OverviewBackground1. Summing Amplifier2. Differentiator3. Integrator4. Logarithmic AmplifierPhysics 375, Laboratory 13Analog MathOverviewThe purpose of these experiments is to use op-amps to perform mathematical operations on inputvoltages. Operations include sum, difference, logarithm, differential and integral. BackgroundAn operational amplifier (op-amp) is a differential amplifier with very high gain, very high inputimpdance, and very low output impedance. The 741 op-amp is an integrated circuit thaat comesin an 8-pin dual in-line package (DIP). The connections for the chip looking down with the notchfacing up is:For large gain, low input current amplifiers used with feedback to the negative input there are tworules to follow:1. I+=I-=0. The input currents are 0.2. v+-v-=0. The input voltage difference is 0.For feedback to the negative input, the general steps for analysis are to find the voltage at the non-inverting input and use rule two to assign that same voltage to the inverting input. The next stepis to find the current flowing at the inverting input from any voltage source. Based on rule one alltis current is assumed to flow into the feedback network, and generate a voltage drop at the outputvoltage.v+voutV+v−V−12345678V−v+v−V+vout+−−+Bal.Bal.Vout−+feedbackVin-Vin+non-inv.invertingPHYS 395 Lab 13, p. 21. Summing AmplifierConnect a 741 op-amp on the Powerace breadboard, and use the +15 V and -15 V supplies topower the chip. Use the variable supply to provide +1 V. Figure 1: Summing AmplifierConnect wires from the +1 V supply to each possible combination of resistors on the invertinginput and measure Voutwith the DMM. How close is the output voltage to the binary value of theinput switches:What is causing the lack of precision?Use a wire to connect the +1 V to Vnon the non-inverting input for no inverting input voltage.What is Vout? Check a couple of inverting input settings to verify the following:Vout+1 VV3−+8.2 kΩ8.2 kΩV2V1V03.9 kΩ2.0 kΩ1.0 kΩ8.2 kΩ 8.2 kΩVnV–outV020V121V222V323+++=2V–outV020V121V222V323Vn20–+++=PHYS 395 Lab 13, p. 32. DifferentiatorBuild the circuit in figure 2 and use sine waves of 0.5 V amplitude for vin.Figure 2: DifferentiatorMeasure vinand voutwith oscilloscope and calculate the gain as a function of frequency, andgraph the gain on a Bode plot (gain in dB vs logf). An ideal differentiator has the relationship:How well does this circuit agree with the ideal? The feedback capacitor should be neglible. Try atriangle wave, and a square wave input and observe the ouput waveforms. Is the slew rate of theop-amp a factor?3.IntegratorBuild the circuit in figure 3 and use sine waves of 0.5 V amplitude for vin.Figure 3: IntegratorMeasure vinand voutwith oscilloscope and calculate the gain as a function of frequency, andgraph the gain on a Bode plot (gain in dB vs log f). An ideal integrator has the relationship:How well does this circuit agree with the ideal? Try a square wave input and observe the ouputwaveforms. Is the slew rate of the op-amp a factor for the integrator?vout−+C =0.1µFRf=10kΩvinCf= 120 pFvoutRfC–vindtd----------=vout−+C =0.1µFR =10kΩvinvout1RC--------– vintd∫=PHYS 395 Lab 13, p. 44. Logarithmic AmplifierUse a 2N2222 transistor and 1N914 diode in parallel in the feedback network of an invertingamplifier.Figure 4: Logarithmic AmplifierThe diode and capacitor are present to protect the circuit, and RBwill conpensate for the bias cur-rent. Use the variable power supply to provide Vin. The current Iinis given byThis current must flow into the collector of the transistor which has the relationshipThe base-emitter voltage is equal to the negative of Voutand the collector current equals the inputcurrent so,Measure VoutforanumberofvaluesofVinand compare to an ideal logarithmic amplifier. If VT=0.026 V, derive a value for I0from the data.Vout−+C =0.1µFR = 100 kΩVinRB= 100 kΩ2N22221N914IinVinR--------=ICI0eVBEVT⁄=VinR--------