CENG4480_A1 Op Amps and Analog Interfacing

1
CENG4480_A1 Op Amps and Analog Interfacing
Week 1

• Analog interfacing techniques

2
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

3
Topics include

• Overall interfacing schemes
• Analog interface circuits, active filters
• Analog/digital conversions
• sensors, controllers
• Control techniques
• Advanced examples

4
Overall view a typical data acquisition and
control system
Timer
Digital control circuit

Sensor
Computer
filter
A/D
Sample Hold
Op-amp
D/A
Power circuit
Mechanical device
5
Analog interface example1 Audio recording systems

• Audio recording systems
• Audio signal is 2020KHz
• 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

6
Analog interface example2 Surround sound audio
systems

• A common two channels audio CD
• Calculate storage size for one hour of recording
  of a CD. 44.1KHz2bytes60sec60min2
  channels633.6Mbytes
• Calculate the play time of a CD.
  700M/(2bytes44.1KHz2channels60sec)61.4
  minutes
• 6 Channels Front R/L,Rear R/L, Middle, Sub
  woofer
• 44.1KHz,
• Calculate the sampling frequency.

7
Analog interface example3Play stations and Wii

• Play station 3, Analog hand held controller
  (http//ryangenno.tripod.com/images/PlayStation3-s
  ystem.gif)
• Wii, http//www.onlinekosten.de/news/bilder/wii_c
  ontroller.jpg
• Driving wheel http//www.bizrate.com/gamecontrolle
  rs/logitech-driving-force-driving-force-wheel--pid
  11297651/

8
Operational Amplifier choices (op amp)

• Why use op amp?
• What kinds of inputs/outputs do you want?
• What frequency responses do you want?

9
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

10
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.

Op- amp
DC Source
Load DC motor
11
Alternating Current (AC) amplifier

• Example Microphone amplifier, signal is AC and
  is changing at a certain frequency range. Current
  is alternating not stable.
• Use capacitors to connect different stages, so no
  need to consider biasing problems.

Op- amp
AC Source
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
Vcc/22.5V
Biased at Vcc/2
12
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
• AC-alternating current amplifier

Op- amp
DC Source
Load
Op- amp
AC Source
Load
13
Input impedance (Rin) and Output impedance (Rout)

• Why do we prefer High Rin and Low Rout?
• Because it is more efficient.
• To maximize Vin2 (input voltage driving stage 2)
  We make Rout1 lower, Rin2 higher.
• Good choice Rin?1M ? or over, Rin? 10?

Stage1(sensor) Vout1 Rout1
Stage 2 Rin2
Vin2
Is equivalent to
Vin2 Vout1Rin2/(Rout1Rin2)
Vout1
Rout1
Rin2
14
Exercise 1.1
Student ID __________________Name
______________________Date_______________
(Submit this at the end of the lecture.)
Sensor Vout_sen 1mV
x1000
Rout_sen
Vout_ amp
Rin_amp

• Sensor (Rout_sen) Vout_sen1mV is sent to an
  amplifier with Rin_amp, gain 1000
• Rout_sen 10 ?, Rin_amp1M, calculate the output
  voltage (out_amp) of the amplifier
• Rout_sen 2K ?, Rin_amp10K, calculate the output
  voltage of the amplifier
• Which above scheme would you prefer and why?

15
Meaning of power gain in dB (Decibel)

• Voutoutput
• Vininput
• Voltage gain Vout/Vin
• Power gain (Vout)2/ (Vin)2
• Power gain in dB10log10(Power gain )
• 20 Log10(Vout/Vin)20Log10G,
• 
  where G Voltage gain
• When power gain(Vout/Vin)21, voltage_gain1,
  power_gain is 0dB
• When power gain(Vout/Vin)20.5,
  voltage_gain(0.5)1/20.707, power_gain is -3dB

