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Kristin Ackerson, Virginia Tech EE Spring 2002

Analog Electronics Tutorial Series Operational Amplifiers. _. +. Kristin Ackerson, Virginia Tech EE Spring 2002. Table of Contents. The Operational Amplifier______________________________slides 3-4 The Four Amplifier Types______________________________slide 5

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Kristin Ackerson, Virginia Tech EE Spring 2002

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  1. Analog Electronics Tutorial Series Operational Amplifiers _ + Kristin Ackerson, Virginia Tech EE Spring 2002

  2. Table of Contents The Operational Amplifier______________________________slides 3-4 The Four Amplifier Types______________________________slide 5 VCVS(Voltage Amplifier) Summary: Noninverting Configuration____________slides 6-9 Inverting Configuration________________slides 10-12 ICIC(Current Amplifier) Summary________________________slide 13 VCIS (Transconductance Amplifier) Summary_____________slides 14-15 ICVS (Transresistance Amplifier) Summary_______________slides 16-18 Power Bandwidth_____________________________________slide 19 Slew Rate____________________________________________slide 20 Slew Rate Output Distortion____________________________ slide 21 Noise Gain___________________________________________slide 22 Gain-Bandwidth Product_______________________________slide 23 Cascaded Amplifiers - Bandwidth________________________slide 24 Common Mode Rejection Ratio__________________________slides 25-26 Power Supply Rejection Ratio___________________________slide 27 Sources_____________________________________________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 A vid 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 1x1010 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 Kristin Ackerson, Virginia Tech EE Spring 2002

  6. VCVS (Voltage Amplifier) Summary Noninverting Configuration i(+) vid = vo/AOL Assuming AOL vid =0 Also, with the assumption that Rin =  i(+) = i(-) = 0 + iO iL vid _ + iF + vO + + vin i(-) vF RF RL vL - _ _ _ + R1 This means that, iF = i1 Therefore: iF = vin/R1 Using the equation to the left the output voltage becomes: vo = vin + vinRF = vin RF + 1 R1 R1 v1 _ Applying KVL the following equations can be found: v1 = vin vO = v1 + vF = vin+ iFRF i1 Kristin Ackerson, Virginia Tech EE Spring 2002

  7. VCVS (Voltage Amplifier) Summary Noninverting Configuration Continued The 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 1 + AOL 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 + RF AF is the amplifier gain with feedback Kristin Ackerson, Virginia Tech EE Spring 2002

  8. VCVS (Voltage Amplifier) Summary Noninverting Configuration Continued Input and Output Resistance Ideally, the input resistance for this configuration is infinity, but the a closer prediction of the actual input resistance can be found with the following 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 a more accurate value: RoF = Ro Where Ro is given for the AOL+ 1specified device. Usually Ro is in the 10s of s range. Kristin Ackerson, Virginia Tech EE Spring 2002

  9. VCVS (Voltage Amplifier) i(+) + iO iL vid _ + iF + vO + + vin i(-) vF RF RL vL - _ _ _ + R1 v1 _ i1 Noninverting Configuration Example Given: vin = 0.6V, RF = 200 k R1 = 2 k , AOL = 400k Rin = 8 M  , Ro = 60  Find: vo , iF , Av ,  , RinF and RoF Solution: vo = vin + vinRF = 0.6 + 0.6*2x105 = 60.6 V iF = vin = 0.6 = 0.3 mA R1 2000 R1 2000 Av = RF + 1 = 2x105 + 1 = 101 = 1 = 1 = 9.9x10-3 R1 2000 AOL 101 RinF = Rin (1 + AOL) = 8x106 (1 + 9.9x10-3*4x105) = 3.1688x1010 RoF = Ro = 60= 0.015  AOL+ 1 9.9x10-3*4x105 + 1 Kristin Ackerson, Virginia Tech EE Spring 2002

  10. VCVS (Voltage Amplifier) Summary Inverting Configuration The same assumptions used to find the equations for the noninverting configuration are also used for the inverting configuration. RF iF R1 i1 + + _ + vin vO _ RL - 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 Resistance Ideally, the input resistance for this configuration is equivalent to R1. However, the actual value of the input resistance is given by the following formula: Rin = R1 + RF 1 + AOL Ideally, the output resistance is zero, but the formula below gives a more accurate value: RoF = Ro 1 + AOL Note:  = R1 This is different from the equation used R1 + RF on the previous slide, which can be confusing. Kristin Ackerson, Virginia Tech EE Spring 2002

