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Lab 7  diode with operational amplifiers by kehali b. haileselassie and kou
Lab 7  diode with operational amplifiers by kehali b. haileselassie and kou
Lab 7  diode with operational amplifiers by kehali b. haileselassie and kou
Lab 7  diode with operational amplifiers by kehali b. haileselassie and kou
Lab 7  diode with operational amplifiers by kehali b. haileselassie and kou
Lab 7  diode with operational amplifiers by kehali b. haileselassie and kou
Lab 7  diode with operational amplifiers by kehali b. haileselassie and kou
Lab 7  diode with operational amplifiers by kehali b. haileselassie and kou
Lab 7  diode with operational amplifiers by kehali b. haileselassie and kou
Lab 7  diode with operational amplifiers by kehali b. haileselassie and kou
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Lab 7 diode with operational amplifiers by kehali b. haileselassie and kou

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Lab 7 diode with operational amplifiers by kehali b. haileselassie and kou

Lab 7 diode with operational amplifiers by kehali b. haileselassie and kou

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  • 1. Diode with operational amplifiers Experiment - #7 Kehali B. Haileselassie and Kou Vue 11/07/2013 ELC ENG 330 – Electronics I Fall 2013
  • 2. Apparatus and Components Agilent multi-meter, DC power Supply, Oscilloscope, Generators, Breadboard, IN914 Diodes, Resistors, Electrolytic Capacitors. Introduction Operational amplifier is one fundamental building block of analog circuits. When usedproperly in negative feedback configurations, the overall closed-loop transfer characteristic can be precisely set by stable passive components such as resistors, capacitors, and diodes, regardless of the potential variation of open-loop parameters. Negative feedback amplifier with op amp operating at its core provides key to highly reliable and stable analog functions. Diodes are semiconductor devices that allow current to flow in only one direction. Diodesare composed of a junction between a P-type semiconductor and an N-type semiconductor.Most diodes are made of silicon. Silicon diodes turn-on strongly when the p-side is madeapproximately 0.7V higher than the n-side. The expression relating current through a diodeto the voltage applied across a diode is given by: (1.1)Id = Is ((exp (λ•vD) – 1) where Id is the diode current, Vd is the voltage from the p-side to the n-side, and Is is thesaturation current which depends on the way the diode is fabricated. Diodes have manyelectronic applications. In this lab we will demonstrate a couple of diodes applications andthe uses of diodes in combination with operational amplifiers.
  • 3. Procedure Part_1: Diode Multipliers 1. Wire up this circuit using the 1N914 diodes, and 47μF electrolytic capacitors. 2. Set up the signal generator to provide a sinusoidal voltage with a peak voltage of 8 volts peak (16 Vpp) and a frequency of 100 Hz. (Use the oscilloscope and you may wish to set the “Out Term” setting on the signal generator to “high–Z”.) 3. Measure the DC voltage across RL using the Agilent multi-meter. 4. Replace RL value with a 100kohms, 10kohms, 1kohms and a 100ohms values, and record the DC output voltage in each case. 6. For the last case in the table (RL = 100 Ω), record Volt as the input frequency for 100Hz, 1kHz, 10kHz and 100kHz. Part_2: Logarithmic Converter 1. Assemble this circuit using the 1N914 diode, wiring it in with the correct polarity! Use a potentiometer (or a variable DC power supply) to generate input voltages which range from 0.25 volts to 5 volts, in increments of 0.25 volts. 2. Generate a plot from the table entries, with input voltage (Vin) on the x-axis and output voltage (Vout) on the y axis. 4. Add another column to the above table. In this column, compute ln(vin). Now, plot the table entries on an x-y graph, with x-axis being ln(vin) and y axis being vout. 5. Drive the diode equation (Id = Is (exp (λ•vD) – 1)) (assuming that the op amp is ideal, Is =10-14 amps, and λ = 40 volts-1.
