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Analog Inverter and Scale Changer:
The circuit of analog inverter is shown in fig. 1. It is same as inverting voltage amplifier.
Assuming OPAMP to be an ideal one, the
differential input voltage is zero.
i.e. vd = 0
Therefore, v1 = v2 = 0
Since input impedance is very high, therefore,
input current is zero. OPAMP do not sink any
current.
\ iin= if
vin / R = - vO / Rf
vo = - (Rf / R) vin
If R = Rf then vO = -vin, the circuit behaves
like an inverter.
Fig. 1
If Rf / R = K (a constant) then the circuit is
called inverting amplifier or scale changer
voltages.
Inverting summer:
The configuration is shown in fig. 2. With three input voltages va, vb & vc. Depending upon the
value of Rf and the input resistors Ra, Rb, Rc the circuit can be used as a summing amplifier,
scaling amplifier, or averaging amplifier.
Again, for an ideal OPAMP, v1 =
v2. The current drawn by OPAMP
is zero. Thus, applying KCL at v2
node
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Fig. 2
This means that the output voltage
is equal to the negative sum of all
the inputs times the gain of the
circuit Rf/ R; hence the circuit is
called a summing amplifier. When
Rf= R then the output voltage is
equal to the negative sum of all
inputs.
vo= -(va+ vb+ vc)
If each input voltage is amplified by a different factor in other words weighted differently at the
output, the circuit is called then scaling amplifier.
The circuit can be used as an averaging circuit, in which the output voltage is equal to the
average of all the input voltages.
In this case, Ra= Rb= Rc = R and Rf / R = 1 / n where n is the number of inputs. Here Rf / R = 1 /
3.
vo = -(va+ vb + vc) / 3
In all these applications input could be either ac or dc.
Noninverting configuration:
If the input voltages are connected to noninverting input through resistors, then the circuit can be
used as a summing or averaging amplifier through proper selection of R1, R2, R3 and Rf. as
shown in fig. 3.
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To find the output voltage expression, v1 is
required. Applying superposition theorem, the
voltage v1 at the noninverting terminal is given
by
Hence the output voltage is
Fig. 3
This shows that the output is equal to the average of all input voltages times the gain of the
circuit (1+ Rf / R1), hence the name averaging amplifier.
If (1+Rf/ R1) is made equal to 3 then the output voltage becomes sum of all three input voltages.
vo = v a + vb+ vc
Hence, the circuit is called summing amplifier.
Example - 1
Find the gain of VO / Vi of the circuit of fig. 4.
Solution:
Current entering at the inveting
terminal
.
Applying KCL to node 1,
Fig. 4
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Applying KCL to node 2,
Thus the gain A = -8 V / VO
Example - 2
Find a relationship between VO and V1 through V6 in the circuit of fig. 5.
Fig. 5
Solution:
Let's consider of V1 (singly) by shorting the others i.e. the circuit then looks like as shown in fig.
6.
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The current flowing through the resistor R
into the i/e.
The current when passes through R, output
an operational value of
Let as now consider the case of V2 with other
inputs shorted, circuit looks like as shown in
fig. 7.
Fig. 6
Now VO is given by
same thing to V4 and V6
net output V" = V2 + V4 + V6
From (2) & (3)
(3)
V' + V" = (V2 + V4 + V6 ) - (V1 + V3 + V5)
So VO = V2 + V4 + V6 - V1 - V3 - V5.
Fig. 7
Example - 3
1. Show that the circuit of fig. 8 has A = VO / Vi = - K (R2 / R1) with K = 1 + R4 / R2
+ R4 / R3, and Ri = R1.
2. Specify resistance not larger than 100 K to achieve A = -200 V / V and Ri = 100
K.
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Fig. 8
Solution:
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Differential Amplifier:
The basic differential amplifier is shown in fig. 1.
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Fig. 1
Since there are two inputs superposition theorem can be used to find the output voltage. When
Vb= 0, then the circuit becomes inverting amplifier, hence the output due to Va only is
Vo(a) = -(Rf / R1) Va
Similarly when, Va = 0, the configuration is a inverting amplifier having a voltage divided
network at the noninverting input
Example - 1
Find vout and iout for the circuit shown in fig. 2. The input voltage is sinusoidal with amplitude of
0.5 V.
Fig. 2
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Solution:
We begin by writing the KCL equations at both the + and � terminals of the op-amp.
