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Lecture - 1: Operational Amplifiers Operational Amplifiers: The operational amplifier is a direct-coupled high gain amplifier usable from 0 to over 1MH Z to which feedback is added to control its overall response characteristic i.e. gain and bandwidth. The op-amp exhibits the gain down to zero frequency. Such direct coupled (dc) amplifiers do not use blocking (coupling and by pass) capacitors since these would reduce the amplification to zero at zero frequency. Large by pass capacitors may be used but it is not possible to fabricate large capacitors on a IC chip. The capacitors fabricated are usually less than 20 pf. Transistor, diodes and resistors are also fabricated on the same chip. Differential Amplifiers: Differential amplifier is a basic building block of an op-amp. The function of a differential amplifier is to amplify the difference between two input signals. How the differential amplifier is developed? Let us consider two emitter-biased circuits as shown in fig. 1 . Fig. 1 The two transistors Q 1 and Q 2 have identical characteristics. The resistances of the circuits are equal, i.e. R E1 = R E2 , R C1 = R C2 and the magnitude of +V CC is equal to the magnitude of –V EE . These voltages are measured with respect to ground. To make a differential amplifier, the two circuits are connected as shown in fig. 1 . The two +V CC and –V EE supply terminals are made common because they are same. The two emitters are also connected and the parallel combination of R E1 and R E2 is replaced by a resistance R E . The two input signals v 1 & v 2 are applied at the base of Q 1 and at the base of Q 2 . The output voltage is taken between two collectors. The collector resistances are equal and therefore denoted by R C = R C1 = R C2 .
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Analog Circuts by Pramod Agrawal From Iit Rookee

Oct 30, 2014

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Lecture - 1: Operational Amplifiers Operational Amplifiers: The operational amplifier is a direct-coupled high gain amplifier usable from 0 to over 1MH Z to which feedback is added to control its overall response characteristic i.e. gain and bandwidth. The op-amp exhibits the gain down to zero frequency. Such direct coupled (dc) amplifiers do not use blocking (coupling and by pass) capacitors since these would reduce the amplification to zero at zero frequency. Large by pass capacitors may be used but it is not possible to fabricate large capacitors on a IC chip. The capacitors fabricated are usually less than 20 pf. Transistor, diodes and resistors are also fabricated on the same chip. Differential Amplifiers: Differential amplifier is a basic building block of an op-amp. The function of a differential amplifier is to amplify the difference between two input signals. How the differential amplifier is developed? Let us consider two emitter-biased circuits as shown in fig. 1. Fig. 1 The two transistors Q1 and Q2 have identical characteristics. The resistances of the circuits are equal, i.e. RE1 = R E2, RC1 = R C2 and the magnitude of +VCC is equal to the magnitude of VEE. These voltages are measured with respect to ground. To make a differential amplifier, the two circuits are connected as shown in fig. 1. The two +VCC and VEE supply terminals are made common because they are same. The two emitters are also connected and the parallel combination of RE1 and RE2 is replaced by a resistance RE. The two input signals v1 & v2 are applied at the base of Q1 and at the base of Q2. The output voltage is taken between two collectors. The collector resistances are equal and therefore denoted by RC = RC1 = RC2. Ideally, the output voltage is zero when the two inputs are equal. When v1 is greater then v2 the output voltage with the polarity shown appears. When v1 is less than v2, the output voltage has the opposite polarity. The differential amplifiers are of different configurations. The four differential amplifier configurations are following: 1.Dual input, balanced output differential amplifier. 2.Dual input, unbalanced output differential amplifier. 3.Single input balanced output differential amplifier. 4.Single input unbalanced output differential amplifier. Fig. 2 These configurations are shown in fig. 2, and are defined by number of input signals used and the way an output voltage is measured. If use two input signals, the configuration is said to be dual input, otherwise it is a single input configuration. On the other hand, if the output voltage is measured between two collectors, it is referred to as a balanced output because both the collectors are at the same dc potential w.r.t. ground. If the output is measured at one of the collectors w.r.t. ground, the configuration is called an unbalanced output. A multistage amplifier with a desired gain can be obtained using direct connection between successive stages of differential amplifiers. The advantage of direct coupling is that it removes the lower cut off frequency imposed by the coupling capacitors, and they are therefore, capable of amplifying dc as well as ac input signals. Dual Input, Balanced Output Differential Amplifier: The circuit is shown in fig. 1, v1 and v2 are the two inputs, applied to the bases of Q1 and Q2 transistors. The output voltage is measured between the two collectors C1 and C2 , which are at same dc potentials. D.C. Analysis: To obtain the operating point (ICC and VCEQ) for differential amplifier dc equivalent circuit is drawn by reducing the input voltages v1 and v2 to zero as shown in fig. 3. Fig. 3 The internal resistances of the input signals are denoted by RS because RS1= RS2. Since both emitter biased sections of the different amplifier are symmetrical in all respects, therefore, the operating point for only one section need to be determined. The same values of ICQ and VCEQ can be used for second transistor Q2. Applying KVL to the base emitter loop of the transistor Q1. The value of RE sets up the emitter current in transistors Q1 and Q2 for a given value of VEE. The emitter current in Q1 and Q2 are independent of collector resistance RC. The voltage at the emitter of Q1 is approximately equal to -VBE if the voltage drop across R is negligible. Knowing the value of IC the voltage at the collector VCis given by VC =VCC IC RC and VCE = VC VE = VCC IC RC + VBE VCE = VCC + VBE ICRC (E-2) From the two equations VCEQ and ICQ can be determined. This dc analysis applicable for all types of differential amplifier. Example - 1 The following specifications are given for the dual input, balanced-output differential amplifier of fig.1:RC = 2.2 k, RB = 4.7 k, Rin 1 = Rin 2 = 50 , +VCC = 10V, -VEE = -10 V, dc =100 and VBE = 0.715V. Determine the operating points (ICQ and VCEQ) of the two transistors. Solution: The value of ICQ can be obtained from equation (E-1).

The voltage VCEQ can be obtained from equation (E-2). The values of ICQ and VCEQ are same for both the transistors. Lecture - 2: Operational Amplifiers Dual Input, Balanced Output Difference Amplifier: The circuit is shown in fig. 1 v1 and v2 are the two inputs, applied to the bases of Q1 and Q2 transistors. The output voltage is measured between the two collectors C1 and C2, which are at same dc potentials. Fig. 1 A.C. Analysis : In previous lecture dc analysis has been done to obtain the operatiing point of the two transistors. To find the voltage gain Ad and the input resistance Ri of the differential amplifier, the ac equivalent circuit is drawn using r-parameters as shown in fig. 2. The dc voltages are reduced to zero and the ac equivalent of CE configuration is used. Fig. 2 Since the two dc emitter currents are equal. Therefore, resistance r'e1 and r'e2 are also equal and designated by r'e . This voltage across each collector resistance is shown 180out of phase with respect to the input voltages v1 and v2. This is same as in CE configuration. The polarity of the output voltage is shown in Figure. The collector C2 is assumed to be more positive with respect to collector C1 even though both are negative with respect to to ground. Applying KVL in two loops 1 & 2. Substituting current relations, Again, assuming RS1 /| and RS2 /| are very small in comparison with RE and re' and therefore neglecting these terms, Solving these two equations, ie1 and ie2 can be calculated. The output voltage VO is given by VO = VC2 - VC1 = -RC iC2 - (-RC iC1) = RC (iC1 - iC2) = RC (ie1 - ie2) Substituting ie1, & ie2 in the above expression Thus a differential amplifier amplifies the difference between two input signals. Defining the difference of input signals as vd = v1 v2 the voltage gain of the dual input balanced output differential amplifier can be given by (E-2) Differential Input Resistance: Differential input resistance is defined as the equivalent resistance that would be measured at either input terminal with the other terminal grounded. This means that the input resistance Ri1 seen from the input signal source v1 is determined with the signal source v2 set at zero. Similarly, the input signal v1 is set at zero to determine the input resistance Ri2 seen from the input signal source v2. Resistance RS1 and RS2 are ignored because they are very small. Substituting ie1, Similarly, The factor of 2 arises because the re' of each transistor is in series. To get very high input impedance with differential amplifier is to use Darlington transistors. Another ways is to use FET. Output Resistance: Output resistance is defined as the equivalent resistance that would be measured at output terminal with respect to ground. Therefore, the output resistance RO1 measured between collector C1and ground is equal to that of the collector resistance RC. Similarly the output resistance RO2 measured at C2 with respect to ground is equal to that of the collector resistor RC. RO1 = RO2 = RC(E-5) The current gain of the differential amplifier is undefined. Like CE amplifier the differential amplifier is a small signal amplifier. It is generally used as a voltage amplifier and not as current or power amplifier. Example - 1 The following specifications are given for the dual input, balanced-output differential amplifier: RC = 2.2 k, RB = 4.7 k, Rin 1 = Rin 2 = 50, +VCC= 10V, -VEE = -10 V, dc =100 and VBE = 0.715V. a.Determine the voltage gain. b.Determine the input resistance c.Determine the output resistance. Solution: (a). The parameters of the amplifiers are same as discussed in example-1 of lecture-1. The operating point of the two transistors obtained in lecture-1 are given below ICQ = 0.988 mA VCEQ=8.54V The ac emitter resistance Therefore, substituting the known values in voltage gain equation (E-2), we obtain b). The input resistance seen from each input source is given by (E-3) and (E-4): (c) The output resistance seen looking back into the circuit from each of the two output terminals is given by (E-5) Ro1 = Ro2 = 2.2 k Example - 2 For the dual input, balanced output differential amplifier of Example-1: a.Determine the output voltage (vo) if vin 1 = 50mV peak to peak (pp) at 1 kHz and vin 2 = 20 mV pp at 1 kHz. b.What is the maximum peal to peak output voltage without clipping? Solution: (a) In Example-1 we have determined the voltage gain of the dual input, balanced output differential amplifier. Substituting this voltage gain (Ad = 86.96) and given values of input voltages in (E-1), we get (b) Note that in case of dual input, balanced output difference amplifier, the output voltage vo is measured across the collector. Therefore, to calculate the maximum peak to peak output voltage, we need to determine the voltage drop across each collector resistor: Substituting IC = ICQ = 0.988 mA, we get This means that the maximum change in voltage across each collector resistor is 2.17 (ideally) or 4.34 VPP. In other words, the maximum peak to peak output voltage with out clipping is (2) (4.34) = 8.68 VPP. Lecture - 3: Difference Amplifiers A dual input, balanced output difference amplifier circuit is shown in fig. 1. Fig. 1 Inverting & Non inverting Inputs: In differential amplifier the output voltage vO is given by VO = Ad (v1 v2) When v2 = 0, vO = Ad v1 & when v1 = 0, vO = - Ad v2 Therefore the input voltage v1 is called the non inventing input because a positive voltage v1 acting alone produces a positive output voltage vO. Similarly, the positive voltage v2 acting alone produces a negative output voltage hence v2 is called inverting input. Consequently B1 is called noninverting input terminal and B2 is called inverting input terminal. 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 v1 and v2 are the two input signals, then the output of a practical op-amp cannot be described by simply v0 = Ad (v1 v2 ) In practical differential amplifier, the output depends not only on difference signal but also upon the common mode signal (average). vd = (v1 vd ) and vC = (v1 + v2 ) The output voltage, therefore can be expressed as vO = A1 v1 + A2 v2 Where A1 & A2 are the voltage amplification from input 1(2) to output under the condition that input 2 (1) is grounded. The voltage gain for the difference signal is Ad and for the common mode signal is AC. The ability of a differential amplifier to reject a common mode signal is expressed by its common mode rejection ratio (CMRR). It is the ratio of differential gain Ad to the common mode gain AC. Date sheet always specify CMRR in decibels CMRR = 20 log CMRR. Therefore, the differential amplifier should be designed so that is large compared with the ratio of the common mode signal to the difference signal. If = 1000, vC = 1mV, vd = 1 V, then It is equal to first term. Hence for an amplifier with = 1000, a 1 V difference of potential between two inputs gives the same output as 1mV signal applied with the same polarity to both inputs. 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. 2. 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.. Fig. 2 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. AC Analysis: The output voltage gain in this case is given by The voltage gain is half the gain of the dual input, balanced output differential amplifier. Since at the output there is a dc error voltage, therefore, to reduce the voltage to zero, this configuration is normally followed by a level translator circuit. 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. Fig. 3, 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. Fig. 3

Example-1 Consider example-1 of lecture-2. The specifications are given again for the dual input, unbalanced-output differential amplifier: RC = 2.2 k, RB= 4.7 k, Rin1 = Rin2= 50, +VCC = 10V, -VEE= -10 V, dc =100 and VBE= 0.715V. Determine the voltage gain, input resistance and the output resistance. Solution: Since the component values remain unchanged and the biasing arrangement is same, the ICQ and VCEQ values as well as input and output resistance values for the dual input, unbalanced output configuration must be the same as those for the dual input, balanced output configuration. Thus, ICQ = 0.988 mA VCEQ = 8.54 V Ri1 = Ri2 = 5.06 k Ro = 2.2 k The voltage gain of the dual input, unbalanced output differential amplifier is given by Example-2 Repeat Example-1 for single input, balanced output differential amplifier. Solution: Because the same biasing arrangement and same component values are used in both configurations, the results obtained in Example-1 for the dual input, balanced output configuration are also valid for the single input, balanced output configuration. That is, ICQ= 0.988 mA VCEQ = 8.54 V Vd = 86.96 Ri = 5.06 k Ro1 = Ro2 = 2.2 k Lecture - 4: Biasing of Differential Amplifiers Constant Current Bias: In the dc analysis of differential amplifier, we have seen that the emitter current IE depends upon the value of |dc. To make operating point stable IE current should be constant irrespective value of|dc. For constant IE, RE should be very large. This also increases the value of CMRR but if RE value is increased to very large value, IE (quiescent operating current) decreases. To maintain same value of IE, the emitter supply VEE must be increased. To get very high value of resistance RE and constant IE, current, current bias is used. Figure 5.1 Fig. 1, shows the dual input balanced output differential amplifier using a constant current bias. The resistance RE is replace by constant current transistor Q3. The dc collector current in Q3 is established by R1, R2, & RE. Applying the voltage divider rule, the voltage at the base of Q3 is Because the two halves of the differential amplifiers are symmetrical, each has half of the current IC3. The collector current, IC3 in transistor Q3 is fixed because no signal is injected into either the emitter or the base of Q3. Besides supplying constant emitter current, the constant current bias also provides a very high source resistance since the ac equivalent or the dc source is ideally an open circuit. Therefore, all the performance equations obtained for differential amplifier using emitter bias are also valid. As seen in IE expressions, the current depends upon VBE3. If temperature changes, VBE changes and current IE also changes. To improve thermal stability, a diode is placed in series with resistance R1as shown in fig. 2. Fig. 2 This helps to hold the current IE3 constant even though the temperature changes. Applying KVL to the base circuit of Q3. Therefore, the current IE3 is constant and independent of temperature because of the added diode D. Without D the current would vary with temperature because VBE3 decreases approximately by 2mV/C. The diode has same temperature dependence and hence the two variations cancel each other and IE3 does not vary appreciably with temperature. Since the cut in voltage VD of diode approximately the same value as the base to emitter voltage VBE3 of a transistor the above condition cannot be satisfied with one diode. Hence two diodes are used in series for VD. In this case the common mode gain reduces to zero. Some times zener diode may be used in place of diodes and resistance as shown in fig. 3. Zeners are available over a wide range of voltages and can have matching temperature coefficient The voltage at the base of transistor QB is Fig. 3 The value of R2 is selected so that I2 ~ 1.2 IZ(min) where IZ is the minimum current required to cause the zener diode to conduct in the reverse region, that is to block the rated voltage VZ. Current Mirror: The circuit in which the output current is forced to equal the input current is said to be a current mirror circuit. Thus in a current mirror circuit, the output current is a mirror image of the input current. The current mirror circuit is shown in fig. 4. Fig. 4 Once the current I2 is set up, the current IC3 is automatically established to be nearly equal to I2. The current mirror is a special case of constant current bias and the current mirror bias requires of constant current bias and therefore can be used to set up currents in differential amplifier stages. The current mirror bias requires fewer components than constant current bias circuits. Since Q3 and Q4 are identical transistors the current and voltage are approximately same For satisfactory operation two identical transistors are necessary. Example - 1 Design a zener constant current bias circuit as shown in fig. 5 according to the following specifications. (a). Emitter current -IE = 5 mA (b). Zener diode with Vz = 4.7 V and Iz = 53 mA. (c). ac = dc = 100, VBE = 0.715V (d). Supply voltage - VEE = - 9 V. Solution: From fig. 6 using KVL we get Fig. 5 Practically we use RE = 820 k Practically we use R2 = 68 The designed component values are: RE = 860 R2 = 68 Fig. 6 Example - 2 Design the dual-input balanced output differential amplifier using the diode constant current bias to meet the following specifications. 1.supply voltage = 12 V. 2.Emitter current IE in each differential amplifier transistor = 1.5 mA. 3.Voltage gain 60. Solution: The voltage at the base of transistor Q3 is Assuming that the transistor Q3 has the same characteristics as diode D1 and D2 that is VD = VBE3, then Practically we take RE = 240 . Fig. 7 Practically we take R2 = 3.6 k. To obtain the differential gain of 60, the required value of the collector resistor is The following fig. 7 shows the dual input, balanced output differential amplifier with the designed component values as RC = 1K, RE = 240 , and R2 = 3.6K. Lecture - 5: The Operational Amplifiers The operation amplifier: An operational amplifier is a direct coupled high gain amplifier consisting of one or more differential (OPAMP) amplifiers and followed by a level translator and an output stage. An operational amplifier is available as a single integrated circuit package. The block diagram of OPAMP is shown in fig. 1. Fig. 1 The input stage is a dual input balanced output differential amplifier. This stage provides most of the voltage gain of the amplifier and also establishes the input resistance of the OPAMP.The intermediate stage of OPAMP is another differential amplifier which is driven by the output of the first stage. This is usually dual input unbalanced output. Because direct coupling is used, the dc voltage level at the output of intermediate stage is well above ground potential. Therefore level shifting circuit is used to shift the dc level at the output downward to zero with respect to ground. The output stage is generally a push pull complementary amplifier. The output stage increases the output voltage swing and raises the current supplying capability of the OPAMP. It also provides low output resistance.

