1. 1 The operational amplifier (“op amp”) Feedback Comparator circuits Ideal op amp Unity-gain voltage follower circuit

2. EE 42/100 Fall 2005Week 8, Prof. White2 The Operational Amplifier The operational amplifier (“op amp”) is a basic building block used in analog circuits. –Its behavior is modeled using a dependent source. –When combined with resistors, capacitors, and inductors, it can perform various useful functions: amplification/scaling of an input signal sign changing (inversion) of an input signal addition of multiple input signals subtraction of one input signal from another integration (over time) of an input signal differentiation (with respect to time) of an input signal analog filtering nonlinear functions like exponential, log, sqrt, etc.

3. EE 42/100 Fall 2005Week 8, Prof. White3 Op Amp Circuit Symbol and Terminals +–+– V + V – non-inverting input inverting input positive power supply negative power supply output The output voltage can range from V – to V + (“rails”) The positive and negative power supply voltages do not have to be equal in magnitude (example: 0V and +3V DC supplies)

4. Week 8, Prof. White4 Op Amp Terminal Voltages and Currents +–+– V + V – All voltages are referenced to a common node. Current reference directions are into the op amp. – V cc + V cc – +vn–+vn– +vp–+vp– common node (external to the op amp) ipip inin ioio +vo–+vo– i c+ i c-

5. 5 Op-Amp Voltage Transfer Characteristic  In the linear region, v o = A (v p – v n ) = A v id where A is the open-loop gain  Typically, V cc  20 V and A > 10 4  linear range: -2 mV  v id = (v p – v n)  2 mV Thus, for an op-amp to operate in the linear region, v p  v n (i.e., there is a “virtual short” between the input terminals.) vovo v id = v p – v n The op-amp is basically a differentiating amplifier: “linear” “positive saturation” “negative saturation” V cc -V cc Regions of operation: slope = A >>1

6. 6 Achieving a “Virtual Short” Recall the voltage transfer characteristic of an op-amp: Q: How does a circuit maintain a virtual short at the input of an op-amp, to ensure operation in the linear region? A: By using negative feedback. A signal is fed back from the output to the inverting input terminal, effecting a stable circuit connection. Operation in the linear region enforces the virtual short circuit. vovo vp–vnvp–vn V cc -V cc slope = A >>1 ~1 mV ~10 V vovo vp–vnvp–vn V cc -V cc slope = A >>1 Plotted using different scales for v o and v p –v n ~10 V Plotted using similar scales for v o and v p –v n

7. 7 Familiar examples of negative feedback: Thermostat controlling room temperature Driver controlling direction of automobile Pupil diameter adjustment to light intensity Familiar examples of positive feedback: Microphone “squawk” in sound system Mechanical bi-stability in light switches Negative vs. Positive Feedback Fundamentally pushes toward stability Fundamentally pushes toward instability or bi-stability

8. 8 Op Amp Operation without Negative Feedback (Comparator Circuits for Analog-to-Digital Signal Conversion) 1. Simple comparator with 1 Volt threshold:  V – is set to 0 Volts (logic “0”)  V + is set to 2 Volts (logic “1”)  A = 100 2. Simple inverter with 1 Volt threshold:  V – is set to 0 Volts (logic “0”)  V + is set to 2 Volts (logic “1”)  A = 100 If V IN > 1.01 V, V 0 = 2V = Logic “1” If V IN < 0.99 V, V 0 = 0V = Logic “0” V0V0 V IN 12 0 1 2 +  V0V0 +  1V V IN 12 0 1 2 V0V0 If V IN > 1.01 V, V 0 = 0V = Logic “0” If V IN < 0.99 V, V 0 = 2V = Logic “1” +  V0V0 V IN +  1V

9. 9 Op Amp Circuits with Negative Feedback Q: How do we know whether an op-amp is operating in the linear region? A: We don’t, a priori. Assume that the op-amp is operating in the linear region and solve for v o in the op-amp circuit. –If the calculated value of v o is within the range from -V cc to +V cc, then the assumption of linear operation might be valid. We also need stability – usually assumed for negative feedback. –If the calculated value of v o is greater than V cc, then the assumption of linear operation was invalid, and the op-amp output voltage is saturated at V cc. –If the calculated value of v o is less than -V cc, then the assumption of linear operation was invalid, and the op-amp output voltage is saturated at -V cc.

