Monday, February 15, 2016

Week 6

Blogsheet week 6
Operational Amplifiers
Explanations of the pin numbers are below:
1: DO NOT USE
8: DO NOT USE
2: Negative input
7: +5 V
3: Positive input
6: output
4: -5 V
5: DO NOT USE

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 -5 V to 5 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?


Figure 1a. Table for Vout vs Vin

Figure 1a'. Graph for Vout vs Vin
          Figure 1a and shows the results we got from the non-inverting input from -5 V to 5 V with 1 V steps with 0 V to the inverting input, and the plot (figure 1a’) is close to the ideal plot. The range is from -5V to 5V since V- for pin 4 is -5V and V+ for pin 7 is +5V. When the non-inverting input from -10V to -0.25V, we got -3.8V; when the non-inverting input is 0, we got 0; when the non-inverting input from 0.25V to 10V, we got 4.56V, and those results are all close to what they suppose to be.


   
b.     Apply 0 V to the non-inverting input. Sweep the inverting input (Vin) from -5 V to 5 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?
   
Figure 1b. Table for Vout vs Vin

Figure 1b'. Graph for Vout vs Vin

     The pictures looks small here...Please click it to see the full size. :)


      Figure 1b and shows the results we got from the inverting input from -5 V to 5 V with 1 V steps with 0 V to the non-inverting input, and the plot (figure 1b’) is close to the ideal plot. The range is also from -5V to 5V since V- for pin 4 is -5V and V+ for pin 7 is +5V. When the inverting input from -10V to -3V, we got 4.53V and when the inverting input from -2V to 10V, we got -3.6V, and those results are all close to what they suppose to be.

2.     Create a non-inverting amplifier. (R2 = 2 kΩ, R1 = 1 kΩ). Sweep Vin from -5 V to 5 V with 1 V steps. Create a table for Vin and Vout. Plot the measured and calculated data together.
Figure 3. Table for Measured Vout vs Vin

Figure 4. Table for calculated Vout vs Vin

Figure 5. Graph for Measured and Calculated Vout vs Vin

Figures above show the calculated and measured values for the non-inverting amplifier. Because R2 = 2 kΩ and R1 = 1 kΩ, so the gain should be 1+2=3. Measured data are close to the calculated data (which they should be), and the with 5V to 10V Vin we got 5.37 which is higher than the maximum range 5V, and that may because the -5V for V- (pin 4) and +5V for V+ (pin 7) are not accurate (higher than 5V).


3.     Create an inverting amplifier. (Rf = 2 kΩ, Rin = 1 kΩ). Sweep Vin from -5 V to 5 V with 1 V steps. Create a table for Vin and Vout. Plot the measured and calculated data together.
Figure 6. Table for measured Vout vs Vin

Figure 7. Table for Measured Vout vs Vin
Figure 8. Graph for Measured and Calculated Vout vs Vin

Figures above show the calculated and measured values for the inverting amplifier. Because Rf = 2 kΩ and Rin = 1 kΩ, so the gain should be -2. Measured data are close to the calculated data (which they should be), and the range this time is also from -5V to 5V.


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?

      In an open loop configuration it does not function as a amplifier, but it compares the V+ and V-. When the V+ is larger than the V- voltage of the opamp the V+ will be the output, and if V- is larger V- will be the output. With an inverting/non-inverting setup we get a small amount of gain because the output is determined by the ratio of the resistors. The larger the ratio the larger the output, the smaller the ratio the smaller the output.





EGR 393 Temperature Controlled LED System

Tips:

1. If something is not working, check your connection first.
2. Check the pins carefully, LM35 is VERY easy to be burned if you connect the wrong pins.
3. Read the datasheet carefully.
4. Before starting to connect the circuit, try to sketch it on a paper first, make sure everything is clear.

