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Operational Amplifier Part 1


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Introduction to operational Amplifier. For A2 level physics (CIE). Discusses characteristics of op amp, inverting and non inverting amplifier, and voltage follower, and transfer characetristics, virtual earth , etc

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Operational Amplifier Part 1

  1. 1. Operational Amplifier Part 1 Mukesh N Tekwani tekwani@
  2. 2. Operational Amplifier <ul><li>Originally an op-amp was an electronic circuit that could carry out mathematical operations of addition, subtraction, differentiation and integration. </li></ul><ul><ul><li>Hence the word “operational” </li></ul></ul><ul><li>Op-amp is used to amplify DC and AC signals. </li></ul>
  3. 3. Operational Amplifier Symbol <ul><li>Circuit Symbol </li></ul>- + +ve supply -ve supply output Inverting i/p V 1 Non-Inverting i/p V 2
  4. 4. Internal Block Diagram
  5. 5. Characteristics of Ideal Op-Amp <ul><li>Infinite input impedance (about 2 Mohm) </li></ul><ul><li>Low output impedance (about 200 ohm) </li></ul><ul><li>Very large voltage gain at low frequency </li></ul><ul><ul><li>Thus, small changes in voltages can be amplified by using an op-amp </li></ul></ul><ul><li>Infinite bandwidth (all frequencies are amplified by same factor </li></ul><ul><li>No slew rate – no delay between change in i/p and changes in o/p </li></ul>
  6. 6. Op Amp Characteristics Explained <ul><li>Infinite input impedance </li></ul><ul><ul><li>no current flows into inputs </li></ul></ul><ul><li>Infinite voltage gain </li></ul><ul><ul><li>a voltage difference at the two inputs is magnified to a very large extent </li></ul></ul><ul><ul><li>in practice, voltage gain ~ 200000 </li></ul></ul><ul><ul><li>means difference between + terminal and  terminal is amplified by 200,000! </li></ul></ul>
  7. 7. Op Amp Characteristics Explained <ul><li>Infinite bandwidth </li></ul><ul><ul><li>In practice, bandwidth limited to few MHz range </li></ul></ul><ul><ul><li>slew rate limited to 0.5–20 V/  s </li></ul></ul>
  8. 8. Op Amp Slew Rate Explained <ul><li>The o/p of an op amp does not change instantaneously. </li></ul><ul><li>The rate of change of o/p of an op amp is limited (about 0.5 V/  sec) </li></ul><ul><li>So, if we want to change the o/p voltage from 0 to 10 V, it would take 20  s </li></ul>
  9. 9. Op Amp Slew Rate Explained
  10. 10. Operational Amplifier Without Feedback <ul><li>The op-amp can be regarded as a device which generates an voltage V o given by: </li></ul><ul><ul><li>V o = A (V 2 – V 1 ) </li></ul></ul><ul><ul><li>A is called as the gain of the amplifier. </li></ul></ul><ul><ul><li>V 1 is the voltage applied at the inverting input, </li></ul></ul><ul><ul><li>V 2 is the voltage applied at the non-inverting input, </li></ul></ul>
  11. 11. Variation of Gain with Frequency <ul><li>The value of gain A depends on the frequency of the i/p signal and is very high at low frequencies. </li></ul><ul><li>At DC, (f = 0 Hz), gain A is about 10 5 . </li></ul><ul><li>But the gain decreases with frequency. </li></ul>
  12. 12. Variation of Output voltage with V 1 <ul><li>V o = A (V 2 – V 1 ) </li></ul><ul><ul><li> When V 2 = 0, V o = -AV 1 </li></ul></ul><ul><li>So, the output voltage is out of phase with the input voltage applied to the inverting input. </li></ul><ul><li>That is why it is called the “ inverting ” input </li></ul>
  13. 13. Variation of Output voltage with V 2 <ul><li>V o = A (V 2 – V 1 ) </li></ul><ul><ul><li> When V 1 = 0, V o = AV 2 </li></ul></ul><ul><li>So, the output voltage is in phase with the input voltage applied to the non-inverting input. </li></ul><ul><li>That is why it is called the “ non-inverting ” input </li></ul>
  14. 14. Variation of Output with Input Voltages <ul><li>V o = A (V 2 – V 1 ) </li></ul><ul><ul><li>If V 2 > V 1 , V o is positive </li></ul></ul><ul><ul><li>If V 2 < V 1 , V o is negative </li></ul></ul><ul><ul><li>If V 2 = V 1 , V o is zero </li></ul></ul>
