computer interfacing Introduction
To learn how to connect the computer to various physical devices.
Some diagrams of this manuscript are taken from the following references:
[1] S.E. Derenzo, Interfacing -- A laboratory approach using the microcomputer for instrumentation, data analysis and control prentice hall.
[2] D.A. Protopapas, Microcomputer hardware design, Prentice hall
[3] G C Loveday, Designing electronic hardware, Addison Wesley
op-amps (v.1a) 3
Topics include:
Overall interfacing schemesAnalog interface circuits, active filtersAnalog/digital conversionssensors, controllersControl techniquesAdvanced examples
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Overall view: a typical data acquisition and control system
Sensor filter A/D
Computer
D/APowercircuit
Mechanicaldevice
Timere.g.
8253
Sample&
Hold
Digital controlcircuit
Op-amp
op-amps (v.1a) 5
Analog interface example1 Audio recording systems
Audio recording systemsAudio signal is 20~20KHzSampling at 40KHz, 16-bit is Hi-fiStereo ADC require to sample at 80KHz.Calculate storage requirement for one hour?Audio recording standards
Audio CD Mini-disk MD MP3
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Analog interface example2 Surround sound audio systems
A common two channels audio CD Calculate storage size for one hour of recording
of a CD. 44.1KHz*2bytes*60sec*60min*2 channels=633.6Mbytes
Calculate the play time of a CD. 700M/(2bytes*44.1KHz*2channels*60sec)=61.4 minutes
6 Channels: Front R/L,Rear R/L, Middle, Sub woofer
44.1KHz, Calculate the sampling frequency.
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Analog interface example3Play stations and Wii
Play station 3, Analog hand held controller (http://ryangenno.tripod.com/images/PlayStation3-system.gif)
Wii, http://www.onlinekosten.de/news/bilder/wii_controller.jpg
Driving wheel http://www.bizrate.com/gamecontrollers/logitech-driving-force-driving-force-wheel--pid11297651/
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Operational Amplifier choices (op amp)
Why use op amp?What kinds of inputs/outputs do you want?What frequency responses do you want?
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Factors for choosing an amplifier Source load requirements:
DC ( static or slow changing, without decoupling capacitors)
AC( for fast changing signal, use decoupling capacitors)
Input range, biased – absolute min, max voltage
Output range, biased – absolute min, max voltage
Frequency: range, allowed attenuation in dB
Noise tolerance Power – output current/output
impedance.
DC-direct current amplifier
AC-alternating current amplifier
Op-
amp
DC
Source Load
Op-
amp
AC
Source Load
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Discussions: what kind of amplifiers should we use? It is an art.
Condenser microphone(+/-10mV)Audio amplifier from MP3 player to speakerUltrasonic sensors (+/-1mV) to ADC (analog to
digital converter) (0-5V)Accelerometers (+/-5V), or (+/-500mV)Temperature sensors to ADC (010mv)Moving coil microphone with 50Hz noise (+/-
0.5mV)
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Suggestions
Condenser microphone(+/-100mV): AC, bandwidth=audio range(20-20KHz, voltage gain ~=100,low power.
Moving coil microphone with 50Hz noise (+/-0.5mV): AC, bandwidth=audio range, voltage gain ~=2000,low power.
Audio amplifier from MP3 player to speaker: AC, high power, voltage gain ~=5, power gain~=100, bandwidth=audio range
Ultrasonic sensors (+/-1mV) to ADC (0-5V): DC-shift needed, voltage gain~=5000. bandwidth=tuned narrow band
Accelerometers (+/-500mV): DC-shift needed, voltage gain ~=10, Temperature sensors to ADC (010mv): DC, voltage gain~=100.