Components:

1. TMP36 Temperature Sensor  2. Lm324 Operational Amplifier  3. OMRON G8QN Relay 4. LED

Procedure:


TMP36 Temperature Sensor: Pin layout – look up characteristics to calculate temperature from datasheet (under Bb/Week6).

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).
    The initial voltage is 0.72V, after covering the sensor with palm then the voltage becomes 0.79V, and if we use hair dryer to heat up the sensor, the voltage becomes 1.3V. Since the temperature that the hair dryer can heat up to is about 75 degrees (Celsius), so the voltage is suppose to increase to about 1.2V according to Figure 6 in the data sheet, so the result we got from the experiment fits the expectation. 


Relay (Manual under Bb/Week6)

Pin 1 – Voltage Source (amount of voltage sent to pins 3 or 4)
Pin 2 – Input Voltage
Pin 3 – Output Voltage when Vin>Vs
Pin 4 – Output Voltage when Vin<Vs
Pin 5 - GND
·       

      schematic view is the bottom view!
1.       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). 
   
      When we applied zero volts to the coil of the relay we get OL from pin 3 to 1, and nearly zero ohms from pin 4 to pin 1. We get these number because where is no voltage applied to the coil of the relay it is in its normal state. When the relay is in its normal state pin four is normally closed contact meaning that it has zero ohms of resistance to pin 1. In the normal state of the relay pin 3 is normally open, which mean it is an open circuit. 
2.       Now sweep your DC power supply from 0V to 8V and back to 0V. What do you observe at the multimeter (resistance measurements similar to #1)? Did you hear a clicking sound? How many times? What is the “threshold voltage values” that cause the “switching?” (EXPLAIN with a VIDEO).

Video 1. Relay with 0V-8V power supply
       The threshold voltage is about 5V, and we heard clicking sounds twice. Because it only clicked when we cross the threshold voltage, so we heard the clicking sound once when we got lost than it, and once again when we decreases past.
3.       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. Measuring V3 and V4

      How relay works depends on the voltage we applied to between two pins, like the video shows.

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)
Figure 9: Schematic of LED on/off by switching of a relay



Video 3. Turn LED on/off by switching the 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)
        
       As you can see from the table below as R2 get larger than R1 there is more of an increase in the output voltage.
Table 1. Three groups with different R1 and R2

2.       Use your temperature sensor as your input. Do you think you can generate enough voltage to trigger the relay? (EXPLAIN)
       
      No with the temperature sensor alone we will not be able to generate enough voltage to engage the relay. The largest amount of voltage we could get out of the temperature sensor was around 1.3 volts, we need about 5.5 volts to engage the relay. We will have to use and operational amplifier to increase the voltage we get from the temperature sensor to engage the relay.

3.       Design a system where LED light turns on when you heat up the temperature sensor. (CIRCUIT schematic and explanation in a VIDEO)
       
      We used a renaissance value of 1200 ohms for R2, and 300 ohms for R1. This gives us a gain of 5, this will allows us to engage the relay using the temperature sensor.

 
Video 4: Explanation of LED controlled by temperature sensor and relay.





4.       BONUS! Show the operation of the entire circuit. (VIDEO)
Video 5. Temperature sensors controls LED 


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8 comments:

  1. everything looks good but some of the tables are going off the screen, it could be my computer but I am not sure!

    ReplyDelete
    Replies
    1. Yes some tables go off the template because we choose to show the original sizes and they looks too big... but we have to do that because even we choose X-large sizes that they are still not clear enough.

      Delete
  2. everything looks good but some of the tables are going off the screen, it could be my computer but I am not sure!

    ReplyDelete
  3. For the question about the temperature sensor where you have to place your finger on it, I like how you compared the output voltage that you received to the expected values.

    ReplyDelete
  4. #4 is not correct. We will talk about it though.
    Video 4: It is not current but voltage that increases at the temperature sensor.
    Great bonus video.
    I like the fact that you acknowledged Stockerfinley but did not agree.

    ReplyDelete