16
Frequency-gain plotWhen power gain(Vout/Vin)21,
voltage_gain1, power_gain is 0dBWhen power
gain(Vout/Vin)20.5, voltage_gain(0.5)1/20.707,
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 -3dB Power gain is 0.5
Gain is 0dB Power gain is 1
0dB -3dB
Slope 20 dB/decade drop
Power Gain (dB)
Log Frequency
One decade  one number is 10 times of the other
number
17
Exercise 1.2 General concept about OP amps
B
A

• Controllable gain
• For DC or AC amplifier
• Not too high frequency responses
• KGain bandwidth gain_bandwidth_product
• Calculate K at A,B,C. Gainpower gain
• At A, GainA________,(?BA)_______,KA___
• At B, GainB________,(?BB)_______,KB___
• At C, GainC________,(?BC)_______,KC___
• What is your conclusion based on the above
  calculation?

C
Power gain in dB 10log10(Power gain )
18
Operational amplifiers (op-amps)
Week 2

• ideal op-amps
• inverting amplifier
• non-inverting amplifier
• voltage follower
• current-to-voltage amplifier
• summing amplifier
• full-wave rectifier
• instrumental amplifier

19
Ideal Vs. realistic op-amp

• Ideal Realistic Rin
• A infinite ? 105-gt108
• Zin infinite ? 106?(bipolar input) ?
  1012?(FET input) output offset exists

2 3
_
V-
6
V0A(V-V-)
LM741
V

20
Exercise 1.3 Inverting amplifier

• Gain(G) -R2/R1
• For min. output offset, set R3 R1 // R2
• RinR1
• Questions
• (i) Derive the gain formula (See appendix)
• (ii) If R11K, R210K, find G and Rin

Virtual-ground,V2
R2
Output
_
V1
R1
V0
A
Input
R3

21
Exercise 1.4 Non-inverting amplifier

• Voltage Gain(G) ? 1 (R2/R1)
• For min. offset output , set R1//R2Rsource
• High input resistance
• Questions
• (i) Derive the gain formula (See appendix)
• (ii) If R11K, R210K, find G and Rin

V1

V0
A
Input
_
Output
R2
V2
R1
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Differential amplifier

• V0(R2/R1)(V2-V1)
• Minimum output offset R1 //R2 R3 //R4

R2
_
R1
V1
V0
A
Input
V2
Output

R3
R4
Exercise proof the gain formula
23
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.

24
Voltage follower (Unit voltage gain, high current
gain, high input impedance)

• Gain1,
• Rinhigh
• For minimum output offset RRsource

V1

V0V1
A
_
R
Exercise proof the gain formula
25
Current to voltage converter Application to
photo detector no loading effect for the light
detector

• V0I R
• I should not be too large otherwise offset
  voltage will be too high.

Photodiode Light detector
R
I
_
V0
A

See http//hyperphysics.phy-astr.gsu.edu/hbase/ele
ctronic/photdet.htmlc1
Exercise proof the formula
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Summing amplifier

• V0 -(V1/R1)(V2/R2)(V3/R3)R

I1
R
V1
R1
_
V2
R2
V0
I1I2I3

Output
V3
R3
Inputs
Exercise proof the gain formula
27
Exercise 1.6Discuss 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)

28
Integrator

Ref http//www.physics.ucdavis.edu/classes/Physics
116/Physics116A04F.html Reading exercise
http//www.electronics-tutorials.ws/opamp/opamp_6.
html
29
Differentiator

Ref http//www.physics.ucdavis.edu/classes/Physics
116/Physics116A04F.html Reading exercise
http//www.electronics-tutorials.ws/opamp/opamp_7.
html
30
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

31
Op-amp dynamic response

• Slew rate -- the maximum rate of output change
  (V/us) for a large input step change.
• ?A741 slew rate0.5V/ ?s. Fast slew rate is
  important in video circuits , fast data
  acquisition etc.
• Gain bandwidth product
• higher gain --gt lower frequency response
• lower gain --gt higher frequency response

32
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_gainGcVo/Vc.

