1 / 35

chapter 18

electronics fundamentals. circuits, devices, and applications. THOMAS L. FLOYD DAVID M. BUCHLA. chapter 18. R out = 0. V in. R in = ∞. A v V in. V out. The ideal op-amp.

isabelle
Download Presentation

chapter 18

An Image/Link below is provided (as is) to download presentation Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author. Content is provided to you AS IS for your information and personal use only. Download presentation by click this link. While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server. During download, if you can't get a presentation, the file might be deleted by the publisher.

E N D

Presentation Transcript

  1. electronics fundamentals circuits, devices, and applications THOMAS L. FLOYD DAVID M. BUCHLA chapter 18

  2. Rout= 0 Vin Rin= ∞ AvVin Vout The ideal op-amp The ideal op-amp is one with optimum characteristics, which cannot be attained in the real world. Nevertheless, actual op-amp circuits can often approach this ideal. The ideal op amp has infinite voltage gain, infinite input resistance (open), and zero output resistance. - +

  3. Rin Vin AvVin Vout Rout The practical op-amp Practical op-amps have limitations including power and voltage limits. A practical op-amp has high voltage gain, high input resistance, and low output resistance. There are two inputs, labeled inverting and non inverting because of the phase relation of the input and output signals. inverting input - non inverting input +

  4. The differential amplifier Most op-amps have a differential amplifier (“diff-amp”) as the input stage. The differential amplifier has important advantages over other amplifiers; for example it can reject common-mode noise. +VCC The signal at the collector of Q2 is not inverted. The signal at the collector of Q1 is inverted. RC1 RC2 Q1 Q2 The input is in single-ended mode. RE At the emitters, the signal is ½ of the input. -VEE

  5. Differential and common-mode signals Signals can be applied to either or both inputs. If two input signals are out of phase, they are in differential-mode. If the signals are in phase, they are in common-mode. When the inputs are out of phase, the outputs are amplified and larger than with one input. +VCC RC1 RC2 When the inputs are in phase, the outputs tend to cancel and are near zero. Q1 Q2 RE Inputs out of phase Inputs in phase -VEE

  6. Common-Mode Rejection Ratio (CMRR) Many times, noise sources will induce an unwanted voltage in a signal line. When the noise is induced in common-mode, the differential amplifier tends to cancel it. (The diff-amp cannot reject any signal that is in differential mode.) The ability to reject common-mode signals is measured with a parameter called the common-mode rejection ratio (CMRR), which is defined as CMRR can be expressed in decibels as

  7. Common-Mode Rejection Ratio (CMRR) Example 1: A certain diff-amp has a differential voltage gain of 500 and a common-mode gain of 0.1. What is the CMRR? Solution: From the defining equation for CMRR: 5000 Expressed in decibels, it is 74 dB

  8. Common-Mode Rejection Ratio (CMRR) Example 2: A certain diff-amp has Ad = 100 and a CMRR of 90 dB. Describe the output if the input is a 50 mV differential signal and a common mode noise of 1.0 V is present. Solution: The differential signal is amplified by 100. Therefore, the signal output is Vout= Av(d)xVin= 100 x 50 mV = 5.0 V The common-mode gain can be found by The noise is amplified by 0.0032. Therefore, Vnoise= AcmxVin= 0.0032 x 1.0 V = 3.2 mV

  9. Op-amp parameters Some important op-amp parameters are: Input bias current: Differential input resistance: Common-mode input resistance: Input offset current: Average of input currents required to bias the first stage of the amplifier: Total resistance between the inverting and non-inverting inputs Total resistance between each input and ground. Absolute difference between the two bias currents:

  10. Op-amp parameters Output resistance: Common-mode input voltage range: Common-mode rejection ratio Slew rate: The resistance when viewed from the output terminal. Range of input voltages, which, when applied to both inputs, will not cause clipping or other distortion. Ratio of the differential gain to the common-mode gain. The differential gain for the op-amp by itself is the same as its open loop gain. The maximum rate of change of the output in response to a step input voltage.

  11. Op-amp parameters Vout (V) Example: 12 10 What is the slew rate for the output signal shown in response to a step input? 0 -10 -12 25 ms Solution: The output goes from -10 V to +10 V in 25 ms. 0.8 V/ms

  12. Negative feedback In 1921, Harold S. Black was working on the problem of linearizing and stabilizing amplifiers. While traveling to work on the ferry, he suddenly realized that if he returned some of the output back to the input in opposite phase, he had a means of canceling distortion. One of the most important ideas in electronics was sketched out on his newspaper that morning. The op-amp has a differential amplifier as the input stage. When a feedback network returns a fraction of the output to the inverting input, only the difference signal (Vin – Vf) is amplified.

  13. A basic configuration is a noninverting amplifier. The difference between Vin and Vf is very small due to feedback. Therefore, + Vin - Rf Vf Feedback network Ri Op amp circuits with negative feedback Negative feedback is used in almost all linear op-amp circuits because it stabilizes the gain and reduces distortion. It can also increase the input resistance. The closed-loop gain for the noninverting amplifier can be derived from this idea; it is controlled by the feedback resistors: Vout

  14. Op amp circuits with negative feedback The inverting amplifier is a basic configuration in which the noninverting input is grounded (sometimes through a resistor to balance the bias inputs). Again, the difference between Vin and Vf is very small due to feedback; this implies that the inverting input is nearly at ground. This is referred to as a virtual ground. The virtual ground looks like ground to voltage, but not to current! Rf Virtual ground The closed-loop gain for the inverting amplifier can be derived from this idea; again it is controlled by the feedback resistors: Ri - Vout Vin +

  15. Vin + Vout - Rf Ri Input resistance for the noninverting amplifier The input resistance of an op-amp without feedback is Rin. For the 741C, the manufacturer’s specified value of Rin is 2 MW. Negative feedback increases this to Rin(NI) = (1 + AolB)Rin. This is so large that for all practical circuits it can be considered to be infinite. Keep in mind that, although Rin(NI) is extremely large, the op-amp is a dc amplifier and still requires a dc bias path for the input.

