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EEE 2302: ANALOG ELECTRONICS III
SIGNAL CONDITIONING
Transducers link up cct to the physical world
...

ii)

Amplification

Transducer producer lower level of voltage level
iii)
iv)
v)
vi)

Conversion CV- conversion
Resistance- voltage conversion
Level translation
Filtering stripping off unwanted frequencies

How do you do signal conditioning? Which device can you use to attain this?
-

By operation amplifier

Precision- differential amplifier (instrumentation)
𝑢0 = 𝑢2 − 𝑢1
1

𝑅4
2𝑅
1+
𝑅2
𝑅1

Advantages: Additional number of features
i)
ii)
iii)

High input impedance
High gain accuracy than use of sing op-amp
High CMRR

Suppose the voltage
𝑢 𝑑 = 2𝑉

i)

𝑢1 = −1
𝑢2 = +1

ii)

𝑢1 = 1
𝑢 𝑑 = 2𝑉
𝑢2 = 3
High gain stability with low temperature coefficient
Low dc off set
Low output impedance

iii)
iv)
v)

One application
Strain Gauge Application

Balanced
𝑅2

Change in small resistance will be measured/translated into 𝑣0
Saying 𝐵1 + 𝐵2 = 1 +∝1 +∝4 making this assumption

2

𝑅3 =

𝑅1

𝑅4

Hence overall equation
−∝2 𝑢2 −∝ 𝑢4 + 𝐵1 𝑢1 + 𝐵2 𝑢2
0
...
7 𝑢2 + 100 𝑢3 −∝ 𝑢4
1 +∝1 +∝4 & 𝐵1 + 𝐵2
= 100
...
7 + 10 = 11
...
1 − 11
...
4

𝐵1 + 𝐵2 𝑔𝑟𝑒𝑎𝑡𝑒𝑟 𝑡𝑕𝑎𝑛 1 +∝2 +∝4

NB: So introduce to 1 +∝2 +∝4 +∝0 & ∝0 = 88
...

But ∝0 should not be connected to any input (because it should not contribute to any output)
with 𝑢0 = 0

Therefore from equation (x) above
𝛽𝑢1 + 𝛽2 𝑢3
= 𝛽1 𝑢1 + 𝛽2 𝑢3
𝛽1 + 𝛽2

1 +∝1 +∝4 +∝0
-

We purposely select ∝0 to get the cancellation
If 1 +∝1 +∝4 is > than 𝛽1 + 𝛽2 put a resistance to ground from positive terminal
...


Simulations of differential amplifiers
𝑑2 𝑢0
𝑑 𝑡2

Historically amplifier have been used as Analog computers in simulation of different
equation
Today also they can be used because op-amps are available as chip components and
therefore this easy to simulate
...

This cct is able to simulate a 2nd order different equation
Integrators and differentiators

4

Already we have seen how to multiply a voltage by a constant (amplifier) add voltage,
subtracting
𝑢0=−𝑖𝑅

Other leaner mathematical operations are;
Integration

capacitor or indictor

= −𝑅𝐶

Differentiation

𝑖=

𝑐𝑑𝑢
− 𝐵𝑎𝑠𝑖𝑐 𝑝𝑟𝑜𝑐𝑒𝑠𝑠 𝑑𝑖𝑓𝑓𝑒𝑟𝑒𝑛𝑡𝑖𝑎𝑡𝑜𝑟
𝑑𝑡

Hence through capacitor is
𝑢𝑐 =

1
𝐶𝑅

𝑐=

𝑢1

𝑅1

𝑢1 𝑑𝑡

𝑢0 = −𝑢 𝑐
=

−1
𝐶𝑅

𝑢1 𝑑1

Meaning in transfer function
From time domain analysis to frequency domain analysis

5

𝑑𝑢1
𝑑𝑡

𝑢0 =
𝑢0

−𝑢1

𝑆𝑅𝐶

-

−1
𝑢 𝑖 = 𝑆𝑅𝐶

Transfer 𝑓𝑥𝑛
1 means integration
𝑠

Differentiation
𝑢0 =

𝑅𝐶𝑑𝑢 𝑖

𝑑𝑐

S= differentiation

= −𝑆𝐶𝑅
i
...

Offset voltage can also drive the integration onto
saturation

Differentiation
𝑢0 = −𝑆𝐶𝑅 = −𝑗𝑤𝐶𝑅
Gain for dc w=0
Advantage of biasing op in differentiation against that of integration
...


- At high frequency the integrator eliminates all the signal
...


-

Integrator trouble (quadrature are implemented using integrator than differentiator
...


Intelligent design & assignment
Question: design a summing amplified using only one op-amp to give an output
𝑢0 = 0
...
7𝑢 𝑖2 + 100𝑢 𝑖3 − 10𝑢 𝑖4

7

Negative coefficient- design inverting amplifier -0
...
7
∝4 = 4
If 𝑅 = 100𝑘
𝑅

100
∝2 =
0
...
They must always be positive
Analysis using sinusoidal response
Consider 𝑢01

𝑢02

𝑎𝑛𝑑

𝑢03

𝑢01 also appear at input hence
𝑢03 =

−𝑢01
𝑆𝐶1 𝑅1

𝑢02 =

−𝑢01
𝑆 2 𝐺𝐶2 𝑅1 𝑅2

−∝0
𝑅𝐶

𝑑 𝑢0
𝑑𝑡

2𝑛𝑑 𝑡𝑖𝑚𝑒 𝑖𝑛𝑡𝑒𝑔𝑟𝑎𝑙

+ 𝑢01 1 −∝01 =

−∝− 𝑢 01
𝑆𝐶1 𝑅1

+ (1−∝)𝑢1 Multiply by a factor of 2 For non-inverting op-

amp gain
Therefore 𝑢01 = 2 1−∝ 𝑢1 − 2

∝𝑢 01

Therefore 𝑢01 = 2 1−∝ 𝑢1 − 2

∝𝑢 01

2 1−∝ 𝑢1 = 𝑢1 1 +

𝑆𝐶1 𝑅1

𝑆𝐶1 𝑅1

− 1+ 𝑅 𝑅 =2
−

𝑢 01
𝑆 2 𝐺𝐶2 𝑅1 𝑅2

2∝
1
+ 2
𝑆𝐺𝑅1
𝑆 𝐺𝐶2 𝑅1 𝑅2

𝑢01
2 1−∝ 𝑆 2 𝐶1 𝐶2 𝑅1 𝑅2
=
𝑢1
𝑆2𝐺𝐶2 𝑅1 𝑅2 + 2 ∝ 𝑆𝐶2 𝑅2 + 1
This transfer 𝑓𝑥𝑛 is called 𝐻𝑃𝐹 transfer 𝑓𝑥𝑛
At high frequency 𝑤 𝑠 2 dominates output = low frequency 𝑤 = 0

𝑢01

𝑢1 = 0

Hence the output is Hp= 𝑢01

Integrating = =

−1

𝑑𝑢

𝑅𝐶

𝑑𝑡

=

−𝑢 01
𝑆𝐶1 𝑅1

𝑢01 =

2(1−∝)𝑆2𝐺 𝐶2 𝑅1 𝑅2
𝑆2𝐺 𝐶2 𝑅1 𝑅2 +2∝𝑆𝐺 𝐶2 𝑅1 𝑅2

9

𝑢03 =

−2(1−∝)𝑆 2 𝐶1 𝐶2 𝑅1 𝑅2

1
𝑆𝐺𝑅1

𝑆2

𝑠 = 𝑗𝑤

== −2 1 ∝ 𝑆𝐶 𝑠 𝑅2
= BPF- transfers 𝑓𝑥𝑛

2
At 𝑤0 =

1
𝐺𝐶2

𝑅1 𝑅2

𝑢03 = 0

𝑤=0

𝐶1 𝐶2 𝑅1 𝑅2 +2∝2 𝐶2 𝑅2 +1

𝑢03 = 0

𝑤=∞

=𝑅𝑒𝑎𝑠𝑛𝑎𝑛𝑡 𝑓𝑟𝑒𝑞𝑢𝑒𝑛𝑐𝑦

Once again integrate 𝑢02 =

−2(1−∝)𝑆 2 𝑅2 𝑅2
𝑆 2 𝑅2 𝐶 2

𝑆 2 𝐶1 𝐶2 𝑅1 𝑅2 +2∝2 𝐶2 𝑅2 +1

𝑠 = 𝑗𝑤
𝑤=0
𝑤=∞

𝑢02 = 𝑖𝑛𝑓𝑖𝑛𝑖𝑡𝑦
𝑢02 = 0

Hpf- transfer 𝑓𝑥𝑛
-

These three outputs are simultaneously available in this cct hence its called universal
active filter
...


