Blogsheet week 6
Operational Amplifiers
Explanations of the pin numbers are below:
1: DO NOT USE
|
8: DO NOT USE
|
2: Negative input
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7: +10V
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3: Positive input
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6: output
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4: -10 V
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5: DO NOT USE
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1. You will use the OPAMP in “open-loop” configuration in
this part, where input signals will be applied directly to the pins 2 and 3.
a.
Apply 0 V to the inverting input. Sweep the
non-inverting input (Vin) from -10 V to 10 V with 1 V steps. Take
more steps around 0 V (both positive and negative). Create a table for Vin
and Vout. Plot the data (Vout vs Vin). Discuss
your results. What would be the ideal plot?
Table 1 Graph 1: Inverting graph in open loop
Table 1 Graph 1: Inverting graph in open loop
Vin (V)
|
Vout (V)
|
-5
|
4.68
|
-4
|
4.68
|
-3
|
4.68
|
-2
|
4.68
|
-1
|
4.68
|
-0.5
|
4.68
|
0
|
0
|
0.5
|
-3.77
|
1
|
-3.77
|
2
|
-3.77
|
3
|
-3.77
|
4
|
-3.77
|
5
|
-3.77
|
In our graph, the slope of the plot should be close to 0. As shown above, the slope of our graph should've been closer to 0, but the outcome of what we have shown in the graph may be from the lack of more data between 0 and +/- 1 V. However, Vout flattens out at 4.68 V at -5V and -3.77 V at 5 V. This occurs because the output can only go so high/low, thus flattening out at the maximum/minimum values. This may also be because of the type of equipment we use. Since it is an inverting input, our output should be the opposite of our input, which our chart clearly shows.
b.
Apply 0 V to the non-inverting input. Sweep the
inverting input (Vin) from -10 V to 10 V with 1 V steps. Take more
steps around 0 V (both positive and negative). Create a table for Vin
and Vout. Plot the data (Vout vs Vin). Discuss
your results. What would be the ideal plot?
Table 2: Graph 2: Non-inverting graph in open loop
Vin (V)
|
Vout (V)
|
-5
|
-3.77
|
-4
|
-3.77
|
-3
|
-3.77
|
-2
|
-3.77
|
-1
|
-3.77
|
-0.5
|
-3.77
|
0
|
0
|
0.5
|
4.68
|
1
|
4.68
|
2
|
4.68
|
3
|
4.68
|
4
|
4.68
|
5
|
4.68
|
Our plot should be similar to the plot above, however the output will be positive and negative respective to the input because the input is non-inverting. The slope should've been closer to 0 also. As shown in our table, when Vin is negative, Vout will also be negative, and same for when Vin is positive.
2. Create
a non-inverting amplifier. (R2 = 2 kΩ, R1 = 1 kΩ). Sweep
Vin from -10 V to 10 V with 1 V steps. Create a table for Vin
and Vout. Plot the measured and calculated data together.
Table 3: Graph 3:Non-inverting with resistor values
Table 3: Graph 3:Non-inverting with resistor values
Vin (V)
|
Vout (V)
|
-5
|
-3.77
|
-4
|
-3.77
|
-3
|
-3.77
|
-2
|
-3.77
|
-1
|
-3.77
|
-0.5
|
-3.77
|
0
|
0
|
0.5
|
4.68
|
1
|
4.68
|
2
|
4.68
|
3
|
4.68
|
4
|
4.68
|
5
|
4.68
|
In our table, Vout are positive and negative respectively since it's non-inverting. When Vin is negative, Vout will also be negative; when Vin is positive, Vout is positive.
3. Create
an inverting amplifier. (Rf = 2 kΩ, Rin = 1 kΩ). Sweep Vin
from -10 V to 10 V with 1 V steps. Create a table for Vin and Vout.
Plot the measured and calculated data together.
Table 4: Graph 4: Inverting with resistor values
Vin (V)
|
Vout (V)
|
-5
|
4.13
|
-4
|
4.13
|
-3
|
4.13
|
-2
|
4.13
|
-1
|
2.91
|
-0.5
|
-1.5
|
0
|
0.8 mV
|
0.5
|
-2.91
|
1
|
-3.5
|
2
|
-3.5
|
3
|
-3.5
|
4
|
-3.5
|
5
|
-3.5
|
When Vin is negative, Vout is positive; when Vin is positive, Vout is negative. This is because it's inverting.
