Signals

1
Signals

• Signals contain information about what is
  happening in the physical world
• Weather information can be contained in signals
  that represent
• Temperature, humidity, pressure, wind speed, etc.
• The voice of an announcer contains information
  about the game
• To extract information from a signal it needs to
  be processed in some manner, most conveniently by
  electronic means.
• But what if the signal is not an electrical
  signal?
• It has to be converted to a current or voltage.
• A transducer converts the signal into an
  electronic form or from and electronic form
• Sound waves from your voice can be converted to
  an electronic signal by a microphone (pressure
  transducer)
• A wide variety of transducers exist
• Transducers will be studied in other courses, in
  this course will be interested in how the
  electronic signals are manipulated

2
Forms of the signal

• The electronic signals can be represented in two
  different but equivalent forms
• Thevenin (voltage) form
• A voltage source and a series source resistance,
  Rs
• Preferred when Rs is low
• and the Norton (current) form
• A current source and a parallel source
  resistance, Rs
• Preferred when Rs is high

3
A time-varying signal

• Information content is contained in the change in
  the magnitude of the signal over time
• In general it is not easy to mathematically
  describe an arbitrary looking waveform, but it is
  very important to have such a description in
  order to design the appropriate signal processing
  circuits to perform the desired functions or
  operations on the given signal

Analog Signal
Amplitude
Continuously time-varying signal
time
4
The Frequency Spectrum of Signals

• An extremely useful way to characterize a signal
  or arbitrary waveform (time function) is in terms
  of its frequency spectrum
• The frequency spectrum description is obtained
  through the use of mathematical tools such as the
  Fourier series and Fourier transform (we will
  only introduce the topic here and come back to it
  in future courses)
• These tools provide the means for representing a
  voltage signal vs(t) or a current signal is(t) as
  the sum of sine-wave signals of different
  frequencies and amplitudes
• This makes the sine-wave a very important signal
  in the analysis and testing of electronic
  circuits so we will briefly review the
  properties of the sinusoid.
• Where Va denotes the peak value or amplitude in
  volts and w denotes the angular frequency in
  radians per second, that is w 2pf radians per
  second, where f is the frequency in hertz, f
  1/T Hz, and T is the period in seconds

T
5
More on the sinusoid

• The sine-wave is completely characterized by
  three quantities
• its peak value, Va
• its frequency, w
• its phase, f with respect to some arbitrary
  reference time. On the previous page the time
  origin was chosen so that the phase angle is
  zero.
• It is common to express the amplitude of the
  sine-wave signal in terms of its root-mean-square
  (rms) value, which is its peak value divided by
  the square root of two. Thus the rms value of the
  sinusoid on the previous page is Va divided by
  the square root of two.
• For example a 120V wall ac power supply is an rms
  value so that the peak value is actually 120
  times the square root of two.

6
A Square-wave signal and its Fourier Series

• The Fourier series representation can also be
  used in the specific case where the signal is a
  periodic function of time
• The waveform is described by the sum of sinusoids
  that are harmonically (multiples of some base
  frequency) related, for instance, take the
  symmetrical (both positive and negative
  amplitude) square wave shown below
• It can be expressed as the following infinite
  Fourier series
• Because the amplitude of the harmonics decrease
  with increasing frequency we can truncate the
  series at some point (make it a finite series)
  with little error

Fundamental (base) frequency
T
7
The Frequency Spectrum

• The sinusoidal components of the square-wave
  signal constitute the frequency spectrum of the
  this particular signal as shown below
• The frequency spectrum of an arbitrary
  non-periodic waveform has a continuous
  distribution of frequencies (all frequencies are
  present in some degree) but much of the
  information is usually confined to a small part
  of the frequency spectrum which can be useful in
  signal processing applications.

frequency
8
Time domain vs. Frequency domain

• The spectrum of audible sounds such as speech and
  music goes from roughly 20Hz to 20kHz (audio
  band)

9
Exercises

• 1.1 - Find the frequencies f, and w of a
  sine-wave signal with a period 1 ms
• 1.2 - What is the period of sine waveforms
  characterized by frequencies of
• a) f 60 Hz
• b) f 10-3 Hz
• c) f 1 MHz

10
Exercise 1.3

• When a symmetrical square-wave signal whose
  Fourier series is given by the following
  equation
  is applied to a resistor the
  total power dissipated may be calculated directly
  from the frequency spectrum using the
  relationship
• The total power can also be found by summing the
  contribution of each of the harmonic components,
  which can be found from using the RMS values,
  that is, P P1 P3 P5
• What fraction of a square waves power is in its
  fundamental?
• What fraction of a square waves power is in its
  first five harmonics (through 5 w0)?