-

Poles of the system are – lie on the negative side hence the systems is stable filter 𝑓𝑥𝑛

-

One can cascade such filter block one can get higher order filter block
...

Pole lie on LH- plane on s-plane – meaning
Zeros can lie anywhere
1
2∝

𝑄=

𝑤0 =

1
𝑅𝐶

Response of the transfer functions above: also called pole-frequency
LPF
𝑢0
𝑢1

=

𝑠2

2(1−∝)
+1 𝑤 𝑄 +1
0
𝑤2

-

0

Poles are given by denominator
1
1
𝑄=
=
2∝ 2∝
𝑤=

1
𝑅𝐶

HPF
2
2(1−∝) 𝑠

𝑢01

𝑢1 = 𝑠 2

2
𝑤0

2
𝑤0

+ 𝑠 𝑤 𝑄+1
0

BP
𝑢03

𝑢01

=

−2(1−∝) 𝑠 𝑤0
𝑠2 2 + 𝑠
𝑤0 𝑄 + 1
𝑤0

Denominator pole- frequency
𝑠1 1,2
𝑤

𝑠

− Normalized frequency- the denominator forms a quadratic equation

−1
1
𝑠1 1,2
4 ± 𝑄2 − 4
=
𝑤
2

𝑤= 𝑥

𝑥2 + 𝑥 𝑄 + 1
𝐴𝑥 2 + 𝐵𝑋 + 𝐶

Will be read if 1 𝑄 > 4 > 2 or 1 𝑄 > 4 >

11

1
2

𝑥=

−𝑏 ±

𝑏2 − 4𝑎𝑐
2𝑎

Poles will be read – this kind of circuit is of no interest
It will be complex conjugate if 1 𝑄 < 4 < 2 𝑜𝑟 𝑄 > 1 2
=

−1
±
𝑄

1

4− 2
𝑄
2

complex conjugate pair if 𝑄 > 1 𝑍

Poles will be
−1
2𝑄

1

Complex conjugate pair gotten

± 𝑗 1 − 4𝑄2

=

Real image part
Actual poles 𝑆1 𝑆2 −

−𝑤 0
2𝑄

1

± 𝑗 1 − 4𝑄2

𝑆1

𝑤

𝑆2

𝑤0

𝑤0

𝑤0 - is called pole frequency and Q is called pole Q
-

As long as the negative real part always

-

Has a negative real part always as along the Q is positive

-

Complex pair of poles in 2

S-Domain Plot
Poles
-

O angle – tells how close is the pole
to imaginary axis

tan 𝜃 =

1 − 1 4𝑄2
1
2𝑄

= 2𝑄 1 −

1
4𝑄2

Θ is directly proposal to Q
...
e
...
And the high pole Q- means high quality factor ( Qquality factor)
-

1

If Q is infinity (when ∝= 0 since 𝑄 = 2∝, meaning there is no real part the poles fall on
imaginary axis
...
e
...
e
...


𝑑2 𝑢0
𝑑 𝑡2

+ 𝛿𝑢0

-

This is second order system of harmonic oscillator

-

When 𝑑𝑢 𝑑𝑡 term is not present this is Harmonic Oscillator
It can by itself without input the output will be close to resonant frequency

-

1−∝ 𝑅 still can exist as R and 𝑢 𝑖 can still exist, but as an oscillator 𝑢1 can as well be
connected to ground

13

𝑑 2 𝑢0
= 𝛿𝑢0
𝑑𝑡 2
-

This cct is famous as Harmonic Oscillator/ Quadrature oscillator analog computer was
used to produce sinusoidal and cosine waves because of phase shift of 90 0 given by
integrator
...


-

This equation always gives poles on the lit side of s-plane
...

A filter 𝑤0 - pole frequency and Q- pole Q

14

𝑤0 =

1

1

𝑅𝐶

𝑄 = 2∝ or 𝑄 > 1 2 to make complex conjugate

Zero location
LPF

2(1−∝) no zeros all are at infinity

HPF

2(1−∝) 𝛺 𝑤0 are two zeros at 𝑆 = 𝑗𝑤0 = 0

BPF

2(1 − 𝑆) 𝑆 𝑤0 one zero at 𝑠 = 𝑗𝑤 = 0

± 𝑗𝑤0 = 0

LP characteristics plot

-

Peaking can be reduced for 𝑄 =
worth filters when 𝑄 =

1
2

1
2

to give maximum flatness hence (MFM)/ Butter

maximally flat magnitude (butter worth filters)

HPF characteristics plot
2(1−∝) 𝑆

HP=

𝑆2

𝑤0 +

𝑆

2
2
𝑤0

𝑤 0 𝑄−1

𝑠 = 𝑗𝑤

𝑤→∞

-

Bandpass filters are frequency selective
...


17

𝐿𝑃 =

𝑢03

𝑢1 = 𝑆 2

1(1−∝)
𝑆
2+
𝑤0 𝑄 + 1
𝑤0

NB: MFM bitter worth says 𝑄 = 1

∝=

2

=1

𝑢0

2∝

2
1
=
2
2

1−

=

1+

𝑤4

𝑤0 = 2𝜋𝑓 = 2𝜋 × 10 = 1 𝑅𝐶
3

Take one value
=

𝑅 = 10𝐾

𝑤0

2
2
𝑤0

𝑤2
+ 2 2
𝑤0

2(1−∝)
2𝑤 2
4 −
2
2
𝑤0
𝑤0 + 2𝑤

1

2

2
𝑤0

2(1−∝)
1
1+ 𝑤

𝐶 = 1𝑠𝑛𝑓

2(1−∝)

𝑢1 =

4
𝑤0

Is an maximal flat magnitude in a polynomial of
𝑤2
2 only the highest polynomials
...
Solution
Define a notch fix- output goes to zero at zone frequency

18

-

Can only be done with zeros
...

Denominator does not change because the loop does
not change

-

𝑓𝑖𝑥𝑒𝑑 𝑙𝑜𝑜𝑝
𝑆2

2
𝑤0

𝑝𝑜𝑙𝑒𝑠 𝑎𝑟𝑒 𝑐𝑜𝑛𝑠𝑡𝑎𝑛𝑡

+ 𝑆 𝑊𝑄 + 1

Because they are characteristics of feedback
...
g
...

All pass used to phase shift without attenuation or with the initial attention presented bit each
frequency is phase shifted
...

