OpAmp Tutorial - Electronics I - Lecture Slides, Slides of Electronics

Download OpAmp Tutorial - Electronics I - Lecture Slides and more Slides Electronics in PDF only on Docsity! Kristin Ackerson, Virginia Tech EE Spring 2002 – VTech –Calvin Project _ + Operational Amplifiers Tutorial Series 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 The Four Amplifier Types Description Gain Symbol Transfer Function Voltage Amplifier or Voltage Controlled Voltage Source (VCVS) Av vo/vin Current Amplifier or Current Controlled Current Source (ICIS) Ai io/iin Transconductance Amplifier or Voltage Controlled Current Source (VCIS) gm (siemens) io/vin Transresistance Amplifier or Current Controlled Voltage Source (ICVS) rm (ohms) vo/iin VCVS (Voltage Amplifier) Summary Noninverting Configuration + _ vin + + - vO vid i(+) i(-) iO iF RF RL R1 i1 vid = vo/AOL Assuming AOL  vid =0 Also, with the assumption that Rin =  i(+) = i(-) = 0 _ vF + _ v1 + _ vL + _ iL Applying KVL the following equations can be found: v1 = vin vO = v1 + vF = vin+ iFRF 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 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 VCVS (Voltage Amplifier) Summary Inverting Configuration + _ RL vO + - vin + _ R1i1 RFiF The same assumptions used to find the equations for the noninverting configuration are also used for the inverting configuration. General Equations: i1 = vin/R1 iF = i1 vo = -iFRF = -vinRF/R1 Av = RF/R1  = R1/RF 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. VCVS (Voltage Amplifier) Summary Inverting Configuration Continued VCVS (Voltage Amplifier) Inverting Configuration Example + _ RL + - vin + _ R1i1 RFiF Given: vin = 0.6 V, RF = 20 k R1 = 2 k , AOL = 400k Rin = 8 M  , Ro = 60  Find: vo , iF , Av ,  , RinF and RoFvO 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 + AOL 1 + 0.09*400,000 Note:  is 0.09 because using different formula than above VCIS (Transconductance Amplifier) Voltage to Current Converter Example + _ Load iL R1i1 vin + _ Given: vin = 2 V, R1 = 2 k vo(max) =  10 V Find: iL , gm and RL(max) 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. VCIS (Transresistance Amplifier) Summary Current to Voltage Converter General Equations: iF = iin vo = -iFRF rm = vo/iin = RF + _ iF iin RF vO + - 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. 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) = (5x10 5 V/s) / (2 * 12) = 6,630 Hz  f is the frequency in Hz Slew Rate Distortion v t desired output waveform actual output because of slew rate limitation  t  v The picture above shows exactly what happens when the slew rate limitations are not met and the output of the operational amplifier is distorted. SR = v/t = m (slope) 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 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 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. 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.