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Introduction to Electronics
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
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Introduction, Page 1.1-1
Introduction to Electronics
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
Rs
vs t 
~ vs t  if R s is low
 and the Norton (current) form
 A current source and a parallel source resistance, Rs
 Preferred when Rs is high
is t 
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Rs
~ is t  if R s is high
Introduction, Page 1.1-2
Introduction to Electronics
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
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Introduction, Page 1.1-3
Introduction to Electronics
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.
va t   Va sin wt 
va
Va
t
T

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
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Introduction, Page 1.1-4
Introduction to Electronics
More on the sinusoid


The sine-wave is completely characterized by three quantities
 it’s peak value, Va
 it’s frequency, w
 it’s 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-meansquare (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.
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Introduction, Page 1.1-5
Introduction to Electronics
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
v
V
t
V
Fundamental (base)
frequency
T

It can be expressed as the following infinite Fourier series
v t  

4V 
1
1





sin
w
t

sin
3
w
t

sin 5w0t   

0
0
p 
3
5

w0 
2p
T
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
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Introduction, Page 1.1-6
Introduction to Electronics
The Frequency Spectrum

The sinusoidal components of the square-wave signal constitute the frequency
spectrum of the this particular signal as shown below
4V
p
1 4V

3 p
w0

3w0
1 4V

5 p
5w0
1 4V

7 p

7w0
w (rad/s)
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.
time
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frequency
Introduction, Page 1.1-7
Introduction to Electronics
Time domain vs. Frequency domain

The spectrum of audible sounds such as speech and music goes from roughly
20Hz to 20kHz (audio band)
1.5
1
f( t ) 0.5
0
0.5
2 10
6
1 10
6
0
1 10
6
2 10
6
t
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Introduction, Page 1.1-8
Introduction to Electronics
Exercises

1.1 - Find the frequencies f, and w of a sine-wave signal with a period 1 ms
1 1 cycle

 1000 Hz or 1000 cycles per second
T 0.001s
w  2p radians per cycle  f  2px103 rad/s
f 

1.2 - What is the period of sine waveforms characterized by frequencies of
 a) f = 60 Hz
T
1
1 cycle

 0.0167 s or 16.7 ms
f 60 cycles per second (Hz)
 b) f = 10-3 Hz
T
1
1 cycle

 1000 s
f 0.001 cycles per second (Hz)
 c) f = 1 MHz
T
1
1 cycle
 6
 106 s or 1s
f 10 cycles per second (Hz)
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Introduction, Page 1.1-9
Introduction to Electronics
Exercise 1.3

When a symmetrical square-wave signal whose Fourier series is given by the following
equation
v t  
where
4V
p
4V 
1
1

 sin w0t   sin 3w0t   sin 5w0t   
p 
3
5

is the peak value of the fundamenta l, the RMS value is
4V
2p
is applied to a resistor the total power dissipated may be calculated directly from
the frequency spectrum using the relationship


1 T  v2 
P    dt
T 0R
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 wave’s power is in its fundamental?
2
  4V  
16V 2




1 T V 
1V T V
2
T
1
8V 2

2
p


2
p
P    dt 

P1   
dt  R  p 2 R
T 0 R
T R 1 R
T 0
R


8


Fraction of power in the first harmonic is 2  0.81 or 81%
p
2

2
2
What fraction of a square wave’s power is in its first five harmonics (through 5 w0)?
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Introduction, Page 1.1-10
Introduction to Electronics
Analog vs. Digital Signals
Analog
Continuous time-varying signal
time
v
Digital - only two discrete levels
Analog
Discrete time-varying signal
 5V
0V
t
1
0
1
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1
0
0
1
time
0
Introduction, Page 1.1-11
Introduction to Electronics


