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
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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