  12. VCVS (Voltage Amplifier) Inverting Configuration Example RF iF Given: vin = 0.6 V, RF = 20 k R1 = 2 k , AOL = 400k Rin = 8 M  , Ro = 60  Find: vo , iF , Av ,  , RinF and RoF R1 i1 + + _ + vin vO _ RL - Solution: vo = -iFRF = -vinRF/R1 = -(0.6*20,000)/2000 = 12 V iF = i1 = vin/R1 = 1 / 2000 = 0.5 mA Av = RF/R1 = 20,000 / 2000 = 10  = R1/RF = 2000 / 20,000 = 0.1 Rin = R1 + RF = 2000 + 20,000 = 2,000.05  1 + AOL 1 + 400,000 RoF = Ro = 60= 1.67 m  1 + AOL1 + 0.09*400,000 Note:  is 0.09 because using different formula than above Kristin Ackerson, Virginia Tech EE Spring 2002

  13. ICIS (Current Amplifier) Summary  Not commonly done using operational amplifiers iL Load _ iin + iin = iL Similar to the voltage follower shown below: 1 Possible ICIS Operational Amplifier Application _ + vin = vo vin _ + + vO - Both these amplifiers have unity gain: Av = Ai = 1 Voltage Follower Kristin Ackerson, Virginia Tech EE Spring 2002

  14. VCIS (Transconductance Amplifier) Summary Voltage to Current Converter iL iL Load Load R1 R1 i1 i1 _ _ 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: RL(max) = vo(max)/iL Kristin Ackerson, Virginia Tech EE Spring 2002

  15. VCIS (Transconductance Amplifier) Voltage to Current Converter Example iL Given: vin = 2 V, R1 = 2 k vo(max) =  10 V Find: iL , gm and RL(max) Load R1 i1 _ + + vin Solution: iL = i1 = vin/R1 = 2 / 2000 = 1 mA gm = io/vin = 1/R1 = 1 / 2000 = 0.5 mS RL(max) = vo(max)/iL = 10 V / 1 mA = 10 k  _ • Note: • If RL > RL(max) the op amp will saturate • The output current, iL is independent of the load resistance. Kristin Ackerson, Virginia Tech EE Spring 2002

  16. VCIS (Transresistance Amplifier) Summary Current to Voltage Converter RF iF _ + + 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 RF iF 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 Bandwidth The 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 2Vo(max) SR = 2fVo(max) where SR is the slew rate Example: Given: Vo(max) = 12 V and SR = 500 kV/s Find: BWp Solution: BWp = 500 kV/s = 6.63 kHz 2 * 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 = 2fVo(max) SR = vo/tmax 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 / 2Vo(max) = (5x105 V/s) / (2 * 12) = 6,630 Hz  f is the frequency in 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 limitation The 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: AN = R1 + RF R1 Example 1: Given a noninverting amplifier with the resistance values, R1 = 2 k and RF = 200 k Find: The noise gain. AN = 2 k + 200 k = 101  Note: For the 2 k noninverting amplifier AN = AV Example 2: Given an inverting amplifier with the resistance values, R1 = 2 k and RF = 20 k Find: The noise gain. AN = 2 k + 20 k = 12  Note: For the 2 k inverting amplifier AN > AV Kristin Ackerson, Virginia Tech EE Spring 2002

  23. Gain-Bandwidth Product In most operational amplifiers, the open-loop gain begins dropping 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 = ANBW Example: For a 741 op amp, a noise gain of 10 k corresponds to a bandwidth of ~200 Hz Find: 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 gain The Total Bandwidth of the Cascaded Amplifiers is: BWT = BWs(21/n – 1)1/2 where n is the number of amplifiers that are being cascaded Example: Cascading 3 Amplifiers with GBW = 1 MHz and AN = 15, Find: The Total Bandwidth, BWT BWS = 1 MHz / 15 = 66.7 kHz BWT = 66.7 kHz (21/3 – 1)1/2 = 34 kHz Kristin Ackerson, Virginia Tech EE Spring 2002

  25. Common-Mode Rejection Ratio The 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 Example Given: A 741 op amp with CMRR = 90 dB and a noise gain, AN = 1 k Find: The common mode gain, Acm Acm = AN = 1000 log-1 (CMRR / 20) log-1 (90 / 20) = 0.0316 It is very desirable for the common-mode gain to be small. Kristin Ackerson, Virginia Tech EE Spring 2002

  27. Power Supply Rejection Ratio One of the reasons op amps are so useful, is that they can be operated from a wide variety of power supply voltages. The 741 op amp can be operated from bipolar supplies ranging 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 Kristin Ackerson, Virginia Tech EE Spring 2002

  29. Sources Dailey, Denton. Electronic Devices and Circuits, Discrete and Integrated. Prentice Hall, New Jersey: 2001. (pp 456-509) 1Table 12.1: Selected Op Amps and Their Open Loop Characteristics, pg 457 Liou, 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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