The Emitter Coupled Pair
The emitter coupled pair or differential pair is shown in figure below. The current source IEE is realized by
current mirror. We assume Q1 and Q2 to be identical transistors and both the collector resistances are fabricated
with equal values. Our objective here is to demonstrate that the differential pair can be utilized as an amplifier
or a switch. To accomplish this we develop the dc transfer characteristics of the circuit.
βπ1 + ππ΅πΈ1 β ππ΅πΈ2 + π2 = 0
πΌπΆ1 = πΌπΌπΈππ ππ΅πΈ1 ππ
πΌπΆ2 = πΌπΌπΈππ ππ΅πΈ2 ππ
πΌπΆ1πΌπΆ2
= π ππ΅πΈ1βππ΅πΈ2 ππ
ππ΅πΈ1 β ππ΅πΈ2 = π1 β π2 = ππ
πΌπΈ1 + πΌπΈ2 = πΌπΈπΈ =πΌπΆ1πΌ
+πΌπΆ2πΌ
πΌπΌπΈπΈπΌπΆ2
=πΌπΆ1πΌπΆ2
+ 1 = 1 + π π£π ππ
πΌπΆ2 =πΌπΌπΈπΈ
1 + π +ππ ππ
A similar analysis gives
πΌπΆ1 =πΌπΌπΈπΈ
1 + π βππ ππ
ππ1 β‘ ππΆπΆ β πΌπΆ1π πΆ
ππ2 β‘ ππΆπΆ β πΌπΆ2π πΆ
ππ = ππ1 β ππ2
Thus it acts as a switch and as an amplifier. When πΌπΆ2 β 0 at ππ β₯ 4ππ,
ππ2 = ππΆπΆ πππ ππ1 = ππΆπΆ β πΌπΌπΈπΈπ πΆ can be made small by taking large π πΆ .
Thus the output of Q1 acts as a closed switch and Q2 as an open switch. By
applying ππ β€ β4ππ, Q1 is open and Q2 is closed. Between β2πππππ 2ππ the
circuit is linear and acts as an amplifier.
The Differential Amplifier
The differential amplifier, emitter-coupled pair, or differential pair, is essential building block in modern IC amplifiers. This
circuit is displayed in Figure below. Both the transistors are matched, i.e., fabricated on a single chip. For a small difference
voltage Vd (4VT > |Vd|), the differential pair behave as a linear amplifier. Included in the figure below is the output resistance
RE of the current source bias network. We assume that the current across RE is negligible as compared to IEE. We assume that
Rs=0.
Differential Mode
For V1 = V2 and assuming π½πΉ β« 1, the collector and emitter currents in
each stage are equal πΌπΆ β |πΌπΈ| . All of these currents have magnitudes
equal to πΌπΈπΈ 2 (approx.) because of the symmetry of the circuit and the
negligible current in RE. Let us now increase V1 by βπ£ 2 and
simultaneously decrease V2 by βπ£ 2 . In effect, we are applying an
incremental signal βπ£ 2 at B1 and applying - βπ£ 2 to B2. The differential
voltage ππ = π1 β π2 increases by βπ£. For βπ£ < 4ππ , the circuit behaves
linearly. Thus IC1 increases by βπΌπΆ , and IC2 decreases by the same amount.
As πΌπΆ β πΌπΈ , the changes in IC1 and IC2 also appears at the emitters.
Consequently, the current in RE remains unchanged (the incremental
current in RE is zero), causing the voltage VE to remain constant.
The situation just described is referred to as the differential mode because the input signals ( βπ£ 2) applied to Q1 and Q2 are
equal and opposite and a difference signal Vd exists. For this mode, the incremental circuit can be redrawn as shown in Figure
(a) below.
(a) The differential-mode small-signal equivalent. (b) The common-mode small signal equivalent of the differential pair
The Common Mode
Now let us consider that both V1 and V2 increase by βπ£ 2. The difference voltage Vd remains zero, and IC1 and
IC2 remain equal. However, because RE is present, both IC1 and IC2 exhibit a small increase πΏπΌπΆ . Again changes in
IC appear at the emitter, and hence the current in RE increases by 2πΏπΌπΆ . The voltage VE is no longer constant but
must increase by 2πΏπΌπΆπ πΈ. This situation, where equal signals are applied to Q1 and Q2 is called common mode.