Binary (base 2) numbers, powers of 2
20=1, 21=2, 22=4, 23=8, 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 8’s place, 4’s place, 2’s place and 1’s place
Binary numbers are listed from MSB to LSB
MSB - Most significant bit (8’s place) and LSB - least significant bit (1’s 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
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Introduction, Page 1.1-12
Introduction to Electronics
Analog to Digital Conversion
Analog
Digital
Continuously
varying signal
1 volt resolution
time
Discrete voltage varying signal
15
14
13
12
11
10
9
8
7
6
5
4
3
2
1
0
Discrete time steps as well
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Introduction, Page 1.1-13
Introduction to Electronics
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
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Introduction, Page 1.1-14
Introduction to Electronics
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
Voltage gain (Av ) =
iI
vI(t)
iO
RL
vO
vO
vI
Av
+
vO(t)
1
0
vI
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Introduction, Page 1.1-15
Introduction to Electronics
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
Power gain (Ap ) =
Load Power PL
Input Power PI
Current gain (Ai ) =
vO i O
vI i I
iO
iI
Ap = Av Ai
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Introduction, Page 1.1-16
Introduction to Electronics
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
Voltage gain in decibels = 20 log Av
Current gain in decibels = 20 log Ai
Recall that P=V2/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
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Introduction, Page 1.1-17
Introduction to Electronics
Amplifier Efficiency and Power Supplies
iI
vI(t)
i1
i2
V1
V+
iO
V+ RL
-V2
+
vO(t)
-
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
Pdc + PI
Some power is consumed by the load and
some is dissapated by the amplifier circuit.
The Efficiency is defined as
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Pdc + PI = PL + Pdissapated
PL x 100
h=
Pdc
Introduction, Page 1.1-18
Introduction to Electronics
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
9
 9V
or Av  20 log 9  19.1dB
V
1
9V
Iˆo 
 9mA
1kW
Iˆ
9
Ai  o 
 90 A
or Ai  20 log 90  39.1dB
A
ˆI i 0.1
Av 
9V 9mA
PL  VoRMS I oRMS 
 40.5mW
2 2
1V 0.1mA
PI  ViRMS I iRMS 
 0.05mW
2
2
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Ap 
PL 40.5mW

 810W
W
PI 0.05mW
Ap  10 log 810  29.1dB
PDC  10Vx9.5mA  10Vx9.5mA  190mW
Pdissapated  PDC  PI  PL  190mW  0.05mW  40.5mW
Pdissapated  149.6mW
The efficiency is then given by
h   PP x100  21.3%
L
DC
Introduction, Page 1.1-19
Introduction to Electronics
Amplifier Transfer Characteristic, Linear to Saturation


L+ is the positive output
saturation level
L- is the negative output
saturation level
vO
Output peaks are
clipped or distorted
L+
Undistorted
Output
Range
vI
0
LThe steeper the slope
of the transfer characteristic
the higher the gain
LAv
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Safe input swing range
L+
Av
Introduction, Page 1.1-20
Introduction to Electronics
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
vO t 
L+
Slope=Av
VO
Q
vI t   VI  vi t 
vO t   VO  vo t 
vo t 
vo t   Avi t 
L-
VI
vi t 
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vI t 
Av =
dvO
dvI at Q
Introduction, Page 1.1-21
Introduction to Electronics
Biasing Example

A transistor amplifier has the following transfer characteristic
11 40v I
O
v  10  10



e
vI  0V and vO  0.3V
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
L  10  1011  10V
vO V 
10
Also find the value of the dc bias voltage VI that results in VO=5V and find
the voltage gain at the corresponding operating point.
 To bias the device so that VO=5V we use the equation and solve for vI .
5
 vI =0.673V
 We can find the gain by taking the derivative at vI =0.673V
dvO
dvI
 40 1011 e 400.673  196
0.3
vI V 
v I 0.673
 Note the negative gain which indicates that we have an inverting
amplifier
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0.673
0.69
Introduction, Page 1.1-22
Introduction to Electronics
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
iA t 

vC t 
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
IB

or
or
VE
Incremental or ac quantities are denoted by a lowercase symbol and a lower case
subscript.
ia t 
or
vc t 
Incremental ac value
ic
Ic
Peak ac value
iC
total
value
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IC
dc value
Introduction, Page 1.1-23
Introduction to Electronics
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.
AV  20 log 100   20 log 102   202   40dB
AI  20 log 1000   20 log 103   203  60dB
AP  10 log  AV AI   10 log 100 x1000   10 log 105   105  50dB

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?
PDC  V I   15 x 0.008  120mW
2
 12 


2
2
  18mW
PLoad  
1000
Pdissapated  120  18  102mW
h
PLoad
18
x100 
x100  15%
PDC
120
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Introduction, Page 1.1-24
Introduction to Electronics
Small-signal approximation Exercise

Lets look at the limitation of the small signal approximation. Consider amplifier with the
following transfer characteristic
vO V 
10

output
swing
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)
11 40v I
1) Let vI=VI+vi and vO=VO+vo=5+vo
O
thus,
v  10  10
5
Input
swing
5  vo  10  1011 e 40VI e40vi  10  1011 e 400.673e 40vi
5  vo  10  5e 40vi
0.3
vI V 
0.673
e
0.69
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vo  5  5e 40vi
using vi  1mV
vo  0.204
V
2) v  200
 vi   2000.001  0.2V
o
V 
Introduction, Page 1.1-25
Introduction to Electronics
Small-signal approximation Exercise continued