Fig. 3 Fig. 4, shows a complete OPAMP circuit having input different amplifiers with balanced output, intermediate stage with unbalanced output, level shifter and an output amplifier. Level Translator: Because of the direct coupling the dc level at the emitter rises from stages to stage. This increase in dc level tends to shift the operating point of the succeeding stages and therefore limits the output voltage swing and may even distort the output signal. To shift the output dc level to zero, level translator circuits are used. An emitter follower with voltage divider is the simplest form of level translator as shown in fig. 2. Thus a dc voltage at the base of Q produces 0V dc at the output. It is decided by R1 and R2. Instead of voltage divider emitter follower either with diode current bias or current mirror bias as shown in fig. 3 may be used to get better results. In this case, level shifter, which is common collector amplifier, shifts the level by 0.7V. If this shift is not sufficient, the output may be taken at the junction of two resistors in the emitter leg. Fig. 2 Fig. 4 Example-1: For the cascaded differential amplifier shown in fig. 5, determine: -The collector current and collector to emitter voltage for each transistor. -The overall voltage gain. -The input resistance. -The output resistance. Assume that for the transistors used hFE = 100 and VBE = 0.715V Fig. 5 Solution: (a). To determine the collector current and collector to emitter voltage of transistors Q1 and Q2, we assume that the inverting and non-inverting inputs are grounded. The collector currents (IC IE) in Q1 and Q2 are obtained as below: That is, IC1 = IC2 =0.988 mA. Now, we can calculate the voltage between collector and emitter for Q1 and Q2 using the collector current as follows: VC1 = VCC = -RC1 IC1 = 10 (2.2k) (0.988 mA) = 7.83 V = VC2 Since the voltage at the emitter of Q1 and Q2 is -0.715 V, VCE1 = VCE2 = VC1 -VE1 = 7.83 + 0715 = 8.545 V Next, we will determine the collector current in Q3 and Q4 by writing the Kirchhoff's voltage equation for the base emitter loop of the transistor Q3: VCC RC2 IC2 = VBE3 - R'E IC3 - RE2 (2 IE3) + VBE= 010 (2.2k) (0.988mA) - 0.715 - (100) (IE3) (30k) IE3 + 10=010 - 2.17 - 0.715 + 10 - (30.1k) IE3 = 0 Hence the voltage at the collector of Q3 and Q4 is VC3 = VC4= VCC RC3 IC3 = 10 (1.2k) (0.569 mA) = 9.32 V Therefore, VCE3 = VVCE4 = VC3 VE3 = 9.32 7.12 = 2.2 V Thus, for Q1 and Q2:ICQ = 0.988 mAVCEQ = 8.545 Vand for Q3 and Q4: ICQ = 0.569 mAVCEQ = 2.2 V [Note that the output terminal (VC4) is at 9.32 V and not at zero volts.] (b). First, we calculate the ac emitter resistance r'e of each stage and then its voltage gain. The first stage is a dual input, balanced output differential amplifier, therefore, its voltage gain is Where Ri2 = input resistance of the second stage The second stage is dual input, unbalanced output differential amplifier with swamping resistor R'E, the voltage gain of which is Hence the overall voltage gain is Ad= (Ad1) (Ad2) = (80.78) (4.17) = 336.85 Thus we can obtain a higher voltage gain by cascading differential amplifier stages. (c).The input resistance of the cascaded differential amplifier is the same as the input resistance of the first stage, that is Ri = 2ac(re1) = (200) (25.3) = 5.06 k (d). The output resistance of the cascaded differential amplifier is the same as the output resistance of the last stage. Hence, RO = RC = 1.2 k Example-2: For the circuit show in fig. 6, it is given that =100, VBE =0715V. Determine -The dc conditions for each state -The overall voltage gain -The maximum peak to peak output voltage swing. Fig. 6 Solution: (a). The base currents of transistors are neglected and VBE drops of all transistors are assumed same. From the dc equivalent circuit, and b) The overall voltage gain of the amplifier can be obtained as below: Therefore, voltage gain of second stage The input impedance of second stage is The effective load resistance for first stage is Therefore, the voltage gain of first stage is The overall voltge gain is AV = AV1 AV2 (c). The maximum peak to peak output votage swing = Vopp = 2 (VC7 - VE7) = 2 x (5.52 - 3.325) = 4.39 V Lecture - 6: Practical Operational Amplifier The symbolic diagram of an OPAMP is shown in fig. 1. 741c is most commonly used OPAMP available in IC package. It is an 8-pin DIP chip. Parameters of OPAMP: The various important parameters of OPAMP are follows: 1.Input Offset Voltage: Input offset voltage is defined as the voltage that must be applied between the two input terminals of an OPAMP to null or zero the output fig. 2, shows that two dc voltages are applied to input terminals to make the output zero. Vio = Vdc1 Vdc2 Vdc1 and Vdc2 are dc voltages and RS represents the source resistance. Viois the difference of Vdc1 and Vdc2. It may be positive or negative. For a 741C OPAMP the maximum value of Vio is 6mV. It means a voltage 6 mV is required to one of the input to reduce the output offset voltage to zero. The smaller the input offset voltage the better the differential amplifier, because its transistors are more closely matched. Fig. 2 2. Input offset Current: The input offset current Iio is the difference between the currents into inverting and non-inverting terminals of a balanced amplifier. Iio = | IB1 IB2 | The Iio for the 741C is 200nA maximum. As the matching between two input terminals is improved, the difference between IB1 and IB2 becomes smaller, i.e. the Iio value decreases further.For a precision OPAMP 741C, Iio is 6 nA 3.Input Bias Current: The input bias current IB is the average of the current entering the input terminals of a balanced amplifier i.e. IB = (IB1 + IB2 ) / 2 For 741C IB(max) = 700 nA and for precision 741C IB = 7 nA 4. Differential Input Resistance: (Ri) Ri is the equivalent resistance that can be measured at either the inverting or non-inverting input terminal with the other terminal grounded. For the 741C the input resistance is relatively high 2 M. For some OPAMP it may be up to 1000 G ohm. 5. Input Capacitance: (Ci) Ci is the equivalent capacitance that can be measured at either the inverting and noninverting terminal with the other terminal connected to ground. A typical value of Ci is 1.4 pf for the 741C. 6. Offset Voltage Adjustment Range: 741 OPAMP have offset voltage null capability. Pins 1 and 5 are marked offset null for this purpose. It can be done by connecting 10 K ohm pot between 1 and 5 as shown in fig. 3. Fig. 3 By varying the potentiometer, output offset voltage (with inputs grounded) can be reduced to zero volts. Thus the offset voltage adjustment range is the range through which the input offset voltage can be adjusted by varying 10 K pot. For the 741C the offset voltage adjustment range is 15 mV. Parameters of OPAMP: 7. Input Voltage Range : Input voltage range is the range of a common mode input signal for which a differential amplifier remains linear. It is used to determine the degree of matching between the inverting and noninverting input terminals. For the 741C, the range of the input common mode voltage is 13V maximum. This means that the common mode voltage applied at both input terminals can be as high as +13V or as low as 13V. 8. Common Mode Rejection Ratio (CMRR). CMRR is defined as the ratio of the differential voltage gain Ad to the common mode voltage gain ACM CMRR = Ad / ACM. For the 741C, CMRR is 90 dB typically. The higher the value of CMRR the better is the matching between two input terminals and the smaller is the output common mode voltage. 9. Supply voltage Rejection Ratio: (SVRR) SVRR is the ratio of the change in the input offset voltage to the corresponding change in power supply voltages. This is expressed in V / V or in decibels, SVRR can be defined as SVRR =A Vio / A V Where A V is the change in the input supply voltage and AVio is the corresponding change in the offset voltage. For the 741C, SVRR = 150 V / V. For 741C, SVRR is measured for both supply magnitudes increasing or decreasing simultaneously, with R3s 10K. For same OPAMPS, SVRR is separately specified as positive SVRR and negative SVRR. 10. Large Signal Voltage Gain: Since the OPAMP amplifies difference voltage between two input terminals, the voltage gain of the amplifier is defined as Because output signal amplitude is much large than the input signal the voltage gain is commonly called large signal voltage gain. For 741C is voltage gain is 200,000 typically. 