The incremental equivalent circuit is displayed in Figure (b) above, in which it is implied that Q1 and Q2 are
represented by their small signal models. We can write 2πΏπΌπΆπ πΈ as πΏπΌπΆ2π πΈ which is shown in the Figure (b)
above showing 2RE. The voltage across each is 2πΏπΌπΆπ πΈ and equals the incremental change in VE; thus the two
resistances are in parallel and 2π πΈ||2π πΈ = π πΈ .
It is evident in the Figures (a) and (b) above that depending on the input signal, the differential amplifier behaves
as either common-emitter stage or a common-emitter stage with emitter resistance. Therefore, the gain of this
stage is significantly higher for differential mode operation than for the common mode operation. Usually,
differential amplifiers are designed so that, for practical purposes, only difference signals are amplified.
As shown in the Figure (a) above, the emitter is at ground for small signal for small signal analysis. Thus it
appears that RE is bypassed. Similarly, the voltage between the two collectors Vo1 β Vo2 is zero in common mode
and is twice the change in Vo1 (Vo2) for the differential mode. Since the applied signal βπ£ can be made positive or
negative, the voltage Vo1 β Vo2 can be positive or negative.
Differential Mode Gain ADM
Consider that a signal VDM is applied at the base of Q1 in Figure below and that βVDM is applied to B2.
For this condition, the circuit in Figure (a) above is valid (with βπ£ 2replaced with VDM). Use of half-circuit concept, that is, analysis of only
one-half of the circuit, results in small signal model in Figure (a) below.
π΄π·π =ππ1ππ·π
=βπ½π πΆππ
= βπππ πΆ
For VDM positive, Vo1 = ADMVDM and as seen in the Equation above, ADM is negative, so that Vo1 is 180o out of phase with VDM
(Vo1 is inverted). Because Q2 is driven by βVDM, Vo2 = -ADMVDM and Vo2 is in phase with VDM (Vo2 is noninverting).
Common-Mode Gain ACM
When the signal VCM is applied to both bases in Figure on the left above, consider the Circuit in Figure (b) to the right above.
For this circuit, the gain ACM is
π΄πΆπ =ππ1ππΆπ
=βπ½π πΆ
2 π½ + 1 π πΈ + ππ
With π½ β« 1 and the division by ππ, reduces to
π΄πΆπ =βπππ πΆ
1+2πππ πΈβ β
π πΆ
2π πΈFor 2πππ πΈ β« 1.
Because the same signal is applied to Q1 and Q2, both Vo1 and Vo2 are 180o out of phase with VCM.
The Common-Mode Rejection Ratio
The differential amplifier is primarily designed to amplify differential signals; hence we
require π΄π·π β« π΄πΆπ. A convenient measure of the differential amplifier performance is the
common-mode rejection ratio or CMRR, defined as
πΆππ π β‘π΄π·π
π΄πΆπ
Combination of equation for ADM and ACM above and substituting in this equation yields
πΆππ π = 1 + 2πππ πΈ β 2πππ πΈ
As seen in this equation, large values of CMRR require large values of RE and often
necessitates the use of current sources having high output resistances. Note that if π πΈ β β,
πΆππ π β β, ACM = 0 and no common-mode component appears at the output.
Output for Arbitrary Input Signals
Consider signals V1 and V2 applied to Q1 and
Q2, respectively. This pair of signals can be
represented as the sum and difference of two
signals VDM and VCM
π1 = ππΆπ + ππ·π π2 = ππΆπ β ππ·π
Solving these equations for VDM and VCM
gives
ππ·π =π1βπ2
2=
ππ
2ππΆπ =
π1+π2
2
The effect of this decomposition is illustrated
in the Figure on the right. Because the circuit
behaves linearly, superposition applies. The
output consists of the two-components, one
attributed to the pair of sources VDM and the
other to the pair of sources VCM. Thus one
component is due to differential input signal,
and the second is due to common-mode input.
