ANALOG CIRCUITS 18EC42 (Module - 3)
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Analog Circuits [18EC42]
Module -3
Feedback Amplifier: General feedback structure: Figure
shows the basic structure of a feedback amplifier.
Figure
shows a signal-flow diagram, where each of the quantities x can represent either a voltage or a current signal.
The
basic amplifier is unilateral and has a gain A A,, known as the open-loop output xo is related to the input x input xi by gain; thus its output x = The The feedback feedback network measures or samples the output signal x signal xo and provides a feedback signal x f that is related to x to xo by by the feedback factor β , x f = βx o The feedback signal x f is subtracted is subtracted from the source signal x s, which is the input to the complete feedback amplifier,1 to produce the signal xi, which is
the input to the basic amplifier, x = x x -x i
s
f
The Closed-Loop Gain: The gain of the feedback amplifier, known as the closed-loop gain or the gaindenoted A , is , with-feedback and denoted A f is defined as
≡ −−
We know that
=
= = Rearranging the above expression +
=
1 By: Mahendra Naik, Department of ECE, PESITM Shivamogga
Analog Circuits [18EC42]
≡ 1+
We get
=
=
+
The quantity Aβ quantity Aβ is called the loop gain.
Properties of negative feedback: 1. Desensitize the gain: that is, make the value of the gain less sensitive to variations in the values of circuit components, such as might be caused by changes in temperature. 2. Reduce nonlinear distortion: that is, make the output proportional to the input (in other words, make the gain constant, independent of signal level). 3. Reduce the effect of noise: that is, minimize the contribution to the output of unwanted electric signals generated, either by the circuit components themselves or by extraneous interference. 4. Control the input and output resistances: that is, raise or lower the input and output resistances by the selection of an appropriate feedback topology. 5. Extend the bandwidth of the amplifier.
The Four Basic Feedback Topologies: Based on the quantity to be amplified (voltage or current) and on the desired form of output (voltage or current), amplifiers can be classified into four categories. 1. Voltage Amplifiers 2. Current Amplifiers 3. Transconductance Amplifiers 4. Transresistance Amplifiers 1. Voltage Amplifiers: Voltage amplifiers are intended to amplify an input voltage signal and provide an output voltage signal. The voltage amplifier is essentially a voltage-controlled voltage- controlled voltage source. The input resistance is required to be high, and the output resistance is required to be low. The most suitable feedback topology for the voltage amplifier is the voltagemixing, voltage-sampling one shown in Fig.
2 By: Mahendra Naik, Department of ECE, PESITM Shivamogga
Analog Circuits [18EC42] Because
of the series connection at the input and the parallel or shunt connection at the output, this feedback topology is also known as series – shunt feedback .
Fig: Block diagram of a feedback voltage amplifier. 2. Current Amplifiers: The input signal in a current amplifier is essentially a current, the output quantity of interest is current; hence the feedback network should sample should sample the output current. The feedback signal should be in current form so that it may be mixed in shunt with the source current. Thus the feedback topology most suitable for a current amplifier is the current- mixing, current-sampling topology illustrated in Fig. Because of the parallel (or shunt) connection at the input, and the series connection at the output, this feedback topology is also known as shunt – series feedback .
Fig: Block diagram of a feedback current amplifier. 3 By: Mahendra Naik, Department of ECE, PESITM Shivamogga
Analog Circuits [18EC42] 3. Transconductan Transconductance ce Amplifiers: In transconductance amplifiers the input signal is a voltage and the output signal is a current. It follows that the appropriate feedback topology is the voltage-mixing , current-sampling topology, illustrated in Fig. The presence of the series connection at both the input and the output gives series feedback . this feedback topology the alternative name series – series
Fig: Block diagram of a feedback transconductance amplifier. 4. Transresistance Amplifiers: In transresistance amplifiers the input signal is current and the output signal is voltage. It follows that the appropriate feedback topology is of the current-mixing, voltage sampling type, shown in Fig. The presence of the parallel (or shunt) connection at both the input and the output makes this feedback topology also known as shunt – shunt shunt feedback.
Fig: Block diagram of a feedback transresistance amplifier.