11. Output voltage Swing: The ac output compliance PP is the maximum unclipped peak to peak output voltage that an OPAMP can produce. Since the quiescent output is ideally zero, the ac output voltage can swing positive or negative. This also indicates the values of positive and negative saturation voltages of the OPAMP. The output voltage never exceeds these limits for a given supply voltages +VCC and VEE. For a 741C it is 13 V. 12. Output Resistance: (RO) RO is the equivalent resistance that can be measured between the output terminal of the OPAMP and the ground. It is 75 ohm for the 741C OPAMP. Example - 1 Determine the output voltage in each of the following cases for the open loop differential amplifier of fig. 4: a.vin 1 = 5 m V dc, vin 2 = -7 Vdc b.vin 1 = 10 mV rms, vin 2= 20 mV rms Fig. 4 Specifications of the OPAMP are given below:A = 200,000, Ri = 2 M , R O = 75, + VCC = + 15 V, - VEE = - 15 V, and output voltage swing = 14V. Solution: (a). The output voltage of an OPAMP is given by Remember that vo = 2.4 V dc with the assumption that the dc output voltage is zero when the input signals are zero. (b). The output voltage equation is valid for both ac and dc input signals. The output voltage is given by Thus the theoretical value of output voltage vo = -2000 V rms. However, the OPAMP saturates at 14 V. Therefore, the actual output waveform will be clipped as shown fig. 5. This non-sinusoidal waveform is unacceptable in amplifier applications. Fig. 5 13. Output Short circuit Current : In some applications, an OPAMP may drive a load resistance that is approximately zero. Even its output impedance is 75 ohm but cannot supply large currents. Since OPAMP is low power device and so its output current is limited. The 741C can supply a maximum short circuit output current of only 25mA. 14. Supply Current : IS is the current drawn by the OPAMP from the supply. For the 741C OPAMP the supply current is 2.8 m A. 15. Power Consumption: Power consumption (PC) is the amount of quiescent power (vin= 0V) that must be consumed by the OPAMP in order to operate properly. The amount of power consumed by the 741C is 85 m W. Parameters of OPAMP: 16. Gain Bandwidth Product: The gain bandwidth product is the bandwidth of the OPAMP when the open loop voltage gain is reduced to 1. From open loop gain vs frequency graph At 1 MHz shown in. fig. 6, It can be found 1 MHz for the 741C OPAMP frequency the gain reduces to 1. The mid band voltage gain is 100, 000 and cut off frequency is 10Hz. Fig. 6 17. Slew Rate: Slew rate is defined as the maximum rate of change of output voltage per unit of time under large signal conditions and is expressed in volts / secs. To understand this, consider a charging current of a capacitor shown in fig. 7. Fig. 6 If 'i' is more, capacitor charges quickly. If 'i' is limited to Imax, then rate of change is also limited. Slew rate indicates how rapidly the output of an OPAMP can change in response to changes in the input frequency with input amplitude constant. The slew rate changes with change in voltage gain and is normally specified at unity gain. If the slope requirement is greater than the slew rate, then distortion occurs. For the 741C the slew rate is low 0.5 V / S. which limits its use in higher frequency applications. 18. Input Offset Voltage and Current Drift: It is also called average temperature coefficient of input offset voltage or input offset current. The input offset voltage drift is the ratio of the change in input offset voltage to change in temperature and expressed in V /C. Input offset voltage drift = ( A Vio /AT). Similarly, input offset current drift is the ratio of the change in input offset current to the change in temperature. Input offset current drift = ( A Iio / AT). For 741C, A Vio / A T = 0.5 V / C.A Iio/ A T = 12 pA / C. Lecture - 7: Parameters of an OPAMP Example - 1 A 100 PF capacitor has a maximum charging current of 150 A. What is the slew rate? Solution: C = 100 PF=100 x 10-12 FI = 150 A = 150 x 10-6 A Slew rate is 1.5 V / s. Example - 2 An operational amplifier has a slew rate of 2 V / s. If the peak output is 12 V, what is the power bandwidth? Solution: The slew rate of an operational amplifier is As for output free of distribution, the slews determines the maximum frequency of operation fmax for a desired output swing. soSo bandwidth = 26.5 kHz. Example - 3 For the given circuit in fig. 1. Iin(off) = 20 nA. If Vin(off) = 0, what is the differential input voltage?. If A = 105, what does the output offset voltage equal? Fig. 1 Solutin: Iin(off) = 20 nA Vin(off) = 0 (i) The differential input voltage = Iin(off) x 1k = 20 nA x 1 k = 20 V (ii) If A = 105 then the output offset voltage Vin(off) = 20 V x 105 = 2 volt Output offset voltage = 2 volts. Example - 4 R1 = 100, Rf = 8.2 k, RC = 10 k. Assume that the amplifier is nulled at 25C. If Vin is 20 mV peak sine wave at 100 Hz. Calculate Er, and Vo values at 45C for the circuit shown in fig. 2. Fig. 2 Solution: The change in temperature T = 45 - 25 = 20C. Error voltage = 51.44 mV Output voltage is 1640 mV peak ac signal which rides either on a +51.44 mV or -51.44 mV dc level. Example - 5 Design an input offset voltage compensating network for the operational amplifier A 715 for the circuit shown in fig. 3. Draw the complete circuit diagram. Fig. 3 Solution: From data sheet we get vin = 5 mV for the operational amplifier A 715. V = | VCC | = | - VEE | = 15 V Now, If we select RC = 10, the value of Rb should beRb = (3000) RC = 30000 = 304 Since R > Rmax, let RS = 10 Rmax where Rmax = Ra / 4. Therefore, If a 124 potentiometer is not available, we may prefer to use to the next lower value avilable, such as 104, so that the value of Ra will be larger than Rb by a factor of 10. If we select a 10 k potentiometer a s the Ra value, Rb is 12 times larger than Ra, Thus Ra = 10 k potentiometer Rb = 30 kRc = 10. The final circuit, which also includes the pin connections for the A 715, shown in fig. 4. Fig. 4 The ideal OPAMP : An ideal OPAMP would exhibit the following electrical characteristic. 1.Infinite voltage gain Ad 2.Infinite input resistance Ri, so that almost any signal source can drive it and there is no loading of the input source. 3.Zero output resistance RO, so that output can drive an infinite number of other devices. 4.Zero output voltage when input voltage is zero. 5.Infinite bandwidth so that any frequency signal from 0 to infinite Hz can be amplified without attenuation. 6.Infinite common mode rejection ratio so that the output common mode noise voltage is zero. 7.Infinite slew rate, so that output voltage changes occur simultaneously with input voltage changes. There are practical OPAMPs that can be made to approximate some of these characters using a negative feedback arrangement. Equivalent Circuit of an OPAMP: Fig. 5, shows an equivalent circuit of an OPAMP. v1 and v2are the two input voltage voltages. Ri is the input impedance of OPAMP. Ad Vd is an equivalent Thevenin voltage source and RO is the Thevenin equivalent impedance looking back into the terminal of an OPAMP. Fig. 5 This equivalent circuit is useful in analyzing the basic operating principles of OPAMP and in observing the effects of standard feedback arrangements vO = Ad (v1 v2) = Ad vd. This equation indicates that the output voltage vO is directly proportional to the algebraic difference between the two input voltages. In other words the OPAMP amplifies the difference between the two input voltages. It does not amplify the input voltages themselves. The polarity of the output voltage depends on the polarity of the difference voltage vd. Ideal Voltage Transfer Curve: The graphic representation of the output equation is shown in fig. 6 in which the output voltage vO is plotted against differential input voltage vd, keeping gain Ad constant. Fig. 6 The output voltage cannot exceed the positive and negative saturation voltages. These saturation voltages are specified for given values of supply voltages. This means that the output voltage is directly proportional to the input difference voltage only until it reaches the saturation voltages and thereafter the output voltage remains constant. Thus curve is called an ideal voltage transfer curve, ideal because output offset voltage is assumed to be zero. If the curve is drawn to scale, the curve would be almost vertical because of very large values of Ad. Lecture - 8: Open loop OPAMP Configuration Open loop OPAMP Configuration: In the case of amplifiers the term open loop indicates that no connection, exists between input and output terminals of any type. That is, the output signal is not fedback in any form as part of the input signal. In open loop configuration, The OPAMP functions as a high gain amplifier. There are three open loop OPAMP configurations. The Differential Amplifier: Fig. 1, shows the open loop differential amplifier in which input signals vin1 and vin2 are applied to the positive and negative input terminals. Fig. 1 Since the OPAMP amplifies the difference the between the two input signals, this configuration is called the differential amplifier. The OPAMP amplifies both ac and dc input signals. The source resistance Rin1 and Rin2 are normally negligible compared to the input resistance Ri. Therefore voltage drop across these resistances can be assumed to be zero. Therefore v1 = vin1 and v2 = vin2. vo = Ad (vin1 vin2 ) where, Ad is the open loop gain. The Inverting Amplifier: If the input is applied to only inverting terminal and non-inverting terminal is grounded then it is called inverting amplifier.This configuration is shown in fig. 2. v1= 0, v2 = vin. vo = -Ad vin Fig. 2 The negative sign indicates that the output voltage is out of phase with respect to input 180 or is of opposite polarity. Thus the input signal is amplified and inverted also. The non-inverting amplifier: In this configuration, the input voltage is applied to non-inverting terminals and inverting terminal is ground as shown in fig. 3. v1 = +vinv2 = 0 vo = +Ad vin This means that the input voltage is amplified by Ad and there is no phase reversal at the output. Fig. 3 In all there configurations any input signal slightly greater than zero drive the output to saturation level. This is because of very high gain. Thus when operated in open-loop, the output