Week 1

CENG4480_A1 Op Amps and Analog Interfacing

Analog interfacing techniques op-amps (v.5f) 1

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.5f) 2

Topics include:

      Overall interfacing schemes Analog interface circuits, active filters Analog/digital conversions sensors, controllers Control techniques Advanced examples op-amps (v.5f) 3



Overall view: a typical data acquisition and control system

Timer Digital control circuit Sensor Mechanical device Op-amp filter Power circuit Sample & Hold A/D Computer D/A op-amps (v.5f) 4

Analog interface example1

     

Audio recording systems

Audio recording systems Audio signal is 20~20KHz Sampling at 40KHz, 16-bit is Hi-Fi Stereo ADC requires to sample at 80KHz.

Calculate storage requirement for one hour?

Audio recording standards  Audio CD  Mini-disk MD  MP3 op-amps (v.5f) 5

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.

op-amps (v.5f) 6

Analog interface example3 Play stations and Wii

   Play station 3, Analog hand held controller ( http://ryangenno.tripod.com/images/PlayStat ion3-system.gif

) Wii, http://www.onlinekosten.de/news/bilder/wii_ controller.jpg

Driving wheel http://www.bizrate.com/gamecontrollers/logit ech-driving-force-driving-force-wheel- pid11297651/ op-amps (v.5f) 7

Operational Amplifier choices (op amp)

   Why use op amp?

What kinds of inputs/outputs do you want?

What frequency responses do you want? op-amps (v.5f) 8

Biasing



{Biasing in electronics is the method of establishing predetermined voltages or currents at various points of an electronic circuit for the purpose of establishing proper operating conditions in electronic components},

from https://en.wikipedia.org/wiki/Biasing op-amps (v.5f) 9

Direct Current (DC) amplifier

    Example: use power op amp (or transistor) to control the DC motor operation.

Need to maintain the output voltage at a certain level for a long time.

All DC (biased) levels must be designed accurately .

Circuit design is more difficult.

DC Source Op amp Load: DC motor op-amps (v.5f) 10

Alternating Current (AC) amplifier

  Example: Microphone amplifier, signal is AC and is changing at a certain frequency range. Current is alternating not stable.

AC Use capacitors to connect different stages, so no need to consider biasing problems.

Biased at Vcc Source Op amp Load Each stage can have its owe biasing level. A capacitor is an isolator, so the circuit is easier to be designed.

Biased at Vcc/2 op-amps (v.5f) Vcc/2=2.5V

11

Factors for choosing an amplifier

      Source DC or AC ?

  DC(static or slow changing input, without decoupling capacitors) AC(for fast changing input, 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 DC Source AC Source Op amp  AC-alternating current amplifier Op amp Load Load op-amps (v.5f) 12

op-amps (v.5f)

Input impedance (Rin) and

 

Output impedance (Rout)

Why do we prefer High Rin and Low Rout?

Because it is more efficient.

Stage1(sensor) Vout1 Rout1 Vin2 Stage 2 Rin2 Is equivalent to  Vout1 Rout1 Rin2 Vin2= Vout1*Rin2/(Rout1+Rin2) To maximize Vin2 (input voltage driving stage 2)  We make Rout1 lower, Rin2 higher. Good choice: Rin  1M  or over, Rin  10  13

Sensor Vout_sen 1mV Rout_sen Rin_amp x1000 Vout_ amp Student ID: __________________ Name: ______________________ Date:_______________ (Submit this at the end of the lecture.)

Exercise 1.1

 Sensor (Rout_sen) Vout_sen=1mV is sent to an a) amplifier with Rin_amp, gain 1000 Rout_sen =10  , Rin_amp=1M, calculate the b) output voltage (out_amp) of the amplifier Rout_sen =2K  , Rin_amp=10K, calculate the output voltage of the amplifier c) Which above scheme would you prefer and why?

op-amps (v.5f) 14

Meaning of power gain in dB

        

(Decibel)

Vout=output Vin=input Voltage gain =Vout/Vin Power gain =(Vout) 2 / (Vin) 2 Power gain in dB=10*log 10 (Power gain ) =20* Log 10 |(Vout/Vin)|=20*Log 10 |G|, where G= Voltage gain When power gain=(Vout/Vin) 2 =1, voltage_gain=1, power_gain is 0dB When power gain=(Vout/Vin) 2 =0.5, voltage_gain=(0.5) 1/2 =0.707, power_gain is -3dB op-amps (v.5f) 15

op-amps (v.5f)  

Frequency-gain plot

When power gain=(Vout/Vin) 2 =1, voltage_gain=1, power_gain is 0dB When power gain=(Vout/Vin) 2 =0.5, voltage_gain=(0.5) 1/2 =0.707,  power_gain is -3dB An amplifier frequency-gain is important to understand its chartered at different frequencies.

