ECE 2006 UMD Circuit AnalysisRev. 1.0 2/04LAB 5 – OPERATIONAL AMPLIFIERSPRE-LAB CALCULATIONS:Use circuit analysis techniques learned in class to analyze the circuit in Figure 5.2. Solve for Vo assuming that the effective resistance of the LED is 50 K-Ohms.OBJECTIVE:1. To introduce the student to the application and analysis of Operational Amplifiers2. To apply circuit analysis techniques to Op Amp circuitsREFERENCES:1. Alexander and Sadiku, Fundamentals of Electric Circuits, 2nd Edition, 2003, McGraw-Hill2. Hambley, ELECTRONICS: A Top-Down Approach to Computer-Aided CircuitDesign, 1994, Prentice-HallEquipment:Digital Multimeter (DMM)DC Power SupplyResistors: 100Ω, 1K, 1K, 10K, 100K, 390K; 5% ToleranceLED741 Op AmpDISCUSSION:Operational Amplifiers or Op Amps are undoubtedly the most versatile analog device in common use. In addition, circuit analysis of Op Amp circuits is a straightforward endeavor. It has become common practice therefore to introduce Op Amp circuits to beginning engineering students as a means to reinforce their newly acquired analysis skills.Without getting into the details of design and construction, an Op Amp can be modeled asshown in Figure 5.0 below:Scott Norr Page 1 1/13/2019ECE 2006 UMD Circuit AnalysisRev. 1.0 2/04Figure 5.0: Operational Amplifier – Equivalent CircuitIt can be seen from the figure above that the difference in voltage across the input terminals, V+ and V-, is multiplied by the gain, A, and is available at the output terminal as Vout (with respect to ground).The ideal Op Amp is characterized by the following parameters: Ri (the input impedance) is Infinite.Ro (the output impedance) is Zero.A (the gain) is Infinite.From this idealization, it is possible to make the following assumptions:Ii (the input current to the Op Amp) is Zero.Vd = (V+) – (V-) = ZeroThus, V+ = V-These conditions make Nodal Analysis of an ideal Op Amp circuit very simple.PROCEDURE:1. Connect the DC circuit shown in Figure 5.1: FIGURE 5.1 – DC VOLTAGE DIVIDERScott Norr Page 2 1/13/2019ECE 2006 UMD Circuit AnalysisRev. 1.0 2/042. Power up the adjustable DC power supply and set it for an output voltage of 6.00 Volts.3. Turn ON the output of the power supply.4. Measure Vo, the voltage drop across the 390 K-Ohm resistor, using the DMM.5. Turn OFF the output of the power supply.6. Vo = ___________________ Volts7. Now connect a Resistor and a Light Emitting Diode (LED) across Vo as shown in Figure 5.2:FIGURE 5.2 – LED CIRCUIT8. Turn ON the output of the power supply.9. Measure again the output voltage, Vo, using the DMM.10. Vo = ___________________ Volts11. Obtain your Instructor’s Signature: ____________________________12. Turn OFF the output of the power supply.13. Answer in the Lab Report: “Why is the value of Vo different?”14. Calculate the “effective resistance” of the LED and 100 Ohm Resistor, by performing nodal analysis at the output node (between the 100 K and 390 K-Ohm resistors). Since Vo is known, solve for RLED.At the Node: Vo – Vs + Vo – 0 . = 0 100 K 390K // RLED15. RLED = _______________ Ohms Scott Norr Page 3 1/13/2019ECE 2006 UMD Circuit AnalysisRev. 1.0 2/0416. A Vo of approximately 1.8 Volts or above is sufficient to make the LED glow, provided that it receives enough current.17. Does the LED turn on (light up) in this circuit? ___________It may be useful to consider the 6-Volt source and the 100 k-Ohm resistor as a Thevenin Pair (i.e. Vth and Rth). If the resistance of the LED were very, very small, say zero ohms, the current delivered by the 6-Volt source would be 60 µ-A. (100 K-Ohm * 60 µ-A = 6 Volts). This is not enough current to make the LED glow. Also, ifRLED were extremely small, Vo would be almost zero.Thus the Load Impedance, RLED, is too small for the resistor bridge and collapses the output voltage, Vo. Or, in other words, the Output Impedance of the source, (Rth), is too high to provide the current necessary to make the LED glow. A practical way to “lower” the output impedance of this voltage-divider circuit is to use an Operational Amplifier (Op Amp).Op Amps have very high input impedance, meaning they don’t draw much current from a source in order to work properly. In addition, they have reasonably low outputimpedance, and can thus supply a fair amount of current to a load.18. Insert an Op Amp into the previous network in order to produce the circuit shown in Figure 5.3 below: (Note that a minus-six volt source is needed to properly bias the op amp)FIGURE 5.3 – VOLTAGE FOLLOWERScott Norr Page 4 1/13/2019ECE 2006 UMD Circuit AnalysisRev. 1.0 2/04The Op Amp circuit above is called a “Voltage Follower”, denoted by the unity feedback loop to the inverting input (i.e. Vout is short-circuited to V-). A voltage follower performs as denoted. Its output follows the input. An ideal voltage followerhas an input of Vo volts and an output of Vo volts. One might argue that a piece of wire also acts as a voltage follower and is much cheaper and easier to use than an Op Amp. The beauty of the Op Amp voltage follower is the current gain of the circuit. The circuit above has an input current of much less than the 60 µ-A available, but the output current of the Op Amp can be much higher. A wire cannot duplicate that. The Op Amp looks like a high impedance load to the voltage-divider source, and also looks like a low impedance output to the LED load.19. Turn ON the output of the power supply.20. Measure Vo and Vout with the DMM. 21. Vo = _______________ Volts Vout = _______________________ Volts22. Turn OFF the output of the power supply. Remove the LED and 100 Ohm resistor from the output.23. Turn ON the power supply and measure Vout now that the load (LED) has been removed.24. Vout (no load) = ___________________ Volts25. Turn OFF the output of the power supply. 26. Did the LED turn on (light up) in this circuit? ___________27. Discussion for the Lab Report: “Describe the impact of putting the Op Amp Voltage Follower between the output voltage, Vo, and the load (the LED). Note the effect on Vo and on source and load impedances.” The Op Amp is probably the most versatile analog chip available. It has a host of applications in a broad range of circuits. The key to making Op Amps do different things is to understand the impact of feedback on Op Amp performance. The first step to such understanding is to analyze the Inverting