of the OPAMP is either negative or positive saturation or switches between positive and negative saturation levels. Therefore open loop op-amp is not used in linear applications. Closed Loop Amplifier: The gain of the OPAMP can be controlled if fedback is introduced in the circuit. That is, an output signal is fedback to the input either directly or via another network. If the signal fedback is of opposite or out phase by 180with respect to the input signal, the feedback is called negative fedback. An amplifier with negative fedback has a self-correcting ability of change in output voltage caused by changes in environmental conditions. It is also known as degenerative fedback because it reduces the output voltage and,in tern,reduces the voltage gain. If the signal is fedback in phase with the input signal, the feedback is called positive feedback. In positive feedback the feedback signal aids the input signal. It is also known as regenerative feedback. Positive feedback is necessary in oscillator circuits. The negative fedback stabilizes the gain, increases the bandwidth and changes, the input and output resistances. Other benefits are reduced distortion and reduced offset output voltage. It also reduces the effect of temperature and supply voltage variation on the output of an op-amp. A closed loop amplifier can be represented by two blocks one for an OPAMP and other for a feedback circuits. There are four following ways to connect these blocks. These connections are shown infig. 4. These connections are classified according to whether the voltage or current is feedback to the input in series or in parallel: -Voltage series feedback -Voltage shunt feedback -Current series feedback -Current shunt feedback Fig. 4 In all these circuits of fig. 4, the signal direction is from input to output for OPAMP and output to input for feedback circuit. Only first two, feedback in circuits are important. Voltage series feedback: It is also called non-inverting voltage feedback circuit. With this type of feedback, the input signal drives the non-inverting input of an amplifier; a fraction of the output voltage is then fed back to the inverting input. The op-amp is represented by its symbol including its large signal voltage gain Ad or A, and the feedback circuit is composed of two resistors R1 and Rf. as shown in fig. 5 Fig. 5 The feedback voltage always opposes the input voltage, (or is out of phase by 180with respect to input voltage), hence the feedback is said to be negative. The closed loop voltage gain is given by The product A and B is called loop gain. The gain loop gain is very large such that AB >> 1 This shows that overall voltage gain of the circuit equals the reciprocal of B, the feedback gain. It means that closed loop gain is no longer dependent on the gain of the op-amp, but depends on the feedback of the voltage divider. The feedback gain B can be precisely controlled and it is independent of the amplifier. Physically, what is happening in the circuit? The gain is approximately constant, even though differential voltage gain may change. Suppose A increases for some reasons (temperature change). Then the output voltage will try to increase. This means that more voltage is fedback to the inverting input, causing vd voltage to decrease. This almost completely offset the attempted increases in output voltage. Similarly, if A decreases, The output voltage decreases. It reduces the feedback voltage vf and hence, vd voltage increases. Thus the output voltage increases almost to same level. Different Input voltage is ideally zero. Again considering the voltage equation, vO = Ad vd or vd = vO / Ad Since Ad is very large (ideally infinite) vd ~ 0. andv1 = v2 (ideal). This says, that the voltage at non-inverting input terminal of an op-amp is approximately equal to that at the inverting input terminal provided that Ad is very large. This concept is useful in the analysis of closed loop OPAMP circuits. For example, ideal closed loop voltage again can be obtained using the results Lecture - 9: Closed Loop Amplifier Input Resistance with Feedback: fig. 1, shows a voltage series feedback with the OPAMP equivalent circuit. Fig. 1 In this circuit Ri is the input resistance (open loop) of the OPAMP and Rif is the input resistance of the feedback amplifier. The input resistance with feedback is defined as Since AB is much larger than 1, which means that Rif is much larger that Ri. Thus Rif approaches infinity and therefore, this amplifier approximates an ideal voltage amplifier. Output Resistance with Feedback: Output resistance is the resistance determined looking back into the feedback amplifier from the output terminal. To find output resistance with feedback Rf, input vin is reduced to zero, an external voltage Vo is applied as shown in fig. 2. Fig. 2 The output resistance (Rof ) is defined as This shows that the output resistance of the voltage series feedback amplifier is ( 1 / 1+AB ) times the output resistance Ro of the op-amp. It is very small because (1+AB) is very large. It approaches to zero for an ideal voltage amplifier. Reduced Non-linear Distortion: The final stage of an OPAMP has non-linear distortion when the signal swings over most of the ac load line. Large swings in current cause the r'e of a transistor to change during the cycle. In other words, the open loop gain varies throughout the cycle of when a large signal is being applied. It is this changing voltage gain that is a source of the non-linear distortion. Noninverting voltage feedback reduces non-linear distortion because the feedback stabilizes the closed loop voltage gain, making it almost independent of the changes in open loop voltage gain. As long as loop gain, is much greater than 1, the output voltage equals 1/B times the input voltage. This implies that output will be a more faithful reproduction of the input . Consider, under large signal conditions, the open loop OPAMP circuit produces a distortion voltage, designated vdist. It can be represented by connecting a source vdist in series with Avd. Without negative feedback all the distortion voltage vdist appears at the output. But with negative feedback, a fraction of vdist is feedback to inverting input. This is amplified and arrives at the output with inverted phase almost completely canceling the original distortion produced by the output stage. The first term is the amplified output voltage. The second term in the distortion that appears at the final output. The distortion voltage is very much, reduced because AB>>1 Bandwidth with Feedback: The bandwidth of an amplifier is defined as the band of frequencies for which the gain remains constant. Fig. 3, shows the open loop gain vs frequency curve of 741C OPAMP. From this curve for a gain of 2 x 105 the bandwidth is approximately 5Hz. On the other hand, the bandwidth is approximately 1MHz when the gain is unity. Fig. 3 The frequency at which gain equals 1 is known as the unity gain bandwidth. It is the maximum frequency the OPAMP can be used for.Furthermore, the gain bandwidth product obtained from the open loop gain vs frequency curve is equal to the unity gain bandwidth of the OPAMP.Since the gain bandwidth product is constant obviously the higher the gain the smaller the bandwidth and vice versa. If negative feedback is used gain decrease from A to A / (1+AB). Therefore the closed loop bandwidth increases by (1+AB). Bandwidth with feedback = (1+ A B) x (B.W. without feedback) ff= fo (1+A B) Output Offset Voltage: In an OPAMP even if the input voltage is zero an output voltage can exist. There are three cause of this unwanted offset voltage. 1.Input offset voltage. 2.Input bias voltage. 3.Input offset current. Fig. 4, shows a feedback amplifier with an output offset voltage source in series with the open loop output AVd. The actual output offset voltage with negative feedback is smaller. The reasoning is similar to that given for distortion. Some of the output offset voltage is fed back to the inverting input. After amplification an out of phase voltage arrives at the output canceling most of the original output offset voltage. When loop gain AB is much greater than 1, the closed loop output offset voltage is much smaller than the open loop output offset voltage. Fig. 4 Voltage Follower: The lowest gain that can be obtained from a non-inverting amplifier with feedback is 1. When the non-inverting amplifier gives unity gain, it is called voltage follower because the output voltage is equal to the input voltage and in phase with the input voltage. In other words the output voltage follows the input voltage. To obtain voltage follower, R1 is open circuited and Rf is shorted in a negative feedback amplifier of fig. 4. The resultant circuit is shown in fig. 5. vout = Avd= A (v1 v2) v1 = vin v2 =vout v1 = v2 if A >> 1 vout = vin. The gain of the feedback circuit (B) is 1. Therefore Af = 1 / B = 1 Fig. 5 Lecture - 10: Voltage Shunt Feedback Voltage shunt Feedback: Fig. 1, shows the voltage shunt feedback amplifier using OPAMP. Fig. 1 The input voltage drives the inverting terminal, and the amplified as well as inverted output signal is also applied to the inverting input via the feedback resistor Rf. This arrangement forms a negative feedback because any increase in the output signal results in a feedback signal into the inverting input signal causing a