Repeat the calculations for input signals of 5mV and 10mV
5mV
 Using the approximation
V 
vo  200 vi   2000.005  1.0V
V 
 Using the full equation
vo  5  5e40vi
using vi  5mV
vo  1.107V

10mV
 Using the approximation
V 
vo  200 vi   2000.01  2.0V
V 
 Using the full equation
vo  5  5e
40vi
vo  2.459V
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using vi  10mV
The larger the signal
the larger the error
when using the approximation
Introduction, Page 1.1-26
Introduction to Electronics
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
Ro
+
vi



Ri
+- A v
vo i
+
vo
-
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
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Introduction, Page 1.1-27
Introduction to Electronics
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.
In order not to lose signal strength at the input,
Ri should be infinitely large or much, much
greater than Rs
vi  vs
Ri
Ri  Rs
vs
is
Rs
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Using voltage division
vo = Avovi
Ro
+
vi
-
Ri
+- A v
vo i
RL
RL + Ro
io
RL
+
vo
-
Introduction, Page 1.1-28
Introduction to Electronics
Comment on the Voltage Amplifier

The overall voltage gain is given by
vo
v v
Ri
RL
 Avo  o i 
vs
vi vs Ri  Rs RL  Ro

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).
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Introduction, Page 1.1-29
Introduction to Electronics
An Example of Cascaded Amplifier Stages
Signal
source
Stage 1
Stage 2
Stage 3
Load
ii
100kW
vi1
1MW
vs


1kW
1kW
vi2
10vi1
100kW
100vi2
10W
vi3
10kW
vi3
100W
1vi3
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
vi1
1MW

 0.909
vs 1MW  100kW
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Introduction, Page 1.1-30
Introduction to Electronics
Example continued
Signal
source
Stage 1
Stage 2
Stage 3
Load
ii
100kW
vi1
1MW
vs


1kW
1kW
vi2
10vi1
100kW
10W
vi3
10kW
vi3
100W
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
Av1 
vi 2
100kW
 10
 9.9V
V
vi1
100kW  1kW
v
100W
Av 3  L  1
 0.909V
V
vi 3
100W  10W
© REP 4/29/2017 EGRE224
Av 2 
Av 
vi 3
10kW
 100
 90.9V
V
vi 2
10kW  1kW
vL
 Av1 Av 2 Av 3  818V  58.3dB
V
vi1
Introduction, Page 1.1-31
Introduction to Electronics
Example continued
Signal
source
Stage 1
Stage 2
Stage 3
Load
ii
100kW
vi1
1MW
vs


1kW
1kW
vi2
vi3
100kW
10vi1
10W
100vi2
10kW
vi3
100W
1vi3
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
vL
v
 Av i1  8180.909 V  743.6
V
vs
vs
or 57.4dB
Find the current gain and the power gain
i
v 100W
Ai  o  L
 104 Av  8.18 x106 A
A
ii vi1 1MW
or 138.3dB

Ap  Av Ai  818V
Ap  6.69 x109 W
V
8.18x10
6
A
W
or 98.3dB
© REP 4/29/2017 EGRE224

A
Introduction, Page 1.1-32
Introduction to Electronics
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).
ii
Vout  Vin _ rms
1MW
Signal
source
(transducer)
vout
2
Vout

105 
_ rms
P

 1011W
10W
10
10W
vs

10W
10
 6
 105  10V rms
1MW  10W 10  10
2
If a unity gain buffer amplifier (AV=1) 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).
Vout  1
Buffer
ii
1MW
10W
vi1
Signal
vs
source
(transducer)
© REP 4/29/2017 EGRE224
1MW
vout
1vi1
10W
Ri
RL
Avo

Ri  Rs
RL  Ro
1MW
10W
1
1MW  1MW 10W  10W
 10.510.5  0.25V
Vout  1
Vout
Introduction, Page 1.1-33
Introduction to Electronics
Exercise Continued
2
Vo
0.252
PL 

