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Showing posts with label ANALOG ELECTRONIC CIRCUITS. Show all posts
ANALOG ELECTRONIC CIRCUITS × Oscillator
• A Crystal Oscillator is basically a tuned circuit Oscillator using a piezoelectric
crystal as a resonant circuit.
• The crystal ( usually quartz) has a greater stability in holding constant at
whatever frequency the crystal is originally cut to operate.
• Crystal Oscillators are used whenever great stability is required, such as
communication transmitters and receivers.
Characteristics of a Quartz Crystal
• A quartz crystal exhibits the property that when mechanical stress is applied
across one set of its faces, a difference of potential develops across the opposite
faces.
• This property of a Crystal is called ‘ Piezoelectric effect’.
• Similarly, a voltage applied across one set of faces of the Crystal causes
mechanical distortion in the Crystal shape.
• When alternating voltage is applied to a crystal, mechanical vibrations are set up
– these vibrations having a natural resonant frequency dependent on the Crystal.
• Although the Crystal has electromechanical resonance, we can represent the
Crystal action by equivalent electrical circuit as shown.
The inductor L and the capacitor C represent electrical equivalents of Crystal mass
and compliance respectively, whereas resistance R is an electrical equivalent of the
crystal structures internal friction. The shunt capacitance CM represents the
capacitance due to mechanical mounting of the crystal. Because the crystal losses,
represented by R, are small, the equivalent crystal Q factor is high – typically 20,000.
Values of Q up to almost 106 can be achieved by using Crystals. The Crystal can
have two resonant frequencies. One resonant condition occurs when the reactances of
the series RLC leg are equal. For this condition, the series – resonant impedance is
very low ( equal to R). The other resonant condition occurs at a higher frequency
when the reactance of the series resonant leg equals the reactance of the capacitor CM.
This is parallel resonance or antiresonance condition of the Crystal,
At this frequency, the crystal offers very high impedance to the external circuit.
To use the crystal properly, it must be connected in a circuit so that its low
impedance in the series resonant operating mode or high impedance in the
antiresonant operating mode is selected.
ANALOG ELECTRONIC CIRCUITS × Oscillator
The Hartley oscillator is similar to the Colpitts. The tank circuit has two inductors and
one capacitor. The calculation of the resonant frequency is the same.
LT= L1 + L2 + 2M
where, M is mutual coupling
ANALOG ELECTRONIC CIRCUITS × Oscillator
• The Colpitts oscillator utilizes a tank circuit (LC) in the feedback loop. The
resonant frequency can be determined by the formula below. Since the input
impedance affects the Q, an FET is a better choice for the active device.
• An Op amp Colpitts Oscillator circuit can also be used wherein the Op amp
provides the basic amplification needed and the Oscillator frequency is set by an
LC feedback network.
ANALOG ELECTRONIC CIRCUITS × Oscillator
• If a transistor is used as the active element of the amplifier stage, the output of the
feedback network is loaded appreciably by the relatively low input resistance
( hie) of the transistor.
• An emitter – follower input stage followed by a common emitter amplifier stage
could be used.If a single transistor stage is desired, the use of voltage – shunt
feedback is more suitable. Here, the feedback signal is coupled through the
feedback resistor R’ in series with the amplifier stage input resistance ( Ri).
f = (1/2pRC)[1/Ö 6 + 4(RC / R)]
hfe > 23 + 29 (R/RC) + 4 (RC / R)
ANALOG ELECTRONIC CIRCUITS × Oscillator
• The amplifier stage is self biased with a capacitor bypassed source resistor Rs and
a drain bias resistor RD . The FET device parameters of interest are gm and rd.
• |A| = gmRL, where RL = (RDrd / RD + rd)
• At the operating frequency, we can assume that the input impedance of the
amplifier is infinite.
• This is a valid approximation provided, the oscillator operating frequency is low
enough so that FET capacitive impedances can be neglected.
• The output impedance of the amplifier stage given by RL should also be small
compared to the impedance seen looking into the feedback network so that no
attenuation due to loading occurs.
ANALOG ELECTRONIC CIRCUITS × Oscillator
• The phase shift oscillator utilizes three RC circuits to provide 180º phase shift that
when coupled with the 180º of the op-amp itself provides the necessary feedback
to sustain oscillations.
• The gain must be at least 29 to maintain the oscillations. The frequency of
resonance for the this type is similar to any RC circuit oscillator:
fr = 1/26RC
ANALOG ELECTRONIC CIRCUITS × Oscillator
• An amplifier with positive feedback results in oscillations if the following
conditions are satisfied:
– The loop gain ( product of the gain of the amplifier and the gain of the
feedback network) is unity
– The total phase shift in the loop is 0°
• If the output signal is sinusoidal, such a circuit is referred to as sinusoidal
oscillator.