decrease in the output signal. The non-inverting terminal is grounded. Resistor R1 is connected in series with the source. The closed loop voltage gain can be obtained by, writing Kirchoff's current equation at the input node V2. The negative sign in equation indicates that the input and output signals are out of phase by 180. Therefore it is called inverting amplifier. The gain can be selected by selecting Rf and R1 (even < 1). Inverting Input at Virtual Ground: In the fig. 1, shown earlier, the noninverting terminal is grounded and the- input signal is applied to the inverting terminal via resistor R1. The difference input voltage vd is ideally zero, (vd= vO/ A) is the voltage at the inverting terminals (v2) is approximately equal to that of the noninverting terminal (v1). In other words, the inverting terminal voltage (v1) is approximately at ground potential. Therefore, it is said to be at virtual ground. Input Resistance with Feedback: To find the input resistance Miller equivalent of the feedback resistor Rf, is obtained, i.e. Rf is splitted into its two Miller components as shown in fig. 2. Therefore, input resistance with feedback Rif is then Fig. 2 Output Resistance with Feedback: The output resistance with feedback Rof is the resistance measured at the output terminal of the feedback amplifier. The output resistance can be obtained using Thevenin's equivalent circuit,shown in fig. 3. iO = ia + ib Since RO is very small as compared to Rf +(R1 || R2 ) Therefore,i.e. iO= ia vO = RO iO + A vd. vd= vi v2 = 0 - B vO Similarly, the bandwidth increases by (1+ AB) and total output offset voltage reduces by (1+AB). Fig. 3 Example - 1 (a).An inverting amplifier is implemented with R1 = 1K and Rf = 100 K. Find the percentge change in the closed loop gain A is the open loop gain a changes from 2 x 105 V / V to 5 x 104 V/V.(b) Repeat, but for a non-inverting amplifier with R1 = 1K at Rf = 99 K. Solution: (a). Inverting amplifier HereRf = 100 K R1 = 1K When, (b) Non-inverting amplifier HereRf = 99 K R1 = 1K Example - 2 An inverting amplifier shown in fig. 4 with R1 = 10 and R2 = 1M is driven by a source v1 = 0.1 V. Find the closed loop gain A, the percentage division of A from the ideal value - R2 / R1, and the inverting input voltage VN for the cases A = 100 V/V, 105 and 105 V/V. Solution: we have when A = 103,

Fig. 4 Example - 3 Find VN, V1 and VO for the circuit shown in fig. 5. Solution: Applying KCL at N or2VN + VN = VO.

Now VO - Vi = 6 as point A and N are virtually shorted. VO - VN = 6 VTherefore,VO = VN + 6 V

Therefore, VN = Vi = 3 V. Fig. 5 Lecture - 11: Applications of Operational Amplifiers 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 = 0Therefore, 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. If Rf / R = K (a constant) then the circuit is called inverting amplifier or scale changer voltages. Fig. 1 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 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) Fig. 2 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. 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, Applying KCL to node 2, Fig. 4 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. 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 (3) From (2) & (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. Fig. 8 Solution:

Lecture - 12: Applications of Operational Amplifiers Differential Amplifier: The basic differential amplifier is shown in fig. 1. 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 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+ = v- Solving these equations, we find vout = 10 vin = 5 sint V Since 2 k 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 Fig. 3 Solution: Since the differential input voltage of OPAMP is negligible, therefore, v1= vxand v2 = vy The input impedance of OPAMP is very large and, therefore, the input current of OPAMP is negligible. ThusAndFrom equation (E-1) or From equation (E-2) 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 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. 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 Lecture - 13: Applications of Operational Amplifiers 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 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> 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. Writing the voltage equation for the input loop. vin = vd + vf But vd since A is very large,therefore, vin = vfvin = 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 equalsiout /|dc. Since the op-amp has to supply this base current iout /|dcmust be less than Iout (max) of the op-amp, typically 10 to 15mA. 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. Fig. 4 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 non-inverting input of the second op-amp. The inverting voltage is vCC- vin 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/ |dc must be less than the Iout(max) of the OPAMP. Furthermore, the load voltage cannot exceed vCC- vin 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. 5 The maximum load current is VCC/ R. In this circuit v in may be positive or negative. Fig. 6 Lecture - 14: Applications Current to voltage converter: The circuit shown in fig. 1, is a current to voltage converter. 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. 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), or, (b). Let the input voltage be sinusoidal of frequency ( / 2) If Z is a capacitor, then Z = 1 / C 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. 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 (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 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. where So when then, 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), or, (b). Let the input voltage be sinusoidal of frequency ( / 2) If Z is a capacitor, then Z = 1 / C 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. 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 (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 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. where So when then, Lecture - 15: Applications 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 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.Low pass filters 2.High pass filter 3.Band pass filter 4.Band reject filter. 5.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. 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. First Order Low Pass Filter: Fig. 2, shows a first order low pass Butter-worth filter that uses an RC network for filtering, opamp is used in non-inverting configuration, R1 and Rf decides the gain of the filter. According to voltage divider rule, the voltage at the non-inverting terminal is: Fig. 2 Thus the low pass filter has a nearly constant gain Af from 0 Hz to high cut off frequency fH. At fH the gain is 0.707 Af and after fH it decreases at a constant rate with an increases in frequency. fH is called cutoff frequency because the gain of filter at this frequency is reduced by 3dB from 0Hz. Filter Design: A low pass filter can be designed using the following steps: 1.Choose a value of high cutoff frequency fH. 2.Select a value of C less than or equal to 1 F. 3.Calculate the value of R using. 4.Finally, select values of R1 and RF to set the desired gain using. Example - 1 Design a low pass filter at a cutoff frequency of 1 kH z with a pass band gain of 2. Solution: Given fH = 1 kHz. Let C = 0.01 F. Therefore, R can be obtained as A 20 k potentiometer can be used to set the resistance R. Since the pass band gain is 2, R1 and RF must be equal. Let R1 = R2 = 10 k. Low pass filter with adjustable corner frequency One advantage of active filter is that it is often quite simple to vary parameter values. As an example, a first-order low-pass filter with adjustable corner frequency is shown in fig. 3. Fig. 3 The voltage at the opamp inputs are given by Setting v + = v -, we obtain the voltage, v1, as follows: where The second opamp acts as an inverting integrator, and Note that we use upper case letters for the voltages since these are functions of s. K is the fraction of V1 sent to the integrator. That is, it is the potentiometer ratio, which is a number between 0 and 1. The transfer function is given by The dc gain is found by setting s = 0 (i.e., j =0) The corner frequency is at KA2 / RC. Thus, the frequency is adjustable and is proportional to K. Without use of the opamp, we would normally have a corner frequency which is inversely propostional to the resistor value. With a frequency proportional to K, we can use a linear taper potentiometer. The frequency is then linearly proportional to the setting of the potentiometer. Example - 2 Design a first order adjustable low-pass filter with a dc gain of 10 and a corner frequency adjustable from near 0 t0 1 KHz. Solution: There are six unknowns in this problems (RA, RF, R1, R2, R and C) and only three equations (gain, frequency and bias balance). This leaves three parameters open to choice. Suppose we choose the following values: C = 0.1 F R = 10 K R1 = 10K The ratio of R2 to R1 is the dc gain, so with a given value of R1 = 10K, R2 must be 100 k. We solve for A1 and A2 in the order to find the ratio, RF / RA. The maximum corner frequency occurs at K = 1, so this frequency is set to 2 x 1000. Since R and C are known, we find A2 = 6.28. Since A2 and A1 are related