6.25mW
RL
10
AV 
Vo 0.25
V

 0.25
Vs
1
V
20 log 0.25  12dB negative means attenuatio n
P
AP  L , Pi  viii
Pi
1MW
1V
 0.5V ; ii 
 0.5A
1MW  1MW
1MW  1MW
Pi  0.5 * 0.5  0.25W
vi  1
6.25mW
 25 * 103
0.25W
10 log AP  44dB
AP 
© REP 4/29/2017 EGRE224
Introduction, Page 1.1-34
Introduction to Electronics
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
Vout  Vamplifier

Rload
Rload  Routput
1  0.2  1
1kW
1kW  Routput
Rout 
1kW  800W
 250W
0.8
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
If
20 log  AV   40 then AV  10
2

RL 


A
v
vo i
2

RL  Ro 
v
2
PL  o  
 2.5vi
RL
RL
40
20
 100
V
V
2
2
v
v
Pi  i  i
Ri 10000
2
P
2.5vi
4W
Ap  L 

2
.
5
x
10
2
W
Pi
vi
10000
© REP 4/29/2017 EGRE224
10 log Ap  44dB
Introduction, Page 1.1-35
Introduction to Electronics
Other Amplifier Types
Current Amp
Voltage Amp
Ro
+
vi
-
Ri
io
+- A v
vo i
+
vo
-
Ri
Ro io
Ri
+- R i
m i
© REP 4/29/2017 EGRE224
Aisii
Ro
+
vo
-
Transconductance Amp
Transresistance Amp
ii
io
ii
io
+
vo
-
+
vi
-
Ri
Gmvi
Ro
+
vo
-
Introduction, Page 1.1-36
Introduction to Electronics
Voltage Amplifier

Voltage in and voltage out, V/V is unit-less
Voltage Amp
Open - Circuit Voltage Gain
Avo 
vo
vi
io  0
V
V
Ideal Characteri stics
Ri   Ro  0
Ro
+
vi
-
Ri
+- A v
vo i
io
+
vo
-
Open - circuit output vol tage is Avo vi
The open - circuit output vol tage of the
v
current amplifier is Aisii Ro and ii  i , so
Ri
Avo  Ais
Ro
Ri
© REP 4/29/2017 EGRE224
Introduction, Page 1.1-37
Introduction to Electronics
Current Amplifier

Current in and current out, I/I is unit-less
Short - Circuit Current Gain
Ais 
io
ii
vo 0
A
A
Ideal Characteri stics
Ri  0 Ro  
Current Amp
io
ii
Ri
Aisii
Ro
+
vo
-
Open - circuit output vol tage is for the
voltage amplifier is Avo vi
The open - circuit output vol tage of the
v
current amplifier is Aisii Ro and ii  i , so
Ri
Avo  Ais
Ro
Ri
© REP 4/29/2017 EGRE224
Introduction, Page 1.1-38
Introduction to Electronics
Transconductance Amplifier

Voltage in and current out, I/V=G
Transconductance Amp
Short - Circuit Transcondu ctance
Gm 
io
vi
vo 0
A
V
Ideal Characteri stics
Ri   Ro  
io
+
vi
-
Ri
Gmvi
Ro
+
vo
-
Avo  Gm Ro
© REP 4/29/2017 EGRE224
Introduction, Page 1.1-39
Introduction to Electronics
Transresistance Amplifier

Current in and a voltage out, V/I =R
Transresistance Amp
ii
Ro io
Open - Circuit Transresis tance
Rm 
vo
ii
io 0
V
A
Ri
+- R i
m i
Ideal Characteri stics
Ri  0 Ro  0
Avo 
© REP 4/29/2017 EGRE224
Rm
Ri
Introduction, Page 1.1-40
+
vo
-
Introduction to Electronics
Bipolar Junction Transistor Example

A bipolar junction transistor (BJT) is a three-terminal device with the following symbol.
C
iB
B
+
vBE


iC
+
vCE
iE
E
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.
vBE = VBE + vbe
iB = IB + ib
iE = IE + ie
iC = IC + ic
vCE = VCE + vce