Horizontal axis is frequency (log scale) in Hz, Vertical axis is gain in dB Gain is 0dB Power gain is 1 Gain is -3dB Power gain is 0.5

0dB -3dB Power Gain (dB) Log Frequency One

decade

= one number is 10 times of the other number Slope: 20 dB/decade drop 16

Exercise 1.2: General concept about OP amps

A         Controllable gain For DC or AC amplifier Not too high frequency responses K=Gain * bandwidth =gain_bandwidth_product  Calculate K at A,B,C. Gain=power gain At A, GainA=________,(  B A )=_______,K A =___ At B, GainB=________,(  B B )=_______,K B =___ At C, GainC=________,(  B C )=_______,K C =___ What is your conclusion based on the above calculation?

op-amps (v.5f) C Power gain in dB= 10*log 10 (Power gain ) 17 B

Week 2

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 op-amps (v.5f) 18

Ideal Vs. realistic op-amp

   Ideal A= infinite Z in = infinite V V + 2 _ + LM741 Realistic Rin    10 5 ->10 8 10 6  (bipolar input) 10 12  (FET input) output offset exists 6 V 0 =A(V + -V ) 3 op-amps (v.5f) 19

Exercise 1.3 Inverting amplifier

   Gain(G) = -R 2 /R 1 For min. output offset, set R 3 = R 1 // R 2 R in =R 1 Virtual-ground,V2  R 2 V1 R 1 _ Input A R 3 + Questions:  (i) Derive the gain formula (See appendix)  (ii) If R1=1K, R2=10K, find G and Rin V0 Output 20 op-amps (v.5f)

Exercise 1.4 Non-inverting amplifier

 Voltage Gain(G)  1 + (R 2 /R 1 )  For min. offset output , set R 1 //R 2 =R source  High input resistance V1 + A V0  Input _ V2 R2 Output  R1 Questions: (i) Derive the gain formula (See appendix)  (ii) If R 1 =1K, R 2 =10K, find G and Rin op-amps (v.5f) 21

Differential amplifier

  V0=(R 2 /R 1 )(V 2 -V 1 ) Minimum output offset R 1 //R 2 =R 3 //R 4 R 2 V 1 Input V 2 R 1 R 3 _ + A Output V 0 R 4 Exercise: proof the gain formula op-amps (v.5f) 22

Exercise 1.5

   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. op-amps (v.5f) 23

Voltage follower (Unit voltage gain, high current gain, high input impedance)

   Gain=1, R in =high For minimum output offset R=R source V 1 + _ A R Exercise: proof the gain formula op-amps (v.5f) V 0 =V 1 24

Current to voltage converter: A

pplication to photo detector – no loading effect for the light detector  V 0 =

I

R 

I

should not be too large otherwise offset voltage will be too high.

Photodiode Light detector

I

_ R A V 0 + See http://hyperphysics.phy-astr.gsu.edu/hbase/electronic/photdet.html#c1 Exercise: proof the formula op-amps (v.5f) 25

Summing amplifier

 V 0 = -{(V 1 /R 1 )+(V 2 /R 2 )+(V 3 /R 3 )}R V V V 1 2 3 R 1 R 2 R 3 I1 I 1 +I 2 +I 3 Inputs Exercise: proof the gain formula op-amps (v.5f) _ + R V0 Output 26

Exercise 1.6

Discuss what kind of amplifiers should we use for the following sensors?       Condenser microphone(+/-10mV) Audio amplifier from MP3 player to speaker Ultrasonic sensors (+/-1mV) to ADC (analog to digital converter) (0-5V) Accelerometers (+/-5V), or (+/-500mV) Temperature sensors to ADC (0  10mv) Moving coil microphone with 50Hz noise (+/ 0.5mV) op-amps (v.5f) 27