11
Analog vs. Digital Signals
Analog
Continuous time-varying signal
time
Digital - only two discrete levels
Analog
Discrete time-varying signal
time
1 0 1 1 0 0 1 0
12

• Binary (base 2) numbers, powers of 2
• 201, 212, 224, 238, etc.
• A single binary digit (or bit as it is called)
  can be either a zero or a one
• Assign each bit a weighting value based on powers
  of two,
• for example 8s place, 4s place, 2s place and
  1s place
• Binary numbers are listed from MSB to LSB
• MSB - Most significant bit (8s place) and LSB -
  least significant bit (1s place)
• A decimal number (base 10) such as 12 has two
  digits one in the tens place and one in the ones
  place. Each digit can be 0-9 in base 10. There
  are 100 or 102 (10 to the number of digits)
  possible combinations (0-99).
• A binary number has two possible values for each
  digit (0 or a 1) so a four bit binary number will
  have 24 or 16 combinations

13
Analog to Digital Conversion
Analog
Digital
Continuously varying signal
Discrete voltage varying signal
time
Discrete time steps as well
14
Amplifiers

• Most transducers provide signals that are in the
  microvolt or millivolt range and have little
  energy
• These signals in their raw state are too weak for
  reliable processing
• Stronger signals are easier to process, therefore
  we need an amplifier
• Amplifiers should be linear so that the
  information being amplified is not being changed
  or distorted in an unwanted way
• A distortion free amplifier can be described by
  the following relationship
• vo(t) Avi(t)
• Where vi and vo are the input and output signals
  and A is a constant which represents the
  amplifier gain. The previous equation describes a
  linear amplifier

15
Voltage Gain

• Triangle of the symbol points in the direction of
  signal flow
• Signals can each have different reference level
  or share a common reference

output
input
Transfer Characteristic
vO
Av
iO
iI
1
vO(t) -
vI
vI(t)
0
RL
16
Power Gain and Current Gain

• We are trying to increase the signal power (its
  ability to do work)
• The additional power is usually externally
  supplied
• The power is to be delivered to the Load ( RL)

iO vO / RL
vO iO vI iI
iO iI
Current gain (Ai )
Ap Av Ai
17
Expressing Gain in Decibels

• Voltage and Current Gain are defined as a ratio
  of quantities with the same dimension so they may
  be written as unit less or as Sedra and Smith
  prefer, V/V or I/I for emphasis
• In many instances a logarithmic expression is
  preferred

Recall that PV2/R or
I2R , the squared term causes the power gain in
decibels to be 1/2
Power gain in decibels 10 log Ap

• Absolute values are used because in some cases
  the gain may be negative
• Negative gain simply means that there is a phase
  change of 180 degrees
• Attenuation (weakening) of a signal is indicated
  by a gain of less than one
• Amplification (strengthening) of a signal is
  indicated by a gain of more than one

18
Amplifier Efficiency and Power Supplies
V1
i1
V
iO
iI
vO(t) -
vI(t)
RL
V
i2
-V2
DC Power delivered to the amplifier is Pdc
V1I1 V2I2
Some power is delivered into the circuit by the
power supplies and some is delivered by the
input signal Some power is consumed by the load
and some is dissapated by the amplifier
circuit. The Efficiency is defined as
Pdc PI
Pdc PI PL Pdissapated
19
Example 1.1

• Consider an amplifier which uses /- 10V
  supplies. The input signal is a 1V peak sinusoid
  while the current is 0.1 mA peak. The output
  delivers a 9V peak sinusoid into a 1 kW load. The
  amplifier draws 9.5 mA from each power supply.
• Find the voltage gain, the current gain, the
  power gain, the power drawn from the dc supplies,
  the power dissipated in the amplifier, in the
  load and the amplifier efficiency

20
Amplifier Transfer Characteristic, Linear to
Saturation
vO

• L is the positive output saturation level
• L- is the negative output saturation level

Output peaks are clipped or distorted
L
Undistorted Output Range
vI
0
L-
The steeper the slope of the transfer
characteristic the higher the gain
L- Av
L Av
Safe input swing range
21
Nonlinear Transfer Characteristics and Biasing

• Bias the circuit so that it operates in a
  linear region of the non-linear characteristic
  for small input signal levels
• The dc operating point or quiescent operating
  point (Q point, from audio, where we have quiet
  or off levels)
• The gain is given by the slope of the transfer
  characteristic at the Q point