The output voltage Vo1 is given as
ππ1 = π΄π·πππ·π + π΄πΆπππΆπ = π΄π·π ππ·π +ππΆπ
πΆππ π
Equation above demonstrate the importance of CMRR if we are to amplify only the difference signals. As the
CMRR is increased, the common-mode output component has diminished significance. The out put voltage Vo2 is
expressed as
ππ2 = βπ΄π·πππ·π + π΄πΆπππΆπ = βπ΄π·π ππ·π βππΆπ
πΆππ π
Further simplifying we get
ππ1 =π΄π·π
2ππ +
π1+π2
πΆππ π
ππ2 =βπ΄π·π
2ππ β
π1+π2
πΆππ π
These equations are an alternative form for the output voltages that appear in the literature. The difference signal
Vd appears explicitly.
OPERATIONAL AMPLIFIER CHARACTERISTICS
The OPAMP is a two-input voltage controlled voltage source whose output voltage is proportional to the difference between
the two input voltages. It is a direct coupled device, high gain amplifier. The schematic diagram of the Op-Amp is displayed in
Figures (a) and (b) below.
The schematic diagram of the OPAMP is displayed in Figure (a) above and its equivalent
circuit in Figure (b) above. As seen in the Figure (b), the OPAMP is a voltage controlled
voltage source. The output voltage vo is the amplified difference signal vi = v1 β v2. The β and
+ symbols at the input of the OPAMP refer to the inverting and noninverting input terminals.
That is, if v2 = 0, vo is 180o out of phased with respect to the input signal vi. When v1 = 0, the
output vo is in phase with vi.
The Ideal OPAMP
The ideal OPAMP has the following characteristics:
1. The input resistance π π β β (open circuited). Consequently, no current enters either input
terminal.
2. The output resistance Ro = 0.
3. The voltage gain π΄π£ β β. The output voltage π£π = βπ΄π£π£π is finite ( π£π < β); thus as
π΄π£ β β, it is required that π£π = 0.4. The amplifier responds equally at all frequencies. (i.e. the bandwidth is infinite).
5. When π£1 = π£2, π£π = 0 and is independent of π£1 . The converse is also true.
The same symbol is used for ideal and practical OPAMP. To distinguish them, we indicate the
finite gain Av in the triangle for practical OPAMPS and omit it in ideal case. This is illustrated in
Figures (a) (b) & (c) above. The circuit in Figure (a) above is an inverting amplifier stage
utilizing an ideal OPAMP. Because the input current is zero, the current I exists in both R1 and
R2. Furthermore, since Vi = 0, it follows that
πΌ =ππ
π 1= β
ππ
π 2ππππ π€βππβ π€π πππ‘ π΄π£ =
π0
ππ = β
π 2
π 1
The OPAMP is used as noninverting amplifier stage in the circuit shown in Figures (b) & (c)
above. Making ππ = 0 requires that
ππ = π1 β π2 =π 1
π 1+π 2ππ β ππ = 0 ππππ£πππ πππ π΄π£ = ππ ππ yields π΄π£ =
π 1+π 2
π 1= 1 +
π 2
π 1
The above equations indicates that the feedback provided by R2 causes Av to depend only on the
resistance ratio R2/R1.
The Voltage Follower
The property of high impedance is very desirable feature of the noninverting configuration. It enables using this
circuit as a buffer amplifier to connect a source with a high impedance to a low-impedance load. In such cases we
make R2 = 0 and R1 = β to obtain the unity gain amplifier shown in the Figure (a) below. This circuit is commonly
referred to as a voltage follower, since the output follows the input. Figure (b) is its equivalent circuit model.
Difference Amplifiers
A difference amplifier is one that responds to difference between the two signals applied at its input and ideally
rejects signals that are common to the two inputs. The representation of signals in terms of their differential and
common mode components are given in the Figure below.