by the dc gain, we determine A1 / A2 = 10 and A1 = 62.8. Now, substituting the expression for A2, we find and since we find RF / RA = 68. RA is chosen to achieve bias balance. The impedance attached to the non-inverting input is 10 K || 100 K = 10 K. Fig. 4 If we assume that RF is large compared with RA ( we can check this assumption after solving for these resistors), the parallel combination will be close to the value of RA. We therefore can choose RA = 10 K. With this choice of RA, RF is found to be 680K and bias balance is achieved. The complete filter is shwn in fig. 4. Lecture - 16: Filters and Precision Diode Second Order Low-Pass Butterworth filter: A stop-band response having a 40-dB/decade at the cut-off frequency is obtained with the second-order low-pass filter. A first order low-pass filter can be converted into a second-order low-pass filter by using an additional RC network as shown in fig. 1. Fig. 1Fig. 2 The gain of the second order filter is set by R1 and RF, while the high cut-ff frequency fH is determined by R2, C2, R3 and C3 as follows: Furthermore, for a second-order low pass Butterworth response, the voltage gain magnitude is given by where, Except for having the different cut off frequency, the frequency response of the second order low pass filter is identical to that of the first order type as shown in fig. 2. Filter Design: The design steps of the second order filter are identical to those of the first order filter as given bellow: 1.Choose a value of high cutoff frequency fH. 2.To simplify the design calculations, set R2 = R3 = R and C2 = C3 = C. Then choose a value of C less than 1 F. 3.Calculate the value of R using. 4.Finally, because of the equal resistor (R2 = R3) and capacitor (C2 = C3) values, the pass band voltage gain AF has to be equal to 1.586. This gain is necessary to guarantee Butterworth response. Therefore, RF = 0.586 R1. Hence choose a value of R1= 100 k and calculate the value of RF. First Order High Pass Butterworth filter: Fig. 3, shows the circuit of first order high pass filter.This is formed by interchanging R and C in low pass filter. The lower cut off frequency is fL. This is the frequency at which the magnitude of the gain is 0.707 times its pass band value. All frequencies higher than fL are pass band frequencies with the highest frequency determined by the closed loop bandwidth of the OPAMP. The magnitude of the gain of the filter is

Fig. 3 If the two filters (high and low) band pass are connected in series it becomes wide band filter whose gain frequency response is shown in fig. 4. Fig. 4 Lecture - 16: Filters and Precision Diode Precision Diodes: If a sinusoid whose peak value is less than the threshold or cut in voltage Vd(-0.6V) is applied to the conventional half-wave rectifier circuit, output will remain zero. In order to be able to rectify small signals (mV), it is necessary to reduce Vd. By placing a diode in the feedback loop of an OPAMP, the cut in voltage is divided by the open loop gain A of the amplifier. Fig. 5, shows an active diode circuit. Fig. 5 Hence VD is virtually eliminated and the diode approaches the ideal rectifying element. If the input Vin goes positive by at least VD/A, then the output voltage (=A vd ) exceeds VD and D conducts and thus, provides a negative feedback. Because of the virtual connection between the two inputs vO= vin-vd=vin- v D / A ~ vin. Therefore, the circuit acts as voltage follower for positive signals (above 60V=0.6 / 1*105) when Vin swing negatively, D is OFF and no current is delivered to the external load. By reversing the diode, the active negative diode can be made. Active Clippers: By slightly modifying the circuit, an active diode ideal clipper circuit is obtained. Fig. 6, shows an active clipper which clips the input voltage below vR. Fig. 6 When vin < VR , then v' is positive and D conducts. Under these conditions, the OPAMP works as a buffer and the output voltage equals the voltage at non-inverting terminal Vout = VR. If vin > VR, then v' is negative and D is OFF and vO = vin RL / (RL + R) ~ Vi if R VR and vO is clamped to VRif vin < VR by about 60 V. Fig. 7, shows the output waveform of clipper circuit.When D is reverse biased a large differential voltage may appear between inputs and the OPAMP must be capable to withstand this voltage. Fig. 7 Lecture - 16: Filters and Precision Diode Active Half Wave Rectifier: The Active half wave rectifier is shown in fig. 8. Fig. 8Fig. 9 If vin is positive then output of the OPAMP becomes negative (the non inverting terminal is grounded). Thus diode D2 conducts and provides a negative feedback. Because of the feedback through D2a virtual ground exists at the input. Thus diode D1 acts as open circuit. The output voltage under this condition is given by vo = v - = 0. If vin goes negative, then output of the OPAMP becomes positive. Thus D1 is conducting and D2 is off. Thus, the circuit behaves as an inverting amplifier. The output of the circuit is given by The resultant output voltage will be positive. If v in is a sinusoid, the circuit performs half wave rectification. The transfer characteristic of the half wave active rectifier is shown in fig. 9. The output does not depend upon the diode forward voltage (vd). Thus, because of the high open loop gain of the OPAMP, the feedback acts to cancel the diode turn-on (forward) voltage. This leads to improved performance since the diode more closely approximates the ideal device. Axis Shifting of the Half Wave Rrectifier: The half wave rectified output waveform can be shifted along the vin axis. This is done by using a reference voltage added to the input voltage of the rectifier as shown in fig. 10. This termed axis shifting. It adds or subtracts a fixed dc voltage to the input signal. This process shifts the diode turn-on voltage point. If a negative reference voltage, VREF, is applied to the circuit, the diode turns on when the input voltage is still positive. This shifts the vout/ vin transfer characteristic to the right. If a positive reference voltage is applied, the vout/ vin transfer characteristic shifts to the left. These shifted characteristics are shown in fig. 10.

Fig. 10Fig. 11 The input-output voltage characteristics can also be shifted up or down. This is termed level shifting and is accomplished by adding a second OPAMP with a reference voltage added to the negative input terminal as shown in fig. 11. Lecture - 17: Applications of Operational Amplifiers Active Full Wave Rectifier: Method 1: A full wave rectifier, or magnitude operator, produces an output which is the absolute value, or magnitude, of the input signal waveform. One method of accomplishing full wave rectification is to use two half wave rectifiers. One of these operates on the positive portion of the input and the second operates on the negative portion. The outputs are summed with proper polarites. Fig. 1 illustrates one such configuration. Note that the resistive network attached to the ouput summing opamp is composed of resistors of higher value than those attached to the opamp that generates v1. This is necessary since for negative vin, v2 follows the curve shown above the node labled v2. That is, as the input increases in a negative direction, v2 increases in a positive direction. Since the input impedance to the non-inverting terminal of the summing opamp is high, the voltage, v+ is simply one half of v2 (i.e., the two 100K resistors form a voltage divider). The voltage at the negative summing terminal, v-, is the same as v+, and therefore is equal to v2 / 2. Now when vin is negative, D2 is open, and the node v1 is connected to the inverting input of the first opamp through a 5 K resistor. The inverting input is a virtual ground since the non-inverting input is tied to ground through a resistor. The result is that the voltage divider formed by the 100 K and 5K resistors. In order to achive a characteristic resembling that shown in the figure, this voltage divider must have a small ratio, on the order of 1 to 20. Fig. 1 Method 2: The method of full wave rectification discussed above requires three separate amplifiers. One simpler circuit or active full wave rectifier, which makes use of only two OPAMPs, is shown in fig. 2. It rectifies the input with a gain of R / R1, controllable by one resistor R1. Fig. 2 When v in is positive then v' = negative, D1 is ON and D2 is virtual ground at the input to (l). Because D2 is non-conducting, and since there is no current in the R which is connected to the non-inverting input to (2), therefore, V1 =0. Hence, the system consists of two OPAMP in cascade with the gain of A1 equal to (-R / R1) and the gain of A2 equal to (-R / R) = -1. The resultant at voltage output is vo = (R / R1 ) vin > 0 (for vin > 0 voltage output of (1) ) Consider now next half cycle when v in is negative. The v' is positive D1 is OFF and D2 is ON. Because of the virtually ground at the input to (2) V2 = V1 = V Since the input terminals of (2) are at the same (ground) potential, the current coming to the inverting terminal of (1) is as indicated in fig. 2. The output voltage is vo = i R + v where i = v / 2R (because input impedance of OPAMP is very high). The sign of vo is again positive because vin is negative in this half cycle. Therefore, outputs during two half cycles are same; and full wave rectified output voltage is