1. EE2301: Basic Electronic Circuit Amplifiers - Unit 1: Amplifier Model1 Block D: Amplifiers

2. Some Applications EE2301: Block D Unit 12 Audio AmplifierKalaok Amplifier Power Amplifier WiFi Repeater

3. Some Applications EE2301: Block D Unit 13 Servo Amplifier

4. Some Applications EE2301: Block D Unit 14 TV and Satellite Antenna Amplifier

5. Some Applications EE2301: Block D Unit 15 Motion Sensor Speed Sensor Temperature Sensor Location Sensor Alcohol Sensor Heart Beat Rate Sensor

6. Amplifiers Need Everywhere EE2301: Block D Unit 16

7. 7 Block D Unit 1 Outline  Concepts in amplifiers > Gain > Input resistance > Output resistance  Introduction to the Operational Amplifier > Characteristics of the Ideal Op Amp > Negative feedback

8. EE2301: Block D Unit 18 Gain: Inverting and Non-inverting  The gain of an amplifier describes the ratio of the output amplitude to that of the input  If the output is an inverted version of the input with a larger amplitude, we say the amplifier is INVERTING > Gain is represented by a negative sign  If the output is simply an amplified version of the input but with no shift in phase, we say the amplifier is NON-INVERTING Non-inverting Inverting

9. EE2301: Block D Unit 19 The decibel  The gain of amplifiers are commonly expressed in decibels (written as dB)  The decibel is a logarithmic unit related to the power gain: Gain in dB = 10 log 10 (P out /P in ) Gain in dB = 20 log 10 (V out /V in )  The dB is extremely useful in finding the overall gain when we cascade amplifiers (connecting the output of one stage to the input of another) like in the figure below: > Note that the overall gain = multiplication of individual gains > Therefore, overall gain (in dB) = sum of individual gains (in dB) A1A1 A2A2 Total gain (in dB) = 20 log 10 (A 1 A 2 ) = 20 log 10 (A 1 ) + 20 log 10 (A 2 )

10. EE2301: Block D Unit 110 Ideal Amplifier We will first consider the amplifier as a black box and see how it connects with the rest of the system. We can see right away that the amplifier is really two port network (REF Block A Unit 3). Input terminals: The input terminals of the amplifier are connected to the source, modeled as a Thevenin circuit. Output terminals: The output terminals of the amplifier are connected to the load.

11. EE2301: Block D Unit 111 Ideal Amplifier INPUT port: Amplifier acts as an equivalent load with respect to the source OUTPUT port: Amplifier acts as an equivalent source with respect to the load Now we take a look at how the amplifier looks like on the inside 1) Input resistance R in : This is seen across the input terminals 2) Dependent source Av in : In the circuit model above, the voltage of the source depends on the voltage drop across R in. A is known as the open loop gain. 3) Output resistance R out : This is seen in series with the dependent source and the positive output terminal.

12. EE2301: Block D Unit 112 Effect of finite resistances From the source to the input terminals, there will be some voltage drop across RS due to the finite value of Rin. Hence V in < V S since: v in is then amplified by a factor A through the dependent source Av in From the dependent source to the output terminals, there will be some voltage drop across R out since Rout is non-zero. Hence V L < Av in since:

13. EE2301: Block D Unit 113 Impedance Dependence  Amplification now dependent on both source and load impedances  Also dependent on input & output resistance of the amplifier  Therefore, different performance using different load/source for same amplifier R in should be very large (ideally infinite), so that v in ≈ v S R out should be very small (ideally zero), so that v L ≈ Av in

14. EE2301: Block D Unit 114 Impedance: Example 1 R S = R in = R out = R L = 50Ω, A = 50 Find v L, (1) without R L, (2) with R L With R L : Without R L :

15. EE2301: Block D Unit 115 Impedance: Example 2 R S = R in = R out = R L = 50Ω, A = 40 If two of the above amplifiers are cascaded, find V L

16. EE2301: Block D Unit 116 Impedance: Example 2 Given: R S = R in = R out = R L = 50Ω, A = 40 +-+- vsvs RsRs R in RLRL R out +-+- +-+- Av

17. EE2301: Block D Unit 117 Impedance: Example 3 R S = R in = R out = R L = 50Ω, A = 20 If it is required for v L = 40v S, how many amplifier stages are needed?