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Operational amplifiers (op-amps)
ideal op-amps inverting amplifier non-inverting amplifier voltage follower current-to-voltage amplifier summing amplifier full-wave rectifier instrumental amplifier
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General concept about OP amps
Controllable gainFor DC or AC amplifierNot too high frequency responses
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Ideal Vs. realistic op-amp
Ideal Realistic RinA= infinite 105->108
Zin= infinite 106(bipolar input) 1012(FET input)
output offset existsV0=A(V+-V-)
V-
V+ +
_
+
2
3
6LM741
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Inverting amplifier
Gain(G) = -R2/R1
For min. output offset, set R3 = R1 // R2
Rin=R1
+
_
R2
R1
R3
V0V1
Virtual-ground,V2
A+
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Non-inverting amplifier
_V0
V1
R1
R2
Gain(G) 1 + (R2/R1)
For min. offset output , set R1//R2=Rsource
High input resistance
+
V2
A
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Differential amplifier
V0=(R2/R1)(V2-V1)
Minimum output offset R1 //R2 =R3 //R4
_V0
R2
+
R1V1
R3V2
R4
A
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Exercise 1
A temperature sensor has an offset of 100mV (produces an output of 100mV at 0 °C-degrees Celsius), and the gradient is 10 mV per °C. The temperature to be measured is ranging from 0 to 50 °C.
The required ADC input range is 0 to 9Volts. Given that the power supply is +/-9V, design a differential amplifier
for this application.
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Voltage follower (Unit voltage gain, high current gain, high input impedance)
Gain=1,Rin=high
For minimum output offset R=Rsource
_V0=V1
V1 +
R
A
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Current to voltage converter: application to photo detector – no loading effect for the light detector
_V0
IR
V0=I R
I should not be too large otherwise offset voltage will be too high.
+A
PhotodiodeLight detector
See http://hyperphysics.phy-astr.gsu.edu/hbase/electronic/photdet.html#c1
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Summing amplifier
_V0
R
+
V0 = -{(V1/R1)+(V2/R2)+(V3/R3)}R
R1
R2
R3
I1
I1+I2+I3
V1
V2
V3
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Op-amp characteristics
Input and output offset voltages It is affected by power supply variations,
temperature, and unequal resistance paths. Some op-amps have offset setting inputs. Unequal resistance paths and bias currents on
inverting and non-inverting inputs
Temperature variations -- read data sheet for operating temperatures
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Op-amp dynamic response
Slew rate -- the maximum rate of output change (V/us) for a large input step change. A741 slew rate=0.5V/ s. Fast slew rate is
important in video circuits , fast data acquisition etc.
Gain bandwidth product higher gain --> lower frequency response lower gain --> higher frequency response
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Common mode gain
If the two inputs (V+,V-) are connected together and is given Vc, output is found to be Vo.
ideal differential amplifier only amplifies the voltage difference between its two inputs, so Vo should be 0.
But in practice it is not.This deficiency can be measured by the Common
mode gain=Vo/Vc.
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Diagram of gain bandwidth product, from [1]
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Instrumental amplifier To make a better DC amplifier from op-amps
Diagram of instrumental amplifier, from [1]
Applications: DSO input amplifiers, amplifiers in medical measurement systems
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Instrumental amplifier
It has all the advantages of an amplifier.Gain(G)=V0/(V+-V-)
=(R4/R3)[1+(2R2/R1)] (typically 10 to 1000)
Even V+=V-= Vc , there is a slight output because of the Common Mode Gain=Gc=V0/Vc
Therefore, V0= G(V+-V-)+GcVc
To measure this imperfection, Common Mode rejection ratio (CMRR)=G/Gc (typically 103 to 107, or 60 to 140 dB)is used , the bigger the better.
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Comparing amplifiers, from [1]
Op Inv. Noninv. Diff. Instu. Amp Amp Amp Amp AmpHigh Rin Yes No Yes No Yes
Diff’tial Yes No No Yes Yes input
Defined No Yes Yes Yes Yesgain
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Operational amplifier selection techniques and keywords National semiconductor is the main manufacturer: See
http://www.national.com/appinfo/milaero/analog/highp.html General Purpose: LM741* High Slew Rate:50V/ ms --> 2000V/ ms (how fast the output can be changed) Follower (high speed):50MHz Low Supply Current: 1.5mA --> 20 µA/Amp Low offset voltage: 100 µV Low Noise Low Input Bias Current: 50pA -->10pA High Power : 0.2A --> 2A Low Drift: 2.5 mV/ _C --> 1.0 mV/ _C Dual/Quad High Power Bandwidth High Power Bandwidth : 300KHz - 230Mhz
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Practical op-amp usage examples
Example 1: Working with one power supply
Example 2: Active filters Example 3: Sample and holdExample 4: Example 4: Voltage
Comparator and schmit triggerExample 5: Power amplifier
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Example 1: Single power supply for op-amps
Small systems usually have 1 power supplyOutput V0 is biased at E/2 rather than 0.
