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Linear ic applications: UNIT-
DIFFERENTIAL AMPLIFIER:
A differential amplifier is a type of that amplifies the difference between two
input but suppresses any voltage common to the two inputs. It is an with two inputs Vin(+) and
Vin(-) and one output Vo in which the output is ideally proportional to the difference between the two voltages
Vo=A[Vin(+)-Vin(-)]
Where, A is the gain of the amplifier.
There are four different types of configuration in differential amplifier which are as follows:
i)Dual input and balanced output ii)Dual input and unbalanced output
iii)Single input and balanced output
iv)Single input and unbalanced output
1) DUAL INPUT, BALANCED OUTPUT DIFFERENTIAL AMPLIFIER
The circuit shown below is a dual-inputbalanced-output differential amplifier.Here in this circuit,the two input signals (dual input), vin1 and vin2, are applied to the bases B 1 and B 2 of transistors Q 1 and Q 2 .The output vo is measured between the two collectors C 1 and C 2 which are at the same dc potential.Because of the equal dc potential at the two collectors with respect to ground,the output is referred as a balanced output.
Circuit Diagram :-
DC Analysis :-
To determine the operating point values (ICQ and VCEQ) for the differential amplifier,we need to obtain a dc equivalent circuit.The dc equivalent circuit can be obtained simply by reducing the input signals vin1 and vin2 to zero.The dc equivalent circuit thus obtained is shown in fig below. Note that the internal resistances of the input signals are denoted by Rinbecause Rin1 = Rin2.Since both emitter biased sections of the differential amplifier are symmetrical (matched in all respects), we need to determine the operating point collector current ICQ and collector to emitter voltage VCEQ for only one section.We shall determine the ICQ and VCEQ values for transistor Q 1 only.These ICQ and VCEQ values can then be used for transistor Q 2 also.
DC EQUIVALENT CIRCUIT FOR DUAL-INPUT BALANCED OUTPUT DIFFERETIAL AMPLIFIER
Applying Kirchhoff’s voltage law to the base-emitter loop of the transistor Q 1 ,
RinIB- VBE - RE(2IE)+VEE = 0 (1)
AC EQUIVALENTCIRCUIT FOR DUAL-INPUT BALANCED OUTPUT DIFFERETIAL
AMPLIFIER
Voltage Gain :-
Before we proceed to derive the expression for the voltage gain Ad the following should be noted about the circuit in the figure above
- Ie1=Ie2; therefore re1=re2. For this reason the ac emitter resistance of transistors Q 1 and Q 2 is simply denoted by re.
- The voltage across each collector resistor is shown out of phase by 180^0 w.r.t the input voltages vin1 and vin2.
Writing Kirchhoff’s voltage eqautions for loops 1 and 2 gives us
vin1 – Rin1ib1 – reie1 – RE(ie1+ie2) = 0 (5)
vin2 – Rin2ie2 – reie2 – RE(ie1+ie2) = 0 (6)
Substituting current relations ib1 = ie1/B (^) ac and ib2 = ie2/B (^) ac yields
vin1 – Rin1ie1/Bac – reie1 – RE(ie1+ie2) = 0
vin2 – Rin2ie2/Bac – reie2 – RE(ie1+ie2) = 0
Generally, Rin1/B (^) ac and Rin2/B (^) ac values are very small therefore we shall neglect them here for simplicity and rearrange these equations as follows:
(re+RE)ie1 + RE2ie2 = vin1 (7)
RE2ie1 + (re+RE)ie2 = vin2 (8)
Eqns (7) and (8) can be solved simultaneously for ie1 and ie2 by using Cramer’s rule:
Ie1 = |(vin1/vin2)(RE/re+RE)|/|{(re+RE)/RE}{RE/(re+RE)}| (9a)
={(re+RE)vin1 – REvin2}/{(re+RE)^2 – (RE)^2 }
Similarly
Ie2 = |(vin1/vin2){(re+RE)/RE}|/|{(re+RE)/RE}{RE/(re+RE)}| (9b)
={(re+RE)vin2 – REvin2}/{(re+RE)^2 – (RE)^2 }
The output voltage is
vo = vc2 – vc
= -RCic2 – (-RCic1) (10)
= RCic1 – RCic
=RC(ie1 – ie2) since ic = ie
Substituting current relations ie1 and ie2 in eqn(10), we get