The Series – Shunt Shunt Feedback Amplifier (The Voltage Amplifier): 4 By: Mahendra Naik, Department of ECE, PESITM Shivamogga
Analog Circuits [18EC42]
Fig: Ideal structure of the series – shunt shunt feedback amplifier The
ideal structure of the series – shunt shunt feedback amplifier is shown in Fig. It consists of a unilateral open-loop amplifier (the A (the A circuit) and an ideal voltage-sampling, voltage-mixing voltage-mixing feedback network (the β circuit). circuit). The The A A circuit has an input resistance R resistance Ri, an open-circuit voltage gain A gain A,, and an output resistance R resistance Ro. It is assumed that the source and load resistances have been absorbed inside the A the A circuit. The β circuit does not load the A the A circuit; that is, connecting the β circuit does
≡ ≡
). not change the value of A of A (defined as circuit of Fig. (a) Exactly follows the ideal feedback model. Therefore the closed-loop voltage gain A gain A f is given by
The
=
+ To determine Input resistance (R iif f ) and Output Resistance (R oof f ): The equivalent circuit model of the series – shunt shunt feedback amplifier is shown in Fig.
5 By: Mahendra Naik, Department of ECE, PESITM Shivamogga
Analog Circuits [18EC42]
Fig: Equivalent circuit of the series – shunt shunt feedback amplifier. Observe that is the open-circuit voltage gain of the feedback amplifier, is its input resistance, and is its output resistance.
Expressions for
:
We know that
Af
= 1 + ≡ = 1 + ×
∴ ∴ ∴
We know that
=
Substituting Vo in above equation gives:
=
=
=
×
1+
1+
1+
≡
Thus the input current becomes
=
Since as
=
is the current drawn from
1+
=
1+
, the input resistance
6 By: Mahendra Naik, Department of ECE, PESITM Shivamogga
can be expressed
Analog Circuits [18EC42]
≡ =
=
Thus
1+
=
=
+
Thus, as expected, the series-mixing feedback results in an increase in the amplifier input resistance by a factor equal to the amount of feedback, + , a highly desirable property for a voltage amplifier.
: Expressions for To determine the output resistance of the feedback amplifier in Fig (a), we set = 0 and apply a test voltage between the output terminals, as shown in Fig.
Fig: Determining the output resistance of the Series-Shunt Seri es-Shunt feedback amplifier If
is output resistance is the ≡ Applying KVL to the output loop, we get the following equations − − = 0 ∴ = − From the input loop we get = − the current drawn from
=
WKT
Thus
Since
=
=
7 By: Mahendra Naik, Department of ECE, PESITM Shivamogga
Analog Circuits [18EC42]
Substituting
in
− − − =
equation we get
=
+ 1+ = =
Substituting
in
≡ =
=
1+
=
∴
1+
+
Thus, as expected, the shunt sampling (or voltage sampling) at the output results in a decrease in the amplifier output resistance by a factor equal to the amount of negative feedback, + , a highly desirable property for a voltage amplifier.
The Series – Series Series Feedback Amplifier (The Transconductan Transconductance ce Amplifier):
Fig: The ideal structure of the series – series series feedback amplifier 8 By: Mahendra Naik, Department of ECE, PESITM Shivamogga
Analog Circuits [18EC42] The
ideal structure of the series – series series feedback amplifier is shown in Fig. The series – series series feedback topology stabilizes and is therefore best suited for transconductance amplifiers. Figure shows the ideal structure for the series – series series feedback amplifier. It consists of a unilateral open-loop amplifier (the A (the A circuit) and an ideal feedback network.
The A The A circuit inputresistance resistance R.i , a short-circuit transconductance resistance R andhas an an output
≡
The circuit samples the short-circuit output current and provides a feedback voltage = that is subtracted from in the series input loop. circuit presents zero resistance to the output loop, and thus does not The load the amplifier output. Since the structure of series – series series feedback amplifier follows the ideal feedback structure of feedback amplifier, we can obtain the closed-loop gain as
≡ =
1+
To determine Input resistance (R iif f ) and Output Resistance (R oof f ): The equivalent circuit model of the series – series series feedback amplifier is shown in Fig.
series feedback amplifier Fig: equivalent circuit of the series – series
Is the short-circuit transconductance.