For practical cases voltage Vo is given by
π£π = π΄ππ£πΌπ + π΄πππ£πΌππ
Where Ad denotes the amplifier differential gain and Acm
denotes the common-mode gain. The efficacy of the
differential amplifier is measured by the degree of its
rejection of common-mode signals in preference to
differential signals. This is usually quantified by a
measure known as Common-Mode Rejection Ratio
(CMRR) defined as
πΆππ π = 20ππππ΄ππ΄ππ
A Single Op-Amp Difference Amplifier
Our first attempt at designing a difference amplifier is motivated by the observation that the gain of the noninverting amplifier
configuration is positive, 1 + π 2 π 1 , while that of the inverting configuration is negative β π 2 π 1 . Combining the two
configurations together is then a step in the right direction β namely, getting the difference between two input signals. We have
to make the two gain magnitudes equal in order to reject common-mode signals. This, however, can be easily achieved by
attenuating the positive input signal to reduce the gain of the positive path from 1 + π 2 π 1 to π 2 π 1 . The resulting
circuit would like as shown in Figure below.
Hence from this Figure
π 4
π 3+π 41 +
π 2
π 1=
π 2
π 1
Orπ 4
π 3+π 4=
π 2
π 1+π 2
This condition is satisfied by selecting
π 4
π 3=
π 2
π 1
To apply superposition, we first reduce vI2 to zero β that is, ground the terminal to which vI2 is applied β and then find the
output voltage, which will be due entirely to vI1. Hence, it is illustrated in Figure (a) below.
π£π1 = βπ 2
π 1π£πΌ1
Next we reduce vI1 to zero and evaluate the corresponding output vo2. The circuit will take the form as in Figure (b) below.
The output voltage vo2 is given by
π£π2 = π£πΌ2π 4
π 3+π 41 +
π 2
π 1=
π 2
π 1π£πΌ2
The superposition principle tells us that the output
voltage vo is equal to the sum of vo1 and vo2. Thus we
have
π£π =π 2
π 1π£πΌ2 β π£πΌ1 =
π 2
π 1π£πΌπ
Thus the circuit acts as a differential amplifier with
the differential gain Ad of π΄π =π 2
π 1.
Let us now consider the circuit with only the
common mode signal applied at the input, as shown
in the Figure on the right.
π1 =1
π 1π£πΌππ β
π 4
π 4+π 3π£πΌππ = π£πΌππ
π 3
π 4+π 3
1
π 1
The output voltage can now be found as
π£π =π 4
π 3+π 4π£πΌππ β π2π 2
π2 =1
π 2
π 4
π 4+π 3π£πΌππ β
1
π 2π£π
Substituting i2 = i1 we get
π£π =π 4
π 4 + π 3π£πΌππ β
π 2π 1
π 3π 3 + π 4
π£πΌππ =π 4
π 3 + π 41 β
π 2π 1
π 3π 4
π£πΌππ
Thus,
π΄ππ β‘π£π
π£πΌππ=
π 4
π 3+π 41 β
π 2
π 1
π 3
π 4for this design if R3 = R1 and R2 = R4 we get Acm = 0.
To find the input resistance between the two input terminals, called the differential input resistance Rid consider the
figure above again.
π ππ β‘π£πΌπ
π1also writing the loop equation
π£πΌπ = π 1π1 + 0 + π 1π1
Thus
π ππ = 2π 1
The Instrumentation Amplifier
It consists of two stages. The first stage is formed by opamps A1 and A2 and their associated resistors, and the second stage is
the difference amplifier formed by opamp A3 and its four associated resistors. A1 and A2 are connected in noninverting
configuration and thus realizes a gain of 1 + π 2 π 1 . It follows that each of vI1 and vI2 is amplified by this factor, and the
resulting amplified signal appear at the outputs of A1 and A2 respectively.
The difference amplifier in the second stage operates
on the difference signal 1 + π 2 π 1 π£πΌ2 β π£πΌ1 =1 + π 2 π 1 π£πΌπ and provides its output as
π£π =π 4
π 31 +
π 2
π 1π£πΌπ
Thus the differential gain realized is
π΄π =π 4
π 31 +
π 2
π 1
The common mode gain will be zero because of the
differencing action of the second stage amplifier. This
circuit has the advantage of very high (ideally infinite)
input resistance and high differential gain.
Voltage-to-Current Converter (Transconductance Amplifier)
The circuit for this configuration is given in Figure below. Thus we get the following equation. Note that iL is
independent of load ZL.