4. Explain
how an OPAMP works. How come is the gain of the OPAMP in the open loop
configuration too high but inverting/non-inverting amplifier configurations
provide such a small gain?
5.
An OPAMP works by amplifying the input voltage, but it can’t
exceed the +/- 5 voltage source. It levels out at a certain point. The
resistors affects the gain in the inverting/non-inverting amplifier while in the open loop, it doesn’t affect the gain. Since open loop is infinite, the gain will be much higher.
5.
Temperature Sensor: Put TMP36
temp sensor on breadboard.
·
Connect the +VS to 5 volts and GND to
ground.
·
Using a voltage meter, measure the output
voltage from the VOUT. Now put your finger (or cover the sensor with
your palm) on the TMP36 temperature sensor for a while, observing how the
output voltage changes. Check Fig. 6 in the data sheet (EXPLAIN).
6. Connect your DC power supply to pin 2 and ground pin 5. Set your power supply to 0V. Switch your multimeter to measure the resistance mode; use your multimeter to measure the resistance between pin 4 and pin 1. Do the same measurement between pin 3 and pin 1. Explain your findings (EXPLAIN).
- The resistance between pin 4 and pin 1 is 0.35 Ω, so basically 0 Ω because voltage is set at 0 V and pin1 and 4 are normally closed.
- The resistance between pin 3 and pin 1 is infinite, because pin 1 and 3 are normally open.
Video 1: Sweeping DC power supply to cause clicking sound
8. How does the relay work? Apply a separate DC voltage of 5 V to pin 1. Check the voltage value of pin 3 and pin 4 (each with respect to ground) while switching the relay (EXPLAIN with a VIDEO).
Video 2: Explanation of relay with basic drawings
LED + Relay
1.
Connect positive end of the LED diode to the pin
3 of the relay and negative end to a 100 ohm resistor. Ground the other end of
the resistor. Negative end of the diode will be the shorter wire.
2.
Apply 3 V to pin 1
3.
Turn LED on/off by switching the relay. Explain
your results in the video. Draw the
circuit schematic (VIDEO)
Video 3: Explanation of turning the LED on/off with relay
Operational
Amplifier (data sheet under Bb/week 6)
1.
Connect the power supplies to the op-amp (+10V
and -10V). Show the operation of LM 124 operational amplifier in DC mode with a
non-inverting amplifier configuration. Choose any opamp in the IC. Method: Use
several R1 and R2 configurations and change your input voltage and record your
output voltage. (EXPLAIN with a TABLE)
R1: 1.2 kΩ R2: 2 kΩ
Vin
|
0 V
|
2 V
|
5 V
|
7 V
|
8 V
|
10V
|
Vout
|
79 mV
|
5.7 V
|
9.1 V
|
9.1 V
|
9.1 V
|
9.1 V
|
R1: 47 Ω R2: 357 Ω
Vin
|
0 V
|
2 V
|
5 V
|
7 V
|
8 V
|
10 V
|
Vout
|
0 V
|
8.5 V
|
8.5 V
|
8.5 V
|
8.5 V
|
8.5 V
|
When using the same Vin values, our Vout values are different when using different resistor values.
1.
Use your temperature sensor as your input. Do
you think you can generate enough voltage to trigger the relay? (EXPLAIN)
Yes, you can
generate enough voltage to trigger the relay. The heat applied to the
temperature sensor increases the output voltage, causing the relay to trigger. Without heat, the voltage at first is low, starting around 1.6 V, but when we applied heat from the hair dryer, the voltage began to increase drastically.
2.
Design a system where LED light turns on when
you heat up the temperature sensor. (CIRCUIT schematic
and explanation in a VIDEO)
Video 4: Schematic of the system and explanation
1.
BONUS! Show the operation of the entire
circuit. (VIDEO)
Video 5: Run through of our circuit
I like the tables, as they are easy to read. Adding some explanations for tables shows that you guys understand the operations of the relay or whatever else you are working with.
ReplyDeleteThanks for the input
ReplyDeleteYour graphs are really easy to read, and I like how you get the light to turn on with such little heating, it felt like ours took forever.
ReplyDelete#2, 3: Plot should have had the theoretical data as well. (-2)
ReplyDelete#4 is not correct. We will talk about it though.
Bonus video! (+2)
Loved the little adjustment on the position of the hair dryer! ;)