L
SlopeAv
Q
L-
22
Biasing Example

• A transistor amplifier has the following transfer
  characteristic
• The characteristic applies for
• Find the limits (L- and L, and the corresponding
  values of vI.)
• The L- limit is obviously 0.3V based on the range
  restriction given above
• By substitution into the transfer characteristic
  equation and solving for vI we get 0.69V
• The L limit is determined by where vI 0, and
  thus
• Also find the value of the dc bias voltage VI
  that results in VO5V and find the voltage gain
  at the corresponding operating point.
• To bias the device so that VO5V we use the
  equation and solve for vI .
• vI 0.673V
• We can find the gain by taking the derivative at
  vI 0.673V
• Note the negative gain which indicates that we
  have an inverting amplifier

10
5
0.3
0.673
0.69
23
Symbol Convention

• Total instantaneous quantities (have a dc and an
  ac component or are arbitrary) are denoted by a
  lower case symbol and an upper case subscript
• Direct-current (dc) quantities are denoted by an
  upper case symbol and an upper case subscript.
  Say to yourself, if they are both uppercase it is
  a dc quantity
• Incremental or ac quantities are denoted by a
  lowercase symbol and a lower case subscript.

Incremental ac value
Peak ac value
total value
dc value
24
Some more Exercises

• An amplifier has a voltage gain of 100 V/V and a
  current gain of 1000 A/A. Express the voltage and
  current gain in decibels and find the power gain
  in dB.
• An amplifier operating from a single 15V power
  supply provides a 12V peak-to-peak sine-wave
  signal to a 1kW load, and draws negligible input
  current from the signal source. The dc current
  drawn from the 15V supply is 8mA. What is the
  power dissipated in the amplifier and what is the
  amplifier efficiency?

25
Small-signal approximation Exercise

• Lets look at the limitation of the small signal
  approximation. Consider amplifier with the
  following transfer characteristic

• A positive input signal of 1mV is superimposed on
  the dc bias voltage VI.
• Find the corresponding output for the following
  two situations
• 1) Use the transfer characteristic of the
  amplifier
• 2) Assume the amplifier gain is linear about the
  bias point (gain of -200)
• 1) Let vIVIvi and vOVOvo5vo
• thus,
• 2)

output swing
10
5
Input swing
0.3
0.673
0.69
26
Small-signal approximation Exercise continued

• Repeat the calculations for input signals of 5mV
  and 10mV
• 5mV
• Using the approximation
• Using the full equation
• 10mV
• Using the approximation
• Using the full equation

The larger the signal the larger the error when
using the approximation
27
Circuit Models for Voltage Amplifiers

• Simple but effective. Input resistance accounts
  for the power drawn from the signal source a the
  output resistance accounts for changes in output
  voltage as current is supplied to the load.
• The model might internally represent from one to
  to 20 or more transistors
• Model values determined by analysis and
  calculation or by direct measurement
• A voltage controlled voltage source which has a
  gain factor of Avo
• An input resistance of Ri that accounts for the
  fact that the amplifier draws an input current
  from the signal source.
• An output resistance Ro that accounts for the
  change in output voltage as the load resistance
  (hence the load current supplied by the
  amplifier) is changed

Ro
vo -
vi -
Ri
Avovi
28
Using Amplifier Circuit Models with a Signal
Source and a Load

• The non-zero output resistance causes only a
  fraction of the Avovi to appear across the output
  (load) as shown on the right below. It follows
  that in order to reduce the losses in coupling
  the amplifier to the load, the output resistance,
  Ro should be as LOW as possible (much lower than
  RL).
• If there is no load, RL is infinitely large (open
  circuit) then the voltage gain AV becomes AVo or
  as it is known, the open-circuit voltage gain.
• When you specify the voltage gain of an amplifier
  you must also specify the value of the load
  resistance the gain is measured at.

Using voltage division
In order not to lose signal strength at the
input, Ri should be infinitely large or much,
much greater than Rs
RL RL Ro
vo Avovi
is
io
Ro
vo -
vi -
Rs
vs
RL
Ri
Avovi
29
Comment on the Voltage Amplifier

• The overall voltage gain is given by
• A buffer amplifier, has high input resistance and
  low output resistance. A buffer has minimal
  voltage gain but can have a significant POWER
  gain (see exercise 1.8).

30
An Example of Cascaded Amplifier Stages
Signal source
Stage 1
Stage 2
Stage 3
Load
ii
100kW
1kW
10W
1kW
vi1
vi2
1MW
vi3
100kW
10kW
vi3
100W
vs
10vi1
1vi3
100vi2

• The amplifier shown is a cascade of three
  separate stages. It is fed by a signal source and
  delivers its output to a load resistance of 100W.
  