S- domain

20

Magnit of ΔPF is unit always of (K)
Phase shift is offered by poles and zero or both
−𝑤 2
𝑆 = 𝑗𝑤

−𝑢2

2
𝑤0

2
𝑤0

+

−

𝑗𝑤

𝑗𝑤

𝑤0 𝑄
𝑊𝑄 + 1

Phase shift variation
𝐺𝑎𝑖𝑛 𝐴 𝑟 =

Syntheses using KHN (state variable filter)

21

1
𝑗

For

∆𝑃𝐹
𝐴𝑑𝑑

∝ 𝑃
2
2(1−∝) 𝑆
2
2(1−∝) 𝑆

𝐵𝑃 =

2
𝑤0

𝑆2

+ 𝑆 𝑊𝑄 + 1

𝑤 = 1 𝑅𝐶 = 2𝜋 × 103
𝑄 = 1 2 ∝ = 100
∝= 1 200
𝑅 = 104
𝐶 = 15 𝑛𝑓

Because of inversion
𝑆2

2
𝑤0

𝑛 𝑆 𝑤 𝑄+1
0

2
𝑤0

+1 𝑤 𝑄+1
0

−2 1−∝
𝑆2

2
𝑤0

+3 𝑤 𝑄+1
0

To get 𝑆 𝑤0 𝑄 from BP we need a
modified coefficient to obtain 1 𝑄
Where;

−𝑠 1−∝

− 𝑆 𝑊 𝑄+1
0

=

2
𝑤0

𝐻𝑃 =
−𝑆 2

𝑊2
0

22

N can be made to be 𝑛 = 1 𝑄 hence this can be

𝑢04

𝑢 𝑖 = −2(1−∝)

OSCILLATORS
Inductor Simulation (Also called Gyrator (rotation))

Important for inductor simulation – important circuit (part of active filter) that is
...

In micro-miniaturization inductor becomes a very component to deal with (i
...
the coil)
...
So all the circuits to be made out
of monolithic ICC require inductor be not exist
...
How
is it possible
...

Meaning a capacitor has a certain relationship with voltage and current
...

Gmesdendinductor 𝐿 = 𝑆𝐶𝑅2 (simulated L)
Use of this in tank network – it stores energy
...

Total admittance
(summance of all admittance)
Re-write using state variable felt to KHN
network
1
𝜔𝑜 =
𝐿𝐶
𝑅𝑠
𝜔𝑜 𝜃 =
𝐿
𝑅𝑠
𝜃=
𝜔𝑜 𝐿

24

𝐿 = 𝑆𝐶𝑅2
1
𝐿𝐶
1
𝐿= 2
𝜔𝑜 𝐶
𝜔 𝑜2 =

𝑅

Q= 𝐿𝜔𝑠 = 𝑅 𝑠 𝜔 𝑜 𝐶
𝑜

Or
𝑅𝑠

𝐿𝐶
𝐿

𝜔 𝑜2 =
=

1
𝐿𝐶

1

=

𝐶𝑅2 𝐶

1
𝑅𝐶
𝐶
= 𝑅𝑠
𝐿

𝑄 = 𝑅𝑠

𝐶
𝐶𝑅2

𝑅𝑠
𝑅
𝑅 𝑠 = 𝑄𝑅

=

2
...


26

Harmonic Oscillator

𝝏𝟐

𝑢𝑖

𝝏𝒕 𝟐

+ 𝜹𝑢 𝑜 = 0 𝑆𝑜𝑙𝑛 = 𝜑 𝑝 𝑠𝑖𝑛(cot 𝜃)

Using L and capacitor C it forms a tank
...


-

It does not tell the amplitude but depends on applied source
...


27

Now
𝜕2 𝜑
+
𝜕𝑡 2

+

𝜑𝑜
𝜕2 𝜑
= 𝑔𝑖𝑣𝑒 𝑎 𝑠𝑖𝑛𝑢𝑠𝑜𝑖𝑑𝑎𝑙 𝑜𝑠𝑐𝑖𝑙𝑙𝑎𝑡𝑖𝑜𝑛 𝑖𝑓
=0
𝐿𝐶
𝜕𝑡

𝛼𝛿𝜑
𝜕𝑡

𝑖𝑓 = 0 𝐻𝑎𝑟𝑚𝑜𝑛𝑖𝑐 𝑂𝑠𝑐𝑖𝑙𝑙𝑎𝑡𝑜𝑟
= -ve poles will ire on the RH side of the S-plane and the oscillation will be growing (ekt)
– growth extra
= +ve decay e-kt it will decay to zero
...

1
𝛿𝜑
𝛿𝜑
𝐶
= 𝛼
𝑅 𝑝 𝜕𝑡
𝜕𝑡
This will cause decay because it is +ve component making 𝑒 𝛼𝑡
...

-

At the end it will be zero/not an oscillator
...


-

The modified differential will be

𝜕2 𝜑
+
𝜕𝑡 2

1
1
+
> 𝑖𝑡 𝑖𝑠 + 𝑣𝑒 − 𝑑𝑒𝑐𝑎𝑦
𝑅𝑝
𝑅

-

So we require –ve resistance to make resistance to infinity; happens only when 𝑅 = 𝑅 𝑝

-

If𝑅 > 𝑅 𝑝
𝑅 < 𝑅𝑝

1
𝑅𝑝

+

1
𝑅

> 𝑖𝑡 𝑖𝑠 + 𝑣𝑒 − 𝐷𝑒𝑐𝑎𝑦

1
1
+
> 𝑖𝑡 𝑖𝑠 − 𝑣𝑒 − 𝐺𝑟𝑜𝑤𝑠
𝑅𝑝
𝑅

𝑅 = 𝑅 𝑝 − 𝐴𝑚𝑝𝑙𝑖𝑡𝑢𝑑𝑒 𝑟𝑒𝑚𝑎𝑖𝑛 𝑐𝑜𝑛𝑠𝑡𝑎𝑛𝑡 𝑖𝑛 𝐹𝐷
In FD +Ve – poles on LH – S-Plane
-Ve – RH – S-Plane
Imaginary axis – then its harmonic oscillator

Building up of oscillation
-

A practical oscillation have a building up of amplitude – this is because when one switches
on the circuit he/she does not expect to see an oscillator
...


-

If oscillation of amplitude is to be build up, the poles on the system should initially lie on the
RHs of S-plane (or 𝑅 < 𝑅 𝑝 R should be initially less than Rp
...

This as theory of amplitude oscillation for any oscillation
...

29

Simulating −𝑹
...

Gyrator circuit thereof (with LC lossy components)

30

-

Why bring in 𝑅 𝑝

-

Up to the circuit above we assuming ideal active parameter gain
𝐴 =∝but𝐴 should be finite 𝐴 𝑜
...


-

𝑅 𝑝 will be equal to compensate for lossy component whose magnitude is equal to 𝑅 𝑝 already
existing in the network
...


-

Because nothing limits the amplitude it will vary as supply voltage
...

This is called gyrator oscillator
Varying C; 𝜔 𝑜 can change/vary
...


-

Is a type of feedback amplifier in which part of the output is fed back to the input via a
feedback circuit
...


-

To visualize the requirements of an oscillator consider the block below
...
(𝑎)
(a) above gives two requirements of oscillation
(i)
Magnitude of the loop gain must at least be 1 (GH)
...
g
...

In addition the frequency of oscillation is determined by the components in the feedback circuit
...

NB: Although there are different types of oscillators they work on the same basic principle
...

34

If the output can be equal to the output in magnitude by cutting the loop can be closed and an
oscillator is realized
If the loop gain is 1 at only 𝜔 𝑜 then it will be a sine wave oscillator at 𝜔 𝑜
...

This is the other way of looking at an oscillator

Loop gain in any loop = 1
Meaning magnitude = 1 (is same as input) and phase shift =0 (meaning the output should be
same as input the loop when broken at certain 𝜔 𝑜 )
This means that initially we apply the voltage 𝜑 𝑖 and 𝜑 𝑜 becomes input at certain frequency
of𝜔 𝑜
...

𝛿2 𝜑
𝛿𝜑 𝑜
+ 𝛼
+ 𝛾𝜑 𝑜 ; 2𝑛𝑑 𝑜𝑟𝑑𝑒𝑟 𝑑𝑖𝑓𝑓𝑒𝑟𝑒𝑛𝑡𝑖𝑎𝑙 𝑤𝑖𝑡𝑕 𝛼 → 0
𝜕𝑡 2
𝜕𝑡
Suppose we have to introduce 𝛼 in the circuit above

35

𝛿2 𝜑
𝛿𝜑 𝑜
+ 𝛼
+ 𝛾𝜑 𝑜 = 0
2
𝜕𝑡
𝜕𝑡
𝛿2 𝜑
𝛿𝜑 𝑜
= −𝛾𝜑 𝑜 − 𝛼
=0
2
𝜕𝑡
𝜕𝑡
To make it to go to oscillation in practice we need a small negative value here for

𝛼

𝛿𝜑 𝑜
𝜕𝑡

which

will go to zero at required amplitude
...

