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Introduction to Op-Amps Swarthmore College Engineering 72: Electronic Circuit Application Lab # 1 September 14, 2004 Heather Jones & David Luong Professor Erik Cheever Procedure The procedures can be found at the following web address: http://www.swarthmore.edu/NatSci/echeeve1/Class/e72/E72L2/Lab2(OpAmp).html Analysis 1. Op-Amp Circuits a. a. Inverting Amplifier Schematic 1 The internal resistances between the inputs (v+ and v-) and ground are infinite. Since v+is v grounded, v - is also grounded. As a result, the current into the node at v – is simply in R2 v by Ohm’s Law. The current across the R1 resistor is also in . By voltage division, R2 v vout in R1 . R2 So the gain of the circuit is vout R 1 . vin R2 Figure 1—Inverting Amplifier with R1 = 10kOhms Figure 2—Inverting Amplifier with R1 = 30kOhm b. Non-Inverting Amplifier Schematic 2 The node between R1 and R2 is vin since v+ = v-. Using a voltage divider, R1 vin vout R1 R2 Rewriting, vout R 1 1 . vin R2 A Multisim produced schematic and simulation are shown below. VCC 12V XFG1 7 1 5 U1 3 6 R1 2 1.00kOhm_1% LF411CH 4 V1 12 V R2 10.0kOhm_1% XSC1 G T B Schematic 3 Figure 3—Multisim Simulation A c. Ideal Integrator Schematic 4 Assume that R1 has no effect on the circuit. Since this is essentially an inverting amplifier, vout Z c , vin R2 where Z c is the impedance of the capacitor. It is equal to 1 vout sC 1 1 . vin R2 sC1 R2 1 . Rewriting, sC1 Figure 4 –Ideal Integrator with R1 having infinite resistance Figure 5–Ideal Integrator with R1 = 1MOhm 2. The resistor labeled R1 in the above diagram allows the integrator to deal with a DC signal. When this resistor is removed, the gain of the circuit is Vo(s)/Vi(s) = -1/(RsC), which goes to infinity as s goes to zero, so any DC signal will cause the amplifier to saturate. 3. Our calculated peak-to-peak value was 500 mV. Referring to the vout/vin relationship of the integrator circuit, we have values for the capacitor and resistor. To determine vout, we place vin on the right side of the equation, and replace it with its Laplace equivalent. Knowing that 1/s^2 is merely –ω^2, vout becomes 1/( ω^2*C1*R2). The time domain equivalent is vout(t) = 1/(C1*R2)*t. Noting that the input frequency is 1 kHz, it takes half that or .5 kHz to reach a peak. Substituting that and values for C1 and R2, vout(t) = 500 mV. Our measured value was 660 mV. Below is a Multisim simulation of the integrator circuit. Figure 6—Multisim Simulation 4. We measured a slew rate of 11.69 V/s for the 411 op-amp, which is close to the rate of 10 V/s listed in the manufacturer’s specifications. 5. An amplifier with a gain of one can be used as a buffer to transfer a voltage between circuits where a direct connection would have allowed current to flow and changed the voltage. 6. In the comparator circuit, if the voltage at the inverting input terminal is greater than the voltage at the non-inverting input terminal, there will be a closed switch inside the comparator, connecting the output to ground. In this case, the resistor R3, which is connected between Vcc and the output, prevents a short circuit between Vcc and ground. When the voltage at the non-inverting input terminal is greater than that at the inverting terminal, the switch is opened, and now no current can flow through R3, and the voltage at the output is pulled up to Vcc. 7. Schmitt Trigger Schematic 5 a. The circuit looks at the inputs at v+ and v- and compares them. The circuit creates two different thresholds, and checks the number of times the input signal crosses the thresholds. Depending on the threshold that is crossed (either high or low), the output changes states—either high or low. Specifically, when the input crosses the high threshold the output is low. When it crosses the low threshold, the output is high. b. When v- is greater than v+, the switch in the circuit, which is connected to ground and the output, is closed. That makes vout = 0 volts, and by voltage R1 k division, v 5 . [R1 is in parallel with R3]. For R1 = R2 = R1 R3 R2 R3 =10 k , v =5/3 volts. When v- is less than v+, the switch is opened. R1 By voltage division, v 5 =2.5 volts. Also vout is also 2.5 volts R1 R2 because no current can flow through R3 as there is no place for it to go. c. From part b, the lower threshold output voltage is 0 volts, and the high is 2.5 volts. d. See the printout below. Figure 7—Oscilloscope Output 8. For the comparator and Schmitt trigger circuits, our output could range from 0-5V, with threshold voltages at 2.5V or 1.67V. To get an input that was centered near this threshold band, and would thus cross through it periodically, we had to add in a DC offset in the function generator. To see this offset on the oscilloscope, we had to turn on DC coupling (AC coupling would have subtracted out the signal mean and shown the signal centered around 0V). 9. Relaxation Oscillator Schematic 6 Figure 8-Oscilloscope Output When the output is high, that means that v+ is greater than v- and the switch is open. This charges up the capacitor and thus increases v- until it becomes greater than v+. This causes the switch to close, pulling the output to a lower voltage than v-, which in turn causes the capacitor to discharge around the feedback loop until v+ is greater than vonce again. This condition causes the switch to open, and the process repeats, making the circuit oscillate. The shape of the output is trivial—it is either high or low depending on the relative magnitudes of the inverting and non-inverting inputs. V+ and v- increase and decrease over time due to the oscillatory nature of the circuit as evidenced by the step triangle waves. The measured frequency of oscillation is 115 kHz, and our calculated one is 1/(R3*C) = 83 kHz. Figure 9-Multisim Simulation