33
Diagram of gain bandwidth product, from 1

Hz
34
Instrumental amplifier To make a better DC
amplifier from op-amps
Applications Digital Oscilloscope DSO input
amplifiers, amplifiers in medical measurement
systems

Diagram of instrumental amplifier, from 1
35
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 VV- Vc , there is a slight output because
  of the Common Mode GainGcV0/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.

36
Comparing amplifiers, from 1

• Op Inv. Noninv. Diff. Instu.
• Amp Amp Amp Amp Amp
• High Rin Yes No Yes No Yes
• Difftial Yes No No Yes Yes
• input
• Defined No Yes Yes Yes Yes
• gain

37
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 Rate50V/ ms --gt 2000V/ ms (how fast
  the output can be changed)
• Follower (high speed)50MHz
• Low Supply Current 1.5mA --gt 20 µA/Amp
• Low offset voltage 100 µV
• Low Noise
• Low Input Bias Current 50pA --gt10pA
• High Power 0.2A --gt 2A
• Low Drift 2.5 mV/ _C --gt 1.0 mV/ _C
• Dual/Quad
• High Power Bandwidth High Power Bandwidth
  300KHz - 230Mhz

38
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

39
Example 1 Single power supply for op-amps

• Small systems usually have a single power supply
• Output V0 is biased at E/2 rather than 0.
• E.g. Inverting amplifier. Gain?-R2/R1

E10V
R2
E
E/25V
_
V1
R1
E/25V
V V-
A
Vo
R3

0-Volt
0Volt
R20K
R20K
E Volts Power supply
E/25V
40
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 2.5V
Biased at around 2.5V
2.5V
Biased at around 4V
Output to Mic-in of power amplifier
2.5V
41
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

42
Types

• 2-1 Low pass
• 2-2 High pass
• 2-3 Band stop (notch) e.g. noise removal
• 2-4 Band pass

Week 3
43
Recall definition of power gain in decibel (dB)

• Output power is P2, input power is P1
• Power Gain in dB10 log10 (P2/P1)
• Or, output voltage is V2, input voltage is V1
• Assume load R is the same, powerV2/R
• Power Gain in dB10 log10 (V22/ V12)
• Power Gain in dB20 log10 (V1/ V2)
• 20 log10 G, where Gvoltage gain

44
Time domain vs. frequency domain

• Time domain we talk about voltage gain against
  time
• Frequency domain we talk about the voltage gain
  against frequency.

Time domain signal plot
1V 0 -1V
VppPeak-to-Peak voltage
Voltage
Time (Seconds, usually linear scale)
Power Gain (dB)
Frequency domain signal plot
0dB -3dB
Frequency (Hz) (can use log scale)
45
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)1/20.707
• I.e. in dB 20log(0.707) -3dB
• Settling time -- time required to rise within 10
  of the final value after a step input.

46
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
47
Diagram for low-pass one pole filter, from
1for simplicity make R2/R11,
Gain G(f) in dB
20 dB/decade drop
3dB
fc
Freq.
Corner frequency fc,

Exercise proof the gain formula based on the
inverting amplifier gain formula (See appendix)
48
Formula
49
2-1aLow pass, one pole filter formulas for
simplicity make R2/R11K

• Corner frequency fc1/(2?R2C)
• The gain drops 6dB/octave or 20 dB/decade

50
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,
  R1R21K , C1uF

51
Diagram for Low-pass two-pole filter, from 1
for simplicity make R3/(R2R1)1
Gain G(f) in dB
40 dB/decade drop
6dB
Where fc(R1//R2)/2? C1 (2 ? R3C2)-1
fc
Freq.