RC will be large and grow to infinity
...

RC goes to infinity as amplitude builds up
...

36

𝛼 -ve terminal
Enables an oscillator to be designed
-

Initially if the oscillator with or RC does not oscillate, put RC and then adjust it to infinity
and the circuit will oscillate locating the poles at the imaginary point
...


37

WIEN BRIDGE

Show that
𝑓𝑜 =

1
2𝜋𝑅𝐶

𝑅 𝑓 = 2𝑅1
1st consider fb circuit of Wien Bridge Oscillator above
...

1+

𝑅𝑓
𝑗𝜔𝐶𝑅 = −𝜔2 𝐶 2 𝑅2 + 𝑗 3𝜔𝐶𝑅 + 1
𝑅1

(i) Real part
−𝜔2 𝐶 2 𝑅2 = −1
𝜔2 =

1
𝑅2 𝐶 2
1

Or 𝑓𝑜 = 2𝜋𝑅𝐶

(ii) Imaginary Part
1+
1+
𝑅𝑓
𝑅1

𝑅𝑓
𝑅1

𝑅𝑓
𝑅1

=3

=2

𝑅𝑓 = 2
𝑅 𝑓 = 2𝑅1

39

𝑗𝜔𝐶𝑅 = 𝑗 3𝜔𝐶𝑅

𝑅𝑏

𝜑𝑜 = 1 +

=

𝑅2

𝑅
𝐴 𝐹 2 1+𝑆𝐶 𝑅
2 2
1+𝑆𝐶2 𝑅2 +

=
=

𝜑𝑖 𝑍 𝑝
𝑍𝑝 + 𝑍𝑠

𝑅𝑎

𝑆𝐶 1 𝑅 1 +1
𝑆 𝐶1

𝐴 𝐹 𝑅2 𝐶1
𝑆2

𝑆𝐶1 𝑅2 +

𝐶1 𝑅1 𝐶2 𝑅2 +𝑆𝐶1 𝑅1 +𝑆𝐶2 𝑅2 +1

1
𝑅
𝐶
𝑆𝐶1 𝑅1 +1+ 1 + 2 +
𝑅2

𝐶1

1
𝑆𝐶 1 𝑅 2

The input will be in phase with the input when the
(i)

Imaginary part is
𝑆𝐶2 𝑅1 =
−𝜔2 =
𝜔=

(ii) Real part

1

1+

𝑆𝐶1 𝑅2
1

𝑅1
𝑅2

+

𝐶2
𝐶1

=0

But 𝐴𝛽 = 1

𝐶1 𝐶2 𝑅1 𝑅2

𝑅
1+ 𝑏 𝑅
𝑎
𝑅
𝐶
1+ 1 + 2

1
𝐶1 𝐶2 𝑅1 𝑅2

𝑅2

=0

𝐶1

If 𝑅1 = 𝑅2, 𝐶1 = 𝐶2
=

1

1+

𝐶𝑅

𝑅𝑏

40

𝑅𝑏

𝑅𝑎 = 3

𝑅𝑎 = 2

OR
Apply that loop gain 𝐴𝛽 = 1, non-investing gain 1 +

1+

𝑅𝑏

𝑅𝑎

𝜑𝑖

1 + 𝑍2 𝑌2
1+
=

𝑅𝑏

𝜑𝑖

𝑅𝑎

1 1
1 + 𝑅2 + 𝑆𝐶 𝑅 + 𝑆𝐶1
2
1
1+

𝜑𝑜 =

𝑅𝑏

𝑅𝑎

𝜑𝑖

𝑅
𝐶
1 + 𝑅1 + 𝐶2 + 𝑆𝐶1 𝑅1 + 𝑆𝐶2 𝑅2
2
1

𝑠=𝑗𝜔

41

𝑅𝑏

𝑅𝑎

Now substitute 𝑠 = 𝑗𝜔
𝑅𝑏

1+
𝑅
𝐶
1 + 𝑅1 + 𝐶2 + 𝑗
2
1

In phase 𝜔 =

𝑅𝑎

1
𝜔𝐶2 𝑅1 + 𝜔𝐶1 𝑅2

1
𝐶1 𝐶2 𝑅1 𝑅2

At that frequency of oscillation
1+

𝑅𝑏

Then connect the loop from input to
output and remove the input signal
...


𝑅𝑎

𝑅
𝐶
1 + 𝑅2 + 𝐶1
1
2

=1

Conditions for oscillation
𝑅2
𝐶1
𝑅 𝑎 = 𝑅1 + 𝐶2

𝑅𝑏

𝑅2 = 𝑅1 = 𝑅
𝐶2 = 𝐶1 = 𝐶

𝑡𝑕𝑒𝑛 3 𝑜 =

1
𝑅𝐶

Attained by setting

𝑅𝑏

𝑅𝑎 = 2

If

𝑅𝑏

at

𝑅𝑏

𝑅𝑏

𝑅 𝑎 > 2 it will build up oscillation
𝑅 𝑎 = 2 poles line on the imaginary – oscillator

𝑅 𝑎 < 𝛼 𝐷𝑒𝑐𝑎𝑦

42

-

Is a common type of oscillator
...


PHASE SHIFT OSCILLATOR

−(𝑆𝐶𝑅)2 𝐾
=1
(1 + 𝑆𝐶𝑅)3

1 + 𝑆𝐶𝑅

3

+ − 𝑆𝐶𝑅

3

𝐾=0

1 + 𝑆 2 𝐶 2 𝑅2 + 3𝑆𝐶𝑅 + 𝑆𝐶𝑅

3

𝐾+1 = 0

𝑠=𝑗𝜔

1 + 3𝜔3 𝐶 2 𝑅2 + 3𝑗𝜔𝐶𝑅 − 𝑗𝜔3 𝐶 3 𝑅3 𝐾 + 1 = 0
If the real part become 0, the imaginary should be =0
...

𝜔=

1
3𝑅𝐶

PHASE OSCILLATOR

Show that;
𝑓𝑜 =

1
2𝜋 6 𝑅𝐶

… … … … …
...


45

RC – network of the phase shift oscillator transform in S-domain
...
9

If 𝑅1 ≫ 𝑅 then the circuit then 𝐼7(𝑠) = 0𝐴, meaning 𝐼5

𝑠 = 𝐼6(𝑠)