E.g. Inverting amplifier. GainR2/R1
+
_
R2
R1
R3Vo
V1
A
R=20K R=20K E Volts Power supply0-VoltE/2
E/2 E/2
Vo
E
0-Volt
V+V-+
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Typical AC amplifier designCondenser microphone amplifier circuit, and the diagram showing theoutput swing around the biased 2.5V. The capacitors isolate the stages.Condenser MIC output impedance is 75 OhmsWhat is the input impedance of the amplifier?
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Example 2: Active filters (analog and using op-amps)
Applications: accept or reject certain signals with specific frequencies. High-pass, low-pass, band-pass etc. E.g. reject noise extract signal after demodulation reject unwanted side effect signals
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Types
2-1: Low pass2-2: High pass2-3: Band stop (notch): e.g. noise removal2-4: Band pass
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definition of power gain in decibel (dB)
Output power is P2, input power is P1
Power Gain in dB=10 log10 (P1/P2)
OrOutput voltage is V2, input voltage is V1
Assume load R is the same, power=V2/RPower Gain in dB=10 log10 (V1
2/ V22)
Power Gain in dB=20 log10 (V1/ V2)
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Important terms for filters: Formulas are not important, remember the frequency-gain curve and concepts
Pass band-- range of frequency that are passed unfiltered
Stop band -- range of frequency that are rejected.Corner frequency -- where amplitude dropped by
2-1/2=0.707= -3dBI.e. in dB: 20log(0.707) = -3dBSettling time -- time required to rise within 10%
of the final value after a step input.
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2-1 Low pass
Only low frequency signal can passone-pole: attenuates slower 20dB/decadetwo-pole: attenuates faster 40dB/decadeApplications:
remove high freq. Noise, remove high freq. before sampling to avoid
aliasing noise
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Diagram for low-pass one pole filter, from [1]for simplicity make R2/R1=1,
-R2/R1
|G| =-------------------
{1+(f/fc)2}1/2
gain
3dB
fc Freq.
20 dB/decadedrop
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Formula, also
22
11 2
2
1 2
2
2
1
2
2 21
// 2 1,
1 2
1( )
// 221 2[ ( )]
2
[1 2 ]
1put
2
1
1 1 1,since ( _ 2)
1 11
cc
c
c
c
c
X RGain X
R j fc
RX R fcj fc
GainR fcR R
j fc
RGain
R j fcR
fcR
RGain
fR j
f
RGain see appendix
R ja af
f
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2-1aLow pass, one pole filter formulas for simplicity make R2/R1=1
-R2/R1
|G| =-------------------
{1+(f/fc)2}1/2
Corner frequency= fc=1/(2R2C)
The gain drops 6dB/octave or 20 dB/decade
op-amps (v.1a) 41
Diagram for Low-pass two-pole filter, from [1] for simplicity make R3/(R2+R1)=1
- R3/(R2+R1)
|G|= ------------------ 1+(f/fc)2
gain
6dB
fc Freq.
40 dB/decadedrop
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2-1bLow-pass two-pole filter formulas for simplicity make R3/(R2+R1)= 1
- R3/(R2+R1)
|G|= ------------------ 1+(f/fc)2
Corner frequency=fc
fc=(R1//R2)/2C1 =(2 R3C2)-1 when gain G drops at -6dB.