52
2-1bLow-pass two-pole filter formulas for
simplicity make R3/(R2R1) 1

• Corner frequencyfc
• fc(R1//R2)/2? C1 (2 ? R3C2)-1 when gain G drops
  at -6dB.
• G is dropping at 40dB/decade

53
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,
  R1R21K ,R32K, C1uF
• Compare the difference between one-pole and
  two-pole low pass filters

54
Matlab, lp42.m

• lp42.m, ceg3480 matlab demo low pass filter-one
  pole
• clear
• f01002000
• Nlength(f)
• fc1000
• for i1N
• -----gain1 , low pass one pole , for
  simplicity make (R2/R1)1
• gv1(i)-1/sqrt(1(f(i)/fc)2)
• db10log_base10(power1/power2) or
  dB20log_base10(voltage1/voltage2
• gain1_db(i)20log10(abs(gv1(i)))
• -----gain2 , low pass two pole , for
  simplicity make R3/(R1R2)1
• gv2(i)-1/(1(f(i)/fc)2)
• db10log_base10(power1/power2) or
  dB20log_base10(voltage1/voltage2
• gain2_db(i)20log10(abs(gv2(i)))
• end
• 
  
• figure(1)
• clf

55
Plotting the comparison of the low pass filters
(one-pole, two-pole)
20dB/decade Slope less steep
40dB/decade Slope more steep
56
2-2High 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.

57
2-2aDiagram for high-pass one-pole filter, from
1For simplicity make R1R2, R3R2 // R1
20 dB/decade drop
Gain G(f) in dB
3dB
fc
Freq.
high freq. Cutoff unintentionally Created by
Op-amp

58
High-pass one-pole filter formulas

• Corner frequency fc1/2 ?(R1C)
• At low f , Glow_freqf/fc
• at high f , Ghigh_freq R2/R1?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.

59
Matlab hp52.m

• hp52.m ceg3480 matlab demo high pass filter-one
  pole
• clear
• f500100100000
• Nlength(f)
• fc1000
• for i1N
• -------------------gain3 , high pass ,one
  pole
• gv3(i)-(f(i)/fc)/sqrt(1(f(i)/fc)2)
• db10log_base10(power1/power2) or
  dB20log_base10(voltage1/voltage2
• gain3_db(i)20log10(abs(gv3(i)))
• end
• 
  
• figure(1)
• clf
• limit_ymin(gain3_db)
• semilogx(f,gain3_db,'k-.')
• hold on
• ------------------------
• semilogx(fc,fc,0,limit_y,'g-.')

60
high pass one pole filter

61
2-3Band stop (notch) filter

• Suppresses a narrow frequency band of signal

62
2-3Band pass filter

• Passes a frequency band of signal.

63
Diagram for Notch filter (band-stop), from 1

64
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

Voltage Gain in dB
frequency
65
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), V0V1 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.

66
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
Analogto-digital converter ADC can have more
time for data conversion

Slight droop may occur
Hold
67
Example 4 Voltage Comparator with hysteresis and
schmit trigger
Comparator gives bad result Unstable region when
V1 and Vref are closed

• E.g. in IR motor speed encoder
• V1IR receiver input

V
V0
V1
comparator
Vref
0V
Better output Using Schmit trigger
V-
IR receiver Signal with noise
Schmit trigger
V
0V
V-
68
Diagram for hysteresis (non-inverting schmit
trigger), see P.420, S. Franco, Design with
operational amplifiers and analog integrated
circuits, McGraw Hill.
Voltage
V0
V1

VTH
VTL
t
Output Voltage
Switch over voltage
10V -10V
VTH -VTL (Vohigh Volow)(R1/R2)2V
VTL -1V
VTH 1V
Input voltage
Vref 0
69
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/R20.1)
Vhigh10V

• When Vo Low(-10Volts), We want to find VTH(low)
  , so that when V1gt VTH(low) Vo will switch from
  low(-10Volts) to high (10Volts).
• At opamp V input, when V1 is close to VTH(low),
  apply current rule
• (V1-0)/R1(V0low-0)/R2 0 (NoteV1? VTH(low))
• (VTH(low)-0)/R1(V0low-0)/R2 0
• (VTH(low)-0)/R1(-10-0)/R2 0
• VTH(low)(R1/R2)(10) (NoteR1/R20.1)
• VTH(low) (0.1)(10)1Volt
• So when V1gt VTH(low), V0 will switch from low
  (-10V) to high(10V)
• Similarly, VTH(high) -1Volt , so that when V1lt
  VTH(low) Vo will switch from high(10volts) to
  low (-10Volts).
• Set R2 gtgt R1, to make a small hysteresis. Ie. for
  Schmit trigger devices R1? 0.1R2, e.g. R11K,
  R210K, so the hysteresis is good enough to
  reject noise. The diodes are used to clamp the
  voltages at /-10V