Integrator

Ref http://www.physics.ucdavis.edu/classes/Physics116/Physics116A04F.html

Reading exercise: http://www.electronics-tutorials.ws/opamp/opamp_6.html

op-amps (v.5f) 28



Differentiator

Ref http://www.physics.ucdavis.edu/classes/Physics116/Physics116A04F.html

Reading exercise: http://www.electronics-tutorials.ws/opamp/opamp_7.html

op-amps (v.5f) 29

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 op-amps (v.5f) 30

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 op-amps (v.5f) 31

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

=G c =Vo/Vc.

op-amps (v.5f) 32

 Diagram of gain bandwidth product, from [1] op-amps (v.5f) Hz 33



Instrumental amplifier

To make a better DC amplifier from op-amps Applications: Digital Oscilloscope DSO input amplifiers, amplifiers in medical measurement systems op-amps (v.5f) Diagram of instrumental amplifier, from [1] 34

Instrumental amplifier

      It has all the advantages of an amplifier.

Gain(G =(R  4 )=V /R 3 0 /(V Even V + =V =

V c

+ -V )[1+(2R 2 ) /R 1 )] (typically 10 to 1000) , there is a slight output because of the

Common Mode Gain

=G c =V 0 /

V c

Therefore, V 0 = G  (V + -V )+G c

V c

To measure this imperfection, Common Mode rejection ratio (

CMRR

)=G  /G c (typically 10 3 to 10 7 , or 60 to 140 dB)is used , the bigger the better.

op-amps (v.5f) 35

Comparing amplifiers, from [1]

   High R in Op Inv.

Noninv.

Diff.

Instu.

Amp Amp Amp Amp Yes No Yes No

Amp Yes

  Diff’tial Yes No input No Yes

Yes

  Defined No gain Yes Yes op-amps (v.5f) Yes

Yes

36

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 op-amps (v.5f) 37

Practical op-amp usage examples

    Example 1: Working with one power supply Example 2: Active filters Example 3: Sample and hold Example 4: Example 4: Voltage Comparator and schmit trigger input ciruit  Example 5: Power amplifier op-amps (v.5f) 38

Example 1: Single power supply for op-amps

 Small systems usually have a single power supply   E/2=5V Output V 0 is biased at E/2 rather than 0.

E.g. Inverting amplifier. Gain  -R2/R1 E=10V R 2 V1 R 3 R 1 _ + A V+ E V 0-Volt Vo E/2=5V 0Volt R=20K E/2=5V R=20K E Volts Power supply op-amps (v.5f) 39

Typical A.C. amplifier design

Condenser a microphone amplifier circuit, and the diagram showing the output swing around the biased (steady state volateg) 2.5V. The capacitors isolate the stages of different biases. Condenser MIC output impedance is 75 Ohms. What is the input impedance of the amplifier?

Answer: See previous notes on inverting amplifier

5V Biased at around 4V Biased at around 2.5V

2.5V

Biased at around 2.5V

Output to Mic-in of power amplifier 2.5V

op-amps (v.5f) 40

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 op-amps (v.5f) 41

Types

    2-1: Low pass 2-2: High pass 2-3: Band stop (notch): e.g. noise removal 2-4: Band pass Week 3 op-amps (v.5f) 42

Recall: definition of power gain in

decibel

(

dB

)

       Output power is P =20 log 10 2 , input power is P Power Gain in dB=10 log Or, output voltage is V 2 Power Gain in dB=20 log 10 10 (P |(V 2 1 /P 1 ) 1 , input voltage is V Assume load R is the same, power=V Power Gain in dB=10 log 10 (V 2 2 / V 1 2 ) / V 2 )| 2 /R |G|, where G=voltage gain 1 op-amps (v.5f) 43

Time domain vs. frequency



domain

Time domain: we talk about voltage gain against time Time domain signal plot +1V Voltage Vpp=Peak-to-Peak voltage 0 Time :(Seconds, usually linear scale) -1V  Frequency domain: we talk about the voltage gain against frequency. Power Gain (dB) 0dB -3dB Frequency domain signal plot op-amps (v.5f) Frequency (Hz) (can use log scale) 44

 

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 (0.5) 1/2 =0.707

I.e. in dB: 20*log(0.707) = -3dB Settling time -- time required to rise within 10% of the final value after a step input. op-amps (v.5f) 45

op-amps (v.5f)