Expressions for
:
9 By: Mahendra Naik, Department of ECE, PESITM Shivamogga
Analog Circuits [18EC42]
≡ ∴ ∴ ∴ We know that
Af
=
=
1+
×
1+
We know that
=
Substituting Vo in above equation gives:
=
=
=
×
1+
1+
1+
≡ ≡
Thus the input current becomes
=
Since as
=
1+
=
1+
, the input resistance
is the current drawn from
can be expressed
=
=
Thus
1+
=
=
Because
+
of the series mixing, the input resistance with feedback,
, will be
larger than the input resistance of the A the A circuit, , by a factor equal to the amount of feedback, + = Expressions for
:
10 By: Mahendra Naik, Department of ECE, PESITM Shivamogga
Analog Circuits [18EC42] To
find the output resistance of the series – series series feedback amplifier, we reduce to zero and break the output circuit to apply a test current I , as shown in Fig.
Fig: Equivalent circuit to determine the output resistance Rof resistance Rof of the series – series feedback amplifier.
≡ − − − − −− ∴ ≡
From the output circuit we get
=
From the input loop we get WKT Thus
=
Since
=
=
=
=
Substituting
in
equation we get
=
=
=
+
=
1+
= 1 +
= 1 +
11 By: Mahendra Naik, Department of ECE, PESITM Shivamogga
Analog Circuits [18EC42] Hence
= 1 +
That
is, in this case the negative feedback increases the output resistance.
The Shunt – Shunt Shunt Feedback Amplifier (Transresistance Amplifier):
Fig: Ideal structure for the shunt – shunt feedback amplifier. – shunt The
ideal structure of the shunt – – shunt shunt feedback amplifier is shown in Fig. The shunt – – shunt shunt feedback topology stabilizes and is thus best suited for transresistance amplifiers.
It
consists of a unilateral open-loop amplifier (the A circuit) and an ideal feedback network. The A circuit has an input resistance , an open-circuit transresistance , and an output resistance . The circuit ssamples amples the open-circuit output voltage and provides a feedback current that is subtracted from the signal-source current at the input nodes. The circuit presents infinite impedance to the amplifier output and thus does not load the amplifier output. The feedback signal = is provided as an ideal current source, and thus the circuit does not load the amplifier input.
≡
A is a transresistance, dimensionless quantity. is a transconductance and thus the loop gain A is a
12 By: Mahendra Naik, Department of ECE, PESITM Shivamogga
Analog Circuits [18EC42] The
closed-loop gain of shunt – shunt – shunt shunt feedback amplifier
≡ =
1+
as
To determine Input resistance (R iif f ) and Output Resistance (R oof f ): The equivalent circuit model of the shunt – – shunt shunt feedback amplifier is shown in
Fig.
Fig: Equivalent circuit of the shunt – – shunt shunt feedback amplifier The
feedback transresistance amplifier can be represented by the equivalent circuit in Fig. Is the open-circuit transresistance.
: Expressions for shunt feedback amplifier causes The shunt connection at the input of shunt – – shunt the feedback current to subtract from resulting in a reduced current into the A the A circuit, =
Substituting
− − =
=
and rearranging, results in = + = 1+ =
=
1+ = 1 + Equation indicates that the shunt mixing reduces the input current by the amount of feedback.
The
input resistance with feedback,
= I 1 + ≡
13 By: Mahendra Naik, Department of ECE, PESITM Shivamogga
Analog Circuits [18EC42] Where
=
Thus,
Substituting in above equation we get = 1 +
as expected, the shunt connection at the input lowers the input resistance by a factor equal to the amount of feedback.
: Expressions for Derivation same as series – shunt shunt amplifier
=
+
The Shunt – Series Series Feedback Amplifier (Current Amplifier):
Fig: Ideal structure for the shunt – series feedback amplifier. – series Figure shows the ideal structure for the shunt – – series series feedback amplifier. It
consists of a unilateral open-loop amplifier (the A circuit) and an ideal feedback network.
