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Lecture 01 The Operational Amplifier Op Amp, A differential amplifier forms the input stage of operational amplifiers The term. differential comes from the amplifier s ability to amplify the difference of two input. signals applied to its inputs Only the difference in the two signals is amplified if there is. no difference the output is zero A basic differential amplifier circuit and its symbol are. shown in Fig 1 3 The transistors Q1 and Q2 and the collector resistors RC1 and RC2 are. carefully matched to have identical characteristics Notice that the two transistors share a. single emitter resistor RE,1 3 Op Amp Operation Modes. The differential amplifier exhibits three modes of operation based on the type of input. and or output signals These modes are single ended double ended or differential and. common Since the differential amplifier is the input stage of the op amp the op amp. exhibits the same modes, Single Ended Input Single ended input operation results when the input signal is. connected to one input with the other input connected to ground Fig 1 4 shows the. signals connected for this operation In Fig 1 4 a the input is applied to the plus input. with minus input at ground which results in an output having the same polarity as the. applied input signal Fig 1 4 b shows an input signal applied to the minus input the. output then being opposite in phase to the applied signal. Lecture 01 The Operational Amplifier Op Amp, Double Ended Differential Input In addition to using only one input it is. possible to apply signals at each input this being a double ended operation Fig 1 5 a. shows an input Vd applied between the two input terminals recall that neither input is at. ground with the resulting amplified output in phase with that applied between the plus. and minus inputs Fig 1 5 b shows the same action resulting when two separate signals. are applied to the inputs the difference signal being Vi1 Vi2. Double Ended Output While the operation discussed so far had a single output the. op amp can also be operated with opposite outputs as shown in Fig 1 6 a An input. applied to either input will result in outputs from both output terminals these outputs. always being opposite in polarity Fig 1 6 b shows a single ended input with a double. ended output As shown the signal applied to the plus input results in two amplified. outputs of opposite polarity Fig 1 6 c shows the same operation with a single output. measured between output terminals not with respect to ground This difference output. signal is Vo1 Vo2 The difference output is also referred to as a floating signal since. neither output terminal is the ground reference terminal Notice that the difference output. is twice as large as either Vo1 or Vo2 since they are of opposite polarity and subtracting. them results in twice their amplitude Fig 1 6 d shows a differential input differential. output operation The input is applied between the two input terminals and the output. taken from between the two output terminals This is fully differential operation. Lecture 01 The Operational Amplifier Op Amp, Common Mode Operation When the same input signals are applied to both inputs.

common mode operation results as shown in Fig 1 7 Ideally the two inputs are equally. amplified and since they result in opposite polarity signals at the output these signals. cancel resulting in 0 V output Practically a small output signal will result. Common Mode Rejection A significant feature of a differential connection is that. the signals which are opposite at the inputs are highly amplified while those which are. common to the two inputs are only slightly amplified the overall operation being to. amplify the difference signal while rejecting the common signal at the two inputs Since. noise any unwanted input signal is generally common to both inputs the differential. connection tends to provide attenuation of this unwanted input while providing an. amplified output of the difference signal applied to the inputs This operating feature. referred to as common mode rejection,1 4 Common Mode Reject Ratio CMRR. One of the more important features of a differential circuit connection as provided in an. op amp is the circuit s ability to greatly amplify signals that are opposite at the two. inputs while only slightly amplifying signals that are common to both inputs An op amp. provides an output component that is due to the amplification of the difference of the. signals applied to the plus and minus inputs and a component due to the signals common. to both inputs Since amplification of the opposite input signals is much greater than that. of the common input signals the circuit provides a common mode rejection as described. by a numerical value called the common mode rejection ratio CMRR. Differential Inputs When separate inputs are applied to the op amp the resulting. difference signal is the difference between the two inputs. Common Inputs When both input signals are the same a common signal element. due to the two inputs can be defined as the average of the sum of the two signals. Output Voltage Since any signals applied to an op amp in general have both. in phase and out of phase components the resulting output can be expressed as. where Ad differential gain and Ac common mode gain of the amplifier. Lecture 01 The Operational Amplifier Op Amp, Having obtained Ad and Ac we can now calculate a value for the common mode. rejection ratio CMRR which is defined by the following equation. The value of CMRR can also be expressed in logarithmic terms as. CMRR dB 20 log 1 5,Exercise 1 1, Calculate the CMRR and express it in decibel for the circuit measurements shown in. Answers 666 7 56 48 dB,Exercise 1 2, Determine the output voltage of an op amp for input voltages of Vi1 150 V. Vi2 140 V The amplifier has a differential gain of Ad 4000 and the value of CMRR. is a 100 and b 105,Answers a 45 8 mV b 40 006 mV,Lecture 01 The Operational Amplifier Op Amp.

1 5 The Inverting Op Amp, Consider the configuration shown in Fig 1 9 a In this very useful application of an. operational amplifier the noninverting input is grounded vin is connected through R1 to. the inverting input and feedback resistor Rf is connected between the output and vi Since. we are using the amplifier in an inverting mode we denote the voltage gain by A. vin vi we define,From Fig 1 9 b, In Eqn 1 6 the gain is negative signifying that the configuration is an inverting. amplifier also the magnitude of vo vin depends only on the ratio of the resistor values The. gain vo vin is a closed loop gain of the amplifier while A is called the open loop gain. Exercise 1 3, Assuming that the operational amplifier in Fig 1 10 is ideal find. a the rms value of vo when vin is 1 5 V rms, b the rms value of the current in the 25 k resistor when vin is 1 5 V rms and. c the output voltage when vin 0 6 V dc,Answers a 8 25 V rms b 60 A rms c 3 3 V dc.