2-1 Low pass

    Only low frequency signal can pass one-pole: attenuates slower 20dB/decade two-pole: attenuates faster 40dB/decade Applications:  remove high freq. Noise,  remove high freq. before sampling to avoid aliasing noise Reading exercise, please read this webpage: Ref : http://www.electronics-tutorials.ws/filter/filter_5.html

46

Diagram for low-pass one pole filter, from [1] for simplicity make

R 2 /R 1 =1,

Gain

G(f) in dB G

(

f

)  

R

2 /

R

1 1    

f f c

   2 3dB 20 dB/decade drop Corner frequency

f c,

f c Freq.

f c

 1 2 

R

2

C

inverting amplifier gain formula (See appendix) op-amps (v.5f) 47

Formula

f

 Frequency.

To proof |

G

| 

V

0

V

1 

R

2

R

1 1     1

f f c

   2 , where

f c

 1 2 

R

2

C

Voltage Gain

G



G



X c

//

R

2

R

1 

X c

//

R

2 , by definition

R

1

R

1

R

2      

R

2     1

j

2 

fC j

1    2 

fC

      

j j X c

2 

fC

2 

fC

 Put

f c

 1 2 

R

2

C G



R

2

R

1    1 

j f f c

   1

j

2 

fC

|

G

| 

R

2

R

1 1     1

f f c

   2 , since 1  1

ja

 1 1 

a

2 see op-amps (v.5f) appendix 48

2-1a Low pass, one pole filter formulas for simplicity make

R 2 /R 1 =1K G

(

f

)  

R

2 /

R

1 1    

f f c

   2   Corner frequency=

f c =1/(2



R 2 C)

The gain drops 6dB/octave or 20 dB/decade op-amps (v.5f) 49

Exercise 1.7

   What is the meaning of -3dB cut off?

What is the meaning of 20dB/decade drop? Plot the power gain(dB) vs frequency diagram of, R1=R2=1K , C=1uF Voltage gain 

G

(

f

)  

R

2 /

R

1 1    

f f c

   2 op-amps (v.5f) 50

Diagram for Low-pass two-pole filter, from [1] for simplicity make

R 3 /(R 2 +R 1 )=1

Gain

G

(

f

)  

R

3 /(

R

2 

R

1 

Where f c =(R 1 //R 2 )/2

   

C 1 f f c

   2

=(2

 1 )

R 3 C 2 ) -1

6dB

G(f) in dB

40 dB/decade drop f c Freq.

op-amps (v.5f)  51

2-1b Low-pass two-pole filter formulas for simplicity make

R 3 /(R 2 +R 1 )= 1 G

(

f

)  

R

3 1  /(

R

2   

f f c

    2

R

1 )   Corner frequency=

f c f c =(R 1 //R 2 )/2



C 1 =(2

 G drops at -6dB.

R 3 C 2 ) -1

when gain  G is dropping at 40dB/decade op-amps (v.5f) 52

Exercise 1.8

   What is the meaning of -6dB cut off?

What is the meaning of 40dB/decade drop? Plot the power gain(dB) vs frequency diagram of, R1=R2=1K ,R3=2K, C=1uF Voltage gain 

G

(

f

)  

R

3 1  /(

R

2   

f f c

    2

R

1 )  Compare the difference between one-pole and two-pole low pass filters op-amps (v.5f) 53

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 simplicity 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 simplicity 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') op-amps (v.5f) 54

Plotting the comparison of the low pass filters (one-pole, two-pole)

20dB/decade Slope less steep 40dB/decade Slope more steep op-amps (v.5f) 55

2-2 High pass

    Only high frequency signal can pass One-pole: attenuates slower 20dB/decade Two-pole: attenuates faster 40dB/decade Applications:  Remove low freq. Noise (50Hz main)  Remove DC offset drift.

op-amps (v.5f) 56

2-2a Diagram for high-pass one-pole filter, from [1] For simplicity make R

1 =R 2, R 3 =R 2 // R 1

G

(

f

)    1   

f f c

 

f f c

  2 , where

f c

 1 2 

R

1

C

20 dB/decade drop Gain

G(f) in dB f

c 3dB Freq.

op-amps (v.5f) high freq. Cutoff unintentionally Created by Op-amp 57 

High-pass one-pole filter formulas

   

G

(

f

)    1   

f f c

 

f f c

  2 , where

f c

 1 2 

R

1

C

Corner frequency=

f c =1/{2



(R 1 C)}

At low

f

,

|G low_freq |=f/f c ; at high f , | G high_freq |=R 2 /R 1



1

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

.