  15. 15. Consequences of Ideal characteristics <ul><li>Infinite input resistance means the current into the inverting input is zero: </li></ul><ul><li>i - = 0 </li></ul><ul><li>Infinite gain means the difference between V 1 and V 2 is zero: </li></ul><ul><li>V 2 – V 1 = 0 </li></ul>
  16. 16. The Basic Inverting Amplifier R 2 V in – + + – V out R 1 + – I 1 I 2 Resistor used to control amplification
  17. 17. How to Calculate the Gain <ul><li>For an Inverting amplifier: </li></ul><ul><li>Gain = -R 2 / R 1 </li></ul><ul><li>Example : if R 2 is 100 kilo-ohm and R 1 is 10 kilo-ohm, Gain = -100 / 10 = -10 If the input voltage is 0.5V then the output voltage would be V in x Gain: V out = 0.5V X -10 = -5V </li></ul>
  18. 18. Inverting Amplifier <ul><li>The i/p voltage to be amplified is fed to the inverting i/p </li></ul><ul><li>A fraction of the o/p signal is fed back to the op-amp through the inverting i/p. </li></ul><ul><li>R 2 is the feedback resistance in this circuit </li></ul><ul><li>Since we have used the inverting i/p, the o/p is out of phase with the i/p signal. </li></ul><ul><li>This process is called negative feedback . </li></ul>
  19. 19. Inverting Amplifier <ul><li>It is called negative feedback because the overall gain of the amplifier reduces . </li></ul><ul><li>So why use negative feedback if gain is reduced? </li></ul><ul><ul><li>The gain is constant over a wide range of input frequencies and input voltages. </li></ul></ul><ul><ul><li>Stability is greater </li></ul></ul><ul><ul><li>Amplification is linear – i.e. distortion of o/p is less </li></ul></ul><ul><ul><li>Gain is independent of the characteristics of op amp. </li></ul></ul>
  20. 20. Solving the Amplifier Circuit <ul><li>Apply KCL at the inverting input: </li></ul><ul><li>i 1 + i 2 + i - =0 </li></ul>– R 1 R 2 i 1 i - i 2
  21. 21. KCL
  22. 22. Solve for V o <ul><li>Amplifier gain: </li></ul>Thus, Gain of an op-amp depends only on the two resistances and not on the op-amp characteristics
  23. 23. Assumptions made in deriving gain equation <ul><li>Each input draws zero current from the signal source. </li></ul><ul><ul><li>Typically, i/p current is 1  A </li></ul></ul><ul><ul><li>That is, input impedances are infinite </li></ul></ul><ul><li>The i/ps are both at the same potential if the op-amp is not saturated. </li></ul>
  24. 24. Transfer Characteristics of Inverting Amplifier V o -V s +V s saturation saturation Vin B A
  25. 25. Transfer Characteristics of a Non-inverting Amplifier V o -V s +V s saturation saturation V 2 – V 1 V 2 > V 1 V 2 < V 1 B A
  26. 26. Transfer Characteristics of an Op-Amp <ul><li>The output (Vo) is directly proportional to the input only within the range AOB. In this region, the op-amp behaves linearly. There is very little distortion of the amplifier output. </li></ul><ul><li>If the inputs are outside this linear range, then saturation occurs. That is output is close to the maximum value it can have i.e. V s or -V s </li></ul>
  27. 27. Transfer Characteristics of an OpAmp V s -V s V o Value V 0 might have for an ac i/p if opamp did not saturate
  28. 28. Transfer Characteristics of an OpAmp <ul><li>Consider an opamp connected to + 9 V supply. </li></ul><ul><li>The o/p voltage can never exceed these values. </li></ul><ul><li> max value of o/p voltage can be +9V or -9V </li></ul><ul><li>Let A = 10 5 (Remember A = V o / V in ) </li></ul><ul><li>So, max i/p voltage is V in = V o / A </li></ul><ul><li> V in = + 9 / 10 5 = + 90  V </li></ul><ul><li>This is the maximum input voltage swing. </li></ul><ul><li>A smaller value of A would allow greater input. </li></ul>
  29. 29. Saturation Effect in Op Amp Suppose gain is -10. Assume the input is a signal of amplitude of 1.4v.  We would expect the output of the amplifier to be a signal of amplitude of 14V because the amplitude of the input is 1.4v and the gain is -10.  But, if you take saturation into account, you will get a signal that is &quot;flattened&quot; at the top and bottom.