Lecture 01 The Operational Amplifier Op Amp,1 6 The Noninverting Op Amp. Fig 1 11 a shows another useful application of an operational amplifier called the. noninverting configuration The input signal vin is connected directly to the noninverting. input and R1 is connected from the inverting input to ground Under the ideal assumption. of infinite input impedance no current flows into the inverting input so i1 if. 0 and where, Eqn 1 7 shows that the closed loop gain of the noninverting amplifier like that of. the inverting amplifier depends only on the values of external resistors Fig 1 11 b. shows a special case of noninverting amplifier used in applications where power gain and. impedance isolation are of primary concern When Rf 0 and R1 so the closed loop. gain is 1 1 This configuration is called a voltage follower because. vo has the same magnitude and phase as vin It has large input impedance and small output. impedance and is used as a buffer amplifier between a high impedance source and a. low impedance load,Exercise 1 4, In a certain application a signal source having 60 k of source impedance RS produces a. 1 V rms signal This signal must be amplified to 2 5 V rms and drive a 1 k load. Assuming that the phase of the load voltage is of no concern design an operational. amplifier circuit for the application, Hint Choose arbitrarily input resistor R1 100 k and find feedback resistor Rf. Since phase is of no concern and the required voltage gain is greater than 1 we can use. either an inverting or noninverting amplifier,Answers Rf inverting 400 k Rf noninverting 150 k.

Lecture 01 The Operational Amplifier Op Amp,1 7 Op Amp Analysis using Feedback Theory. We have seen that we can control the closed loop gain vo vin of an operational amplifier by. introducing feedback through external resistor combinations We wish now to examine the. feedback mechanism in detail and discover some other important consequences of its use. Feedback theory is widely used to study the behavior of electronic components as well as. complex systems in many different technical fields so it is important to develop an. appreciation and understanding of its underlying principles. 1 7 1 Feedback in the Noninverting Op Amp, Fig 1 12 shows the noninverting configuration along with an equivalent block diagram on. which we can identify the signal and feedback paths A represents the amplifier and its. open loop gain is called the feedback ratio and represents the output voltage that is. fed back to the input ve vin vf ve is often called the error voltage The feedback voltage. vf vo corresponds to vi in the amplifier circuit Since the feedback voltage subtracts. from the input voltage the amplifier is said to have negative feedback. With reference to Fig 1 12 b we see that,Form a voltage divider across vo. noninverting op amp 1 9,Lecture 01 The Operational Amplifier Op Amp. Substituting Eqn 1 9 into Eqn 1 8 we find, Eqn 1 11 is exactly the same result we obtained from Eqn 1 7.

Negative feedback improves the performance of an amplifier in several ways In the. case of the noninverting amplifier it can be shown that the input resistance seen by the. signal source looking directly into the terminal is. where rid is the differential input resistance of the amplifier. The closed loop output resistance of the noninverting amplifier is also improved by. negative feedback, where ro is the open loop output resistance of the amplifier. Exercise 1 5, Find the closed loop gain of the amplifier in Fig 1 13 when a A b A 106 and. Answers a 10 b 9 9990 c 9 90099,Exercise 1 6, An operational amplifier has open loop gain A 104 Compare its closed loop gain with. that of an ideal amplifier when a 0 1 and b 0 001,Answers a 9 99 b 909 09. Exercise 1 7, A noninverting op amp has open loop gain A 105 feedback ratio 0 01 differential.

input resistance rid 20 k and open loop output resistance ro 75 Find the closed. loop input rif and output rof resistances of the amplifier. Answers 20 M 0 075,Lecture 01 The Operational Amplifier Op Amp. 1 7 2 Feedback in the Inverting Op Amp, To investigate the effect of open loop gain A and feedback ratio on the closed loop gain. of the inverting amplifier let us recall Fig 1 9 b. Once again when A we see that the closed loop gain reduces to the ideal. amplifier value Rf R1 Eqn 1 6 By the superposition principle we can analyze the. contribution of the feedback source by grounding all other signal sources When this is. done as shown in Fig 1 14 we see that the feedback voltage in both configurations is. developed across R1 and Rf by voltage divider and R1 R1 Rf in both cases In view. of this fact we can write Eqn 1 14 as, Towards developing a feedback model for the inverting amplifier consider the. block diagram shown in Fig 1 15 It is quite similar to Fig 1 12 b for the noninverting. amplifier except that we now denote the open loop gain by A v represents an arbitrary. input voltage rather than vin As shown in the figure. Multiplying the right side by the factor we would obtain. Lecture 01 The Operational Amplifier Op Amp, Eqn 1 16 gives us exactly the same result Eqn 1 15 with vin v that we obtain. for the inverting amplifier Therefore we modify the block diagram model in Fig 1 15 by. adding a block that multiplies the input by The complete feedback model is. shown in Fig 1 16 As can be seen the loop gain for the inverting amplifier is A the. same as that for the noninverting amplifier, It can be shown that the input resistance seen by the signal source driving the.

inverting amplifier is, As with the noninverting amplifier the output resistance of the inverting amplifier is. decreased by the negative feedback In fact the relationship between output resistance and. loop gain is the same for both,Lecture 01 The Operational Amplifier Op Amp. In closing our discussion of feedback theory we should note once again that the. same relationship between actual and ideal closed loop gain applies to inverting and. noninverting amplifiers This relationship is, where ideal closed loop gain is the closed loop gain vo vin that would result if the. amplifier were ideal A We saw this relationship in Eqn 1 10 and Eqn 1 15. repeated here,noninverting op amp,inverting op amp. In both cases the numerator is the closed loop gain that would result if the amplifier. were ideal Also in both cases the greater the value of the loop gain A the closer the. Lecture 01 The Operational Amplifier Op Amp 3 Double Ended Differential Input In addition to using only one input it is possible to apply signals at each input this being a double ended operation

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