The A
circuit has an input resistance
, a short-circuit current gain
=
,
and an output resistance . The circuit samples the short-circuit output current and provides a feedback current that is subtracted from the signal-source current at the input node. The feedback signal = is provided as an ideal current source, and thus the circuit does not load the amplifier input. the closed-loop current gain of shunt – – series series feedback amplifier is
≡ = 1 +
14 By: Mahendra Naik, Department of ECE, PESITM Shivamogga
Analog Circuits [18EC42] To determine Input resistance (R iif f ) and Output Resistance (R oof f ): The equivalent circuit model of the shunt – – series series feedback amplifier is shown in Fig.
Fig: Equivalent circuit of the shunt – – series series feedback amplifier Expressions for
:
=
Expressions for
:
1+
= 1 +
For Derivation refer previous topologies
15 By: Mahendra Naik, Department of ECE, PESITM Shivamogga
Analog Circuits [18EC42]
OUTPUT STAGES AND POWER AMPLIFIERS: Classification of Output Stages: Output stages are classified according to the collector current waveform that
results when an input signal is applied. Class A Stage They 1. are The 2. The Class B Stage 3. The Class AB Stage 4. The Class C Amplifier Stage
The Class A Stage: The class A stage, whose associated waveform is shown in Fig. It is biased at a current greater than the amplitude of the signal current, The transistor in a class A stage conducts for the entire cycle of the input signal; that is, the conduction angle is 360°.
The Class B Stage: The class B stage, whose associated waveform is shown in Fig. It is biased at zero dc current. Thus a transistor in a class B stage
conducts for only half the cycle of the input sine wave, resulting in a conduction angle of 180°.
16 By: Mahendra Naik, Department of ECE, PESITM Shivamogga
.
Analog Circuits [18EC42] The Class AB Stage: The class AB stage, whose associated waveform is shown in Fig. An intermediate class between A and B, appropriately named class AB, involves biasing the transistor at a nonzero dc current much smaller than the peak current of the sine-wave signal. the transistor conducts for an interval slightly greater than half a cycle, as
illustrated in Fig. resulting conduction angle is greater than 180° but much less than 360°.
The
The Class C Amplifier Stage Figure shows the collector-current waveform for a transistor operated as a class C amplifier. The transistor conducts for an interval shorter than that of a halfcycle; that is, the conduction angle is less than 180°. The result is the periodically pulsating current waveform shown. To obtain a sinusoidal output voltage, this current is passed through a parallel LC parallel LC circuit, tuned to the frequency of the input sinusoid.
17 By: Mahendra Naik, Department of ECE, PESITM Shivamogga
Analog Circuits [18EC42] CLASS A OUTPUT STAGE: The transistor in a class A stage conducts for the entire cycle of the input
signal; that is, the conduction angle is 360°. Transfer characteristic of class an output stage:
Figure shows an emitter Q1 biased with a constant current I current I supplied by transistor Q2follower .
In the above circuit:
= + The bias current I current I must be greater than the largest negative load current; otherwise, Q1 cuts off and class A operation will no longer be maintained. The transfer characteristic of the emitter follower of above Fig. is described by
=
− Where depends on the emitter current and thus on the load current If.we neglect the relatively small changes in , the linear transfer curve shown in waveform results. 1
1
1
1
The
thus
positive limit of the linear region is determined by the saturation of Q1;
−
= 1 In the negative direction, depending on the values of I of I and and RL RL,, the limit of the linear region is determined either by Q1 turning off, = OR −
18 By: Mahendra Naik, Department of ECE, PESITM Shivamogga
Analog Circuits [18EC42] By
saturating Q2 we get
−
= + 2 The absolutely lowest (most negative) output voltage is that given g iven by above Eq. and is achieved provided the bias current I current I is greater than the magnitude of the corresponding load current, + 2
≥ −
Signal Waveforms: Consider the operation of the emitter-follower circuit of below Fig
19 By: Mahendra Naik, Department of ECE, PESITM Shivamogga
Analog Circuits [18EC42] For
a sine-wave input Neglecting V CE CE sat, sat, Selecting the proper bias current I, The output voltage can swing from −VCC − VCC to +VCC + VCC with the quiescent value being zero is as shown in Fig. (a).
Figure
Now,
(b) shows the corresponding waveform of
1=
V CC CC −
.
assuming that the bias current I is selected to allow a maximum negative load current of , that is
=
The
collector current of Q1 will have the waveform shown in Fig. (c).
Fig.