• Find the fraction of the source signal appearing
  at the input terminals of the amplifier
• Use voltage division

31
Example continued
Signal source
Stage 1
Stage 2
Stage 3
Load
ii
100kW
1kW
10W
1kW
vi1
vi2
1MW
vi3
100kW
10kW
vi3
100W
vs
10vi1
1vi3
100vi2

• Find the gain of each of the three stages and the
  three in cascade,
• For the first stage consider the input resistance
  of the second stage to be the load resistance of
  the first stage

32
Example continued
Signal source
Stage 1
Stage 2
Stage 3
Load
ii
100kW
1kW
10W
1kW
vi1
vi2
1MW
vi3
100kW
10kW
vi3
100W
vs
10vi1
1vi3
100vi2

• Find the gain from source to load. Here we simply
  realize that some of the signal is lost due to
  voltage division between the source and the first
  stage and we already calculated that only 90.9
  gets to the input, so
• Find the current gain and the power gain

33
Exercise to Determine the Benefits of Using a
Buffer

• 1.8 - A transducer produces 1 Volt rms and has a
  resistance of 1MW and its output is to drive a
  10W load. If the transducer is connected
  directly to the load what voltage and power
  levels are seen at the load (draw a circuit
  diagram, it always helps).
• If a unity gain buffer amplifier (AV1) with a
  1MW input resistance and 10W output resistance is
  connected between the transducer and the load,
  what are the output voltage and power levels now?
  Find the gain (in dB) from source to load and the
  ac power gain (also in dB).

ii
1MW
Signal source (transducer)
vout
10W
vs
Buffer
ii
1MW
10W
vi1
vout
1MW
10W
Signal source (transducer)
vs
1vi1
34
Exercise Continued
35
More Exercises

• 1.9 - The output voltage of an amplifier is found
  to decrease by 20 when a total load resistance
  of 1kW is connected. What is the value of the
  amplifier output resistance?
• We can view this as a voltage division problem
  between the amplifier output resistance and the
  load resistance
• 1.10 - An amplifier with a voltage gain of 40dB,
  an input resistance of 10kW, and an output
  resistance of 1kW is used to drive a 1kW load.
  What is the value of the voltage gain AV? Find
  the value of power gain in dB.
• Since the load resistance is equal to the output
  resistance the voltage gain will be the same as
  the amplifier gain

36
Other Amplifier Types
Current Amp
Voltage Amp
ii
Ro
vo -
vo -
vi -
Ro
Ri
Aisii
Ri
Avovi
Transconductance Amp
Transresistance Amp
ii
Ro
vo -
vo -
vi -
Ro
Ri
Ri
Rmii
Gmvi
37
Voltage Amplifier

• Voltage in and voltage out, V/V is unit-less

Voltage Amp
Ro
vo -
vi -
Ri
Avovi
38
Current Amplifier

• Current in and current out, I/I is unit-less

Current Amp
ii
vo -
Ro
Ri
Aisii
39
Transconductance Amplifier

• Voltage in and current out, I/VG

Transconductance Amp
vo -
vi -
Ro
Ri
Gmvi
40
Transresistance Amplifier
Transresistance Amp

• Current in and a voltage out, V/I R

ii
Ro
vo -
Ri
Rmii
41
Bipolar Junction Transistor Example

• A bipolar junction transistor (BJT) is a
  three-terminal device with the following symbol.
• The terminals are called the emitter (E), the
  collector (C) and the base (B). The device is
  basically non-linear (relationships between the
  terminal voltages and terminal currents). We can
  use the small-signal linear approximation about a
  quiescent bias point as we discussed in section
  1.4.
• The total instantaneous terminal voltage and
  current quantities can be expressed using the
  conventions presented earlier.
• The device itself may be represented by one of
  several equivalent circuit models

C
iC
vCE -
iB
B
vBE -
iE
E
vBE VBE vbe
iB IB ib
iE IE ie
iC IC ic
vCE VCE vce
42
Two Equivalent (p) Circuit Models for Bipolar
Junction Transistors

• Notice that the components of the two models are
  the same but that they use different dependencies
  for the current source (one a current and the
  other a voltage) .

ic b ib
ib
iC
C
B
vbe -
rp
bib
Current Amplifier
ie ib ic(b 1)ib
ie
E
ib
iC
ic gmvbe
B
vbe -
C
Trans-conductance Amplifier
rp
gmvbe
b
P model name comes from the fact that the circuit
configuration looks like an upside down greek
letter p
gm
rp
ie
E
43
Why so many different models of the same device?