When the switch at the amplifier input is open, there are no oscillations. Imagine that a
voltage Vi is fed to the circuit and the switch is closed. This results in Vo = AV Vi and
bVo = Vf is fed back to the circuit. If we make Vf = Vi, then even if we remove the input
voltage to the circuit, the output continues to exist.
Vo = AV Vi
bVo = Vf
b AV Vi = Vf
If Vf has to be same as Vi, then from the above equation, it is clear that, b AV =1.
Thus in the above block diagram, by closing the switch and removing the input, we are
able to get the oscillations at the output if b AV =1, where b AV is called the Loop gain.
Positive feedback refers to the fact that the fed back signal is in phase with the input
signal. This means that the signal experiences 0° phase shift while traveling in the loop.
The above condition along with the unity loop gain needs to be satisfied to get the
sustained oscillations. These conditions are referred to as ‘Barkhausen criterion’.
Another way of seeing how the feedback circuit provides operation as an oscillator is
obtained by noting the denominator in the basic equation
Af = A / (1+bA).
When bA = -1 or magnitude 1 at a phase angle of 180°, the denominator becomes 0 and
the gain with feedback Af becomes infinite.Thus, an infinitesimal signal ( noise voltage)
can provide a measurable output voltage, and the circuit acts as an oscillator even without
an input signal.
ANALOG ELECTRONIC CIRCUITS × Power Amplifiers
Ø The transformer can step up or step down a voltage applied to primary coil .
Transformer coupled class A PA
• A form of class A amplifier having maximum efficiency of 50% uses
transformer to couple the output signal to the load.
Impedance transformation
Load resistance reflected to the primary side as,
Transformer coupled amplifier
• Drawing DC and AC load line
• Signal swing and output AC power
Power across the load can be expressed as
• with the output ac power then calculated using
PL=IL
2(rms) RL
ANALOG ELECTRONIC CIRCUITS × Power Amplifiers
It is a fixed bias circuit
DC bias operation
The DC bias set by Vcc and Rb
Collector current IC=βIB
Collector –emitter voltage
VCE=VCC-ICRC
Load line
Power considerations
Ø The power into an amplifier is provided by the power supply
Ø With no input supply, current drawn is collector bias current ICq.
pi(dc)=VCCICq
Output power
Ø The output voltage and current varying around the bias point provide ac
power to the load.
Using rms signals
Ø P0(ac)=VCE(rms)IC(rms)
=I2
C(rms) Rc
=V2
C(rms)/Rc
Ø Using peak signals
The ac power delivered to the load is
p0={VCE(p) IC(p)} / 2
or = {I2
c(p)/2} Rc
={V2
CE(p)}/2Rc
Ø Using peak-peak signals
P0(ac)={VCE(p-p) IC(p-p)}/8
= {I2
c(p-p)/8} Rc
={V2
CE(p)}/8Rc
Efficiency
Ø Efficiency of an amplifier represents the amount of ac power delivered from
dc source. It can be calculated using
Maximum Efficiency
Ø Maximum voltage swing VCE(p-p)=VCC
Ø Maximum current swing IC(p-p)=VCC/RC
Ø Maximum power
Ø The maximum power input evaluated using dc bias current set to half of the
maximum value….
Ø Maximum Pi(dc)=VCC(maximum IC)
maximum Pi(dc)} x100
= 25%
Maximum efficiency
Ø The maximum efficiency of a class A series fed amplifier is thus seen to be
25%.
Ø The maximum efficiency occurs only for ideal conditions of both voltage
and current swing .thus practical circuits will have less than this percentage.
ANALOG ELECTRONIC CIRCUITS × Power Amplifiers
• Class D amplifiers are much more efficient than Class AB power amplifiers.
As such, Class D amplifiers do not need large transformers and heavy
heatsinks, which means that they are smaller and lighter in weight than an
equivalent Class AB amplifier. All power devices in a Class D amplifier are
operated in on/off mode.
• These amplifiers use pulse width modulation,
ANALOG ELECTRONIC CIRCUITS × Power Amplifiers
• Class C amplifiers conduct less than 50% of the input signal and the
distortion at the output is high, but high efficiencies (up to 90%) are
possible. Some applications (for example, megaphones) can tolerate the
distortion. A much more common application for Class C amplifiers is in RF
transmitters, where the distortion can be vastly reduced by using tuned loads
on the amplifier stage.
• The input signal is used to roughly switch the amplifying device on and off,
which causes pulses of current to flow through a tuned circuit.
ANALOG ELECTRONIC CIRCUITS × Power Amplifiers
• A practical circuit using Class B elements is the complementary pair or
"push–pull" arrangement. Here, complementary or quasi-complementary
devices are used to each amplify the opposite halves of the input signal,
which is then recombined at the output. This arrangement gives excellent
efficiency, but can suffer from the drawback that there is a small mismatch
at the "joins" between the two halves of the signal..