10. 10 Op Amp Circuit Model (Linear Region) vpvp vnvn RiRi +–+– RoRo vovo A(vp–vn)A(vp–vn) R i is the equivalent resistance “seen” at the input terminals, typically very large (>1M , as large as 10 12  for FET input op- amps ), so that the input current is usually very small: i p = – i n  0 Note that significant output current (i o ) can flow when i p and i n are negligible! ipip inin ioio

11. 11 Ideal Op-Amp Assumptions: –R i is large (  10 5  ) –A is large (  10 4 ) –R o is small (<100  ) Simplified circuit symbol: –power-supply terminals and dc power supplies not shown +–+– +vn–+vn– +vp–+vp– ipip inin ioio +vo–+vo– i p = – i n = 0 v p = v n Note: The resistances used in an op-amp circuit must be much larger than R o and much smaller than R i in order for the ideal op-amp equations to be accurate.

12. 12 Unity-Gain Voltage-Follower Circuit V0V0 +  V IN V 0 (V) V IN (V) 12 1 2 Note that the analysis of this simple (but important) circuit required only one of the ideal op-amp rules. Q: Why is this circuit important (i.e., what is it good for)? A: A “weak” source can drive a “heavy” load; in other words, the source V IN only needs to supply a little power (since I IN = 0), whereas the output can drive a power-hungry load (with the op-amp providing the power). v p = v n  V 0 = V IN ( valid as long as V –  V 0  V + ) vnvn vpvp I IN

13. 13 What’s Inside an Op-Amp?

14. 14 Lecture Week 8 (continued) Op-Amp circuits continued: examples: Inverting amplifier circuit Summing amplifier circuit Non-inverting amplifier circuit Differential amplifier circuit Current-to-voltage converter circuit Reading (Note: amplifiers are discussed in great detail in Ch. 11)

15. 15 Negative feedback is used to “linearize” a high-gain differential amplifier. V0V0 +  V+V+ VV V 0 (V) 5 55 Without feedback V0V0 +  V IN 1 2 3 V0V0 1 2 3 4 5 4 5 Review: Negative Feedback With feedback

16. 16 Gain vs. Frequency of the Basic Op-Amp without and with Feedback (Hambley, Sec. 14.5) Facts: 1.The open-loop gain of an op--amp (no feedback from output to inverting input) is constant from DC to a frequency f BOL, after which the open-loop gain drops at a rate of -20dB/decade of frequency (Fig. 14.20 – next slide). 2.As the negative feedback is made stronger, the gain decreases but the bandwidth of the amplifier increases. (Bandwidth of an amplifier is the frequency range over which it amplifies.) 3.One can show (Hambley) that the product of the dc gain and the bandwidth is a constant that is independent of the amount of negative feedback. This is called the gain-bandwidth product for the op-amp. Example: The op-amp you’ll use in the lab (National Semiconductor LMC6482) is rather like that shown in Hambley Fig. 14.22, with a DC open-loop gain of 100 dB (voltage amplification of 10 5 ) out to about only 40 Hz! With negative feedback, however, the amplifier has an increasing bandwidth but with decreasing gain. The unity gain (0 dB) bandwidth is 4 MHz; one can get 40 dB of gain (V out /V in = 100) from DC to about 40 kHz.

17. 17 Op-Amp Frequency Response with and without Negative Feedback

18. 18 Application of Voltage Follower: Sample and Hold Circuit

19. 19 Inverting Amplifier Circuit +–+– +vo–+vo– –+–+ vsvs RsRs RfRf inin i n = 0  i s = -i f v p = 0  v n = 0 +vp–+vp– +vn–+vn– isis ifif

20. 20 Analysis using Realistic Op-Amp Model In the analysis on the previous slide, the op-amp was assumed to be ideal, i.e. R i =  ; A =  ; R o = 0 In reality, an op-amp has finite R i, finite A, non-zero R o, and usually is loaded at its output terminals with a load resistance R L.