18. EE2301: Block D Unit 118 Impedance: Example 3 solution From previous example 2: To achieve V L = 40V S Need an integer number of stages, so therefore 2 stages required.

19. EE2301: Block D Unit 119 Block D Unit 1 Outline  Concepts in amplifiers > Gain > Input resistance > Output resistance  Introduction to the Operational Amplifier > Characteristics of the Ideal Op Amp > Negative feedback Let’s con’t in next lecture Much Simpler to Use

20. EE2301: Block D Unit 120 Block D Unit 1 Outline  Concepts in amplifiers > Gain > Input resistance > Output resistance  Introduction to the Operational Amplifier > Characteristics of the Ideal Op Amp > Negative feedback

21. EE2301: Block D Unit 121 Operational Amplifier vS+vS+ vS-vS- v out + _ v+v+ v-v- v + Non-inverting input v - Inverting input v S + Positive power supply v S - Negative power supply v out Output  The operational amplifier (or op-amp for short) behaves much like an ideal difference amplifier  It amplifies the difference between two input voltages v + and v - v out = A V(OL) (v + - v - )

22. EE2301: Block D Unit 122 Op-amp Model v+v+ v-v- General amplifier model Op-amp model + v in - R out R in +-+- i in Av in + v in - R out R in A V(OL) v in +-+- i in + v out -

23. EE2301: Block D Unit 123 Ideal Op-amp Characteristics + v in - R out R in v+v+ v-v- A V(OL) v in +-+- i in + v out - There are 3 main assumptions we make for an ideal op amp: 1)Infinite input resistance (i.e. R in → ∞) 2)Zero output resistance (i.e. R out = 0) 3)Infinite open loop gain (i.e. A V(OL) → ∞) Given that R in → ∞, this means that no current flows into or out of any of the input terminals (inverting as well as non-inverting) Given that R out = 0, this means that v out = Av in But what about the effect of infinite open loop gain?

24. Negative feedback EE2301: Block D Unit 124 In this course, we will focus on the op amps being used in negative feedback. What this means is that we introduce a direct electrical path between the output and the inverting input terminals. What this does is to take some of the output and feed it back to the input in the opposite sense. This provides stability to the circuit and is commonly used to build amplifiers and filters. + - V in V out We can see that: V out = A(V in – V out ) Re-arranging the terms: Now if A is infinite, this will then force V out to equal V in. In other words, an infinite A has the effect of forcing the input terminals to be at the same voltage.  V + = V -

25. Before we continue, let’s go through some op-amp applications EE2301: Block D Unit 125

26. Amplifiers - Unit 2: The operational amplifier26 Inside the Op Amp Model 741 Made up of many transistors and circuits…

27. Circuit Example 1 - Timer Amplifiers - Unit 2: The operational amplifier27

28. Circuit Example 2 - Ramp Generator Amplifiers - Unit 2: The operational amplifier28

29. Circuit Example 3 – Optical Receiver Amplifiers - Unit 2: The operational amplifier29

30. Circuit Example 5 – Active Filter Amplifiers - Unit 2: The operational amplifier30

31. Circuit Example 6 – Dual Power Regulator Amplifiers - Unit 2: The operational amplifier31

32. Circuit Example 7 – Radio Receiver Amplifiers - Unit 2: The operational amplifier32

33. EE2301: Basic Electronic Circuit If you want to do these designs, you have to start from the basic first Feel Confusion! Amplifiers - Unit 2: The operational amplifier33

34. EE2301: Block D Unit 234 Block D Unit 2 Outline  Op-amp circuits with resistors only > Inverting amplifier > Non-inverting amplifier > Summing amplifier > Differential amplifier > Instrumentation amplifier  Op-amp circuits with reactive components > Active filters (Low pass, High pass, Band pass) > Differentiator & Integrator  Physical limits of practical op-amps

35. EE2301: Block D Unit 235 Source follower + - V in V out RsRs RLRL Find the gain of the above circuit The key features of the source follower are: 1)Large input resistance 2)Small output resistance 3)Unity gain (i.e. gain of close to one) It is therefore commonly used as a buffer between a load and source where the impedances are not well matched How to prove it?