G is dropping at 40dB/decade
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Matlab, lp42.m %lp42.m, ceg3480 matlab demo %low pass filter-one pole clear f=[0:100:2000] N=length(f) fc=1000 for i=1:N % -----gain1 , low pass one pole , for simplifcity make (R2/R1)=1 gv1(i)=-1/sqrt(1+(f(i)/fc)^2); %db=10*log_base10(power1/power2); or dB=20*log_base10(voltage1/voltage2; gain1_db(i)=20*log10(abs(gv1(i))); % -----gain2 , low pass two pole , for simplifcity make {R3/(R1+R2)}=1 gv2(i)=-1/(1+(f(i)/fc)^2); %db=10*log_base10(power1/power2); or dB=20*log_base10(voltage1/voltage2; gain2_db(i)=20*log10(abs(gv2(i))); end % %======================================================== figure(1) clf limit_y=min(gain2_db) semilogx(f,gain1_db,'g-.') hold on semilogx(f,gain2_db,'b--') %------------------------ semilogx([fc,fc],[0,limit_y],'k-.') legend('one pole','two pole','fc=1000',4); %------------------------------------------------------- semilogx(f,ones(N)*-3,'r') text(f(N),-3,'- 3d B line') %------------------------ semilogx(f,ones(N)*-6,'r') text(f(N),-6,'- 6d B line') ylabel('low pass filter gain in d B') xlabel('freq Hz f in log scale')
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Plotting the comparison of the low pass filters (one-pole, two-pole)
op-amps (v.1a) 45
2-2High pass
Only high frequency signal can passone-pole: attenuates slower 20dB/decadetwo-pole: attenuates faster 40dB/decadeApplications:
Remove low freq. Noise (50Hz main) Remove DC offset drift.
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2-2aDiagram for high-pass one-pole filter, from [1]For simplicity make R1=R2
(f/fc)
|G|=------------------- {1+(f/fc)2}1/2
fc=1/{2 (R1C)}
gain3dB
Freq.
20 dB/decadedrop
high freq. CutoffunintentionallyCreated by Op-amp
fc
op-amps (v.1a) 47
High-pass one-pole filter formulas
(f/fc)
|G|=------------------- {1+(f/fc)2}1/2
Corner frequency= fc=1/{2 (R1C)}
At low f , |G|=f/fc; at high f , |G|=R2/R11
Since op-amp has a certain gain-bandwidth, so at high frequency the gain drops. So all op-amp high-pass filters are actually band-pass.
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Matlab hp52.m %hp52.m ceg3480 matlab demo %high pass filter-one pole clear f=[500:100:100000] N=length(f) fc=1000 for i=1:N % -------------------gain3 , high pass ,one pole gv3(i)=-(f(i)/fc)/sqrt(1+(f(i)/fc)^2); %db=10*log_base10(power1/power2); or dB=20*log_base10(voltage1/voltage2; gain3_db(i)=20*log10(abs(gv3(i))); end % %======================================================== figure(1) clf limit_y=min(gain3_db) semilogx(f,gain3_db,'k-.') hold on %------------------------ semilogx([fc,fc],[0,limit_y],'g-.') legend('one pole','fc=1000',4); %------------------------------------------------------- semilogx(f,ones(N)*-3,'r') text(f(N),-3,'- 3d B line') ylabel('high pass filter gain in d B') xlabel('f freq in log scale')
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high pass one pole filter
op-amps (v.1a) 50
2-3Band stop (notch) filter
Suppresses a narrow frequency band of signal
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2-3Band pass filter
Passes a frequency band of signal.
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Diagram for Notch filter (band-stop), from [1]
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Notch filter (band-stop) formulas
Rejects a narrow band of frequencies and passes all others. Say reject the 60Hz main noise for noise removal.
High Q,Fc1=(4 RC)-1
Low Q, Fc2=( RC)-1
Voltagegain
frequency
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Example 3: Sample and hold amplifier
For a fast changing signal, if you want to know the voltage level of a snap shot (e.g. using a slow AD converter to view a short pulse), you need a sample and hold device, e.g. AD582, AD389 etc.
At Sample(S), V0=V1; at Hold(H) the output is held at the level just before switching to H. It is like taking a photograph of a signal.
Some AD converter has this circuit incorporated inside.
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Diagram for Sample and hold amplifier, from [1]
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Example 4: Voltage Comparator with hysteresis and schmit trigger
E.g. in IR motor speed encoder V1=IR receiver input
Vref
V1comparator
Comparator gives bad resultUnstable region when V1 and Vref are closed
V0
0V
0V=
V+
V-
V-
V+
IR receiverSignal with noise
Better outputUsing Schmit trigger
Schmit trigger
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Diagram for hysteresis (non-inverting schmit trigger), see P.420, S. Franco, Design with operational
amplifiers and analog integrated circuits, McGraw Hill.