vo = RC[{(re+RE)vin1 - REvin2}/{(re+RE)^2 – (RE)^2 } - {(re+RE)vin2 - REvin1}/{(re+RE)^2 – (RE)^2 }]
= RC[{(re+RE)(vin1 – vin2)+(RE)(vin1 – vin2)}/{(re+RE)^2 – (RE)^2 }]
=RC[(re+2RE)(vin1 – vin2)/re(re+2RE)]
=(RC/re)(vin1 – vin2) (11)
Thus a differential amplifier amplifies the difference between two input signals as expected,the figure below shows the input and output waveforms of the dual-input balanced-output differential amplifier. By defining vid = vin1 as the difference in input voltages, we can write the voltage-gain equation of the dual-input balanced-output differential amplifier as follows:
Ad = vo/vid = RC/re
(12)
Similarly, the input resistance Ri2 seen from the input signal source vin2 is defined as
Ri2 = |vin2/ib2|Vin1=
=|vin2/(ie2/Bac)|Vin1=
Substituting the value of ie2 from eqn(9b), we get
Ri2 = Bacvin2/[{(re+RE)vin2 – RE(0)}/{(re+RE)^2 – (RE)^2 }] (15)
=[Bac(re^2 +2reRE)]/(re+RE)
=[Bac re(re+2RE)]/(re+RE)
However, (re+2RE) and (re+RE) = RE if RE>>re. Therefore eqn(15) can be rewritten as
Ri2 = Bacre(2RE)/RE = 2Bacre (16)
Output Resistance :-
Output resistance is defined as the equivalent resistance that would be measured at either output terminal w.r.t ground.
Ro1 = Ro2 = RC (17)
The current gain of the differential amplifier is undefined; therefore, the current-gain equation will not be derived for any of the four differential amplifier configurations.
Common mode Gain:-
A common mode signal is one that drives both inputs of a differential amplifier equally. The common mode signal is interference, static and other kinds of undesirable pickup etc.
The connecting wires on the input bases act like small antennas. If a differential amplifier is operating in an environment with lot of electromagnetic interference, each base picks up an unwanted interference voltage. If both the transistors were matched in all respects then the balanced output would be theoretically zero. This is the important characteristic of a differential amplifier. It discriminates against common mode input signals. In other words, it refuses to amplify the common mode signals.
The practical effectiveness of rejecting the common signal depends on the degree of matching between the two CE stages forming the differential amplifier. In other words, more closely are the currents in the input transistors, the better is the common mode signal rejection e.g. If v 1 and v 2 are the two input signals, then the output of a practical op-amp cannot be described by simply
v 0 = Ad (v 1 -v 2 )
In practical differential amplifier, the output depends not only on difference signal but also upon the common mode signal (average).
vd = (v 1 -v 2 )
andvC = ½ (v 1 + v 2 )
The output voltage, therefore can be expressed as
vO = A 1 v 1 + A 2 v 2
Where A 1 & A 2 are the voltage amplification from input 1(2) to output under the condition that input 2 (1) is grounded.
2)DUAL INPUT, UNBALANCED OUTPUT DIFFERENTIAL
AMPLIFIER:
In this case, two input signals are given however the output is measured at only one of the two- collector w.r.t. ground as shown in fig. 1. The output is referred to as an unbalanced output because the collector at which the output voltage is measured is at some finite dc potential with respect to
ground. In other words, there is some dc voltage at the output terminal without any input signal applied. DC analysis is exactly same as that of first case.
DC Analysis :
The dc analysis procedure for the dual input unbalanced output is identical to that dual input balanced output because both configuration use the same biasing arrangement. Therefore the emitter current and emitter to collector voltage for the dual input unbalanced output differential amplifier are determined from equations.