36. EE2301: Block D Unit 236 Inverting op amp Note that the inverting input (node X) is at ground R 1 and R 2 are in series + - V in V out R2R2 R1R1 Applying KCL at XClosed-loop gain Negative sign indicates a 180 0 shift in the phase Gain is set by the resistor values

37. EE2301: Block D Unit 237 Non-inverting amplifier R + _ +_+_ + v out - RFRF RSRS v in + v RS - i in Note that since i in = 0: 1)v + = v in (no voltage drop across R) 2)We can apply voltage divider rule to R S & R F But we also note that A is infinite, so: 1)v in = v RS 2)Hence we then obtain:

38. EE2301: Block D Unit 238 Summing Amplifier + - + v out - +_+_ +_+_ v S1 v S2 R S1 R S2 RFRF Apply NVA at the inverting input terminal: Hence the form of the gain relation can be described by: V out = -(A 1 v S1 + A 2 v S2 ) A 1 and A 2 are set by the resistor values chosen. We can extend this result to write a general expression for the gain:

39. EE2301: Block D Unit 239 Differential amplifier + - V1V1 V out R1R1 V2V2 R2R2 R1R1 R2R2 One way of analyzing this circuit is to apply superposition (find V out for V 1 or V 2 only) If we short V 2 first, we obtain: + - V1V1 V out1 R1R1 R2R2 If we short V 1 now, we obtain: + - [R 2 /(R 1 +R 2 )]V 2 R2R2 R1R1 V out2 Hence, finally: V out = (R 2 /R 1 )(v 2 – v 1 ) Output is the difference between the inputs amplified by a factor set by the resistor values.

40. EE2301: Block D Unit 240 Difference amplifier In the case whereby all the resistors are different, as shown in the circuit below, while v out1 remains unchanged, v out2 now becomes: Hence the overall gain expression is given by: This is similar to the form of the summing amplifier except that we take the difference between the two inputs: V out = A 2 V 2 – A 1 V 1 A 2 and A 1 are set by the resistor values + - V1V1 V out R1R1 V2V2 R2R2 R3R3 R4R4

41. EE2301: Block D Unit 241 Diff Amp: Example 1 Find v out if R 2 = 10kΩ and R 1 = 250Ω + - V1V1 V out R1R1 V2V2 R2R2 R1R1 R2R2

42. EE2301: Block D Unit 242 Diff Amp: Example 2 For the same circuit in the previous example, given v 2 = v 1 : Find v out if v 1 and v2 have internal resistances of 500Ω and 250Ω respectively + - V1V1 V out R1R1 V2V2 R2R2 R1R1 R2R2 500Ω 250Ω Now: R 2 = 10kΩ, R 1 = 750Ω, R 3 = 500Ω, R 4 = 10kΩ v out = 13.65 v 2 – 13.33v 1 We can see that as a result of the source resistances: 1)Differential gain has changed 2)v out is not zero for v 2 = v 1

43. EE2301: Basic Electronic Circuit Recap in last lecture EE2301: Block B Unit 243

44. EE2301: Block D Unit 244 Block D Unit 2 Outline  Ideal Op-Amp Characteristic > Three basic assumptions 1.Infinitive input impedance 2.Zero output impedance 3.Infinitive open-loop gain  V+ = V-  Op-amp circuits with resistors only > Source Follower > Inverting amplifier > Non-inverting amplifier > Summing amplifier > Differential amplifier (Difference amplifier) > Instrumentation amplifier

45. EE2301: Block D Unit 245 Source follower + - V in V out RsRs RLRL Find the gain of the above circuit The key features of the source follower are: 1)Large input resistance 2)Small output resistance 3)Unity gain (i.e. gain of close to one) It is therefore commonly used as a buffer between a load and source where the impedances are not well matched

46. EE2301: Block D Unit 246 Inverting op amp + - V in V out R2R2 R1R1 Negative sign indicates a 180 0 shift in the phase Gain is set by the resistor values

47. EE2301: Block D Unit 247 Non-inverting amplifier R + _ +_+_ + v out - RFRF RSRS v in + v RS - i in

48. EE2301: Block D Unit 248 Summing Amplifier + - + v out - +_+_ +_+_ v S1 v S2 R S1 R S2 RFRF We can extend this result to write a general expression for the gain:

49. EE2301: Block D Unit 249 Differential amplifier + - V1V1 V out R1R1 V2V2 R2R2 R1R1 R2R2 + - V1V1 R1R1 V2V2 R2R2 R3R3 R4R4

50. EE2301: Block D Unit 250 Diff Amp: Example 2 For the same circuit in the previous example, given v 2 = v 1 : Find v out if v 1 and v2 have internal resistances of 500Ω and 250Ω respectively + - V1V1 V out R1R1 V2V2 R2R2 R1R1 R2R2 500Ω 250Ω Now: R 2 = 10kΩ, R 1 = 750Ω, R 3 = 500Ω, R 4 = 10kΩ v out = 13.65 v 2 – 13.33v 1 We can see that as a result of the source resistances: 1)Differential gain has changed 2)v out is not zero for v 2 = v 1

51. EE2301: Block D Unit 251 Instrumentation Amplifier + - R2R2 RXRX R3R3 R1R1 R3R3 v o1 v o2 The instrumentation amplifier takes care of this problem by including a non- inverting amplifier (which possesses an infinite input resistance) between the differential amplifier an the inputs. - + R2R2 R1R1 + - V1V1 V out V2V2

52. EE2301: Block D Unit 252 Instrumentation Amplifier 2 nd stage is a differential amplifier: V out = (R 3 /R 2 ) (V o2 - V o1 ) R3R3 R3R3 + - V out V o1 V o2 + - R X /2 R1R1 V1V1 1 st stage comprises a pair of non-inverting amplifiers: V o1 = (2R 1 /R x +1)V 1 V o1 Closed-loop gain: V out = (R 3 /R 2 )(2R 1 /R x + 1)(V 2 - V 1 )

53. EE2301: Block D Unit 253 Op amp example 1 Problem 8.5 Find v 1 in the following 2 circuits What is the function of the source follower?

54. EE2301: Block D Unit 254 Op amp example 1 solution This slide is meant to be blank Fig (a): 6Ω || 3Ω = 2Ω v 1 = {2 / (2+6)}*V g = 0.25V g Fig (b): Voltage at the non-inverting input = 0.5V g Voltage at output = Voltage at non-inverting input = 0.5V g

55. EE2301: Block D Unit 255 Op amp example 2 Problem 8.7 Find the voltage v 0 in the following circuit Transform to Thevenin

56. EE2301: Block D Unit 256 Op amp example 2 solution This slide is meant to be blank Thevenin equivalent circuit: R th = 6 + 2||4 = 22/3 kΩ V th = {4/(2+4)}*11 = 22/3 V Closed loop gain expression: A = - 12kΩ / R th (With V th as input source) V out = 12/(22/3) * (22/3) = 12V

57. EE2301: Block D Unit 257 Op amp example 3 + - + v out - R2R2 R1R1 v in RBRB Find the closed-loop gain Consider: Current through R B = 0A (Infinite input resistance of op amp)  Voltage across R B = 0V  Voltage at inverting input = 0V Same as inverting op amp: A = -R 2 / R 1

58. EE2301: Block D Unit 258 Op amp example 4 + - + v out - R2R2 Find the closed-loop gain + v in - Consider: Current through R 2 = 0A (Infinite input resistance of op amp)  Voltage across R 2 = 0V  V out = V -  V out = V in (Infinite open loop gain of op amp)

59. Common and differential mode EE2301: Block D Unit 259 A differential amplifier should ideally amplify ONLY DIFFERENCES between the inputs. That is to say identical inputs should give an output of zero. Ideal differential amplifier: V out = A(V 2 – V 1 ) In reality this is not the case as we have seen in a previous example. The output of the amplifier is more accurately described by: V out = A 2 V 2 + A 1 V 1 A 2 : Gain when V 1 = 0 (V 2 is the only input) A 1 : Gain when V 2 = 0 (V 1 is the only input) It is then useful to describe the performance of a differential amplifier in terms of the gain when both input are identical (common mode) and when the inputs are equal and out-of-phase (differential) We would like to express A 1 and A 2 by A cm and A dm