  30. 30. Problem 1: <ul><ul><li>In this circuit, we want a gain of ten.  If R 1 is 5 K ohm, what is the value you need to use for R 0 ?  Give your answer in ohms. </li></ul></ul>50,000 ohm
  31. 31. Problem 2: <ul><ul><li>In this circuit, you have it set up for a gain of -10.  The input voltage is 0.24v.  What is the output voltage? </li></ul></ul>Gain = - Vo / Vi Vo = Gain x Vi Vo = (-10) x 0.24 Vo = -2.4 V
  32. 32. Problem 3: <ul><ul><li>In this circuit, Ro and R1 values are shown. The input signal is also shown. Sketch the o/p signal. </li></ul></ul>10 K ohm 2 . 7 K ohm
  33. 33. Problem 3:
  34. 34. Problem 3: <ul><li>Gain A = Ro / R1 </li></ul><ul><ul><li>So, A = - 10 K / 2.7 K = -3.7 </li></ul></ul><ul><li>Amplitude of i/p signal is 4 V </li></ul><ul><li>So max o/p voltage is Vo = A x Vin </li></ul><ul><li> Vo = 3.7 x 4 = 14.8 V </li></ul><ul><li>But power supply is only + 9V </li></ul><ul><li>So 9V is the max o/p the amplifier can provide. </li></ul>
  35. 35. Problem 3: <ul><li>Amplifier is saturated </li></ul><ul><li>It will remain saturated as long as size of i/p </li></ul><ul><li>voltage is greater than 9V / 3.7 = 2.4 V </li></ul><ul><li>That is why we observe that the o/p gets clipped as soon as the i/p rises above 2.4 V </li></ul>
  36. 36. Concept of virtual earth R 2 V in P Q – + + – V out R 1 + – I 1 I 2 V Q V P
  37. 37. Virtual earth <ul><li>In the previous figure, V Q = 0 and  V P = 0 </li></ul><ul><li>P is called a virtual earth or ground point even though it is not connected to the ground. </li></ul>
  38. 38. Non-inverting Amplifier
  39. 39. Non-inverting op amp – + V i V o R f R i
  40. 40. Non-inverting Amplifier <ul><li>The output (Vo) is in phase with the input. </li></ul><ul><li>R f and R i form a voltage divider circuit. </li></ul><ul><li>A fraction of o/p voltage (Vo) developed across R f is fed back to the inverting i/p </li></ul><ul><li>This fraction is called feedback factor and is given by </li></ul><ul><ul><li> = Ri / (Ri + Rf) </li></ul></ul><ul><li>Gain of this amplifier is: </li></ul><ul><li> </li></ul><ul><li> A = 1 + R f </li></ul><ul><li> R i </li></ul><ul><li>There is no virtual earth at the non-inverting i/p terminal. </li></ul>
  41. 41. Voltage Follower – + V i V o
  42. 42. Voltage Follower <ul><li>This is a special case of the non-inverting amplifier. </li></ul><ul><li>In case of non-inverting amplifier, gain </li></ul><ul><li>A = 1 + R f </li></ul><ul><li>R i </li></ul><ul><li>If we set R f = 0, A = 1 ( unity gain ) </li></ul><ul><li>This is called voltage follower because the o/p voltage is locked to the i/p voltage (both are same) </li></ul><ul><li>Advantage: op amp has very high i/p impedance so it can measure V i without drawing any current. </li></ul>
  43. 43. Characteristics of Voltage Follower <ul><li>This is a special case of the non-inverting amplifier. </li></ul><ul><li>Gain A = 1 </li></ul><ul><li>The o/p voltage “follows” the i/p voltage </li></ul><ul><li>Op amp has very high i/p impedance and very low i/p impedance </li></ul>
  44. 44. Voltage Follower used for measuring charge ? Test Plate
  45. 45. Voltage Follower used for measuring charge <ul><li>This circuit uses a capacitor to make a charge-measuring device. </li></ul><ul><li>If a charged object touches the test plate, it will transfer charge to the capacitor. </li></ul><ul><li>The p.d. between the plates of the capacitor rises </li></ul><ul><li>If the capacitor is connected directly to a voltmeter, this charge will drain away through the meter and incorrect reading would be obtained. </li></ul><ul><li>Op-amp has very high i/p impedance and so practically no charge is removed from the capacitor and yet measured by the voltmeter </li></ul>
  46. 46. Thank You