(d) shows the waveform of the instantaneous power dissipation in Q1, 1 1 1
≡
20 By: Mahendra Naik, Department of ECE, PESITM Shivamogga
Analog Circuits [18EC42] Power Dissipation: Below waveform indicates that the maximum instantaneous power dissipation in Q1 is
=
This
power dissipation is equal to the power dissipation in Q1 with no input signal applied, that is, the quiescent power dissipation. The emitter-follower transistor dissipates the largest amount of power when
= 0. The power dissipation in Q depends on the value of R of R . When = ∞( output open circuit): = I is constant and the instantaneous power dissipation in In this case, i = I 1
L
C 1
Q1 will depend on the instantaneous value of vO. The maximum power dissipation will occur when vO = −V CC CC , for in this case vCE 1 is a maximum of 2V 2V CC CC I . CC and P D 1 = 2V CC With an open-circuit load, the average power dissipation in Q1 is V CC I . CC When = 0 ( output short circuit): With an output short circuit, a positive input voltage would theoretically result in an infinite load current.
A very large current may flow through Q1, and if inthe short-circuit condition persists; the resulting large power dissipation Q1 can raise its junction temperature beyond the specified maximum, causing Q1 to burn up. To guard against such a situation, outputs stages are usually equipped with short-circuit protection. The power dissipation in Q2 also must be taken into account in designing an emitter follower output stage. Since Q2 conducts a constant current I current I , and the maximum value of vCE 2 is 2V CC CC , the maximum instantaneous power dissipation in Q2 is PD2=V CC CC I . A significant quantity for design purposes is the average power dissipation in Q2, which is V CC I . CC I
21 By: Mahendra Naik, Department of ECE, PESITM Shivamogga
Analog Circuits [18EC42] Power-Conversion Power-Conversi on Efficiency: The power-conversion efficiency of an output stage is defined as
≡ For
Load power
Supply power
the emitter follower circuit, assuming that the output o utput voltage is a
, the average load power will be 1 2 = = 2
sinusoid with the peak value
2
Since
2
the current in Q2 is constant ( I ), ), the power drawn from the negative
supply is V CC I . CC The
average current in Q1 is equal to I to I , and thus the average power drawn
from the positive supply is V CC I . CC
Thus the total average supply power is = + + = 2
− ∴ ≤ ≤
Substituting
and
in we get
1 2 = 2
Rearranging
=
2
1 4
1
4
, maximum efficiency is obtained when
=
The
2
the above equation we get
=
Since
=
maximum efficiency attainable is =
1
4
= 25%.
22 By: Mahendra Naik, Department of ECE, PESITM Shivamogga
Analog Circuits [18EC42] TRANSFORMER-COUPLED POWER AMPLIFIERS: In
the previous section we have seen transformer-less Class-A power amplifiers. Transformer-less power amplifier is also known as RC – coupled coupled power amplifiers. The efficiency of RC-coupled Class-A power amplifier is very less, i.e 25%. To improve the efficiency we prefer transformer coupled amplifiers. TRANSFORMER-COUPLED CLASS-A POWER AMPLIFIERS:
(a)
(b)
The
schematic diagram of a transformer-coupled class A power amplifier is as shown in the figure (a).
In this amplifier a By adjusting turn
transformer is used to couple ac power to the load. ratio of the primary windings to the secondary windings, one can match the source and load impedance for a maximum power transfer. This makes transformer coupled power amplifiers more efficient as compared to RC-coupled power amplifiers, as maximum power transfer can take place. The impedance matching of the transformer can be
1
1
=
2
2
Where
1
2
,
1
2
, and
1
2
1
2
=
2
1
are the number of turns, voltages, and
currents respectively, in the primary (secondary) coil of the transformer.
23 By: Mahendra Naik, Department of ECE, PESITM Shivamogga
Analog Circuits [18EC42]
∴ ′ 2
1
2
1
=
and represents the Since represents the effective load resistance output load resistance , then = ′ used to bias the transistor for class A The resistor R and R are 1
1
2
2
1
2
2
2
1
2
1
2
operation. The dc and ac load lines for the amplifier are shown in the figure (b). The dc load line is vertical to VCC, with infinite slope. The ac load line is -(1/R aacc), where R aacc is the ac resistance of the primary windings. The intersection of dc and ac load lines gives the operating point of the amplifiers. Due to counter- emf effect of the transformer, the output signal will swing from 0v to 2VCC as shown in fig (b). Efficiency (η) (η) of of transformer-coupled class-A power amplifier: a mplifier:
η
=
ac power delivered to the load Pac = Pdc dc power supplied
Pdc = Pac =
=
2 2
From the load line graph, the peak values of output voltage and current are equal to =VCC and =ICQ , respectively.