• Different models are used in different circuit
  configurations in order to make the derivation of
  the circuit behavior easier. For example, the p
  models on the previous page are easy to use when
  the input signal source resistance appears in
  series with rp in the model since they can be
  combined by adding.
• Other circuit configurations become easier to
  analyze when other models are used
• It takes a little experience to pick the right
  models to simplify your design work. The
  alternative is to use computer algorithms to
  solve the complex circuits regardless of the
  choice of models, but the designer loses some of
  the intuitive feel for the behavior of the
  circuit when they do that.
• Another issue is conversion between model types.
  What if someone gives me parameters for one model
  and I find that a different model simplifies the
  analysis?
• Equations for each model can usually be linked
  and solved such that the parameter values for one
  model can be found in terms of the other model
  parameters allowing conversion between models

44
Low Frequency T Model for common base
configuration

• We can derive expressions that allow us to
  convert from one model form to another

ib
iC
ic a ie
B
C
vbe -
a ie
ieib ic therefore ib (1-a )ie
re
But what are a and re? From the p model we can
write
ie
E
Therefore
v1 -
E
C
T
re
The T model is useful in the Common base
configuration
iC
gmv1
B
45
Input Resistance in an active device (BJT)

• 1.14 - Find the input resistance between the base
  and ground in the circuit shown below. The
  voltage vx is a test voltage and the input
  resistance is defined as Rin vx/ix.

• Using KCL at the emitter node we get
• Now using KVL around the input loop we get
• Note We cant see any resistances on the other
  side of the current source since the current
  source output is constant no matter what voltage
  appears across it.

ib
iC
B
ix
ieib ic or ic (b1 )ib (b1 )ix
C
vbe -
rp
bib
RL
vx
vxvbevE(ixrp) (b1 )ixRe or vx/ix Rinrp
(b1 )Re
ie
E
Re
G
Rin
46
Frequency Response of Amplifiers

• We can make up any signal from a sum of different
  frequency and amplitude sinusoidal signals
  (Fourier series), so it is important to know how
  our amplifier responds to various different
  sinusoidal frequencies.
• Such a characterization of amplifier performance
  is called the amplifier frequency response.
  Various test instruments can assist in the
  characterization process as we shall see in lab.
• How can we apply a signal and measure the
  response?
• The gain is given by Vo/Vi and the phase
  difference or phase angle is given by f.
• The gain and phase angle measured can change
  depending on the test frequency w used. An
  oscilloscope can be used to measure these
  quantities (see next page)
• The amplifier transmission or transfer function,
  T(w) versus frequency is found by making
  measurements over a wide range of frequencies.

linear amplifier
-
47
Measuring Frequency Response with an Oscilloscope

• If the output is one half cycle behind the input
  then the phase difference is 180 degrees
• In this example it appears that the gain (Vo/Vi)
  goes up with increasing frequency and the phase
  difference (referenced to the input) goes down
  (gets smaller
• If we took data at many more frequencies we could
  construct a detailed plot of these two responses
  versus frequency
• This is a tedious process so the Gain-Phase
  Analyzer was invented to do this for us and plot
  the results

Low Frequency
f
input
output
w w1
Higher Frequency
input
f
output
w w2
48
Single Time Constant (STC) Networks

• Lets consider a Low Pass filter
• The transfer function T(s) of an STC (single time
  constant) low-pass circuit can always be written
  in the form of the complex frequency variable, s
• For physical frequencies, where sjw, we get
• Where K is the magnitude of the transfer
  function at w0 (dc) and w0 is defined being 1/t
  .
• Tau is the time constant, which in this case is
  RC.
• On the next page we will look at the magnitude
  response of the transfer function for various
  frequencies and plot the results

49
Sample Calculations

• Let R100,000W and C equal 0.1mF, tRC is 0.01,
  w01/t100 so the transfer function is
• Lets pick a frequency and calculate the magnitude
  of the response, LET w1
• Plot the denominator as a complex number

jw
The magnitude of the denominator is given by the
length of the complex vector. Using trigonometry
we know that c2 a2b2 so that c is
The phase angle of the denominator is given by
the angle the complex vector makes with the
positive real axis. Using trigonometry we know
that this angle is given by the inverse tangent
of imaginary part over the real part
0.01
-Re
Re
1
-jw
50
Continuing with the example

• We looked at the denominator so the total
  magnitude response is the magnitude of the
  numerator divided by the magnitude of the
  denominator, or 1/1 1
• The total phase response is the phase angle on
  the numerator (zero) minus the phase angle of the
  denominator (since it is division), or -f
• Now lets consider a different frequency w10.
  Remember w0 is 100.
• And w100

jw
0.1
jw
-Re
Re
1
1
-jw
-Re
Re
1
-jw
51
and at higher frequencies

• Finally lets consider a frequency of w1000.
  Remember w0 is 100.
• What we are observing is this capacitor gradually
  changing its behavior from an open circuit at low
  frequencies (wltltw0) to a short circuit at higher
  frequencies (wgtgtw0) which shorts out the output
  response

-jw
10
-Re
Re
1
-jw
52
The General Expression (Low Pass)

• Thus the low pass magnitude response in general
  is given by
• The general low pass phase response is given by