obtained also shown in fig. 2. Lecture - 17: Applications of Operational Amplifiers Clampers: Fig. 3, shows an active positive clamper circuit. Fig. 3 The first negative half cycle produces a positive OPAMP output, which turns ON the diode. This capacitor charges to the peak of the input with the polarity shown in fig. 3. Just beyond the negative peak the diode turns off, the feedback loop opens, and the virtual ground is lost. Therefore, vout = vin + VP Since VP is being added to a sinusoidal voltage, the final output waveform is shifted positively through VP volts. The output wave form swing from 0 to 2VP as shows in fig. 4. Again the reduction of the diode-offset voltage allows clamping with low-level inputs. During most of the cycle, the OPAMP operates in negative saturation. Right at the negative input peak, the OPAMP produces a sharp positive going pulse that replaces any change lost by the clamping capacitor between negative input peaks. Fig. 4 Lecture - 17: Applications of Operational Amplifiers Comparators: An analog comparator has two inputs one is usually a constant reference voltage VR and other is a time varying signal vi and one output vO. The basic circuit of a comparator is shown in fig. 5. When the noninverting voltage is larger than the inverting voltage the comparator produces a high output voltage (+Vsat). When the non-inverting output is less than the inverting input the output is low (-Vsat). Fig. 5, also shows the output of a comparator for a sinusoidal. Fig. 5 vO = -Vsat if vi > VR = + Vsat if v i < VR If VR = 0, then slightest input voltage (in mV) is enough to saturate the OPAMP and the circuit acts as zero crossing detector as shown in fig. 6. If the supply voltages are 15V, then the output compliance is from approximate 13V to +13V. The more the open loop gain of OPAMP, the smaller the voltage required to saturate the output. If vdrequired is very small then the characteristic is a vertical line as shown in fig. 6. Fig. 6 If we want to limit the output voltage of the comparator two voltages (one positive and other negative) then a resistor R and two zener diodes are added to clamp the output of the comparator. The circuit of such comparator is shown in fig. 7, The transfer characteristics of the circuit is also shown in fig. 7. Fig. 7 The resistance is chosen so that the zener operates in zener region. When VR= 0 then the output changes rapidly from one state to other very rapidly every time that the input passes through zero as shown in fig. 8. Fig. 8 Such a configuration is called zero crossing detector. If we want pulses at zero crossing then a differentiator and a series diode is connected at the output. It produces singl Lecture - 18: Applications of Operational Amplifiers Schmitt Trigger: If the input to a comparator contains noise, the output may be erractive when vin is near a trip point. For instance, with a zero crossing, the output is low when vin is positive and high when vin is negative. If the input contains a noise voltage with a peak of 1mV or more, then the comparator will detect the zero crossing produced by the noise. Fig. 1, shows the output of zero crossing detection if the input contains noise. Fig. 1Figure 19.2 This can be avoided by using a Schmitt trigger, circuit which is basically a comparator with positive feedback. Fig. 2, shows an inverting Schmitt trigger circuit using OPAMP. Because of the voltage divider circuit, there is a positive feedback voltage. When OPAMP is positively saturated, a positive voltage is feedback to the non-inverting input, this positive voltage holds the output in high stage. (vin< vf). When the output voltage is negatively saturated, a negative voltage feedback to the inverting input, holding the output in low state. When the output is +Vsat then reference voltage Vref is given by If Vin is less than Vref output will remain +Vsat. When input vin exceeds Vref = +Vsat the output switches from +Vsat to Vsat. Then the reference voltage is given by The output will remain Vsat as long as vin > Vref. Fig. 3Fig. 4 If vin < Vref i.e. vin becomes more negative than Vsat then again output switches to +Vsat and so on. The transfer characteristic of Schmitt trigger circuit is shown in fig. 3. The output is also shown infig. 4 for a sinusoidal wave. If the input is different than sine even then the output will be determined in a same way. Lecture - 18: Applications of Operational Amplifiers Positive feedback has an unusual effect on the circuit. It forces the reference voltage to have the same polarity as the output voltage, The reference. voltage is positive when the output voltage is high (+vsat) and negative when the output is low (vsat). In a Schmitt trigger, the voltages at which the output switches from +vsat to vsat or vice versa are called upper trigger point (UTP) and lower trigger point (LTP). the difference between the two trip points is called hysteresis. Fig. 5 The hysteresis loop can be shifted to either side of zero point by connecting a voltage source as shown in fig. 5. When VO= +Vsat , the reference. Voltage (UTP) is given by When VO= -Vsat , the reference. Voltage (UTP) is given by If VR is positive the loop is shifted to right side; if VR is negative, the loop is shifted to left side. The hysteresis voltage Vhys remains the same. Non-inverting Schmitt trigger: In this case, again the feedback is given at non-inverting terminal. The inverting terminal is grounded and the input voltage is connected to non-inverting input. Fig. 6, shows an non-inverting schmitt trigger circuit. Fig. 6 To analyze the circuit behaviour, let us assume the output is negatively saturated. Then the feedback voltage is also negative (-Vsat). Then the feedback voltage is also negative. This feedback voltage will hold the output in negative saturation until the input voltage becomes positive enough to make voltage positive. When vin becomes positive and its magnitude is greater than (R2 / R1) Vsat, then the output switches to +Vsat. Therefore, the UTP at which the output switches to +Vsat, is given by Simillarly, when the output is in positive saturation, feedback voltage is positive. To switch output states, the input voltage has to become negative enough to make. When it happens, the output changes to the negative state from positive saturation to negative saturation voltage negative. When vin becomes negative and its magnitude is greater than R2 / R1 vsat, then the output switches to -vsat. Therefore, The difference of UTP and LTP gives the hysteresis of the Schmitt trigger. In non inverting Schmitt trigger circuit, the is defined as Lecture - 18: Applications of Operational Amplifiers Example - 1 Design a voltage level detector with noise immunity that indicates when an input signal crosses the nominal threshold of 2.5 V. The output is to switch from high to low when the signal crosses the threshold in the positive direction, and vice versa. Noise level expected is 0.2 VPP, maximum. Assume the output levels are VH = 10 V and VL = 0V. Solution: For the triggering action required an inverting configuration is required. Let the hysteresis voltage be 20% larger that the maximum pp noise voltage, that is, Vhys = 0.24V. Thus, the upper and lower trigger level voltages are -2.5 0.12, or UTP = 2.38 V and LTP = -2.62 V Since the output levels are VH and VL instead of +Vsat and Vsat, therefore, hysteresis voltage is given by orand The reference voltage V R can be obtained from the expression of LTP. Given that VL = 0, and LTP= -262, we obtain VR = (1 + R2 / R1) LTP = (1 + 1 / 40.7) (-2.62) = - 2.68 V We can select any values for R2 and R1 that satisfy the ratio of 40.7. It is a good practice to have more than 100 k for the sum of R1 and R2 and 1 k to 3k for the pull up resistor on the output. The circuit shown in fig. 7 shows a possible final design. The potentiometer serves as a fine adjustment for VR, while the voltage follower makes VR to appear as an almost ideal voltage source. Fig. 7 Lecture - 19: Schmitt Trigger and Relaxation Oscillator Example - 1 The Schmitt trigger circuit of fig. 1 uses 6V zener diodes with VD = 0.7 V. if the threshold voltage V1 is zero and the hysteresis is VH = 0.2V. Calculate R1 / R2 and VR. Fig. 1 Solution: The normal output voltage of Schmitt trigger circuit will be either +VO or VO,Where, VO = VZ + VD = 6.7 V Let the output voltage be +VO. The voltage V1 can be obtained from the voltage divider circuit consisting of R1 and R2. When vin > V1 then vo = -VOTherefore, upper trigger point voltage will be given by, Similarly, the lower trigger point voltage will be given by, Therefore, the hysteresis voltage is Since, the threshold voltage v1 is zero, Therefore, Relaxation Oscillator: With positive feedback it is also possible to build relaxation oscillator which produces rectangular wave. The circuit is shown in fig. 2. Fig. 2 In this circuit a fraction R2/ (R1 +R2) = | of the output is feedback to the non-inverting input terminal. The operation of the circuit can be explained as follows: Assume that the output voltage is +Vsat. The capacitor will charge exponentially toward +Vsat. The feedback voltage is +|Vsat. When capacitor voltage exceeds +|Vsat the output switches from +Vsat to -Vsat. The feedback voltage becomes -Vsat and the output will remain Vsat. Now the capacitor charges in the reverse direction. When capacitor voltage decreas