∴η
=
Hence
Pac Pdc
=
η VCC
2 × ICQ
=
Pac
Pdc
2
× 100 =
= 50%
24 By: Mahendra Naik, Department of ECE, PESITM Shivamogga
1 2
× 100 100
Analog Circuits [18EC42] CLASS B OUTPUT STAGE: Figure
shows a class B output stage. It consists of a complementary pair of transistors (an npn and a pnp pnp)) connected in such a way that both cannot conduct simultaneously.
Circuit Operation: When the input voltage v I is zero: Both transistors are cut off and the output voltage vO is zero. As v I goes positive and exceeds about 0.5 V: Q N conducts and operates as an emitter follower. In this case vO follows v I (i.e., vO = v I − v BEN ) and Q N supplies the load current. Meanwhile, the emitter – base junction of Q P will be reverse-biased by the V BE of Q N , which is approximately 0.7V.
Thus Q P will be cut off. negative by more than about 0.5 V: Q P turns on and acts as an emitter follower. Again vO follows v I (i.e., vO = v I + v EBP ), but in this case Q P supplies the load current and Q N will be cut off. The circuit operates in a push – pull pull fashion: Q N pushes (sources) current into the load when v I is positive, and Q P pulls (sinks) current from the load when v I is negative.
v I goes As
25 By: Mahendra Naik, Department of ECE, PESITM Shivamogga
Analog Circuits [18EC42] Transfer Characteristic: A sketch of the transfer characteristic of the class B stage is shown in Fig. Note that there exists a range of v I centered around zero where both transistors are cut off and vO is zero.
Crossover distortion:
The
dead band where both transistors are cut off vO is zero results in the crossover distortion illustrated in Fig. for the case of an input sine wave. The effect of crossover distortion will be most pronounced when the amplitude of the input signal is small. Crossover distortion in audio power amplifiers gives rise to unpleasant sounds. 26 By: Mahendra Naik, Department of ECE, PESITM Shivamogga
Analog Circuits [18EC42] POWER-CONVERSION EFFICIENCY: The power-conversion efficiency of an output stage is defined as
≡
Load power
Supply power To calculate the power-conversion power-conversion efficiency, η, of the class B stage, we neglect the crossover distortion and consider the case of an output sinusoid of peak amplitude . The average load power will be
2
=
The
2
=
1
2
2
current drawn from each supply will consist of half-sine waves of peak
amplitude Thus
.
the average current drawn from each of the two power supplies will be
.
It
follows that the average power drawn from each of the two power supplies will be the same, 1 = = +
the
− − − ≅
total supply power will be
=
Substituting
and
=
in we get
=
The
++
2
1 2
2
2
=
4
maximum efficiency is obtained when is at its maximum. This maximum is limited by the saturation of Q N and Q P to = At this value of peak output voltage, the power-conversion efficiency is
= = 78.5% 4 The maximum average power available from a class B output stage is obtained by substituting = = Substituting in we get 2 1 = 2 27 By: Mahendra Naik, Department of ECE, PESITM Shivamogga
Analog Circuits [18EC42] Power Dissipation:
The
quiescent power = 0 dissipation of the class B stage is zero. When an input signal is applied, the average average power power dissipated in the class B stage is given by
=
for− = results in we get Substituting for = and 1 2 = − 2 2
2
1
2
2
From
symmetry we see that half of P of P D is dissipated in Q N and the other half
in Q P . Thus
Q N and Q P must be capable of safely dissipating
Since
1 2
watts.