53
Plots of the Frequency Response
Magnitude
1
0
1 10 100 1k
Radian Frequency
1 10 100 1k
Phase Angle of the transfer function
54
Bode (Bo-duh) Plots of the Frequency Response
(Low Pass)

• A decade is a factor of 10 and an octave is a
  doubling
• Note
• and

Normalized Magnitude
0 -10 -20 -30
0.1 1 10
55
Single Time Constant (STC) High Pass Network

• Lets consider a High Pass filter now
• The transfer function T(s) of an STC (single time
  constant) low-pass circuit can always be written
  in the form of the complex frequency variable, s
• For physical frequencies, where sjw, we get
• Where K is the magnitude of the transfer function
  at w0 (dc) and w0 is defined as being 1/t .
• Tau is the time constant, which in this case is
  1/RC.

56
The General Expression (High Pass)

• Thus the high pass magnitude response in general
  is given by
• The general high pass phase response is given by
• In this case the capacitor is and open at low
  frequencies and the output is zero, but at higher
  frequencies the capacitor becomes a short circuit
  and the output is finite

57
Bode (Bo-duh) Plots of the Frequency Response
(High Pass)
Normalized Magnitude
0 -10 -20 -30
0.1 1 10
Normalized Frequency
slope
0.1 1 10
58
Bode Plots of Factored Transfer Functions
z2
z1
z1
p2
p1
p3
z2
I expected everyone to do the problem
graphically and by calculation
p1
T(jw) in dB
p2
83.5
80
43.5
p3
43.5
40
107
108
1
103
10
104
106
102
105
z1
z2
p3
p1
p2
59
Example 1.5

• The circuit below shows a voltage amplifier
  having an input resistance Ri, an input
  capacitance Ci, a gain factor m, and an output
  resistance Ro. The amplifier is fed with a
  voltage source Vs having a source resistance Rs,
  and a load resistance RL is connected to the
  output.
• Derive an expression for the amplifier voltage
  gain Vo/Vs as a function of frequency. From
  this find expressions for the dc gain and the
  3-dB frequency.

60
Example 1.5 contd
Using the voltage divider rule
,
Thus,
Low pass STC network
At the output side of the amplifier
Amplifier transfer function
61
Example 1.5 contd

• By inspection, we see that the input circuit is a
  STC network. We can find the time constant by
  reducing Vs to zero, resulting in the resistance
  seen by Ci is Ri // Rs.

(from previous page)
dc gain
3-dB frequency
62
Example 1.5 contd

• Calculate the values of the dc gain, the 3-db
  frequency, and the frequency at which the gain
  becomes 0 db (i.e., unity) for the case Rs
  20kW, Ri 100KW, Ci 60pF, m 144V/V, Ro
  200W, and RL 1kW.
• Since the gain falls off at a rate of -20
  db/decade, the gain will reach 0 dB in two
  decades (starting at w0). Unity-gain frequency
  108 rad/s or 15.92 MHz

dB
V/V
rad/s
63
Example 1.5 contd

• Find vo for vi 0.1sin102t,V
• Find vo for vi 0.1sin105t,V
• Find vo for vi 0.1sin106t,V

V
V
V
64
Classification of Amplifiers based on Frequency
Response

• Capacitively coupled amplifier
• Directly coupled amplifier
• Tuned or bandpass amplifier

mid-band frequencies
decibels (dB)
bandwidth
decibels (dB)
bandwidth
decibels (dB)
bandwidth
65
Frequency Dependent Gain

• INTERNAL device (transistor) capacitances (in th
  pF range) cause the gain of amplifiers to fall
  off at high frequencies.
• The gain at low frequencies usually falls off at
  low frequencies due to coupling capacitors.
• Coupling capacitors are used to couple (connect)
  one amplifier stage to another. Coupling
  capacitors simplify the design of each stage by
  allowing the designer to separately bias each
  stage.
• The coupling capacitors are usually quite large
  (0.1mF - 50mF typically). This value range is
  chosen so that the coupling capacitors will have
  a low resistance (be short circuits) at the
  frequencies of interest. The internal capacitance
  will still be open circuits at mid-band
  frequencies.
• Coupling capacitors cause the gain to be zero at
  dc
• A directly coupled amplifier has gain down to dc
  and, integrated circuit fabrication techniques do
  not easily allow for the fabrication of the large
  value capacitors required for capacitively
  coupling amplifiers, so most are direct coupled.
• Tuned amplifiers are often needed in the design
  of radio and television circuits and are used in
  the tuner part of the receiver (allows you to
  select the channel). The center frequency is
  adjusted to match the broadcast frequency. Other
  frequencies (channels) outside of the pass band
  are not amplified or received.