∴Using voltage divider rule
𝜑𝑓
𝜑2 =

𝑅𝐶 𝑠 + 1
𝜑 𝑓………………………………………
𝑆𝐶𝑅

𝑠

=

𝑅
𝑅+1

𝜑2

𝑠

𝑆𝐶

10

Substitute 𝜑2 in 8
𝜑𝑖 =

𝑆𝐶𝑅 + 1 𝜑 𝑓
𝑆𝐶𝑅 𝜑 𝑜
+
… … … … … … … … …
...
12

Equate (11) and (12) and simplify for

𝜑𝑓

𝜑𝑜

46

𝜑𝑓

𝑆 3 𝐶 3 𝑅3
𝜑 𝑜 = 𝑆 3 𝐶 3 𝑅3 + 6𝑆 2 𝐶 2 𝑅2 + 5𝑆𝐶𝑅 + 1 … … … … … … 13

47

Consider the part of op-amp of the phase oscillator

𝐴∪ =

𝜑𝑜
𝜑𝑓

𝑠

=−

𝑠

𝑅𝑓
𝑅

For oscillation to occur
𝐴𝛽 = 1
=−

∴ 𝑒𝑞𝑢𝑎𝑡𝑖𝑛𝑔 13 𝑎𝑛𝑑 14

𝑅𝑓
𝑆 3 𝐶 3 𝑅3
=1
𝑅 𝑆 3 𝐶 3 𝑅3 + 6𝑆 2 𝐶 2 𝑅2 + 5𝑆𝐶𝑅 + 1

Substitute 𝑠 = 𝑗𝜔
𝑅𝑓
𝑗𝜔3 𝐶 3 𝑅3 = −𝑗𝑅3 𝐶 3 𝜔3 − 6𝜔2 𝐶 2 𝑅2 + 5𝐶𝑅𝜔 + 1
𝑅

(i) Real Part

(ii) Imaginary part
𝑅𝑓

−6𝜔2 𝐶 2 𝑅2 + 1 = 0
𝜔=

𝑅

1

𝑅𝑓

6 𝑅𝐶

𝑅

𝑗𝜔3 𝐶 3 𝑅3 = −𝑗𝑅3 𝐶 3 𝜔3 + 5𝐶𝑅𝜔
= 1 − 5 𝜔 2 𝐶 2 𝑅2 substitute for 𝜔2 in

the preceding equation
1

𝜔2 = 6𝑅 2 𝐶 2
𝑅𝑓
= 29
𝑅

48

49

Example
1
...
1 below, the gain of the forward amplifier A is frequency/dependent
6
3
and given by 𝐴 = 9 × 10 𝑗𝜔
...

Solution:
The loop gain
−9 × 106 × 6 × 103
𝐴𝛽 =
𝑗𝜔 3 × 103 + 𝑗𝜔 3 × 103 + 𝑗𝜔
=

…………
...

j 𝜔 × 9 × 106 − 𝜔3 = 0
or𝜔 = 3 × 103 𝑟𝑎𝑑 𝑠
Substituting this in 1 we have

−54×10 9

Aβ = −6×10 3 ×9×10 6 = 1
The condition for oscillation are satisfied
...
7hz
...
7
2
...
1 the gain of the forward amplifier is given by A= −4 × 107 𝑗𝜔
If the feedback fraction 𝛽 is 3
...
The gain of the forward amplifier A in the circuit fig
...
1 i−4 × 107 𝑗𝜔
...


FEEDBACK SINUSOIDAL OSCILLATOR

50

Oscillators which are formed by linear amplifiers terminating in their own impedances,
producing unity amplification with zero loop phase shift are called feedback sinusoidal
oscillators
...


Basic positive feedback oscillator circuit with no external ac exaltation
...

𝐴 < 00
𝐴𝐵 =

𝐵 < 00 = 1 < 00

𝑎𝑛𝑑 < 𝜃 + 𝜑 = 0 … … … … … … … … … … … … … …
...
(𝑖𝑖𝑖)

For sustained oscillation at a frequency 𝑓𝑜 , the loop gain
𝐿 𝑗𝑤 = 𝐴 𝑗𝑤 𝐵 𝑗𝑤 = 1
𝐴 𝑟 𝑗𝑤 = 1
𝐵 𝑟 𝑗𝑤 … … … … … … … … … … … … (𝑖𝑣)
𝐵𝑖 𝑤 = 0

𝑠𝑖𝑛𝑐𝑒 𝐴 𝑟𝑜 ≠ 0 … … … … …
...

Since ac power required by the input of an amplifier is lees than output power, it is possible
to make the amplifier supply its own input and hence generate self-sustained oscillations
...

The initiating ac triggering signal is provided by the abrupt flow of forward bias device
current from its zero value
...


Note the overall fain 𝐴 𝑓 =

𝐴
1−𝐴𝐵

, under the conditions above, it tends to be infinite and hence

the amplitude must be self-limited
...
However, with the help of linear analysis, one can
obtain the frequency of oscillations and the conditions for starting of the oscillations
...

Frequency determining network can be RC-type and LC type for frequency upto few look HZ
...


𝑢2 = 𝑢2 +
𝑢2 +

𝑢2
𝑗𝑤𝐶𝑅

𝑎𝑠

3
𝑗𝑤𝑐

𝑢2
… … … … … … … … …
...
(𝑖𝑖)
𝑅
𝑢3 = 𝑢4 +

= 𝑢2 +

𝑢4
𝑅

−𝑗2

𝑗𝑤𝐶

𝑢2
2𝑢2
𝑢
+
− 2 𝑤 2 𝐶 2 𝑅2 … … … … … … … … … … …
...
(𝑖𝑣)

𝑢3 +

−4𝑢2

𝑗𝑤𝐶𝑅 +

𝑅

𝑙1

𝑗𝑤𝐶

𝑤 2 𝐶 2 𝑅2 −

𝑢2

𝑗𝑤 3 𝐶 3 𝑅3

6𝑢2
5𝑢2
𝑢2
− 2 2 2 − 3 3 3 … … … … … …
...
(𝑣𝑖)

The imaginary part of 𝐵(𝑗𝑤) come from the first and third term in denominator namely odd
power terms
...
(𝑣𝑖𝑖)
𝑗𝑤𝐶𝑅
𝑤0 =

𝑓0 =

1
2𝜋𝑅𝐶 6

1
𝑅𝐶 6

… … … … … … … … (𝑣𝑖𝑖𝑖)

54

All this frequency oscillation the phase shift network produces a phase shift of 1800
From equation the real part of 𝐵(𝑗𝑤) at 𝑤0 is
𝐵𝑟 𝑤 =

1
5
𝑗𝑤0 𝑅𝐶

2

+1

From (viii)
𝐵𝑟 𝑤 =

1
−1
=
5 −6 + 1
29

For oscillations to occur we must have 𝐴𝐵 ≥ 1 and hence using this expression
𝐴𝐵 ≥ 1
𝐴 ≥ 29
Is the condition for oscillation to occur

If the op-amp (inverting portion) had the above components the resistors 𝑅 𝑓 and 𝑅1 provide A
the inverting gam
𝐴 =
𝑅𝑓

𝑢0

𝑢𝑓 =

−𝑅𝐹
𝑅1

𝑅1 = 29 also the start of oscillation condition is satisfied if 𝑅1 𝑎𝑛𝑑 𝑅2 selected provide

𝐴 𝑟 > 29
...
It follows that for such an oscillator the frequency of oscillation is
given by 𝑤0 =

6
𝑅6

And gain condition is 𝐴 𝑟 ≥ 29
55

𝐴𝑟 =

𝑢0

0
𝑢1

= −𝑔𝑚𝑅 𝐿

Assuming we are using ideal op amp (𝑅1 - infinite) i
...
𝑙5 𝑙6
𝐼4 𝑍2 = 𝐼5 𝑍1 + 𝑍2
=

𝑢𝑓
𝑍 + 𝑍2 = 𝑢 𝑓 𝐾 + 1 … … … (𝑖)
𝑍2 1

Where 𝐾 =

𝑍1

𝑍2 further using equation (i) we have

𝐼3 = 𝐼4 + 𝐼5
=

𝑢 𝑓 (𝐾 + 1) 𝑢 𝑓 (𝐾 + 1)
+
𝑍2
𝑍1 + 𝑍2

=

𝑢 𝑓(𝐾+2)
(𝐾 + 2)
= 𝑢 (𝐾+1)
𝑓
𝑍2
𝑍1 + 𝑍2
𝑍2

Now 𝐼2 𝑍2 = 𝐼3 𝑍1 + 𝐼4 𝑍2
𝐼2 =

𝐾𝐼3 + 𝐼4

Using equation (ii) and (i) we get
𝐼2 =

𝑢𝑓

𝑍2 𝐾 𝐾 + 1 +

𝑢𝑓

𝑍2 𝐾 + 1 … … …
...
(𝑖𝑖)

Now using equation (ii) and (iii) from the figure above we obtain
𝑢0 = 𝐼1 𝑍1 + 𝐼2 𝑍2
= 𝐼2 𝐼3 𝑍1 + 𝐼2 𝑍2
= 𝐼2 𝑍1 + 𝑍2 + 𝐼3 𝑍1
=

𝑢𝑓
𝑍2

𝐾 2 + 3𝐾 + 1
𝑢𝑓
=
𝑢0

𝐵=

𝐾3

+

𝑍1 + 𝑍2 +

𝑢𝑓
𝑍2

𝐾 + 2 𝑍1 … …
...
(𝑣𝑖)

The phase shift introduced by the network would be 1800 when the j-component of B is zero i
...

when
𝑥 3 − 6𝑥 = 0
𝑥 = ± 6 0𝑟 0 … … … … … … … … … …
...
Further, negative value of 𝑥 is
also admissible since 𝑤 is a positive quantity
...
Putting 𝑥 = 6 in equation (6) we get 𝐵 =

−1
29

For oscillation to occur we must have 𝐴𝐵 ≥ 1 and hence using it 𝐴𝐵 ≥ 1
𝐴 ≥ 29 is the condition for oscillations to occur
...


