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- 1. _+ Kristin Ackerson, Virginia Tech EE Spring 2002
- 2. Table of ContentsThe Operational Amplifier______________________________slides 3-4The Four Amplifier Types______________________________slide 5VCVS(Voltage Amplifier) Summary: Noninverting Configuration____________slides 6-9 Inverting Configuration________________slides 10-12ICIC(Current Amplifier) Summary________________________slide 13VCIS (Transconductance Amplifier) Summary_____________slides 14-15ICVS (Transresistance Amplifier) Summary_______________slides 16-18Power Bandwidth_____________________________________slide 19Slew Rate____________________________________________slide 20Slew Rate Output Distortion____________________________ slide 21Noise Gain___________________________________________slide 22Gain-Bandwidth Product_______________________________slide 23Cascaded Amplifiers - Bandwidth________________________slide 24Common Mode Rejection Ratio__________________________slides 25-26Power Supply Rejection Ratio___________________________slide 27Sources_____________________________________________slide 28 Kristin Ackerson, Virginia Tech EE Spring 2002
- 3. The Operational Amplifier• Usually Called Op Amps• An amplifier is a device that accepts a varying input signal and produces a similar output signal with a larger amplitude.• Usually connected so part of the output is fed back to the input. (Feedback Loop)• Most Op Amps behave like voltage amplifiers. They take an input voltage and output a scaled version.• They are the basic components used to build analog circuits.• The name “operational amplifier” comes from the fact that they were originally used to perform mathematical operations such as integration and differentiation.• Integrated circuit fabrication techniques have made high- performance operational amplifiers very inexpensive in comparison to older discrete devices. Kristin Ackerson, Virginia Tech EE Spring 2002
- 4. The Operational Amplifier +VS i(-) _ Inverting RO vid A Output Ri vO = Advid Noninverting i(+) + -VS• i(+), i(-) : Currents into the amplifier on the inverting and noninverting lines respectively• vid : The input voltage from inverting to non-inverting inputs• +VS , -VS : DC source voltages, usually +15V and –15V• Ri : The input resistance, ideally infinity• A : The gain of the amplifier. Ideally very high, in the 1x10 10 range.• RO: The output resistance, ideally zero• vO: The output voltage; vO = AOLvid where AOL is the open-loop voltage gain Kristin Ackerson, Virginia Tech EE Spring 2002
- 5. The Four Amplifier Types Gain Transfer Description Symbol Function Voltage Amplifier or Av vo/vinVoltage Controlled Voltage Source (VCVS) Current Amplifier or Ai io/iinCurrent Controlled Current Source (ICIS) Transconductance Amplifier gm or io/vinVoltage Controlled Current Source (VCIS) (siemens) Transresistance Amplifier rm or vo/iinCurrent Controlled Voltage Source (ICVS) (ohms) Kristin Ackerson, Virginia Tech EE Spring 2002
- 6. VCVS (Voltage Amplifier) Summary Noninverting Configuration i(+) + iO iL vid = vo/AOL vid Assuming AOL _ iF + + + + vO vid =0vin i(-) vF RF RL vL - _ _ _ Also, with the assumption that Rin = + i(+) = i(-) = 0 v1 R1 This means that, Applying KVL the _ i1 iF = i1 following equations Therefore: iF = vin/R1 can be found: Using the equation to the left the output v1 = vin voltage becomes: vO = v1 + vF = vin+ iFRF vo = vin + vinRF = vin RF + 1 R1 R1 Kristin Ackerson, Virginia Tech EE Spring 2002