  16. Vin + Vout - Rf Ri Output resistance for the noninverting amplifier The output resistance of an op-amp without feedback is Rout. Negative feedback decreases this by a factor of (1 + AolB). This is so small that for all practical circuits it can be considered to be zero. The low output resistance implies that the output voltage is independent of the load resistance (as long as the current limit is not exceeded).

  17. Example: What are the input and output resistances and the gain of the noninverting amplifier? Assume the op amp has Aol = 100,000, Rin = 2 MW, and Rout = 75 W. Solution: Vin + Vout The gain is - Rf 25 36 kW Ri The feedback fraction is 1.5 kW The input resistance is 8 GW Solution continued on next slide…

  18. Solution: (continued) Vin + Vout The last result illustrates why it is rarely necessary to calculate an exact value for the input resistance of a noninverting amplifier. For practical circuits, you can assume it is ideal. - Rf 36 kW Ri 1.5 kW The output resistance is 0.019 W This extremely small resistance is close to ideal. As in the case of the input resistance, it is rarely necessary to calculate an exact value for the noninverting amplifier.

  19. Input resistance for the inverting amplifier Recall that negative feedback forces the inverting input to be near ac ground for the inverting amplifier. For this reason, the input resistance of the inverting amplifier is equal to just the input resistor, Ri. That is, Rin(I) = Ri. Rf The low input resistance is usually a disadvantage of this circuit. However, because the Rin(I) is equal to Ri, it can easily be set by the user for those cases where a specific value is needed. Ri Vin - Vout +

  20. Output resistance for the inverting amplifier The equation for the output resistance of the inverting amplifier is the essentially the same as the noninverting amplifier: Rf Although Rout(I) is very small, this does not imply that an op-amp can drive any load. The maximum current that the op-amp can supply is limited; for the 741C, it is typically 20 mA. Ri Vin - Vout +

  21. Example: What is the input resistance and the gain of the inverting amplifier? Rf 36 kW Ri Vin - Vout 1.5 kW Solution: + The gain is -24 The input resistance = Ri = 1.5 kW The output resistance is nearly identical to the noninverting case, where it was shown to be negligible.

  22. Voltage-follower The voltage-follower is a special case of the noninverting amplifier in which Acl = 1. The input resistance is increased by negative feedback and the output resistance is decreased by negative feedback. This makes it an ideal circuit for interfacing a high-resistance source with a low resistance load. Vin + Vout -

  23. Selected Key Terms A special type of amplifier exhibiting very high open-loop gain, very high input resistance, very low output resistance, and good rejection of common-mode signals. Operational amplifier Differential amplifier Common-mode rejection ratio (CMRR) An amplifier that produces an output proportional to the difference of two inputs. A measure of a diff-amp's or op-amp's ability to reject signals that appear the same on both inputs; the ratio of differential voltage gain or open-loop gain (for op-amps) to common-mode gain.

  24. Selected Key Terms Open-loop voltage gain Closed-loop voltage gain Noninverting amplifier Inverting amplifier The internal voltage gain of an op-amp without feedback. The overall voltage gain of an op-amp with negative feedback. An op-amp closed-loop configuration in which the input signal is applied to the noninverting input. An op-amp closed-loop configuration in which the input signal is applied to the inverting input.

  25. Quiz 1. When two identical in-phase signals are applied to the inputs of a differential amplifier, they are said to be a. feedback signals. b. noninverting signals. c. differential-mode signals. d. common-mode signals.

  26. Quiz 2. Assume a differential amplifier has an input signal applied to the base of Q1 as shown. An inverted replica of this signal will appear at the • emitter terminals. • collector of Q1 • collector of Q2 • all of the above. RC1 RC2 Q1 Q2 RE -VEE

  27. Quiz 3. A differential amplifier will tend to reject • noise that is in differential-mode. • noise that is in common-mode. • only high frequency noise. • all noise.

  28. Quiz 4. The average of two input currents required to bias the first stage of an op-amp is called the a. input offset current. b. open-loop input current. c. feedback current. d. input bias current.

  29. Quiz 5. The slew rate illustrated is a. 0.5 V/ms b. 1.0 V/ms c. 2.0 V/ms d. 2.4 V/ms Vout (V) 12 10 0 -10 -12 10 ms

  30. + Vin - Rf Vf Feedback network Ri Quiz 6. For the circuit shown, Vf is approximately equal to a. Vin b. Vout c. ground. d. none of the above. Vout

  31. Quiz 7. For the inverting amplifier shown, the input resistance is closest to a. zero b. 10 kW c. 2 MW d. 8 GW Rf 150 kW Ri Vin - Vout 10 kW +

  32. Quiz 8. For the inverting amplifier shown, the output resistance is closest to a. zero b. 10 kW c. 150 kW d. 8 GW Rf 150 kW Ri Vin - Vout 10 kW +

  33. Quiz 9. The gain of the inverting amplifier shown is a. -1 b. -10 c. -15 d. -16 Rf 150 kW Ri Vin - Vout 10 kW +

  34. Quiz 10. A voltage follower has a. current gain. b. voltage gain. c. both of the above. d. none of the above.

  35. Quiz Answers: 1. d 2. b 3. b 4. d 5. c 6. a 7. b 8. a 9. c 10. a

More Related