IE=ICQ= (VEE.VBE) / (2RE+βdc)
VCE=VCEQ=VCC+VBE-RCICQ
AC Analysis:
The output voltage gain in this case is given by
INPUT RESISTENCE:
The only difference between the circuits is the way output voltage is measured. The input resistance seen from either input source does not depend on the way the output voltage is measured.
Ri1=RI2=2βacre
OUTPUT RESISTENCE:
The output resistance R 0 measured at collector C 2 with respect to ground is equal to the collector resistor RC.
R 0 =RC
3)SINGLE INPUT, BALANCED OUTPUT DIFFERENTIAL AMPLIFIER:
From the figure of single input balanced output differential amplifier, input is applied to the base of transistor Q1 and the output is measured between 2 collectors which are at the same dc potential. Therefore ,the output is said to be a balanced output
DC Analysis:
The dc analysis procedure and bias equations for the single input balanced output differential amplifier are identical to those of the 2 previous configurations is the same.Thus the bias equations are
IE=ICQ=(VEE-VBE)/(2RE + Rinβdc)
VCE=VCEQ=VCC+VBE-RCICQ
AC Analysis :
The ac equivalent circuit of this differential amplifier with a small input T-equivalent model substituted for transistors
From input and output waveforms,
During the positive half cycle of the input signal, the base-emitter voltage of the transistor Q1 is positive and that of transistor Q2 is negative. This means that the collector current in Q1 increases and that in transistor Q2 decreases from the operating point value ICQ. This change in collector currents during the positive half cycle of the input signal is indicated in figure in which the currents of both the sources ic1 and ic2 are shown to be in the same direction. In fact, during the negative half cycle of the input signal, the opposite action takes place that is; the collector current of transistor Q1 decreases and that in transistor Q2 increases.
DIFFERENTIAL AMPLIFIER WITH SWAMPING RESISTORS
By using external resistors R'E in series with each emitter, the dependence of voltage gain on variations of r'e can be reduced. It also increases the linearity range of the differential amplifier shows the differential amplifier with swamping resistor R'E. The value of R'E is usually large enough to swamp the effect of r'E.
4)DUAL INPUT, BALANCED OUTPUT DIFFERENTIAL AMPLIFIER USING CONSTANT
CURRENT BIAS
The collector current IC3 in transistor Q3 is fixed and must be invariant signal is injected into either the emitter or the base of Q3.thus the transistor Q3 is a source of constant emitter current for transistor Q1 and Q2 of the differential amplifier.
Recall that in the analysis of differential amplifier circuit with emitter bias we required that Rb>>IC.Besides supplying constant emitter current the constant current bias also provides a very high source resistance since the ac equivalent of the dc current source is ideally a open circuit.therefore the
performance equations obtained for the differential amplifier configuration using emitter base are also applicable to differential amplifier using constant current bias.
To improve the thermal stability of constant bias replace R1 by diodes D1 and D2. Note that high to flows to the node at the base of Q3 and then divides paths IB3 if the temperature Q3 increases the emitter voltage VBE.
In silicon units VBEdecreases 2mv/c and in germanium units VBE decreases 1.6mv/c.this decreased VBE tends to raisethe voltage drop across R2and in turn current IE.for better performance of transistor CA3086 have been used a constant current bias.
R2 =(VEE - 1.4V)/IE
VE3 = -VEE + VZ–VBE
IE3 =( VE3 – (-VEE))/RE
IE3 =( VZ – VBE3)/RE
connected Vcc is the dc supply given to the circuit. The external resistances Re' and Re' are connected in series with each emitter. The dependence of the voltage gain of the differential amplifier or variations in Re can be reduced. Re also increases the linearity range of the differential amplifier. Generally value of Re'is large enough to swamp the effect of Re.for this reason the Re' is referred to as the swamping resistance.Vee the supply at the emitter resistance
Advantages of the sawmpling resistors is:
Input resistance is high Increase the linearity range of the differential amplifier Minimization of the changes in the transistor parameters due to the temperature