  • 4. Results and Analysis The DC output voltage after replacing RL with the given values RL (Ohms) Vout (DC, volts) 100k 14.309 10k 12.246 1k 5.76 100 0.831 The DC output voltage as the input frequency increases in the table shown bellow Frequency(Hz) Vout (DC, volts) 100 0.831 1k 2.570 10k 2.806 100K 2.826
  • 5. Input Voltage(volts DC) Input Current Output voltage lnVin (computed) 0.25 0.000025 -0.436 -1.386 0.50 0.00005 -0.468 -0.693 0.75 0.000075 -0.488 -0.287 1.00 0.0001 -0.501 0 1.25 0.000125 -0.512 0.223 1.50 0.00015 -0.519 0.405 1.75 0.000175 -0.527 0.559 2.00 0.0002 -0.533 0.693 2.25 0.000225 -0.537 0.810 2.50 0.00025 -0.542 0.916 2.75 0.000275 -0.547 1.0116 3.00 0.0003 -0.551 1.098 3.25 0.000325 -0.554 1.178 3.50 0.00035 -0.558 1.252 3.75 0.000375 -0.562 1.321 4.00 0.0004 -0.566 1.386 4.25 0.000425 -0.567 1.446 4.50 0.00045 -0.570 1.504 4.75 0.000475 -0.574 1.558 5.00 0.0005 -0.576 1.609
  • 6. "Input voltage (Vin) VS output voltage (Vout)” 1.2 1 0.8 0.6 0.4 Figure_1 input voltage(Vin) Versus output voltage(Vout) 0.2 0 -0.2 -0.4 -0.6 -0.8 Figure_1 Natural Logarithm input voltage (lnVin) versus output voltage(Vout) Natural Logarithm input voltage (lnVin) versus output voltage(Vout) 0 -2 -1 -0.1 0 1 2 -0.2 -0.3 -0.4 Natural Logarithm input voltage (lnVin) versus output voltage(Vout) -0.5 -0.6 -0.7 Figure_2
  • 7. Stimulation Results Part_A: Diode Multiplier
  • 8. Part_B: Logarithmic Converter
  • 9. Response to the questions that stated in the lab manual 1) Measure the DC voltage across RL using the Agilent multi-meter. Is the DC output voltage higher, lower, or the same as the peak voltage of the input sine wave? The DC output voltage that we measured across the RL using Agilent multi-meter is 14.309v which is higher than the sinusoidal peak voltage of 8v. 2) Does the output voltage remain the same as the load resistance decreases? Can you explain why? After we replace the RLwith the given 4 values, the output voltage was not remaining the same as the load resistance decreases. That’s due to the reduction of the values of the resistors and also the fact thatthe waveform from the oscilloscope varies in each case. 3) Generate a plot from the table entries, with input voltage (vin) on the x-axis and output voltage (vout) on the y axis. Is voutvsvin a linear function? Yes, it is a linear function. 4) Add another column to the above table. In this column, compute ln(vin). Now, plot the table entries on an x-y graph, with x-axis being ln(vin) and y axis being vout. Is voutvsln(vin) a linear function? Yes, it is a linear function. 5) Using the diode equation (iD = Is (exp (λ•vD) – 1)) and assuming that the op amp is ideal, can you derive an expression for vout as a function of vin?
  • 10. Yes Vin= Iin* Rin Since id = iin Vin = 10k* Is (exp (λ•vD) – 1)) Conclussion Diodes are very useful in building circuits for various instrumentation applications and also for a power supply applications. In this lab, we explored and analyzed the applications of diodes and their utilities in combination with operational amplifiers.Logarithmic converters and Diode multipliers are used in many electrical and electronic circuit applications.it is necessary to have a very high DC voltage generated from a relatively low AC voltage. In conclusion, by using diode multipliers and logarithmic converter, we can simply build circuits for various applications.Thesetype of rectifiers (diode multipliers and logarithmic converter) produce a high DC voltage multiplier and converter inorder to make easier the electric and electronic circuit applications.

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