For the negative terminal,
Therefore,
15 v- = vout
For the positive terminal,
This yields two equations in three unknowns, vout, v+ and v-. The third equation is the
relationship between v+ and v- for the ideal OPAMP,
v+ = vSolving these equations, we find
vout = 10 vin = 5 sinωt V
Since 2 kΩ resistor forms the load of the op-amp, then the current iout is given by
Example - 2
For the different amplifier shown in fig. 3, verify that
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Fig. 3
Solution:
Since the differential input voltage of OPAMP is negligible, therefore,
v1= vx
and v2 = vy
The input impedance of OPAMP is very large and, therefore, the input current of OPAMP is
negligible.
Thus
And
From equation (E-1)
or
From equation (E-2)
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or
The OPAMP3 is working as differential amplifier, therefore,
Integrator:
A circuit in which the output voltage waveform is the integral of the input voltage waveform is
called integrator. Fig. 4, shows an integrator circuit using OPAMP.
Fig. 4
Here, the feedback element is a capacitor. The current drawn by OPAMP is zero and also the V2
is virtually grounded.
Therefore, i1 = if and v2 = v1 = 0
Integrating both sides with respect to time from 0 to t, we get
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The output voltage is directly proportional to the negative integral of the input voltage and
inversely proportional to the time constant RC.
If the input is a sine wave the output will be cosine wave. If the input is a square wave, the
output will be a triangular wave. For accurate integration, the time period of the input signal T
must be longer than or equal to RC.
Fig. 5, shows the output of integrator for square and sinusoidal inputs.
Fig. 5
Example - 3
Prove that the network shown in fig. 6 is a non-inverting integrator with
.
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Solution:
The voltage at point A is vO / 2 and it is also the
voltage at point B because different input voltage is
negligible.
vB = VO / 2
Therefore, applying Node current equation at point
B,
Fig. 6
Differentator:
A circuit in which the output voltage waveform is the differentiation of input voltage is called
differentiator.as shown in fig. 1.
Fig. 1
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The expression for the output voltage can be obtained from the Kirchoff's current equation
written at node v2.
Thus the output vo is equal to the RC times
the negative instantaneous rate of change
of the input voltage vin with time. A cosine
wave input produces sine output. fig. 1
also shows the output waveform for
different input voltages.
The input signal will be differentiated
properly if the time period T of the input
signal is larger than or equal to Rf C.
T ³ Rf C
As the frequency changes, the gain
changes. Also at higher frequencies the
circuit is highly susceptible at high
frequency noise and noise gets amplified.
Both the high frequency noise and
problem can be corrected by additing, few
components. as shown in fig. 2.
Fig. 2
Voltage to current converter:
Fig. 3, shows a voltage to current converter in which load resistor RL is floating (not connected
to ground).
The input voltage is applied to the non-inverting input terminal and the feedback voltage across
R drives the inverting input terminal. This circuit is also called a current series negative
feedback, amplifier because the feedback voltage across R depends on the output current iL and
is in series with the input difference voltage vd.
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Writing the voltage equation for the
input loop.
vin = vd + vf
But vd » since A is very large,therefore,
vin = vf
vin = R iin
iin = v in / R.
and since input current is zero.
iL = iin = vin ./ R
The value of load resistance does not
appear in this equation. Therefore, the
output current is independent of the
value of load resistance. Thus the input
voltage is converted into current, the
source must be capable of supplying this
load current.
Fig. 3
Grounded Load:
If the load has to be grounded, then the above circuit cannot be used. The modified circuit is
shown in fig. 4.
Since the collector and emitter currents are equal to a
close approximation and the input impedance of
OPAMP is very high,the load current also flows
through the feedback resistor R. On account of this,
there is still current feedback, which means that the
load current is stabilized.
Since vd= 0
\ v2 = v1 = vin
\ iout = (vCC� vin ) / R
Thus the load current becomes nearly equal to iout.
There is a limit to the output current that the circuit
can supply. The base current in the transistor equals
iout / bdc. Since the op-amp has to supply this base
current iout / bdc must be less than Iout (max) of the op-
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amp, typically 10 to 15mA.
Fig. 4
There is also a limit on the output voltage, as the load
resistance increases, the load voltage increases and
then the transistor goes into saturation. Since the
emitter is at Vin w. r. t. ground, the maximum load
voltage is slightly less than Vin.
In this circuit, because of negative feedback VBEis automatically adjusted. For instance, if the
load resistance decreases the load current tries to increase. This means that more voltage is
feedback to the inverting input, which decreases VBE just enough to almost completely nullify
the attempted increase in load current. From the output current expression it is clear that as Vin
increases the load current decreases.
Another circuit in which load current increases as Vin increases is shown in fig. 5.