66
The complex frequency plane

• Complex frequency is a unifying concept. It ties
  together
• resistive circuit analysis
• steady state sinusoidal analysis
• transient analysis
• forced response
• complete response and the analysis of of circuits
  excited by exponential forcing functions and
  exponentially damped sinusoidal forcing functions
• They all become special cases of one general
  technique which is associated with the complex
  frequency concept
• Sigma is the damping level (real axis), is
  growing and - is decreasing
• Omega is the frequency
• sigma 0 and omega 0 is a dc signal
• sigma and omega 0 is an increasing
  exponential
• sigma - and omega 0 is a decreasing exponential
• See page 420 of Hayt and Kemmerly

67
The Inverter

• An inverter is the fundamental circuit in digital
  logic as the amplifier is in analog circuitry, in
  fact, an inverter is an amplifier
• We have looked at amplifier transfer
  characteristics and determined that by proper
  biasing (choosing a Q point) and use of a small
  ac signal we can use a non-linear characteristic
  to create a linear amplifier. In digital logic
  we make use of the gross non-linearity of the
  characteristic since we are only interested in
  the two (binary) saturation levels and not in the
  transition region in between.
• One saturation region is called a 1, or high or
  true. The signal is usually a voltage level but
  current signals are sometimes used as well
• The other saturation region is called a 0, or
  low or false
• In positive logic a 1 is a more positive voltage
  than a 0, in a negative logic system the more
  negative voltage is defined to be a 1. Negative
  logic is commonly used with logic circuits made
  up of only p-channel metal oxide semiconductor
  field effect transistors (MOSFETs).
• The voltage transfer functions of several
  inverters are shown on the next page. Some key
  points to observe are
• The slope at any point on the VTC is the gain of
  the inverter
• If the slope is greater than one (steep) the
  gain is high and if the slope is less than one we
  have attenuation
• A slope of -1 on the VTC is a 45 degree line
  (provided both axes are scaled the same). The
  minus sign come from the fact that it is an
  inverter.
• The point at which the output voltage is equal to
  the input voltage is called Vinvert and is of
  interest in digital logic (ideally Vinvert should
  be 1/2 the supply voltage).

68
The Voltage Transfer Characteristic (VTC) of an
Inverter

• The gain is the change in Vout divided by the
  change in Vin or the slope of the VTC

VOUT (Volts)
5
Ideal Inverter
HIGH, 1 5V
BJT Common- Emitter Inverter
4
GAINSLOPE
3
2
CMOS Inverter
1
LOW, 0 0V
0
1
2
4
5
3
0
VIN (Volts)
LOW, 0 0V
HIGH, 1 5V
69
n-channel MOS Transistors

• Enhancement Mode Transistors. A normally open
  switch. At zero volts on the gate no current
  flows ( a positive Voltage must be applied to the
  gate to enhance a channel of electrons)
• Depletion Mode Transistors. A normally closed
  switch. At zero volts on the gate a current flows
  ( a negative Voltage must be applied to the gate
  to deplete the channel of electrons)

70
p-channel MOS Transistors

• Enhancement Mode Transistors. A normally open
  switch. At zero volts on the gate no current
  flows ( a negative Voltage must be applied to the
  gate to enhance a channel of holes)
• Depletion Mode Transistors. A normally closed
  switch. At zero volts on the gate a current flows
  ( a positive Voltage must be applied to the gate
  to deplete the channel of electrons)

71
Inverter operation

• The figure on the left shows that when the
  switch is open (transistor is off) there is no
  connection between Vout and ground and there is
  no path for current flow through the resistor. If
  there is no current flow through the resistor
  then both ends must be at the same potential and
  Vout must be five volts.
• When the switch is closed (transistor is on) a
  low resistance path is formed between Vout and
  ground through the transistor. The output will be
  pulled towards ground since the on resistance of
  the transistor is less than the other resistance
  value and Vout will be close to zero volts. A
  current will flow from the 5 volt supply to
  ground when the transistor is on.
• Resistors take up too much space (area)

5
5
say R 10k Ohms
Vout5V Logic 1
Vout 0.5 Volts Logic 0
Vin0V Logic 0
Vin5V Logic 1
closed say Ron 1k Ohms
open
72
The Complementary MOS (CMOS) Inverter

• Complementary means both nMOS and pMOS
  transistors are used
• A polite tug OWar! Only one device pulls at a
  time
• A High Voltage on Vin turns On the NMOS device
  and turns Off the PMOS device
• A Low Voltage on Vin turns off the NMOS and turns
  On the PMOS
• Power is dissipated only when the output is
  switching from low to high or high to low

p channel (n well)
n channel (p wafer)
The source is where charge carriers come from and
the drain is where they flow to, holes come from
the higher voltage and flow towards a more
negative terminal, electrons come from the more
negative terminal and flow towards the positive
73
Noise Margins