- 7. VCVS (Voltage Amplifier) Summary Noninverting Configuration ContinuedThe closed-loop voltage gain is symbolized by Av and is found to be: Av = vo = RF + 1 vin R1 The original closed loop gain equation is: Av = AF = AOL AF is the amplifier gain with 1 + AOL feedback Ideally AOL , Therefore Av = 1 Note: The actual value of AOL is given for the specific device and usually ranges from 50k 500k. is the feedback factor and by assuming open-loop gain is infinite: = R1 R1 + R F Kristin Ackerson, Virginia Tech EE Spring 2002
- 8. VCVS (Voltage Amplifier) Summary Noninverting Configuration Continued Input and Output ResistanceIdeally, the input resistance for this configuration is infinity, but the acloser prediction of the actual input resistance can be found with thefollowing formula: RinF = Rin (1 + AOL) Where Rin is given for the specified device. Usually Rin is in the M range.Ideally, the output resistance is zero, but the formula below gives amore accurate value: RoF = Ro Where Ro is given for the AOL + 1 specified device. Usually Ro is in the 10s of s range. Kristin Ackerson, Virginia Tech EE Spring 2002
- 9. VCVS (Voltage Amplifier) i(+) Noninverting Configuration Example + iO iL vid Given: vin = 0.6V, RF = 200 k _ iF + + + + vO R1 = 2 k , AOL = 400kvin i(-) vF RF RL vL - _ _ _ Rin = 8 M , Ro = 60 + Find: vo , iF , Av , , RinF and RoF v1 R1 _ i1Solution:vo = vin + vinRF = 0.6 + 0.6*2x105 = 60.6 V iF = vin = 0.6 = 0.3 mA R1 2000 R1 2000Av = RF + 1 = 2x105 + 1 = 101 = 1 = 1 = 9.9x10-3 R1 2000 AOL 101RinF = Rin (1 + AOL) = 8x106 (1 + 9.9x10-3*4x105) = 3.1688x1010 RoF = Ro = 60 = 0.015 Kristin Ackerson, Virginia Tech EE AOL + 1 9.9x10-3*4x105 + 1 Spring 2002
- 10. VCVS (Voltage Amplifier) Summary Inverting Configuration RF The same iF assumptions used to R1 find the equations for i1 the noninverting + configuration are + _ + also used for thevin vO inverting _ RL - configuration. General Equations: i1 = vin/R1 iF = i1 vo = -iFRF = -vinRF/R1 Av = RF/R1 = R1/RF Kristin Ackerson, Virginia Tech EE Spring 2002
- 11. VCVS (Voltage Amplifier) Summary Inverting Configuration Continued Input and Output ResistanceIdeally, the input resistance for this configuration is equivalent to R 1.However, the actual value of the input resistance is given by thefollowing formula: Rin = R1 + RF 1 + AOLIdeally, the output resistance is zero, but the formula below gives amore accurate value: RoF = Ro 1 + AOLNote: = R1 This is different from the equation used R1 + R F on the previous slide, which can be confusing. Kristin Ackerson, Virginia Tech EE Spring 2002
- 12. VCVS (Voltage Amplifier) Inverting Configuration Example iF RF Given: vin = 0.6 V, RF = 20 k i1 R1 R1 = 2 k , AOL = 400k + + _ + Rin = 8 M , Ro = 60 vin vO _ RL - Find: vo , iF , Av , , RinF and RoFSolution:vo = -iFRF = -vinRF/R1 = -(0.6*20,000)/2000 = 12 ViF = i1 = vin/R1 = 1 / 2000 = 0.5 mAAv = RF/R1 = 20,000 / 2000 = 10 = R1/RF = 2000 / 20,000 = 0.1Rin = R1 + RF = 2000 + 20,000 = 2,000.05 1 + AOL 1 + 400,000RoF = Ro = 60 = Note: is 0.09 because using 1.67 m Kristin Ackerson, Virginia Tech EE different formula than above Spring 2002
- 13. ICIS (Current Amplifier) Summary Not commonly done using operational amplifiers iL Load _iin + iin = iL Similar to the voltage1 Possible follower shown below: ICISOperational Amplifier _ + vin = voApplication vin _ + + vO -Both these amplifiers have unity gain: Voltage Follower Av = A i = 1 Kristin Ackerson, Virginia Tech EE Spring 2002