P D depends on we must find the worst-case power dissipation, . Eq. with respect to and equating the derivative to zero Differentiating that results in maximum average power dissipation as gives the value of
We get
− − − ∴
= 0
2
1 2
2
= 0
2
2
1 2
1 × 2 × 2
2
=
2
= 0
= 0
When maximum average power
=
Substituting
=
2
in
2
we get
28 By: Mahendra Naik, Department of ECE, PESITM Shivamogga
Analog Circuits [18EC42]
=
2
2
2
Thus
2
= = At the point of maximum power dissipation, the efficiency can be evaluated we get = 50%. by substituting for = into = . of P versus the peak output voltage Figure shows a sketch of P 2
2
4
D
From
2
π decreases the power dissipated in the class B stage while increasing the load
the above graph we can observe that: Increasing VO beyond VCC
power.
29 By: Mahendra Naik, Department of ECE, PESITM Shivamogga
Analog Circuits [18EC42] CLASS AB OUTPUT STAGE:
Crossover
distortion can be virtually eliminated by complementary output transistors at a small nonzero current.
biasing
the
The class AB output is shown in the Fig.bases of Q and Q . bias voltage V BB is stage applied between N P For v I = 0, vO = 0, and a voltage 2 appears across the base – emitter emitter junction of each of Q N and Q P . Assuming matched devices, 2 = = = The value of V q uiescent current current I I Q. BB is selected to yield the required quiescent
A
Circuit Operation: When v I goes
positive by a certain amount, the voltage at the base of Q N increases by the same amount and the output becomes positive at an almost equal value,
− =
+
2 The positive vO causes a current i L to flow through R through R L, and thus i N must increase; that is, = + The increase in i N will be accompanied by a corresponding increase in v BEN (above the quiescent value of V BB/2). However, since the voltage between the two bases remains constant at V BB, the increase in v BEN will result in an equal decrease in v EBP and hence in i P . The relationship between i N and i P can be derived as follows: + =
ln
+
ln
= 2
ln
30 By: Mahendra Naik, Department of ECE, PESITM Shivamogga
Analog Circuits [18EC42]
−
= 2 Thus, as i N increases, i P decreases by the same ratio while the product remains constant. Substituting = + in = 2 we get = 2 2 = 2 2 2 = 0 From the equations above, we can see that for positive output voltages, the load current is supplied by Q N , which acts as the output emitter follower. Meanwhile, Q P will be conducting a current that decreases as vO increases; for large vO the current in Q P can be ignored altogether. For negative input voltages the opposite occurs: The load current will be supplied by Q P , which acts as the output emitter follower, while Q N conducts a current that gets smaller as v I becomes becomes more negative.
∴ −−−
Transfer characteristic of the class AB stage:
Figure
shows the transfer characteristic of the class AB stage. For small v I , both transistors conduct, and as v I is increased or decreased, one of the two transistors takes over the operation. Since the transition is a smooth one, crossover distortion will be almost totally eliminated.
31 By: Mahendra Naik, Department of ECE, PESITM Shivamogga
Analog Circuits [18EC42] CLASS C OUTPUT STAGE: In
the class C amplifier, the transistor is biased such that it remains off for no-signal conditions and operates in the saturation region when an input signal is present. When the transistor is off, the current through it is very small and hence the
transistor dissipates negligible power. when transistor operates in saturation Similarly,
region, the voltage across it is very small, and again the power dissipation is small. Therefore, in the class C amplifier, as the transistor dissipates less power, its efficiency is higher than that of class A amplifier. However, drawback of the class C amplifier is, it is highly non-linear and produces distorted output. The drawback is overcome by connecting a low-pass filter at the output. The Class C amplifier:
Fig (a): THE CLASS C AMPLIFIER
Fig (b): Input and waveforms at the collector terminal of the class C amplifier 32 By: Mahendra Naik, Department of ECE, PESITM Shivamogga
Analog Circuits [18EC42] The
schematic diagram of class C amplifier is as shown in the figure (a). The figure (b) shows the input and the waveforms at the collector terminal. When the input signal is positive and above the cut-in voltage of the transistor, the transistor operates in saturation region. o During this period, the output voltage is equal to the saturation voltage of the transistor remains constant as long as the input voltage is above the cut-in and voltage. When the input voltage is less than the cut-in voltage, the transistor remains off, while the induced emf in the inductor provides the collector voltage as shown in the graph.
33 By: Mahendra Naik, Department of ECE, PESITM Shivamogga
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