Block D: Amplifiers

1. Block D: Amplifiers Amplifiers - Unit 1: Amplifier Model

2. Some Applications WiFi Repeater Audio Amplifier Kalaok Amplifier Power Amplifier EE2301: Block D Unit 1

3. Some Applications Servo Amplifier EE2301: Block D Unit 1

4. Some Applications TV and Satellite Antenna Amplifier EE2301: Block D Unit 1

5. Some Applications Alcohol Sensor Motion Sensor Temperature Sensor Speed Sensor Location Sensor Heart Beat Rate Sensor EE2301: Block D Unit 1

6. Amplifiers Need Everywhere EE2301: Block D Unit 1

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 EE2301: Block D Unit 1

8. 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 EE2301: Block D Unit 1

9. 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 log10(Pout/Pin) Gain in dB = 20 log10(Vout/Vin) • 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) A1 A2 Total gain (in dB) = 20 log10(A1A2) = 20 log10(A1) + 20 log10(A2) EE2301: Block D Unit 1

10. 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. EE2301: Block D Unit 1

11. Ideal Amplifier Now we take a look at how the amplifier looks like on the inside 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 1) Input resistance Rin: This is seen across the input terminals 2) Dependent source Avin: In the circuit model above, the voltage of the source depends on the voltage drop across Rin. A is known as the open loop gain. 3) Output resistance Rout: This is seen in series with the dependent source and the positive output terminal. EE2301: Block D Unit 1

12. 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 Vin < VS since: vin is then amplified by a factor A through the dependent source Avin From the dependent source to the output terminals, there will be some voltage drop across Rout since Rout is non-zero. Hence VL < Avin since: EE2301: Block D Unit 1

13. 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 Rin should be very large (ideally infinite), so that vin ≈ vS Rout should be very small (ideally zero), so that vL ≈ Avin EE2301: Block D Unit 1

14. Impedance: Example 1 RS = Rin = Rout = RL = 50Ω, A = 50 Find vL, (1) without RL, (2) with RL Without RL: With RL: EE2301: Block D Unit 1

15. Impedance: Example 2 RS = Rin = Rout = RL = 50Ω, A = 40 If two of the above amplifiers are cascaded, find VL EE2301: Block D Unit 1

16. Impedance: Example 2 Rout Rout Rs +- +- +- Rin Av Rin Av RL vs Given: RS = Rin = Rout = RL = 50Ω, A = 40 EE2301: Block D Unit 1

17. Impedance: Example 3 RS = Rin = Rout = RL = 50Ω, A = 20 If it is required for vL = 40vS, how many amplifier stages are needed? EE2301: Block D Unit 1

18. Impedance: Example 3 solution From previous example 2: To achieve VL = 40VS Need an integer number of stages, so therefore 2 stages required. EE2301: Block D Unit 1

19. 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 Much Simpler to Use Let’s con’t in next lecture EE2301: Block D Unit 1

20. 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 EE2301: Block D Unit 1

21. Operational Amplifier vS+ v+ Non-inverting input v- Inverting input vS+ Positive power supply vS- Negative power supply vout Output v+ + vout _ v- vout = AV(OL)(v+ - v-) vS- • 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- EE2301: Block D Unit 1

22. + vout - Op-amp Model iin Rout iin v+ Rout + vin - + vin - Rin Rin + - Avin + - v- AV(OL)vin General amplifier model Op-amp model EE2301: Block D Unit 1

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

24. Negative feedback 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. We can see that: Vout = A(Vin – Vout) Re-arranging the terms: Vin + Vout - Now if A is infinite, this will then force Vout to equal Vin. In other words, an infinite A has the effect of forcing the input terminals to be at the same voltage. V+ = V- EE2301: Block D Unit 1

25. Before we continue,let’s go through some op-amp applications EE2301: Block D Unit 1

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

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

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

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

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

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

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

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

34. 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 EE2301: Block D Unit 2

35. Source follower + Vin Rs Vout - RL Find the gain of the above circuit • The key features of the source follower are: • Large input resistance • Small output resistance • 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? EE2301: Block D Unit 2