The current through the first
transistor is
i = vin / R
This current produces a collector
voltage of vC = vCC� i R = vCC �
vin
Since this voltage drives the noninverting input of the second opamp. The inverting voltage is vCCvin to a close approximation. This
implies that the voltage across the
final R is
vCC - (vCC - vin ) = vin
and the output current .
iout = vin / R
As before, this output current must
satisfy the condition,that Iout/ bdc
must be less than the Iout(max) of the
OPAMP. Furthermore, the load
voltage cannot exceed vCC- vin
Fig. 5
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because of transistor saturation,
therefore Iout R must be less than
vCC- vin.This current source produces
unidirectional load current. fig. 6,
shows a Howland current source,
that can produce a bi-directional
load current.
Fig. 6
The maximum load current is VCC/ R. In this
circuit v in may be positive or negative.
Current to voltage converter:
The circuit shown in fig. 1, is a current to voltage converter.
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Fig. 1
Due to virtual ground the current through R is zero and the input current flows through Rf.
Therefore,
vout =-Rf * iin
The lower limit on current measure with this circuit is set by the bias current of the inverting
input.
Example �1:
For the current to current converter shown in fig. 2, prove that
Fig. 2
Solution:
The current through R1 can be obtained from the current divider circuit.
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Since, the input impedance of OPAMP is very large, the input current of OPAMP is negligible.
Thus,
Example - 2
(a). Verify that the circuit shown in fig. 3 has input impedance.
(b). If Z is a capacitor, show that the system behaves as an inductor.
(c). Find the value of C in order to obtain a 1H inductance if R1 = R2= 1K.
Fig. 3
Solution:
Let the output of OPAMP (1) be v and the output of OPAMP (2) be vo. Since the differential
input voltage of the OPAMP is negligible, therefore, the voltage at the inverting terminal of
OPAMP (1)will be vi.
For the OPAMP (2),
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or,
(b). Let the input voltage be sinusoidal of frequency (ω / 2π)
If Z is a capacitor, then Z = 1 / ωC
Let L = C R1 R2
Therefore,
and the system behaves as an inductor.
(c). Given R1 = R2 = 1 K, L = 1 H
Example - 3
Show that the circuit of fig. 4 is a current divider with io = ii / ( 1 + R2 + R1) regardless of the
load.
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Fig. 4
Solution:
since non-inverting and inverting terminals are virtually grounded.
Therefore, V1 = VL.
Now applying KCL to node (1)
Now current is
Example - 4
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(a). For the circuit shown in fig. 5 prove that
.
(b). Verify that if R3 / R4 = R1 / R2, the circuit is an instrumentation amplifier with gain A = 1 +
R2 / R1.
Solution:
Here
Fig. 5
(b).
So in this condition circuits as an instrument amplifier with gain
.
Example - 3
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Obtain an expression of the type iO = Vi / R - VO / RO for the circuit shown in fig. 6. Hence verify
that if R4 / R3 = R2 / R1 the circuit is a V-I converter with RO =∞ and R = R1 R5 / R2 .
Fig. 6
Solution:
Here
iO = current through the resistor.
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where
So when
then,
Filters:
A filter is a frequency selective circuit that, passes a specified band of frequencies and blocks or
attenuates signals of frequencies out side this band. Filter may be classified on a number of ways.
1. Analog or digital
2. Passive or active
3. Audio or radio frequency
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Analog filters are designed to process only signals while digital filters process analog signals
using digital technique. Depending on the type of elements used in their consideration, filters
may be classified as passive or active.
Elements used in passive filters are resistors, capacitors and inductors. Active filters, on the other
hand, employ transistors or OPAMPs, in addition to the resistor and capacitors. Depending upon
the elements the frequency range is decided.
RC filters are used for audio or low frequency operation. LC filters are employed at RF or high
frequencies.
The most commonly used filters are these:
1.
2.
3.
4.
5.
Low pass filters
High pass filter
Band pass filter
Band reject filter.
All pass filter
Fig. 1, shows the frequency response characteristics of the five types of filter. The ideal response
is shown by dashed line. While the solid lines indicates the practical filter response.
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Fig. 1
A low pass filter has a constant gain from 0 Hz to a high cutoff frequency fH. Therefore, the
bandwidth is fH. At fH the gain is down by 3db. After that the gain decreases as frequency
increases. The frequency range 0 to fH Hz is called pass band and beyond fH is called stop band.
Similarly, a high pass filter has a constant gain from very high frequency to a low cutoff
frequency fL. below fL the gain decreases as frequency decreases. At fL the gain is down by 3db.
The frequency range fL Hz to ∞ is called pass band and bleow fL is called stop band.
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