The device itself may be represented by one of several equivalent circuit models
© REP 4/29/2017 EGRE224
Introduction, Page 1.1-41
Introduction to Electronics
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) .
ib =
vbe
rp
ib
+
vbe
-
B
rp
Current Amplifier
ic = b ib
C
iC
b ib
ie
ie = ib+ ic=(b + 1)ib
E
ib
B
rp
iC
+
vbe
-
C
gmvbe
gm =
ie
E
© REP 4/29/2017 EGRE224
ic = gmvbe
Trans-conductance Amplifier
b
rp
P model name comes from
the fact that the circuit
configuration looks like an
upside down greek letter p
Introduction, Page 1.1-42
Introduction to Electronics
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
© REP 4/29/2017 EGRE224
Introduction, Page 1.1-43
Introduction to Electronics
Low Frequency “T” Model for common base configuration

ib
iC
B
+
vbe
-
re
a ie
C
ie =
vbe
re
But what are a and re?
From the p model we can write
ic
=
ie
E
Therefore
+ v1 re
ic = a ie
ie=ib +ic therefore ib= (1-a )ie
ie
E
We can derive expressions that allow us to convert
from one model form to another
T
gmv1 iC
B
© REP 4/29/2017 EGRE224
C
b
b +1
b
a =
b +1
ie =
b +1
rp
re =
vbe
rp
b +1
The “T” model is useful in the
Common base configuration
Introduction, Page 1.1-44
Introduction to Electronics
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.
ix
ib
B
rp
vx
iC
+
vbe
-
Rin
G
© REP 4/29/2017 EGRE224
Using KCL at the emitter node we get
ie=ib +ic or
ic= (b+1 )ib= (b+1 )ix
C
b ib
ie
E


RL
Now using KVL around the input loop
we get
vx=vbe+vE=(ixrp)+ (b+1 )ixRe
or
vx/ix= Rin=rp + (b+1 )Re
Re

Note: We can’t “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.
Introduction, Page 1.1-45
Introduction to Electronics
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?
linear amplifier
+
vo  Vo sin w t  f 
vi  Vi sin w t



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.
T w  
© REP 4/29/2017 EGRE224
Vo
Vi
T w   f
Introduction, Page 1.1-46
Introduction to Electronics
Measuring Frequency Response with an Oscilloscope

Low Frequency
f
input
w= w1

output

Higher Frequency
w= w2
f
© REP 4/29/2017 EGRE224
input
output

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
Introduction, Page 1.1-47
Introduction to Electronics
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
T s  

K
R
vI
K
T w  
1  j  w 
 w0 


C
1   s 
 w0 
For physical frequencies, where s=jw, we get
Where K is the magnitude of the transfer function at w=0
(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
© REP 4/29/2017 EGRE224

vo

T s  
1
sC
R
T w  
1
sC

1
sRC  1
1
jw RC  1
Introduction, Page 1.1-48
Introduction to Electronics
Sample Calculations
Let R=100,000W and C equal 0.1F, t=RC is 0.01, w0=1/t=100 so the transfer function is
T w  
1

1 j w
100

Lets pick a frequency and calculate the magnitude of the response, LET w=1
T 1 
1
1

1  j 0.01
1 j 1
100


Plot the denominator as a complex number
+jw
-Re
The magnitude of the denominator is given by the length of the
complex vector. Using trigonometry we know that c2 =a2+b2
so that c is
c  a 2  b2  12  0.012  1
1

100 0.01
1
-jw
© REP 4/29/2017 EGRE224
+Re
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
1
f  tan
1100  0.57 degrees  0
Introduction, Page 1.1-49
Introduction to Electronics
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 w=10. Remember w0 is 100.
And w=100
+jw
T w  100  
+jw
100
1
100
1
1.41
T w  100   0.707
-Re
10
 0.1
100
T w  10 
1
1
c  12  0.12  1
1
+Re


fdenom  tan 1 10100 
c  12  12  2  1.41
+Re
1
-Re


fdenom  tan 1 100100  45 degrees
-jw
© REP 4/29/2017 EGRE224
-jw
fdenom  5.7 degrees  0
fT  5.7
fT  45
Introduction, Page 1.1-50
Introduction to Electronics
and at higher frequencies

Finally lets consider a frequency of w=1000. Remember w0 is 100.
-jw
1000
 10
100
T w  1000 
c  12  102  10
-Re
1
1
10
T w  1000  0.1 ~ 0
+Re


fdenom  tan 1 1000100  84.3 degrees  90
-jw

fT  90
What we are observing is this capacitor gradually changing its behavior from an open
circuit at low frequencies (w<<w0) to a short circuit at higher frequencies (w>>w0) which
shorts out the output response
R
vI
C

vo

© REP 4/29/2017 EGRE224
Introduction, Page 1.1-51
Introduction to Electronics
The General Expression (Low Pass)
Thus the low pass magnitude response in general is given by
T w  
K
1   w 
 w0 
2
The general low pass phase response is given by
f w    tan 1 w w 