58

𝑍1 =

𝑢𝑖
=
𝑍

𝑢𝑖

𝑢𝑖

𝑖
𝑅 + 𝑗𝑤𝐶
𝑗𝑤𝑐𝑢 𝑖
1 + 𝑗𝑤𝐶𝑅

=

𝑢0

=

𝑗𝑤𝐶𝑅𝑢 𝑖
1 + 𝑗𝑤𝐶𝑅

𝑢0

1−

𝑗

→

𝑥(𝑥 − 𝑗𝑅)
𝑅2 + 𝑥 2

𝑥− 𝑗
𝑥− 𝑗

= 𝑢 𝑢1

𝜃 = tan − −𝑅 𝑥

1 + 𝑗𝑥
𝑥2 + 1

= − tan − 𝑅 𝑋
𝑢0

𝑢1 =

Method 2:
1
𝑗𝑤𝐶

𝑢 𝑖 𝑅 = 𝑢0 𝑅 − 𝑗𝑥𝐶
𝑢0

𝑅2 + 𝑥 2

𝑥

𝑔𝑎𝑖𝑛

𝑅 − 𝑗𝑥0 =

𝑥

𝐺𝑎𝑖𝑛 =

𝑗
𝑥+ 𝑗

−𝑗𝑥
𝑅 − 𝑗𝑥

−𝑗𝑥(𝑅 + 𝑗𝑥)
𝑅2 + 𝑥 2

𝑥

=

𝑅

𝑢1 =

𝑗𝑤𝐶𝑅
𝑢 𝑖 = 1 + 𝑗𝑤𝑅𝐶
𝑗

𝑅+

1
𝑗𝑤𝐶

1
= 𝑖𝑗𝑥
𝑗𝑤𝐶𝑅

𝑢0 = 𝑙 𝑖 𝑅

=

𝑅+

−𝑗𝑥𝑢 𝑖 = 𝑢0 𝑅𝑗𝑥

Voltage across R has a value

𝑢 𝑖 𝑅 = 𝑢0

1
= 𝑢0
𝑗𝑤𝐶

𝑢𝑖

𝑅 𝑅 + 𝑗𝑥 𝑐
𝑢
= 0 𝑢𝑖
2 + 𝑥2
𝑅
59

1
𝑥2 + 1

Gain

𝑢0

𝑢1 =

𝜃 = tan −

𝑅
𝑅 2 +𝑋 2

𝑥𝑐
=
𝑅

1
𝑤𝐶𝑅

To analyze the operation of the transfer 𝑓𝑥𝑛 of the feedback network must be deduced
...
(2)
𝑅

2𝑅 − 𝑗𝑥
𝑢 𝑖 … … … … … …
...

𝑓0 =

1
2𝜋𝑅𝐶 6

At this frequency if the magnitude of the shift of 1800 at a frequency 𝑓0
At this frequency the magnitude of the small signal amplitude gain is arranged to be precisely 29
...


Biasing comprise of Admittance because the feedback at to be made only resonate at certain
frequency
...

60

Yb- load and admittance e
...
c
Yb- load and out
Ya- biasing cct and input reaction
∇𝑌 =

𝑌𝑎 +

1
+ 𝑌𝑐
𝑟𝑐 (𝐵 + 1)

𝑌𝑏 𝑌𝑐 +

1
+ 𝑌𝑐 𝑔𝑚 − 𝑌𝑐 = 0
𝑟𝑐 𝑒

This equation gives the condition as well as the frequency of oscillation
...

Select up to possible combinations
...

𝑌𝑐 =

1
𝑗𝑤𝐿

𝑌 𝑎 = 𝑗𝑤𝐶1

-

Feedback element takes one part/one type and other two
opposite

𝑌𝑏 = 𝑗𝑤𝐶2
Two capacitative 𝑌 𝑎 𝑌𝑐
One inductive 𝑌𝑐

Two oscillators produced

Two inductive 𝑌 𝑎 𝑌𝑏
One capacitative 𝑌𝑐

Real part

61

𝑌 𝑎 𝑌𝑏 + 𝑌𝑐 𝑌 𝑎 +𝑌𝑏 +

−𝑤 2 𝐶1 𝐶2 +

1
=0
𝑟𝑒 (𝐵 + 1)𝑟𝑐 𝑒

𝐶1 + 𝐶2
1
+
=0
𝐿
𝑟𝑐 (𝐵 + 1)𝑟𝑐 𝑒

𝑤2 =

𝐶1 + 𝐶2
+
𝐿𝐶1 𝐶2

𝑤0 ≅

𝐶1 + 𝐶2
𝐿𝐶1 𝐶2

1
𝑟𝑐 (𝐵 + 1)𝑟𝑐 𝑒
𝐶1 𝐶2

≈

1
𝐿𝐶1 𝐶2
𝐶1 + 𝐶2

Frequency of oscillation takes care of real part completely
...

𝑌𝑐 - Feedback reactance
...
It is not a feedback but a biasing
circuit
...
(Takes the active device to oscillation)
Objective




To know whether such circuit can be made into an oscillator?
Analysis of such a phenomenon
...

Matrix of parameters
...

= ∆𝑌 =

𝑌 𝑝 𝑌𝑜 − 𝑌𝑓 𝑌𝛾 = 0

Condition for oscillation is that ∆𝑌 = 0 at 𝜔 = 𝜔 𝑜 which is a single resonant frequency
...
At certain frequency it
should be = 0
...

The equation must have the real and imaginary part and they should be made to go to zero
...

Real part =0
Im=0
Choose the possible combination for this to work
...


𝑌 𝑎 = 𝑗𝜔𝐶1
𝑌𝑏 = 𝑗𝜔𝐶2
(i)
(ii)

Two capacitive 𝑌 𝑎 𝑌𝑐
One inductive 𝑌𝑐
Two inductive 𝑌 𝑎 , 𝑌𝑏
One capacitive 𝑌𝑐

Two oscillators produced

65

Real Part
...

Condition for oscillation
1
1
1
𝐶2
𝐶1
𝑔𝑚 +
+
= 𝜔
+
𝜔𝐿
𝛾 𝑐𝑒
𝛾 𝑐𝑒
𝛾 𝑒 𝛽 +1
𝛾 𝑒 𝛽 +1

1
𝜔2 𝐿

=

𝑡𝑎𝑘𝑖𝑛𝑔 𝜔 𝑜 =

𝐶1 +𝐶2
𝐿𝐶1 𝐶2

66

𝐶1 𝐶2
𝑛𝑒𝑔𝑙𝑖𝑎𝑏𝑙𝑒
𝑔𝑚 +
𝑠𝑚𝑎𝑙𝑙 =
𝐶1 + 𝐶2
𝐶1
𝑔𝑚 =
𝐶1 + 𝐶2
𝐶1
=
𝐶1 + 𝐶2

𝐶2
𝐶1
+
𝑟𝑐 (𝐵 + 1) 𝑟𝑐 𝑒

1
𝐶1 + 𝐾2
+
𝑟𝑐 (𝐵 + 1)
𝑟𝑐 𝑒

1
𝐶1 𝐾2
+
𝐵+1
𝑔𝑚𝑟𝑐 𝑒

For condition for oscillation
𝐵+1 =1+
𝐵=

𝐶2

𝐶2
𝐶1

𝐶1

Case II
𝑌𝑐 =

1
𝑗𝑤𝑐

𝑌𝑎 =

1
𝑗𝑤𝐿1

𝑌𝑏 =

𝑌3
𝑌1

1

𝑌2

𝑗𝑤 𝐿2

In the cct shown let 𝑌𝑐 = 𝑗𝑤𝐶, 𝑌 𝑎 =
Show that the frequency of oscillation 𝑤0 =

1
𝐿1 +𝐿2 𝐶

Determine the relationship/ the condition for oscillation in terms of 𝐿1 𝑎𝑛𝑑 𝐿2 and B
...

