Figure 1: Operational Amplifier DiagramECE 2006 University of Minnesota Duluth Lab 5OPERATIONAL AMPLIFIERSINTRODUCTIONThe student will be introduced to the application and analysis of operational amplifiers in this laboratory experiment. The student will apply circuit analysis techniques to study circuits containing operational amplifiers.BACKGROUNDOperational 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 introduceOp 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 as shown in Figure 1.It can be seen in Figure 1 that thedifference in voltage across theinput terminals, V+ and V-, ismultiplied by the gain, A, and isavailable at the output terminal asVout (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.Scott Norr Page 1 January 2005Figure 2: DC Voltage DividerFigure 3: LED CircuitECE 2006 University of Minnesota Duluth Lab 5THEORETICAL PROCEDUREUsing circuit analysis techniques, analyze thecircuit in Figure 3 to solve for Vo. Assume thatthe effective resistance of the LED is 50 KΩ.Vo = ___________________ VoltsEXPERIMENTAL PROCEDUREConnect the DC circuit shown in Figure 2 Power up the adjustable DC power supply and set it for an output voltage of 6.00 Volts.Turn ON the output of the power supply.Measure Vo, the voltage drop across the 390 K-Ohm resistor, using the DMM.Vo = ___________________ VoltsTurn OFF the output of the power supply.Now connect a Light Emitting Diode (LED) across Vo as shown in Figure 3.Turn ON the output of the power supply.Measure again the output voltage, Vo, using the DMM.Vo = ___________________ VoltsScott Norr Page 2 January 2005ECE 2006 University of Minnesota Duluth Lab 5Obtain your Instructor’s Signature: ____________________________Turn OFF the output of the power supply.Why is the value of Vo different? Answer question in lab report.Calculate the “effective resistance” of the LED, by performing nodal analysis at the output node (between the 100 and 390 K-Ohm resistors). Since Vo is known,solve for RLED.At the Node: Vo – Vs + Vo – 0 . = 0 100 K 390K // RLEDRLED = _______________ OhmsA Vo of approximately 2.0 Volts or above is sufficient to make the LED glow, provided that it receives enough current.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, if RLED 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 ofthe 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 bridge 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 output impedance, and can supply a fair amount of current to a load.Insert an Op Amp into the previous network in order to produce the circuit shown in Figure 4 below: (Note that a minus-six volt source is needed to properly bias the op amp)Scott Norr Page 3 January 2005Figure 4: Voltage FollowerECE 2006 University of Minnesota Duluth Lab 5The Op Amp circuit above is called a “Voltage Follower” circuit, 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 follower has an input of Vo volts and an output of Vo volts. One mightargue that a piece of wire also acts as a voltage follower and is much cheaperand easier to use than an Op Amp. The beauty of the Op Amp voltage follower is the 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 canbe much higher. A wire cannot duplicate that. The Op Amp looks like a high impedance load to the resistor bridge source, and also looks like a low impedance source to the LED load.Scott Norr Page 4 January 2005Figure 5: Inverting AmplifierECE 2006 University of Minnesota Duluth Lab 5Turn ON the output of the power supply.Measure Vo and Vout with the DMM. Vo = _______________ Volts Vout = ___________________ VoltsTurn OFF the output of the power supply. Remove the LED.Turn ON the power supply and measure Vout now that the load (LED) has been removed.Vout (no load) = ___________________ VoltsTurn OFF the output of the power supply. Compare Vo for Figure 4 with Vo for Figures 2 and 3. Which is it closer to? ______________________Does the LED turn on (light up) in this circuit? ___________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. Include Discussion in lab report. The Op Amp is probablythe most versatile analogchip available. It has ahost of applications in abroad range of circuits.The key to making OpAmps do different things isto understand the impactof feedback on Op Ampperformance. The firststep to suchunderstanding is toanalyze the InvertingAmplifier circuit.Connect the Op Amp circuit shown in Figure 5.Scott Norr Page 5 January 2005ECE 2006 University of Minnesota Duluth Lab 5Nodal analysis at the node labeled V- (between resistors R1 and R2) produces the following results: (V-) – Vs + (V-) – (Vout) = 0 ; and (V-) ≈ (V+) = 0 100 K 390 KVout = - 390 K * Vs = -3.9 Vs ;