- 14. VCIS (Transconductance Amplifier) Summary Voltage to Current Converter iL iL Load Load i1 R1 i1 R1 _ _ + OR +vin + vin + _ _ + vin _ General Equations: iL = i1 = v1/R1 v1 = vin The transconductance, gm = io/vin = 1/R1 Therefore, iL = i1 = vin/R1 = gmvin The maximum load resistance is determined by: Kristin Ackerson, Virginia Tech EE RL(max) = vo(max)/iL Spring 2002
- 15. VCIS (Transconductance Amplifier) Voltage to Current Converter Example iL Load Given: vin = 2 V, R1 = 2 k i1 R1 vo(max) = 10 V _ + Find: iL , gm and RL(max)vin + _ Solution: iL = i1 = vin/R1 = 2 / 2000 = 1 mANote: gm = io/vin = 1/R1 = 1 / 2000 = 0.5 mS• If RL > RL(max) the op amp will saturate RL(max) = vo(max)/iL = 10 V / 1 mA• The output current, iL is independent of the load = 10 k resistance. Kristin Ackerson, Virginia Tech EE Spring 2002
- 16. VCIS (Transresistance Amplifier) Summary Current to Voltage Converter iF RF _ + iin + vO - General Equations: iF = iin vo = -iFRF rm = vo/iin = RF Kristin Ackerson, Virginia Tech EE Spring 2002
- 17. VCIS (Transresistance Amplifier) Summary Current to Voltage Converter • Transresistance Amplifiers are used for low-power applications to produce an output voltage proportional to the input current. • Photodiodes and Phototransistors, which are used in the production of solar power are commonly modeled as current sources. • Current to Voltage Converters can be used to convert these current sources to more commonly used voltage sources. Kristin Ackerson, Virginia Tech EE Spring 2002
- 18. VCIS (Transresistance Amplifier) Current to Voltage Converter Example iF RF Given: iin = 10 mA _ RF = 200 Find: iF , vo and rm +iin + vO - Solution: iF = iin = 10 mA vo = -iFRF = 10 mA * 200 = 2 V rm = vo/iin = RF = 200 Kristin Ackerson, Virginia Tech EE Spring 2002
- 19. Power BandwidthThe maximum frequency at which a sinusoidal output signal can be produced without causing distortion in the signal. The power bandwidth, BWp is determined using the desired output signal amplitude and the the slew rate (see next slide) specifications of the op amp. BWp = SR 2π Vo(max) SR = 2π fVo(max) where SR is the slew rateExample:Given: Vo(max) = 12 V and SR = 500 kV/sFind: BWpSolution: BWp = 500 kV/s = 6.63 kHz 2π * 12 V Kristin Ackerson, Virginia Tech EE Spring 2002
- 20. Slew Rate A limitation of the maximum possible rate of change of the output of an operational amplifier.As seen on the previous slide, This is derived from:SR = 2π fVo(max) SR = vo/ tmax f is the frequency in Hz Slew Rate is independent of the closed-loop gain of the op amp.Example:Given: SR = 500 kV/s and vo = 12 V (Vo(max) = 12V)Find: The t and f.Solution: t = vo / SR = (10 V) / (5x105 V/s) = 2x10-5 s f = SR / 2π Vo(max) = (5x105 V/s) / (2π * 12) = 6,630 Hz Kristin Ackerson, Virginia Tech EE Spring 2002
- 21. Slew Rate Distortion v desired output waveform SR = v/ t = m (slope) v t t actual output because of slew rate limitationThe picture above shows exactly what happens when the slew rate limitations are not met and the output of the operational amplifier is distorted. Kristin Ackerson, Virginia Tech EE Spring 2002
- 22. Noise Gain The noise gain of an amplifier is independent of the amplifiers configuration (inverting or noninverting) The noise gain is given by the formula: A N = R 1 + RF R1Example 1: Given a noninverting amplifier with the resistance values, R1 = 2 k and RF = 200 kFind: The noise gain.AN = 2 k + 200 k = 101 Note: For the 2 k noninverting amplifier AN = AVExample 2: Given an inverting amplifier with the resistance values, R1 = 2 k and RF = 20 kFind: The noise gain.AN = 2 k + 20 k = 12 Note: For the Kristin Ackerson, Virginia Tech EE 2 k inverting amplifier2002 N > AV Spring A