The analysis is valid only that in FET we replace open cct for 𝑟𝑒 (𝐵 + 1) and for 𝑟𝑐 𝑒 )the rest
remain the same)
...

Hartley
Culprit

22-

inductors

inductor

1- capacitor

2- capacitors
68

The same are most important used oscillators
...


It has series capacitor- it can resonate with 𝐶 𝑠 (series capacitor) in terms of series resonance
...

Crystal
Can be used as LC block for the oscillator
...

Hence the cct can be used as a feedback cct (structure where it can take the inductive or cap
actative (reactance) at the frequency of resonance and therefore will always oscillate at
crystal frequency
...
It can replace LC circuit because
it is a combination of series capacitance cct and parallel capacitance
...

Thus, the equivalent cct consist of two capacitors giving a pair of closely spaced series and
parallel resonant frequencies
...
05 pF and 3H respectively
...

1
1
𝑗𝑤𝐿 + 𝑗𝑤𝐶 ×
−𝑗
𝑗𝑤𝐶 1
𝑗𝑥 =
=
1
1
𝑤𝐶 1
𝑗𝑤𝐿 + 𝑗𝑤𝐶 +
1
𝑗𝑤𝐶

1
𝑤 2 − 𝐿𝐶
1 1 1
𝑤2 − 𝐿 𝐶 + 𝐶

1
𝑓2 − 2
−𝑗
4𝜋 𝐿𝐶
=
;
1 1
2𝜋𝐹𝐶 𝑓 2 − 1
2 𝐿 𝐶+ 𝐶
4𝜋
=

𝑗 2 − 𝑓𝑠2
𝑓 2 − 𝑓𝑝2

−𝑗
2𝜋𝑓𝐶 1

The above expression shows that the crystal can have both series a parallel resonance
...


Positive feedback occurs when the feedback signal is in phase with input signal and under the
proper conditions, oscillation is possible

The oscillation conditions
-

The basic oscillator is shown below

1
...
The phase of 𝑢 𝑓 determines if it
0
adds or subtracts to 𝑢1
...
1 above
...
However, in many oscillators, at the
frequency of oscillation, the amplifier is operating in its mid band region where 𝐴 𝑟 (𝑗𝑤) is
a real constant 𝐴 𝑟 𝑗𝑤 = 𝐴 𝑢𝑜

If 𝐴 𝑟 𝑗𝑤 = 𝐴 𝑢𝑜 and it is positive, the phase shift through the amplifier is 00 and for a positive
feedback the phase through 𝐵(𝑗𝑤) should be 00 or a multiple of 3600
If 𝐴 𝑢𝑜 is negative number the phase shift through the amplifier is ±1800

73

From above
𝑢0 = 𝐴 𝑟 𝑗𝑤 𝑢 𝑑 … … … … … … … … …
...
(2)

From (1) (2) & (3), the closed loop voltage gain 𝐴 𝑟𝑓 (𝑗𝑤) is given by
𝐴 𝑟𝑓 𝑗𝑤 =

𝑢0
0
𝑢1

= 1−𝐵

𝐴 𝑟 (𝑗𝑤 )
𝑗𝑤 𝐴∝ (𝑗𝑤 )

… … … … … … (4)

The quantity 𝑏 𝑗𝑤 𝐴 𝑟 (𝑗𝑤) is called the loop gam
0
For oscillation to occur an output signal must exist with no input signal applied
...
e
...
This relation
is known as Barkhawsen criterion
...
(6)
𝐵𝑟(𝑤)

𝐵𝑖 𝑤 = 0

𝑆𝑖𝑛𝑐𝑒

𝐴 𝑢𝑜 ≠ 0 … … … … … … …
...
(8)
And 𝐵 𝑗𝑤 𝐴 𝑟 𝑗𝑤 = ±𝑛3600 … … … … … … … … … … … … … … … … …
...


-

Also for 𝐴 𝑟 𝑗𝑤 = 𝐴 𝑟 and with 𝐵 𝑖 𝑗𝑤 = 0 (8) reduces to (6)
...


Current amplifier feedback network

The current form of the basic feedback network
𝐴1 (𝑗𝑤)- Current gain of the amplifier
∝ 𝑗𝑤 − Current feedback factor
∝ 𝑗𝑤 =

𝑗𝑓
𝑗0

For oscillation to take place the loop gain must be unity ∝ 𝑗𝜔 𝐴 𝑖 𝑗𝜔 = 1
Types of oscillators
Types of components used
RC
LC
Crystal oscillators

Frequency of oscillation
Audio frequency (AF)
Radio frequency (RF)

75

Types of waveform
Sinusoidal
Square wave
Triangular wave, Saw tooth
wave e
...
c

Classification of oscillators based on the generated frequency ranges
Oscillators class

Frequency range

Audio frequency (AF) oscillators

A few HZ – 20KHZ

Radio frequency (RF) oscillators

20KHZ- 30MHZ

Very high frequency (VHF) oscillators

30MHZ- 300MHZ

Ultra-high frequency (UHF) oscillators

300MHZ – 3GHZ

Microwave oscillators

3GHZ-several GHZ

-

-

L and C confine the application of LC feedback oscillators mainly for generation of radio
frequency signals
...
Examples
– phase shift, Wien bridge oscillators
...
These are example of crystal oscillators
...


- 11

0 < 𝐴𝐵 < 1 … … … … … … … … … … 12
𝐴𝐵

𝑛

→ 0 and the input voltage to the amplifier settles down to a value

𝑢𝑠
(1 − 𝐴𝐵)

Hence under this condition a positive feedback amplifier delivers a steady output
...

76

-

When 𝐴𝐵 > 1 we find from equation (11), the amplifier input voltage goes on increasing
and becomes infinitely large as 𝑛 →∝ hence instability results
...


-

In fact, non-linearities in the amplifier limit the build up of signal and a steady state is
reached within a very short time
...
Then the signal feed back to the
input is equal to that present in the preceding cycle
...


Question: what are basic requirements of a feedback oscillator (page 313)
BASIC OSCILLATOR CIRCUIT

Voltage gain of oscillator
𝑢0
𝐴(𝑗𝑤)
= 𝐴 𝑓 𝑗𝑤 =
𝑢𝑠
1 − 𝐴 𝑗𝑤 𝑊
𝑢𝑠 = 0
For sustained oscillation at a frequency 𝑓0 , the loop gain 𝐿 𝑗𝑤 = 𝐴 𝑗𝑤 𝐵 𝑗𝑤 = 1
Without any input how can we get 𝑢0 ? The starting point is random electric noise
...


Using inverting terminal
The cct will oscillate at a frequency 𝑓0 at which the phase shift of the feedback network is 1800
...