36. Inverting op amp R2 Note that the inverting input (node X) is at ground R1 and R2 are in series R1 - Vin Applying KCL at X Closed-loop gain Vout + Negative sign indicates a 1800 shift in the phase Gain is set by the resistor values EE2301: Block D Unit 2

37. R + +_ vin + vout- _ RF RS Non-inverting amplifier iin • Note that since iin = 0: • v+ = vin (no voltage drop across R) • We can apply voltage divider rule to RS & RF iin + vRS- • But we also note that A is infinite, so: • vin = vRS • Hence we then obtain: EE2301: Block D Unit 2

38. Summing Amplifier RS1 RF Apply NVA at the inverting input terminal: +_ vS1 - RS2 + vout- + +_ vS2 Hence the form of the gain relation can be described by: Vout = -(A1vS1 + A2vS2) A1 and A2 are set by the resistor values chosen. We can extend this result to write a general expression for the gain: EE2301: Block D Unit 2

39. Differential amplifier R2 One way of analyzing this circuit is to apply superposition (find Vout for V1 or V2 only) R1 V1 - Vout If we short V2 first, we obtain: R1 + V2 R2 R1 V1 - R2 Vout1 + If we short V1 now, we obtain: [R2/(R1+R2)]V2 + Hence, finally: Vout = (R2/R1)(v2 – v1) Output is the difference between the inputs amplified by a factor set by the resistor values. Vout2 - R2 R1 EE2301: Block D Unit 2

40. Difference amplifier In the case whereby all the resistors are different, as shown in the circuit below, while vout1 remains unchanged, vout2 now becomes: R2 Hence the overall gain expression is given by: R1 V1 - Vout R3 + V2 This is similar to the form of the summing amplifier except that we take the difference between the two inputs: Vout = A2V2 – A1V1 A2 and A1 are set by the resistor values R4 EE2301: Block D Unit 2

41. Diff Amp: Example 1 Find vout if R2 = 10kΩ and R1 = 250Ω R2 R1 V1 - Vout R1 + V2 R2 EE2301: Block D Unit 2

42. Diff Amp: Example 2 For the same circuit in the previous example, given v2 = v1: Find vout if v1 and v2 have internal resistances of 500Ω and 250Ω respectively R2 500Ω R1 V1 - Now: R2 = 10kΩ, R1 = 750Ω, R3 = 500Ω, R4 = 10kΩ Vout R1 + V2 250Ω R2 vout = 13.65 v2 – 13.33v1 • We can see that as a result of the source resistances: • Differential gain has changed • vout is not zero for v2 = v1 EE2301: Block D Unit 2

43. Recap in last lecture EE2301: Block B Unit 2

44. Block D Unit 2 Outline • Ideal Op-Amp Characteristic • Three basic assumptions • Infinitive input impedance • Zero output impedance • 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 EE2301: Block D Unit 2

45. Source follower + Vin Rs Vout - RL Find the gain of the above circuit • The key features of the source follower are: • Large input resistance • Small output resistance • 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 EE2301: Block D Unit 2

46. Inverting op amp R2 R1 - Vin Vout + Negative sign indicates a 1800 shift in the phase Gain is set by the resistor values EE2301: Block D Unit 2

47. R + +_ vin + vout- _ RF RS Non-inverting amplifier iin iin + vRS- EE2301: Block D Unit 2

48. Summing Amplifier RS1 RF +_ vS1 - RS2 + vout- + +_ vS2 We can extend this result to write a general expression for the gain: EE2301: Block D Unit 2

49. Differential amplifier R2 R1 V1 - Vout R1 + V2 R2 R2 R1 V1 - R3 + V2 R4 EE2301: Block D Unit 2

50. Diff Amp: Example 2 For the same circuit in the previous example, given v2 = v1: Find vout if v1 and v2 have internal resistances of 500Ω and 250Ω respectively R2 500Ω R1 V1 - Vout R1 + Now: R2 = 10kΩ, R1 = 750Ω, R3 = 500Ω, R4 = 10kΩ V2 250Ω R2 vout = 13.65 v2 – 13.33v1 • We can see that as a result of the source resistances: • Differential gain has changed • vout is not zero for v2 = v1 EE2301: Block D Unit 2