ππΏ π‘ =π£π (π‘)
π
Current βto-Voltage Converter (Transresistance Amplifier)
Photocells and Photomultiplier tubes give an output current which is independent of the load. The circuit in Figure below
shows an Op-Amp used as a current-to-voltage converter. Due the virtual ground at the amplifier input, the current through Rs
is zero and it flows through the feedback resistor Rβ. Thus the output vo = -isRβ. It is common to parallel Rβwith a Capacitance
Cβ to reduce high frequency noise and the possibility of oscillations. The current to voltage converter makes excellent current-
measuring instrument since it is an ammeter with zero voltage across the meter.
Logarithmic Amplifiers Using Matched Transistors
Consider the circuit of logarithmic amplifier given below.
We now derive the logarithmic expression for vo. For the
present discussion, ignore the high resistance balancing
potentiometer arrangements. For matched transistors and
with ππ΅ βͺ ππΆ, the positive input to A2 is at a voltage
π£ β‘ ππ΅πΈ2 β ππ΅πΈ1 = ππ ln ππΆ2 β ππππ ππΆ1 = βππππππΆ1
ππΆ2
Since v equals the small difference in the base-emitter
voltages of Q2 and Q1, we neglect v compared with the
reference voltage VR. Then since ππ΅2 βͺ ππΆ2 and because of
the virtual ground at the input of A1, it follows that
ππΆ2 =ππ
π 2and ππΆ1 =
π£π
π 1Since A2 is noninverting Op-
Amp,
π£π = π£ π 3 + π 4 /π 3, Combining these equations we get
π£π = βπ 3 + π 4
π 3ππ
π£π π 1
π 2ππ
Experimentally, it is found that the above equation is satisfied over a dynamic range of four decades., from input voltages of 2
mV to 20 V.
Exponential (Antilog) Amplifier
This amplifier is depicted in the given Figure. In the exponential amplifier the feedback current iC1 is constant and is derived
from the reference voltage VR, whereas iC2 depends upon the input signal. In logarithmic amplifier the converse of this is true.
Because of the virtual ground at the inputs to A1 and A2, the collector and base of Q1 are at the same potential βπ£ = ππ΅πΈ1 βππ΅πΈ2. Neglecting v relative to VR, we obtain
ππΆ1 =ππ
π 2and ππΆ2 =
π£π
π 1From the input attenuator it is clear that βπ£ =
π 3π£π
π 3+π 4= ππππ
ππΆ1
ππΆ2. Substituting the currents iC1
and iC2 from the above equations we obtain
π£π = βπππ 3+π 4
π 3ππ
π£π
π 1
π 2
ππ
Hence,
π£0 =π 1ππ
π 2ππ₯π β
π£π
ππ
π 3
π 3+π 4
The system is calibrated for mismatch and
offset voltages by setting vs = 0 and then
adjusting the potentiometer P until
π£π = π 1ππ π 2.
Comparator
An analog comparator, or simply a comparator, has two input voltages v1 and v2 and one output voltage vo. Often, the input (v2)
is constant reference voltage VR, and the other is time-varying signal. The ideal comparator is shown in Figure below. The
voltage transfer characteristic shown in Figures to the right below. It has the constant output voltage vo = V(0) if v1 β v2 = vi < 0
and a different constant voltage vo = V(1) if vi > 0. Hence, if v2 = VR, a reference voltage, vo = V(0) when v1 < VR and vo =
V(1) if v1 > VR.
Square Wave Generator From a Sinusoid
The comparator performs highly nonlinear wave shaping because the output bears no resemblance to the input waveform. It is
often used to transform a signal which varies slowly with time to another which exhibits an abrupt change. One such
application is the generation of square wave from sinusoidal signal. If VR is set to zero, the output will change from one stage
to the other very rapidly (limited by slew rate) every time the input passes through zero. Such a configuration is called zero-
crossing detector. If the input to a comparator is a sine wave, the output is square wave. If a zero crossing detector is used, a
symmetrical square wave results, as shown in the Figures below.