- 23. Gain-Bandwidth Product In most operational amplifiers, the open-loop gain beginsdropping off at very low frequencies. Therefore, to make the op amp useful at higher frequencies, gain is traded for bandwidth. The Gain-Bandwidth Product (GBW) is given by: GBW = ANBWExample: For a 741 op amp, a noise gain of 10 k corresponds to a bandwidth of ~200 HzFind: The GBW GBW = 10 k * 200 Hz = 2 MHz Kristin Ackerson, Virginia Tech EE Spring 2002
- 24. Cascaded Amplifiers - Bandwidth Quite often, one amplifier does not increase the signal enough and amplifiers are cascaded so the output of one amplifier is the input to the next.The amplifiers are matched so:BWS = BW1 = BW2 = GBW where, BWS is the bandwidth of all AN the cascaded amplifiers and AN is the noise gainThe Total Bandwidth of the Cascaded Amplifiers is:BWT = BWs(21/n – 1)1/2 where n is the number of amplifiers that are being cascadedExample: Cascading 3 Amplifiers with GBW = 1 MHz and A N = 15,Find: The Total Bandwidth, BWTBWS = 1 MHz / 15 = 66.7 kHzBWT = 66.7 kHz (21/3 – 1)1/2 = 34 kHz Kristin Ackerson, Virginia Tech EE Spring 2002
- 25. Common-Mode Rejection RatioThe common-mode rejection ratio (CMRR) relates to the ability of the op amp to reject common-mode input voltage. This is very important because common-mode signals are frequently encountered in op amp applications. CMRR = 20 log|AN / Acm| Acm = AN log-1 (CMRR / 20)We solve for Acm because Op Amp data sheets list the CMRR value.The common-mode input voltage is an average of the voltages that are present at the non-inverting and inverting terminals of the amplifier. vicm = v(+) + v(-) 2 Kristin Ackerson, Virginia Tech EE Spring 2002
- 26. Common-Mode Rejection Ratio ExampleGiven: A 741 op amp with CMRR = 90 dB and a noise gain, AN = 1 kFind: The common mode gain, AcmAcm = AN = 1000 log-1 (CMRR / 20) log-1 (90 / 20) = 0.0316It is very desirable for the common-mode gain to be small. Kristin Ackerson, Virginia Tech EE Spring 2002
- 27. Power Supply Rejection RatioOne of the reasons op amps are so useful, is that they canbe operated from a wide variety of power supply voltages. The 741 op amp can be operated from bipolar suppliesranging from 5V to 18V with out too many changes to the parameters of the op amp.The power supply rejection ratio (SVRR) refers to the slight change in output voltage that occurs when the power supply of the op amp changes during operation. SVRR = 20 log ( Vs / Vo) The SVRR value is given for a specified op amp. For the 741 op amp, SVRR = 96 dB over the range 5V to 18V. Kristin Ackerson, Virginia Tech EE Spring 2002
- 28. Open-Loop Op Amp Characteristics Table 12.11 Device LM741C LF351 OP-07 LH0003 AD549K HybridTechnology BJT BiFET BJT BiFET BJT AOL(typ) 200 k 100 k 400 k 40 k 100 k Rin 2 M 1012 8 M 100 k 1013 || 1 pF Ro 50 30 60 50 ~100 SR 0.5 V/µs 13 V/µs 0.3 V/µs 70 V/µs 3 V/µs CMRR 90 dB 100 dB 110 dB 90 dB 90 dB Kristin Ackerson, Virginia Tech EE Spring 2002
- 29. SourcesDailey, Denton. Electronic Devices and Circuits, Discrete and Integrated. Prentice Hall, New Jersey: 2001. (pp 456-509) 1 Table 12.1: Selected Op Amps and Their Open Loop Characteristics, pg 457Liou, J.J. and Yuan, J.S. Semiconductor Device Physics and Simulation. Plenum Press, New York: 1998.Neamen, Donald. Semiconductor Physics & Devices. Basic Principles. McGraw-Hill,Boston: 1997. (pp 351-357) Web Sources www.infoplease.com/ce6/sci/A0803814.html http://www.infoplease.com/ce6/sci/A0836717.html http://people.msoe.edu/~saadat/PSpice230Part3.htm Kristin Ackerson, Virginia Tech EE Spring 2002

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