© REP 4/29/2017 EGRE224
0

Introduction, Page 1.1-52
Introduction to Electronics
Plots of the Frequency Response
w0
Magnitude
1
0
w (log scale)
1
10
100
1k
Radian Frequency
1
10
100
1k
w (log scale)
Phase Angle
5 .7 
of the
 45
transfer function
 90
© REP 4/29/2017 EGRE224
5 .7 
f w 
Introduction, Page 1.1-53
Introduction to Electronics
Bode (Bo-duh) Plots of the Frequency Response (Low Pass)
T  jw 
20 log
K
dB  Normalized Magnitude

0
3 dB
-10
Slope of
 6 dB
octave
 20 dB
decade
-20
w
-30 0.1
f w 

1
w0
20log (0.707)  3dB
or
(log scale)
10
A decade is a factor of 10 and
an octave is a doubling
Note

and
20log (2)  6dB
20log (10)  20dB
Normalized Frequency
0.1
5 .7 
 45
1
10
w
w0 (log scale)
 45 / decade
5 .7 
 90
© REP 4/29/2017 EGRE224
Introduction, Page 1.1-54
Introduction to Electronics
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
T s   K



R

jw
T w   K
jw  w 0
1
1
vI

vo
s
s  w0
For physical frequencies, where s=jw, we get
T w   K
C
w0
jw
T w   K
1
w 
1  j 0 
w 
T s  
R
R
1
sC

sRC
sRC  1
T s  
s
s  RC
jw
T w  
jw  RC
Where K is the magnitude of the transfer function at w=0
(dc) and w0 is defined as being 1/t .
Tau is the time constant, which in this case is 1/RC.
© REP 4/29/2017 EGRE224
Introduction, Page 1.1-55
Introduction to Electronics
The General Expression (High Pass)
Thus the high pass magnitude response in general is given by
T w  
K
w
1   0 
 w
2
The general high pass phase response is given by
f w   tan 1 w 0 w 


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
C
vI
R

vo

© REP 4/29/2017 EGRE224
Introduction, Page 1.1-56
Introduction to Electronics
Bode (Bo-duh) Plots of the Frequency Response (High Pass)
20 log
Slope of
 6 dB
octave
 20 dB
decade
T  jw 
K
0
dB  Normalized Magnitude
3 dB
20log (0.707)  3dB
or
-10
w
-20
5.7 f w 
-30 0.1
90
1
10
Normalized Frequency
20log (2)  6dB
20log (10)  20dB
slope
 45 / decade
45
5 .7 
0
0.1
w0 (log scale)
1
© REP 4/29/2017 EGRE224
w
w0 (log scale)
10
Introduction, Page 1.1-57
Introduction to Electronics
Bode Plots of Factored Transfer Functions
z2
z1 




s 1  s 
100
,
000
 K @ s  jw  0  20 log 150  43.5dB


T s   150 
s 
s 
s


 1 
1 
 
1 
  100  p 1,000  1,000,000  
z1
z2
p1
p1
p3
2


1,0001  1,000 


100,000 


  15,000
T s  jw  1000   150
  1,000  1,000 
1,000  
 1 

 1 
 1 
100  1,000  1,000,000  

T(jw) in dB
I expected everyone to
do the problem graphically
and by calculation
K @ s  jw  1,000  20 log 15,000  83.5dB
p2
83.5
80
43.5
40
p3
dB
dec.
43.5
dB
slope  20
dec.
1
z1
© REP 4/29/2017 EGRE224
slope  20
slope  20
dB
dec.
10 102 103 104 105 106 107 108
z2 p3
p1 p2
Introduction, Page 1.1-58
Introduction to Electronics
Example 1.5

The circuit below shows a voltage amplifier having an input resistance Ri, an input
capacitance Ci, a gain factor , 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.
Rs
Vs

Ro
+
Vi
-
Ri
Ci
+-
Vi
RL
+
Vo
-
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.
© REP 4/29/2017 EGRE224
Introduction, Page 1.1-59
Introduction to Electronics
Example 1.5 cont’d
Using the voltage divider rule:
V V
i
Z
s
Z R
i
Thus, V
V
i
s
,
i

s
1  ( Rs
Y
i
 1
Z
i
i
V
V
1
Ri)  s C i Rs
At the output side of the amplifier:
Amplifier transfer function:
© REP 4/29/2017 EGRE224
1
V V
V
V
o
s