• In positive logic a high is more positive than
  a low
• High is 5V or High is 3.3V or 12V, etc.
• Low is zero
• In negative logic a high is more negative than
  a low
• High is -10V
• Low is zero
• In a real binary (2 level) logic system we have
  to assign a high to a range of voltages and a
  low to another range of voltages
• For example, let a high be 4V lt Vout
  lt 5V and
• let a low be
  0V lt Vout lt 1V
• The output of logic gates must fall within these
  ranges in order to clearly (solidly) convey to
  the outside world what value the gate is
  producing
• If Vout is 4.2V it is a high. If Vout is 0.3V it
  is a low. If Vout is 2.1V then the output is said
  to be in an illegal or undertermined logic state.
• Another problem is that we would like to be able
  to transmit these solid high and low signals to
  remote locations. In the process these signals
  might get degraded. So, we must be able to accept
  these degraded inputs (within reason) and still
  recognize them.
• For example, A 4.05V signal generated by a
  logic gate is a solid high, but if it degrades to
  a 3.7V signal when it arrives, we must still
  recognize it as a marginal high.
• Noise margins are ranges adjacent to strong
  signal levels, which are also recognized as
  having the same level. If 92 and above is a
  strong A, a grade of 88 might be a marginal
  A.

74
How are Noise Margins Determined?

• The slope of the voltage transfer characteristic
  of an inverter is the gain.
• There are three key points on the gain plot
• The point at which the magnitude of the gain is
  first equal to unity (one, 45 degrees)
• The point at which the magnitude is maximum
• The point second point at which the gain is again
  equal to unity

slope 1
5
VOH
slope Maximum 5
4
VOUT (Volts)
3
2
slope 1
1
VOL
VIN (Volts)
0
4
1
2
3
5
VIH
VIL
gain slope
max
5
1
VIN (Volts)
75
What Noise Margins really mean

• On the previous page VIL was equal to 1.2V and
  VIH was equal to 3V
• VOL was equal to 0.7V and VOH was equal to 4.9V
• The Noise Margins are defined as follows
• NML VIL - VOL in our case
  1.2 - 0.7 0.5 Volts
• NMH VOL - VIL
  4.9 - 3.0 1.9 Volts

76
How does an Inverter (with gain) restorea poor
signal level?

• Assume that we have two identical inverters in
  series and that they both have the same voltage
  transfer characteristic given below.
• Lets say that the input to the first inverter is
  3.1 Volts, which is about as marginal a high
  signal as will be recognized as a high by the
  inverter.

• The output of the first inverter will be 0.65
  volts. If we take that as the input to the second
  inverter the output of the second inverter will
  be 5 volts which is a solid high, and is much
  improved over the 3.1 volts we saw on the
  original input signal.
• The inverter supplies (5 and ground) and the
  gain drive illegal and marginal signals towards
  solid levels

77
Capacitors

• A capacitor is a circuit element that stores
  charge. The energy is not stored chemically as in
  a battery but rather the energy is stored in the
  form of an electric field.
• The Electric Field (Volts per cm) is proportional
  to the voltage across the capacitors terminal.
• The charge stored is proportional to the voltage
  (and the area), and the proportionality constant
  is called the capacitance
• QCV dQ/dt C DV/dt but dQ/dt is
  current, I so that
• I C (dV/dt) the units of capacitance are
  Farads Coulomb/Volt
• The bathtub analogy. The voltage across a
  capacitor can not change instantaneously just
  like the water level in a bathtub can not change
  instantaneously, instead the rate at which it
  changes depends on the value of the current
  flowing into or out of the capacitor. I C
  (dV/dt) Note the level can change pretty fast
  if the current is large enough

Partially charged capacitor
Fully charged capacitor


Electric field
Stronger electric field
- - -
- - - - - - - -
78
The RC time constant

• Consider the following circuit with a battery,
  switch, resistor and capacitor. The capacitor
  initially has no charge stored on it.
• Charging a capacitor
• Discharging a capacitor
• Note RC has units of time. Ohms are Volts divided
  by Coulombs per second and capacitance is Farads
  which are Coulombs per Volt, which leaves
  seconds. We use the greek letter t (tau) to
  represent the RC product.
• The behavior of the circuit is found by solving a
  differential equation. The solution uses the base
  e 2.72
• e-1 0.3737 e-2 0.13514 e-3
  0.055

VC
t 0
V0 10V
10V
R
90
C
i
V V0 (1 - e-t/t)
6.3V
VC 0
time
RC
2.3RC
VC
t 0
10V
R
V V0(e-t/t)
C
i
VC 10
3.6V
10
time
RC
2.3RC
79
Propagation Delay and Rise and Fall Timesof a
signal