• Class AB sacrifices some efficiency over class B in favor of linearity, so will
always be less efficient (below 78.5%). It is typically much more efficient
than class A.
ANALOG ELECTRONIC CIRCUITS × Power Amplifiers
• Class B amplifiers only amplify half of the input wave cycle. As such they
create a large amount of distortion, but their efficiency is greatly improved
and is much better than Class A. Class B has a maximum theoretical
efficiency of 78.5% (i.e., π/4). This is because the amplifying element is
switched off altogether half of the time, and so cannot dissipate power.
• A single Class B element is rarely found in practice, though it can be used in
RF power amplifier where the distortion levels are less important. However
Class C is more commonly used for this.
ANALOG ELECTRONIC CIRCUITS × Power Amplifiers
• Class A amplifying devices operate over the whole of the input cycle such
that the output signal is an exact scaled-up replica of the input with no
clipping. Class A amplifiers are the usual means of implementing smallsignal
amplifiers. They are not very efficient. a theoretical maximum of 50%
is obtainable with inductive output coupling and only 25% with capacitive
coupling.
• In a Class A circuit, the amplifying element is biased so the device is always
conducting to some extent, and is operated over the most linear portion of its
characteristic curve Because the device is always conducting, even if there is
no input at all, power is drawn from the power supply. This is the chief
reason for its inefficiency.
ANALOG ELECTRONIC CIRCUITS × Power Amplifiers
Ø Amplifier classes represent the amount the output signal varies over one
cycle of operation for a full cycle of input signal
Ø So the following classes of PA are defined
Ø Class A
Ø Class B
Ø Class AB
Ø Class C
Ø Class D
ANALOG ELECTRONIC CIRCUITS × Feedback Amplifier:
• Gain without feedback is A
• Feedback factor β
• Gain with feedback is (1+A β)
Voltage series feedback
• With zero feedback then Vf=0 the voltage gain of amplifier stage is
A=Vo/Vs=Vo/Vi
• If feedback of Vf is connected then,
Vi=Vs-Vf
Vo=AVi=A(Vs-Vf)=AVs-AVf=A(Vs-A(βVo)
Then, (1+ βA)Vo=AVs
Overall gain with feedback is
Af=Vo/Vi=A/(1+A β)
This shows that gain of feedback has reduced by factor (1+A β)
Voltage shunt feedback
• Af=Vo/Is=A Ii / (Ii+If)=AIi/(Ii+ βAIi)
• Af=A/(1+ βA)
Input impedance with FB
• Ref to fig(1)
Ii=Vi/Zi=(Vs-Vf) / Zi = (Vs- βVo) / Zi
Ii Zi= Vs- βAVi
Vs=Ii Zi+ β A Vi = Ii Zi+ β A Ii Zi
Zif = Vs/Ii=Zi+(βA)Zi=Zi(1+ βA)
ANALOG ELECTRONIC CIRCUITS × Feedback Amplifier:
Feedback connection types
• Voltage series feedback
• Voltage shunt feedback
• Current series feedback
• Current shunt feedback
Ø Here voltage refers to small part of voltage as input to the feedback network
Ø Current refers to tapping some part of output current through feedback
network.
Ø Series refers to connecting feedback signal in series with the input signal
voltage.
Ø Shunt refers to connecting feedback signal in shunt with the input signal
voltage
Ø Series feedback connections increases the input resistance
Ø Shunt feedback connections decreases the input resistance.
Voltage series feedback Af=Vo/Vs
Voltage shunt feedback
Current series feedback
Current shunt feedback
ANALOG ELECTRONIC CIRCUITS × Feedback Amplifier:
• Depending on the relative polarity of fed back signal in to the circuit, there
are two types of feedback
> Negative feedback
> Positive feedback
Negative feedback results in Reduced gain
Positive feedback are used in oscillators.
Feedback amplifier
Negative feedback circuits
• Reduces the gain
• Increases input impedance
• Better stabilized frequency response
• Lower output impedance
• Reduced noise
• More linear operation
Feedback connection types
• Voltage series feedback
• Voltage shunt feedback
• Current series feedback
• Current shunt feedback
Ø Here voltage refers to small part of voltage as input to the feedback network
Ø Current refers to tapping some part of output current through feedback
network.
Ø Series refers to connecting feedback signal in series with the input signal
voltage.
Ø Shunt refers to connecting feedback signal in shunt with the input signal
voltage
Ø Series feedback connections increases the input resistance
Ø Shunt feedback connections decreases the input resistance.
ANALOG ELECTRONIC CIRCUITS × General Amplifiers:
• Main feature of the Darlington connection is that the composite acts as a
single unit with a current gains of individual transistors.
• Darlington connection provides a current gain of βo= β1+ β2
• If β1= β2= β then βo= β2
• This configuration provides a transistor having a very large current gain,
typically a few thousands