V
o
s
1  Rs Y i

i
s
1
1  Rs [(1
R )  sC ]
i
1
i
1
R ) 1  s C [( R R ) /( R  R )]
Low pass STC network
1  ( Rs
 V i
1
1  ( Rs
V s
i
R
s
i
i
s
i
L
RL  Ro
1
1
R ) 1  ( R R ) 1  s C [ R R ) /( R  R )
i
o
L
i
s
i
s
i
Introduction, Page 1.1-60
Introduction to Electronics
Example 1.5 cont’d

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.
t  C RR
s
i
R R
s
V
V
o
s

 C i ( Rs // Ri )
i
i
1
1  ( Rs
1
1
(from previous page)
R ) 1  ( R R ) 1  s C [ R R ) /( R  R )
i
dc gain:
o
K V
V
3-dB frequency:
L
o
i
( s  0)  
s
s
i
i
1
1  ( Rs
1
1
R ) 1  (R R
i
o
L
1
w  t  C ( // )
R R
o
i
© REP 4/29/2017 EGRE224
s
s
i
Introduction, Page 1.1-61
Introduction to Electronics
Example 1.5 cont’d

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.
K V
V
o
( s  0)  
s
1
1  ( Rs
1
R ) 1  (R R
i
o
L
K  144 1  (201 100) 1  (2001 1000  100 V/V
1
1
w  t  C ( // )
R R
o
i
fo 

s
i

60  10
12
 40 dB
1
 10 6 rad/s
3
 (20  100 /( 20  100))  10
10 6
 159.2kHz
2p
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
© REP 4/29/2017 EGRE224
Introduction, Page 1.1-62
Introduction to Electronics
Example 1.5 cont’d

Find vo for vi = 0.1sin102t,V
T ( jw )  V o ( jw ) 
V
T  100
s
f   tan 1 10 4  0
v (t )  10 sin 10
o

100
1  j (w 10 6 )
2
tV
Find vo for vi = 0.1sin105t,V
T  99.5
f   tan 1 0.1  5.7
v (t )  9.95sin( 10 t  5.7) V
5
o

Find vo for vi = 0.1sin106t,V
T  100
2  70.7
f  45
v (t )  7.07 sin( 10 t  45) V
6
o
© REP 4/29/2017 EGRE224
Introduction, Page 1.1-63
Introduction to Electronics
Classification of Amplifiers based on Frequency Response

Capacitively coupled amplifier
decibels (dB)
20 log T w 
mid-band frequencies
bandwidth

Directly coupled amplifier
20 log T w 
wL
wH
w
decibels (dB)
bandwidth

wH
Tuned or bandpass amplifier
20 log T w 
w
decibels (dB)
bandwidth
wC
© REP 4/29/2017 EGRE224
w
Introduction, Page 1.1-64
Introduction to Electronics
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 (0.1F - 50F 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 midband 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.
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Introduction, Page 1.1-65
Introduction to Electronics
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
© REP 4/29/2017 EGRE224
Introduction, Page 1.1-66
Introduction to Electronics
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 nonlinearity 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).
© REP 4/29/2017 EGRE224
Introduction, Page 1.1-67
Introduction to Electronics
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
HIGH, “1”
5V
Ideal Inverter
BJT
CommonEmitter
Inverter
4
3
GAIN=SLOPE
2
LOW, “0”
0V
CMOS Inverter
1
0
0
LOW, “0”
0V
© REP 4/29/2017 EGRE224
1
2
3
VIN (Volts)
4
5
HIGH, “1”
5V
Introduction, Page 1.1-68
Introduction to Electronics
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)
Source
Drain
Drain
Gate
Gate
n
p substrate
Substrate
n
Source
n-channel
Source - where electrons come from (-)
Drain - where electrons flow to (+)
IDS
Channel is enhanced (resistive)

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)
-VGS
Channel is off
+VGS
0
Source
Gate
Drain
Drain
Gate
n
n
p substrate
IDS
n-channel
Substrate
Source
Source - where electrons come from (-)
Drain - where electrons flow to (+)
Channel is made stronger
Channel is depleted (Off)
-VGS
© REP 4/29/2017 EGRE224
0
+VGS
Introduction, Page 1.1-69
Introduction to Electronics
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)
Source
Drain
Drain
Gate
Gate
p
Substrate
p
Source
p-channel
n substrate
0
-VGS
Source - where holes come from (+)
Drain - where holes flow to (-)
+VGS
Channel is enhanced