Vo
Vo
0V
Vlow-10V
The op-amp uses V,V_ power supplies.Output is
clamped to Vlow or Vhigh, setVref 0 to make
the math easier
70
Extra informationExample 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.
• V0V1-1.2 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/we
  b/en/catalog/sense_power/FM123/SC1592)

71
Power transistors, from 1

From http//www.st.com/stonline/books/pdf/docs/41
36.pdf
72
From http//www.fairchildsemi.com/ds/TI/TIP41C.p
df

73
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

74
Appendix

• To be discussed in class

75
Appendix 1, To prove

• 1/(1ja)1/(1ja)(1-ja)/(1-ja)
• (1-ja)/(12-(ja)2)(1-ja)/(1a2), since j2 -1
• 1/(1a2)(-ja)/(1a2)real imaginary
• so
• 1/(1ja)real2 imaginary21/2
• 1 /(1a2)2(-ja)/(1a2)21/2
• 1a22-(1a2)a2/1a221/2
• 12a2a4-a2-a4/1a221/2
• 1a2/1a221/2
• 1/1a21/2, proved!

76
ANS Solution for Exercise 1.5

• Gain Vout/Vin9V/(10mV50 C )18, set
  R2/R1R4/R318
• How to solve the offset problem.
• Sensor ? V2
• Offset of 100mV at V1, 9Rb/(RaRb)100mV (make
  R4gtgt Ra) why?
• Add a small variable resistor Rc between 9V Ra
  for offset trimming.

9V
R2
V1
Ra
9V
_
R1
V0
A
Sensor
Rb

R3
V2
-9V
0V
R4
77
Appendix 2aDerive 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 -I30 (rule 2) . So
  using by Kirchhoff current law (sum of all
  currents going to a point is 0), I1I2I30.
• So (V1-0)/R1 (V0-0)/R200, hence
• (V1-0)/R1-(V0-0)/R2, therefore the
  amplificationV0/V1-R2/R1

R2
Virtual-ground,V-
I2
Output
_
V1
R1
I1
I3
Input
V0
A

R3
78
Appendix 2bDerive 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 V2V1. The current I3 is 0
  (rule2) . So using by Kirchhoff current law at V2
  (sum of all currents going to a point is 0),
  I1I2I30.
• So (0-V2)/R1(V0-V2)/R20, or (0-V1)/R1(V0-V1)/R2
  0 hence
• (V0-V1)/R2V1/R1, therefore the
  amplificationV0/V11(R2/R1)

Output
V1

V0
A
Input
_
R2
I3
I2
V2
R1
I1
79
Appendix 2b Sketch the Bode plot (frequency
response) of a first-order low pass filter Power
gain in dB (20log10(Gv(f)dB ) VS. frequency
(log10 scale) plot, Assume R2R1 1K , C1uF

• From 0 Hz to a point much lower than 159Hz (e.g.
  130 Hz), a horizontal line (0dB), because
  f/159ltlt1, 20log(Gv(f)) 20log10(1)2000
• At ffccorner frequency1/(2?R2C)159Hz,
  f/fc159/1591, hence power gain
  20log(1/sqrt(11))20log (1/1.414) -3dB (half
  power)
• When fgtgtfc, so (f/ fc)2 gtgt1, hence power gain
  20log(fc/f), on the log scale frequency plot, it
  is a line of -20dB per decade gradient.
• Meaning that, it decreases 20dB for each 10 times
  of frequency increment.

Gain G(f) in dB
0
3dB
fc
Freq.(f)
20 dB/decade drop line