op-amps (v.5f) 58

                         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') op-amps (v.5f) 59



high pass one pole filter

op-amps (v.5f) 60

2-3 Band stop (notch) filter

 Suppresses a narrow frequency band of signal op-amps (v.5f) 61

2-3 Band pass filter

 Passes a frequency band of signal.

op-amps (v.5f) 62

Diagram for Notch filter (band-stop), from [1]  op-amps (v.5f) 63

Notch filter (band-stop) formulas

   Rejects a narrow band of frequencies and passes all others. Say reject the 60Hz main noise for noise removal.

Voltage Gain in dB High Q,  F c1 =(4  Low Q,  F c2 =(  RC) RC) -1 -1 frequency 64 op-amps (v.5f)

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.

op-amps (v.5f) 65

Diagram for Sample and hold amplifier, from [1]  Sample: sampling Hold: When the switch is at H, Vo keeps unchanged for a long time. So the Analog–to-digital converter ADC can have more time for data conversion op-amps (v.5f) Hold Slight droop may occur 66

Example 4: Voltage Comparator with hysteresis and schmit trigger

  E.g. in IR motor speed encoder V1=IR receiver input V+ Comparator gives bad result Unstable region when V 1 and V ref are closed V 0 IR receiver Signal with noise 0V V 1 comparator V ref V Schmit trigger Better output Using Schmit trigger V+ 0V= op-amps (v.5f) V 67

Diagram for hysteresis (non-inverting schmit trigger), see P.420,

S. Franco, Design with operational amplifiers and analog integrated circuits, McGraw Hill.

Voltage V1 V0  V TH V TL t Output Voltage Switch over voltage 10V V TH -V TL = (V ohigh –V olow )(R1/R2)=2V -10V V TL =-1V V op-amps (v.5f) TH =1V V ref =0 Input voltage 68

   Example 4: Schmit trigger using dual-power supply non inverting op amp : A small amount of hysteresis is used to stabilize the output when V 1 is near to V ref .(set R1/R2=0.1) When Vo =Low(-10Volts), We want to find V TH(low) that when V1> V TH(low) Vo will switch from low( 10Volts) to high (10Volts).

, so V high =10V        At opamp V+ input, when V1 is close to V TH(low) , apply current rule : (V1-0)/R1+(V0 low -0)/R2 =0; (Note:V1  V TH(low) ) (V TH(low) -0)/R1+(V0 low -0)/R2 =0 (V TH(low) -0)/R1+(-10-0)/R2 =0 V TH(low) =(R1/R2)(10); (NoteR1/R2=0.1) V TH(low) = (0.1)(10)=1Volt So when V1> V TH(low) , V0 will switch from low ( 10V) to high(+10V) 0V= Similarly, V TH(high) =-1Volt , so that when V1< V TH(low) Vo will switch from high(+10volts) to low (-10Volts).

Set R 2 >> R 1 , to make a small hysteresis. Ie. for Schmit trigger devices R 1  0.1*R 2 , e.g. R1=1K, R2=10K, so the hysteresis is good enough to reject noise. The diodes are used to clamp the voltages at +/-10V Vo op-amps (v.5f) V low The op-amp uses V+,V_ power supplies.Output is clamped to V low or V high , setV ref =0 to make the math easier 69 Vo =-10V

 Extra information:  Example 5: Power Transistors 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.

  V 0 =V 1 -1.2 Volts; TIP3055 type transistors can drive current up to 15A.

(Note: Transistor TIP3055 is not an op amp, but there are power op amps , see http://www.st.com/web/en/catalog/sense_power/FM123/S C1592 ) op-amps (v.5f) 70

Power transistors, from [1]

 From: http://www.st.com/stonline/books/pdf/docs/4136.pdf

op-amps (v.5f) 71

From : http://www.fairchildsemi.com/ds/TI/TIP41C.pdf

 op-amps (v.5f) 72

Summary

 Studied  Basic digital data acquisition systems  Low-pass, high-pass and band-pass filter design  The configuration of operational amplifier circuits and their applications op-amps (v.5f) 73