21. 21 Summing Amplifier Circuit +–+– +vo–+vo– –+–+ vcvc RcRc RfRf inin +vp–+vp– +vn–+vn– icic ifif ibib iaia –+–+ vbvb –+–+ vava RbRb RaRa i n = 0  i a + i b + i c = -i f v p = 0  v n = 0 superposition !

22. EE 42/100 Fall 2005Week 8, Prof. White22 A DAC can be used to convert the digital representation of an audio signal into an analog voltage that is then used to drive speakers -- so that you can hear it! 00 1 01 00 11 0 1 00 1.5 2 0 1 1 0 0 111 1 000 2.5 3 3.5 4 1 00 1 1 0 1 0 11 11 00 11 0 1 111 0 1111 4.5 5 5.5 6 6.5 7 7.5 Binary number Analog output (volts) 0000 000 1 0.5 MSB LSB S1 closed if LSB =1 S2 " if next bit = 1 S3 " if " " = 1 S4 " if MSB = 1 4-Bit D/A 8V  + V0V0 5K 80K 40K 20K 10K S1 + -+ - S3 S2 S4  + Application: Digital-to-Analog Conversion “Weighted-adder D/A converter” (Transistors are used as electronic switches)

23. 23 Digital Input 0 1 2 3 4 5 6 7 8 0246810121416 Analog Output (V) 1111 1000 0100 0000 0001 Characteristic of 4-Bit DAC

24. 24 Noninverting Amplifier Circuit +–+– +vo–+vo– –+–+ vgvg RgRg RfRf i p = 0  v p = v g  v n = v g i n = 0  R s & R f form a voltage divider + vpvp – RsRs + vnvn – inin ipip

25. 25 Differential Amplifier Circuit +–+– +vo–+vo– –+–+ RbRb inin +vp–+vp– +vn–+vn– vbvb –+–+ vava RcRc RaRa i n = 0  i p = 0  RdRd ipip

26. 26 More usual version of differential amplifier (Horowitz & Hill, “Art of Electronics”, 2 nd Ed., p. 184-5 (part of their “smorgasbord” of op-amp circuits): Let R a = R c = R 1 and R b = R d = R 2 Then v 0 = (R 2 / R 1 )(v b – v a ) To get good “common-mode rejection” (next slide) you need well-matched resistors (R a and R c, R b and R d ), such as 100k  0.01% resistors Differential Amplifier (cont’d)

27. 27 Differential Amplifier – Another Perspective Redefine the inputs in terms of two other voltages: 1. differential mode input v dm  v b – v a 2. common mode input v cm  (v a + v b )/2 so that v a = v cm – (v dm /2) and v b = v cm + (v dm /2) Then it can be shown that “common mode gain”“differential mode gain”

28. 28 Differential Amplifier (cont’d) An ideal differential amplifier amplifies only the differential mode portion of the input voltage, and eliminates the common mode portion. –provides immunity to noise (common to both inputs) If the resistors are not perfectly matched, the common mode rejection ratio (CMRR) is finite:

29. 29 Op-Amp Current-to-Voltage Converter

30. 30 Summary Voltage transfer characteristic of op-amp: A feedback path between an op-amp’s output and its inverting input can force the op-amp to operate in its linear region, where v o = A (v p – v n ) An ideal op amp has infinite input resistance R i, infinite open-loop gain A, and zero output resistance R o. As a result, the input voltages and currents are constrained: v p = v n and i p = -i n = 0 vovo vp–vnvp–vn V cc -V cc slope = A >>1 ~1 mV ~10 V

31. EE 42/100 Fall 2005Week 8, Prof. White31

32. Week 8, Prof. White32

33. Week 8, Prof. White33

34. 34 Differentiator (from Horowitz & Hill)

35. 35 Op-Amp Imperfections Discussed in Hambley, pp. 651-665 (of interest if you need to use an op-amp in a practical application!) Linear range of operation: Input and output impedances not ideal values; gain and bandwidth limitations (including gain-bandwidth product); Nonlinear limitations: Output voltage swing (limited by “rails” and more); output current limits; slew-rate limited (how fast can the output voltage change: 0.5V/  s for the 741 op-amp, to 6000V/  s for high slew-rate op-amp); full-power bandwidth DC imperfections: Offset current (input currents don’t sum exactly to zero); offset voltage; There are pins (denoted offsets) for inputs to help control this.