Regenerative Comparator (Schmitt Trigger)
The regenerative comparator of Figure (a) below is commonly referred to as a Schmitt trigger. The input voltage is applied to
the inverting terminal 2 and the feedback voltage to the noninverting terminal 1. Assuming that the output resistance of the
comparator is negligible compared with R1 + R, we obtain π£1 =π 2
π 1+π 2π£π
Since v1 = vi with v2 = 0, vo = Avvi and use of small signal analysis gives the return ratio as
π =βπ 2π΄π£
π 1+π 2Clearly, with Av > 0, T < 0 and the feedback is positive (Regenerative). Let Vo = VZ + VD and assume
that v2 < v1 so that vo = +Vo. From the Figures above we find that the voltage at the noninverting terminal is given
by
π£1 = ππ΄ +π 2
π 1+π 2ππ β ππ΄ β‘ π1
If v2 is now increased, vo remains constant at Vo, and v1 = V1 = constant until v2 = V1. At this threshold, critical, or
triggering voltage, the output regeneratively switches to vo = -Vo and remains at this value as long as v2 > V1. The
transfer characteristics is indicated in Figure (b) above. The voltage at the noninverting terminal for v2 > V1 is
π£1 = ππ΄ βπ 2
π 1+π 2ππ + ππ΄ β‘ π2
Note that V2 < V1, and the difference between theses two values is called the hysteresis VH
ππ» = π1 β π2 =2π 2ππ
π 1+π 2
Square Wave Generator
The inverting Schmitt trigger can be used to obtain a free-running square wave generator by connecting an RC feedback
network between the output and the inverting input. The circuit is displayed in Figure below.
In Figure (a) above the differential input voltage vi is given by
π£π = π£π β π£1 = π£π βπ 2
π 1+π 2π£π = π£π β π½π£π
From the ideal-comparator characteristics π£π = ππ· + ππ = ππ if vi < 0 and vo = - Vo if vi > 0. Consider an instance
of time when vi < 0 or vc < Ξ²vo = Ξ²Vo. The capacitor C charges exponentially toward Vo, through the RC
combination. The output remains at Vo until vc equals +Ξ²Vo at which time the comparator output reverses to βVo.
Now vc charges exponentially toward βVo. The output voltage vo and the capacitor voltage vc waveforms are shown
in Figure (b) above. If we let t = 0 when vc = -Ξ²Vo for the first half cycle, we have
π£π π‘ = ππ 1 β (1 + π½)πβ π‘ π πΆ
Since at t = T/2 vc(t) = +Ξ²Vo, we find T solving equation above is given by
π = 2π πΆππ1+π½
1βπ½= 2π πΆππ 1 +
2π 1
π 2Note that T is independent of Vo
This square wave generator is particularly useful in the frequency range of 10 Hz to 10 KHz. At higher frequencies
the slew rate of the OPAMP limits the slope of the output square wave.
Triangular Wave Generator
The triangular wave generator is illustrated in the Figure (a) below and the triangular wave is illustrated in the Figure (b) below.
To find the maximum value of the triangular waveform assume that output voltage vo of the Schmitt trigger is at its negative
value, -(VZ + VD) = -Vo. With a negative input, the output vβ(t) of the integrator is at an increasing ramp.
The voltage at the noninverting comparator input
v1 is obtained by the use of superposition and is
π£1 = βπππ 2
π 1+π 2+
π£πβ²π 1
π 1+π 2
When v1 rises to VR, the comparator changes
state, vo = +Vo and vβo(t) starts decreasing
linearly. Hence the peak Vmax of the triangular
waveform occurs for v1 = VR. From the above
equation
ππππ₯ = ππ π 1+π 2
π 1+ ππ
π 2
π 1
By similar argument it is found that
ππππ = ππ π 1+π 2
π 1β ππ
π 2
π 1
The peak to peak swing is ππππ₯ β ππππ = 2πππ 2
π 1
We now calculate the sweep time T1 and T2 for Vs = 0. The capacitor charging current is given
as
ππ = πΆππ£π
ππ‘= βπΆ
ππ£πβ²
ππ‘
Where vc = π£πβ² is the capacitor voltage. For vo = -Vo, i = -Vo/R, and the positive sweep speed is
ππ£πβ² ππ‘ = ππ π πΆ . Hence
π1 =ππππ₯βππππ
ππ π πΆ=
2π 2π πΆ
π 1
Since the negative sweep speed has the same magnitude as calculated above, T2 = T1 = T/2 =
1/2f, where the frequency f is given by
π =π 1
4π 2π πΆ
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