Appendix

To be discussed in class op-amps (v.5f) 74

Appendix 1, To prove

          1/(1+ja)=1/(1+ja)*[(1-ja)/(1-ja)] = (1-ja)/(1 2 -(ja) 2 )=(1-ja)/(1+a 2 ), since j 2 = -1 =1/(1+a 2 )+(-ja)/(1+a 2 )=real + imaginary so |1/(1+ja)|={real 2+ imaginary 2 } 1/2 ={[1 /(1+a 2 )] 2 +[(-ja)/(1+a 2 )] 2 } 1/2 ={[1+a 2 ] 2 -(1+a 2 )a 2 /[1+a 2 ] 2 } 1/2 ={[1+2a 2 +a 4 -a 2 -a 4 /[1+a 2 ] 2 } 1/2 ={[1+a 2 ]/[1+a 2 ] 2 } 1/2 =1/[1+a 2 ] 1/2 , proved! 1  1

ja

 1 1 

a

2 op-amps (v.5f) 1  1

ja

 1 1 

a

2 75

ANS:

Solution for Exercise 1.5

  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.

9V R 2 Ra Rb V 1 Sensor V 2 R 1 R 3 _ + A 9V -9V V 0 0V R 4 op-amps (v.5f) 76

Appendix 2a Derive the gain formula for the inverting amplifier  To proof: Gain(G) = -R2/R1  Op-amp calculation rules:  (1) Assume ‘+’ and ‘–’ inputs are at the same voltage potential.

 (2) The current going into ‘–’ or ‘+’ input of an Op-amp are assumed to very small (approaching 0).

 Kirchhoff current law: The sum of currents entering a point is 0.

 Potential at the ‘-’ input is V =0 (virtual ground, because ‘+’ is also at 0 and they should be the same using rule (1) above, and the current going into the ‘-’ input is -I 3 =0 (rule 2) . So using by Kirchhoff current law (sum of all currents going to a point is 0), I 1 +I 2 +I 3 =0.   So (V1-0)/R1+ (V0-0)/R2+0=0, hence (V1-0)/R1=-(V0-0)/R2, therefore the amplification=V0/V1=-R2/R1 Virtual-ground,V V1 Input R 3 R 1 I 2 I 1 I 3 _ + A R 2 op-amps (v.5f) V0 Output 77

Appendix 2b Derive the gain formula for the non-inverting amplifier  To proof: Gain(G) = 1+(R2/R1)  Op-amp calculation rules:  (1) Assume ‘+’ and ‘–’ inputs are at the same voltage potential.

 (2) The current going into ‘–’ or ‘+’ input of an Op-amp are assumed to very small (approaching 0).

 Kirchhoff current law: The sum of currents entering a point is 0.

 The potential at the ‘-’ input is V2, it is the same as V1 (rule1), so V2=V1. The current I 3 is 0 (rule2) . So using by Kirchhoff current law at V2 (sum of all currents going to a point is 0), I 1 +I 2 +I 3 =0.   So (0-V2)/R1+(V0-V2)/R2=0, or (0-V1)/R1+(V0-V1)/R2=0 hence (V0-V1)/R2=V1/R1, therefore the amplification=V0/V1=1+(R2/R1) Input V1 + A _ I 3 V2 R2 I 2 op-amps (v.5f) I 1 V0 R1 Output 78

Appendix 2b

Sketch the Bode plot (frequency response) of a first-order low pass filter

Power gain in dB (

20*log 10 (|G v (f)|

dB ) VS. frequency (log 10 Assume

R 2 =R 1

=1K , C=1uF scale) plot,

G v

(

f

)  

R

2 /

R

1 1   

f f c

  2  1   1

f

159 2 , hence

G v

(

f

)  1  1

f

159 2     From 0 Hz to a point much lower than 159Hz (e.g. 130 Hz), a horizontal line (0dB), because

f/159<<1, 20*log(|G v (f)|) =20*log 10 (1)=20*0=0

At

f=f c

=corner frequency=

1/(2



R 2 C)=159Hz, f/f c =159/159=1,

hence power gain=

20*log(1/sqrt(1+1))=20*log (1/1.414) =-3dB (half power) When f>>fc,

so (

f/ fc) 2 >>1, hence

power gain=

20*log(f c /f),

on the log scale frequency plot, it is a line of -20dB per decade gradient. Gain 0

G(f) in dB

drop line

f c

20 dB/decade 3dB Freq.

(f)

79 Meaning that, it decreases 20dB for each 10 times of frequency increment.

op-amps (v.5f)