60. Common mode rejection ratio EE2301: Block D Unit 260 Common mode gain (A cm ): Gain when both inputs are exactly the same (V 1 = V 2 = V in ) V ocm = (A 2 + A 1 )V in Voltage output for common input V odm = (A 2 - A 1 )V in Voltage output for differential input Differential mode gain (A dm ): Gain when both inputs are equal but out-of-phase (V 2 = -V 1 = V in ) The common mode rejection ratio (CMRR) is simply the ratio of the differential mode gain (A dm ) over the common mode gain (A cm ): CMRR = A dm /A cm A large CMRR is therefore desirable for a differential amplifier Divide by 2 since the difference of a pair of differential inputs is twice that of each input

61. EE2301: Block D Unit 261 Block D Unit 2 Outline  Op-amp circuits with resistors only > Inverting amplifier > Non-inverting amplifier > Summing amplifier > Differential amplifier > Instrumentation amplifier  Op-amp circuits with reactive components > Active filters (Low pass, High pass, Band pass) > Differentiator & Integrator  Physical limits of practical op-amps

62. EE2301: Basic Electronic Circuit Let’s con’t in this lecture EE2301: Block B Unit 262

63. Active filters EE2301: Block D Unit 263 Range of applications is greatly expanded if reactive components are used Addition of reactive components allows us to shape the frequency response Active filters: Op-amp provides amplification (gain) in addition to filtering effects Substitute R with Z now for our analysis: Z F and Z S can be arbitrary (i.e. any) complex impedance

64. Active low pass filter EE2301: Block D Unit 264 Z S = R S Shaping of frequency response Amplification

65. Active high pass filter EE2301: Block D Unit 265 Z F = R F Shaping of frequency response Amplification ω→∞, V out /V S → -R F /R S

66. Active band pass filter EE2301: Block D Unit 266 Z S : C S blocks low frequency inputs but lets high frequency inputs through  High pass filter Z F : C F shorts R F at high frequency (reducing the gain), but otherwise looks just like a high pass filter at lower frequencies  Low pass filter

67. Active band pass filter EE2301: Block D Unit 267 ω HP = 1/(C S R S ) – Lower cut-off frequency ω LP = 1/(C F R F ) – Upper cut-off frequency For ω HP < ω LP : Frequency response curve is shown below

68. Second-order Low Pass Filter EE2301: Block D Unit 268 R 1, R 2, C and L are specially chosen so that: ω 0 = 1/(CR 2 ) = R 1 /L. The frequency response function then simplifies to:

69. Second-order Low Pass Filter EE2301: Block D Unit 269 R 1, R 2, C and L are specially chosen so that: ω 0 = 1/(CR 2 ) = R 1 /L. Above ω 0, H v is reduced by a factor of 100 for a ten fold increase in ω (40dB drop per decade) First-order filter: H v is reduced by a factor of 10 for a ten fold increase in ω (20dB drop per decade)

70. Ideal integrator EE2301: Block D Unit 270 KCL at inverting input: i S = -i F Virtual ground at inverting input: i S = V S /R S For a capacitor: i F = C F [dV out /dt] Output is the integral of the input

71. Ideal differentiator EE2301: Block D Unit 271 KCL at inverting input: i S = -i F Virtual ground at inverting input: i F = V out /R F For a capacitor: i S = C S [dV S /dt] Output is the time- differential of the input

72. Physical limit: Voltage supply limit EE2301: Block D Unit 272 The effect of limiting supply voltages is that amplifiers are capable of amplifying signals only within the range of their supply voltages. R S = 1kΩ, R F = 10kΩ, R L = 1kΩ; V S + = 15V, V S - = -15V; V S (t) = 2sin(1000t) Gain = -R F /R S = -10 V out (t) = 10*2sin(1000t) = 20sin(1000t) But since the supply is limited to +15V and -15V, the op-amp output voltage will saturate before reaching the theoretical peak of 20V.

73. Physical limit: Frequency response limit EE2301: Block D Unit 273 So far we have assumed in our ideal op-amp model that the open loop gain A V(OL) is infinite or at most a large constant value. In reality, A V(OL) varies with a frequency response like a low pass filter: ω 0 : frequency when the response starts to drop off The consequence of a finite bandwidth is a fixed gain-bandwidth product  If closed loop gain is increased, -3dB bandwidth is reduced  Increasing the closed loop gain further results in a bandwidth reduction till the gain-bandwidth produce equals the open-loop gain  Gain bandwidth product = A 0 ω 0