OutputVoltage
Input voltage
Switch overvoltage
t
VTH
VTL
V1 V0Voltage
VTH -VTL=(Vohigh –Volow)(R1/R2)=2V
Vref =0
VTL
=-1VVTH
=1V
10V
-10V
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Example 4: Schmit trigger using dual-power supply non-inverting op amp : A small amount of hysteresis is used to stabilize the output when V1 is near to Vref.(set R1/R2=0.1)
To make Vo from low to high V1 >VTH, at V1=VTH
(VTH-0)/R1=(0-Vlow)/R2
V1>VTH= -R1/R2(Vlow)=1V
V1>VTH= -(0.1)(-10)=1V
To make Vo from high to low V1 <VTL
(Vhigh-0)/R2=(0-VTL)/R1
V1<VTL= -R1/R2(VHigh)= -1V
If R2 >> R1, hysteresis is small. For schmit trigger devices R1 0.1*R2, e.g. R1=1K, R2=10K, so
hysteresis is good enough to reject noise.
0V= Vo
The op-amp uses V+,V_ power supplies.Output is clamped to Vlow
or Vhigh, setVref =0 to make the math easier
Vhigh=10V
Vlow=-10V
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Example 5: Power amplifier
Most op-amps can drive outputs with low currents, we need transistors to raise the power to drive heavy loads, e.g. mechanical relays, motors or speakers.
V0=V1-1.2 Volts;
TIP3055 type transistors can drive current up to 15A.
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Power amplifier, from [1]
From: http://www.st.com/stonline/books/pdf/docs/4136.pdf
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From : http://www.fairchildsemi.com/ds/TI/TIP41C.pdf
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Summary
Studied Basic digital data acquisition systems The configuration of operational amplifier
circuits and their applications
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Appendix
To be discussed in class
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Appendix 1Exercise—problem setting
A magnetic sensor is used to detect the magnetic flux density (in K Gauss) of an environment. The resistance of the sensor is proportional to the flux density detected with a gradient of 4 KΩ per K Gauss, and when there is no magnetic flux the resistance is 2K. The range of the magnetic flux density in the environment is between 0 and 5 K Gauss, and the flux density changes at a rate of not more than 10 K Gauss per second. The smallest change in flux density detectable is required to be 5 Gauss. The system uses an Analogue-to-Digital Converter (ADC) to convert the magnetic flux density into digital data, and the power supply used for this system is 5 Volt.
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Exercise — Question Draw the circuit diagram of the bridge circuit and the operational
amplifier circuit needed to transform the flux density detected to an output voltage. The output voltage is proportional to the flux density. When the flux density is 0 the output is 0 Volt; when the flux density is 5 K Gauss the output is 5 Volts.
It is found that the temperature of the environment affects the resistance of the magnetic sensor at a rate of RT (in KΩ per degree Celsius). Discuss how you can use two or more sensors to reduce the effect of temperature change. Draw the circuit of your scheme and explain with the help of formulas of how the system can be freed from the effect of temperature change.
op-amps (v.1a) 66
Appendix 2, To prove
1/(1+ja)=1/(1+ja)*[(1-ja)/(1-ja)] = (1-ja)/(12-(ja)2)=(1-ja)/(1+a2), since j2= -1 =1/(1+a2)+(-ja)/(1+a2)=real + imaginary so |1/(1+ja)|={real2+ imaginary2}1/2
={[1 /(1+a2)]2+[(-ja)/(1+a2)]2}1/2
={[1+a2]2-(1+a2)a2/[1+a2]2}1/2
={[1+2a2+a4-a2-a4/[1+a2]2}1/2
={[1+a2]/[1+a2]2}1/2
=1/[1+a2]1/2, proved! 2
1 1
1 1ja a
2
1 1
1 1ja a
op-amps (v.1a) 67
Solution for Exercise 1
Gain =Vout/Vin=9V/(10mV*50 °C )=18, set R2/R1=R4/R3=18 How to solve the offset problem.
Sensor V2 Offset of 100mV at V1, 9*Rb/(Ra+Rb)=100mV (make R4>> Ra) why? Add a small variable resistor Rc between 9V & Ra for offset trimming.
_V0
R2
+
R1
V1
R3
Sensor
R4
A
Ra
Rb
9V
0V-9V
9V
V2
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