The imaginary part of 𝐵(𝑗𝑤) will vanish when
3

+6

𝑤 = 𝑤0 =

1
=0
𝑗𝑤𝑅𝐶

________________(2)

1

𝑓0 =

6𝑅𝐶

1

(3)

2𝜋𝑅𝐶 0

At this frequency of oscillation the phase shift network produces a phase shift of 1800
From equation (1) the real part 𝐵 𝑗𝑤 at 𝑤0 is
𝐵 𝑟 𝑗𝑤 =

1
5
𝑗𝑤𝑐𝑅𝐶

… … … … … … … … … … …
...
(5)
5 −6 + 1
29

From the gain condition of oscillator
𝐴𝑟 𝐵 = 1
𝐴 𝑟0 =

1
𝐵

= 29

=

1
𝐵 𝑟 (𝑤 0 )

… … … … … … … … … … … … (6)

From the figure above the resistors 𝑅1 and 𝑅2 provide an inverting gain of
𝐴 𝑟𝑜 =

𝑢0 −𝑅2
=
𝑢𝑓
𝑅1

Hence, the start of oscillator condition is satisfied if 𝑅1 and 𝑅2 selected to provide 𝐴 𝑢𝑜 > 29
...


81

NB: the phase shift oscillator in figure above have been implemented by inter-changing R and C
in the phase shift sections
...


FEEDBACK THEORY
-

Important for circuit design
...

Advantages and disadvantages
...
X can be any variable
...

𝐼𝑓 𝑡𝑕𝑒 𝑒𝑟𝑟𝑜𝑟 𝑖𝑠 ε Xi – HX0 𝑖𝑡 𝑤𝑖𝑙𝑙 𝑏𝑒 𝑝𝑜𝑠𝑖𝑡𝑖𝑣𝑒 𝑓𝑒𝑒𝑑𝑏𝑎𝑐𝑘
𝑋0
𝐺
=
𝑋 𝑖 1 − 𝐺𝐻

Discussing negative feedback:
-

To remove distortion used to improve the performance of circuit (Used in eplepsy)
...

Reduction in active parameter sensitivity
...
X0 can be voltage, current, resistance, phase
...
𝑃𝑕𝑎𝑠𝑒 𝑓𝑜𝑙𝑙𝑜𝑤𝑒𝑟 𝑓𝑟𝑒𝑞𝑢𝑒𝑛𝑐𝑦 𝑓𝑜𝑙𝑙𝑜𝑤𝑒𝑟
𝑋 𝑖 ∅1
𝜔0

H – is decided by positive devices – R, C, (linear)
The negative gain advantage:- reduction of active parameter sensitivity
...

BJT = 𝑔 𝑚 ∞𝐼 and dependent on temperature

Feedback situation makes an amplifier with gain more than 1, using positive, attenuators,
resistors, capacitors, inductors
...
𝑅, 𝐶 𝑎𝑛𝑑 𝑙𝑖𝑛𝑒𝑎𝑟
The resultant gain is 𝐺 𝑓
𝜕𝐺 𝑓 /𝐺 𝑓
=
𝜕𝐺/𝐺

𝐺𝑓
𝐺

(𝑆𝑒𝑛𝑠𝑖𝑡𝑖𝑣𝑖𝑡𝑦 𝑓𝑎𝑐𝑡𝑜𝑟 𝑜𝑓 𝐺 𝑓 𝑟𝑒𝑙𝑎𝑡𝑖𝑜𝑛 𝑡𝑜 𝐺
...
If GH >> very high 1+𝐺𝐻 ≪ 1 hence sensitivity factor gets to zero
...
𝑖𝑓 𝐺 = 10, 𝐺𝐻 = 0
...
9 (𝑙𝑒𝑠𝑠 𝑡𝑕𝑎𝑛 1) to get higher
gain
...

FEEDBACK AMPLIFIERS
-

Negative feedback – inverse/degenerate feedback
...


Block diagram of Feedback Amplifier
Has internal amplifier/basic and feedback network

85

-

Feedback network has a passive element such as resistors, capacitors or inductors
...

The feedback network extracts a portion of the output 𝜇 𝑜 voltage of the amplifier
...

𝜇 𝑖 = 𝜇 𝑠 + 𝜇 𝑓 ; 𝑖𝑛𝑑𝑖𝑐𝑎𝑡𝑖𝑛𝑔 𝑏𝑜𝑡𝑕 𝑛𝑒𝑔𝑎𝑡𝑖𝑣𝑒 𝑎𝑛𝑑 𝑝𝑜𝑠𝑖𝑡𝑖𝑣𝑒 𝑓𝑒𝑒𝑑𝑏𝑎𝑐𝑘
...

The whole system is feedback amplifier with overall gain 𝐴 𝑓 =
Transfer gain of an amplifier with feedback
...

(1 + 𝛽𝐴)is called return difference
...

(i)

If 1 + 𝛽𝐴 > 1, 𝐴 𝑓 < 𝐴 𝑎𝑛𝑑 𝑓𝑒𝑒𝑑𝑏𝑎𝑐𝑘 𝑖𝑠 𝑛𝑒𝑔𝑎𝑡𝑖𝑣𝑒
...
e
...

If 1 + 𝛽𝐴 > 1, 𝐴 𝑓 > 𝐴 ; and the feedback is positive since the overall gain of an
amplifier with the positive feedback, there is a great chance of oscillations to occur
...
Making it undesirable
...

N = 𝜕𝛽 of feedback

=

20log10

=

20log10

For negative feedback
1 + 𝛽𝐴 > 1
N = is –ve

𝐴𝑓
𝐴
1
1+𝐴𝛽

Forpositive feedback
1 + 𝛽𝐴 < 1
N = is +ve number

If 𝛽𝐴 = 1 𝐴 𝑓 =∝ meaning there is an output without an input
...
What actually is experienced practically is that the amplifier becomes an
oscillator (unstable) supplying its own input
...
e
...

What is an oscillator?
Maybe defined as follows
...

It is an electronic source of alternating current or voltage having Sine, Square or Saw-tooth
or pulse shapes
...

It is unstable amplifier
Signal input

Amplifier

Output Signal

Oscillator

DC power
input

Output Signal

DC power
input

Classification of oscillators
Two broad groups
i)
ii)

-

Sinusoidal/harmonics oscillators – sine waveforms
...

- Find use in timing and control application
...

- Ramposc – are found in the horizontal sweep circuit oscilloscope and TV sets
...
g
...


Objectives
-

Feedback approach to oscillator design is discussed
...

Rectifiers AC-DC
Oscillators DC – AC sine waves

88

Theory of sinusoidal oscillation
-

To build a sinusoidal oscillation we need to use an amplifier with positive feedback
...
e
...
This is
called an oscillator
...

 -ve feedback cause an amplifier to be used for amplification
...

The condition 𝐴𝛽 = 1 is called Bakhausen criterion
...


-

Is 𝐴𝛽 𝑖𝑠 𝑔𝑟𝑒𝑎𝑡𝑒𝑟 𝑡𝑕𝑎𝑛 1, 𝐴𝛽 − 𝑉𝑖𝑛 is greater than 𝜇 𝑖 and the output voltagebuilds up as
shown
...


Phase requirements
In many amplifiers we have 1800 phase shift between the input and the output (transistor circuit
amplifier)
...

89

NB: RC network produces/introduces definite phase shift
...

∅ = tan−

𝑋𝑐
𝑅

= −tan−

𝑋𝑐
𝑅


...
Remember,
1

since X1, is frequency sensitive𝐴1 = 2𝜋 f , the 1800 phase lift also can be achieved only at a
𝑐

particular frequency
...

Such as oscillator is called phase shift oscillator
...

-

Moderately stable in frequency and amplification
...

Used – guitar

Advantage
-

Investing
Non-investing

Buffered phase shift oscillator

This buffer prevents the RC section from loading each other, hence the buffered phase-shift
oscillator performs closer to the calculated frequency and gain
...
of three such networks

Phase difference due to one item
...

The output voltage is real at a frequency making factor (Resonant frequency)
...
Hence the gain of the amplifier must be greater than 29 since 𝐴𝛽 must be 1
𝐴𝛽 = 𝐴 ×

1
=1
29

𝐶 = 0
...
𝜕𝛽

93



---