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)
IDS
Source
Drain
Drain
Gate
Gate
p
p
p-channel
n substrate
-VGS
Substrate
Source
Source - where holes come from (+)
Drain - where holes flow to (-)
0
+VGS
Channel is depleted
IDS
© REP 4/29/2017 EGRE224
Introduction, Page 1.1-70
Introduction to Electronics
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
Vin=0V
Logic 0
Vout=5V
Logic 1
open
© REP 4/29/2017 EGRE224
Vin=5V
Logic 1
Vout ~ = 0.5 Volts
Logic 0
closed
say Ron = 1k Ohms
Introduction, Page 1.1-71
Introduction to Electronics
The Complementary MOS (CMOS) Inverter


Complementary means both nMOS and pMOS transistors are used
A polite “tug O’War”! 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
+5 V
gate
Vgsp= Vgp-Vsp
= 5-5 = 0
source of holes
p channel (n well)
drain for holes
Vin
+5 V
g=5
s=5
open
Vin = + 5V
Vout = 0 V
Vout
gate
drain for electrons
n channel (p wafer)
source of electrons
closed
g=5
s=0
Vgsn= Vgn-Vsn
= 5-0 = 5
+5 V
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
© REP 4/29/2017 EGRE224
Vgsp= Vgp-Vsp
= 0-5 = -5
g=0
s=5
closed
Vin = 0
Vgsn= Vgn-Vsn
= 0-0 = 0
Vout = + 5V
g=0
open
s=0
Introduction, Page 1.1-72
Introduction to Electronics
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 < Vout < 5V
and

let a low be
0V < Vout < 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”.
© REP 4/29/2017 EGRE224
Introduction, Page 1.1-73
Introduction to Electronics
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
VOH
VOUT
4
3
(Volts)
2
VOL
|slope| = 1
5
|slope| = Maximum = 5
|slope| = 1
1
0
|gain|
slope
5
1
© REP 4/29/2017 EGRE224
1
VIL
2
3
VIH
4
5
VIN (Volts)
max
VIN (Volts)
Introduction, Page 1.1-74
Introduction to Electronics
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
What the output
produces
High VOH =
(5- 4.9) V
“solid”
signals
Low VOL =
(0.7- 0) V
© REP 4/29/2017 EGRE224
What the input
accepts
solid high
VIH = 3V
and up
VIL = 1.2V
solid low
+5V
Marginal High
NMH = (5-3)-(5-4.9)
=1.9V
Marginal Low
NML = 1.2-9.7=0.5V
0V
Introduction, Page 1.1-75
Introduction to Electronics
How does an Inverter (with gain) restore
a “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
VOH
input to the second
inverter the output of the
second inverter will be 5
volts which is a solid
VOUT
high, and is much
improved over the 3.1
volts we saw on the
(Volts)
original input signal.
The inverter supplies (+5
and ground) and the gain
VOL
drive illegal and marginal
signals towards solid
Output of
levels
inverter # 1
= 0.65V
© REP 4/29/2017 EGRE224
5
0.65
3.1
5
4
3
2
|slope| = 1
1
0
2
1
VIL
Input to
INV #2
3
VIH
4
VIN (Volts)
5
Input to
inverter # 1
= 3.1V
Introduction, Page 1.1-76
Introduction to Electronics
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.
Partially charged
+
capacitor
Electric field


+
+
-
-
Fully charged
capacitor
Stronger
electric
field
++++++++
--------
The Electric Field (Volts per cm) is proportional to the voltage across the capacitor’s
terminal.
The charge stored is proportional to the voltage (and the area), and the proportionality
constant is called the capacitance
 Q=CV
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
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Introduction, Page 1.1-77
Introduction to Electronics
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
VC
t=0
V0 =
10V
R
+
-
C
i
VC = 0
10V
90%
6.3V
V = V0 (1 - e-t/t)
RC
 Discharging a capacitor
2.3RC
time
VC
t=0
10V
R
C
i
VC = 10
3.6V
V = V0(e-t/t)
10%
RC



2.3RC
time
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 = 0.37~37%
e-2 = 0.135~14% e-3= 0.05~5%
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Introduction, Page 1.1-78
Introduction to Electronics
Propagation Delay and Rise and Fall Times
of a signal
vI
Input Signal
VOH
90% of (VOH-VOL)
50% 1/2(VOL+VOH)
10% of (VOH-VOL)
VOL
tr
vO
tf
tPHL
Time
tPLH
VOH
VOL
tTHL
Output Signal
© REP 4/29/2017 EGRE224
tTLH
Time
Introduction, Page 1.1-79