Lab 5 - EE 420L 

Authored by Shada Sharif,

sharifs@unlv.nevada.edu  

6 March 2015

Pre-lab work:

• Watch the video by Dr. Baker.
• Read lab 5 details before coming to lab. 

• This lab is about learning how the integrator op-amp work. Knowing how to design an integrator with a specified frequency and input voltage, as well as being able to pick the right capacitor and resistor values appropriately. 

• Hand calculation for the integrator values needed.
• Simulation for the circuits built. 
• Experimentally verifying all the hand calculation and simulate it.

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Experiment #1

• Using the LM324.pdf op-amp again for this lab with a V+ = 5V, V- = 0V. The frequency response of the circuit was derived in the hand calculation's pictures attached below, and then we had to construct the actual circuit and test it to see if the results match the hand calculation. 
• To simplify the calculation the big 100k ohms resistor can be ignored in calculation so that only the capacitor can be in the feedback of the op-amp, but during lab the resistor had to be included so that it discharges the C1 capacitor and the offset doest cause any problems in the output. 
• The circuit would not work properly without the resistor in the feedback in parallel with the capacitor because in DC the capacitor acts as an open and if there is no resistance in the feedback then the op-amp will have infinite gain. An ideal op-amp should have an output of zero in DC without any offset. But real op-amps have an offset so the output without a resistor in DC doest have a feedback loop will go to infinity. This will cause the output to go to VCC+/- depending on the input and we will not get a triangle wave. So ideally we can ignore it, but in reality it has to be included.
• In the frequency response the 100k ohms resistor does not affect much as shown in the calculation below it will only cause high magnitude frequencies rather than smaller ones.

 

• So using the equations below and to verify experimentally we picked to have an input signal at 1k ohms with the same value for resistors and capacitors so plugging that into the magnitude equation the gain should be 0.159 so Vout = Vin * 0.159. 

• From scope wave above input is 680mV and output measured in the scope is a lot larger than what it actually is. This is because the pk-pk measured the small impulses seen in the signal. So it is hard to see the the output is 0.159 of the input.
• Shown below is the input and output signal at the unity frequency ~159 Hz, and the phase is 95º which is really close to the calculated 90º phase. So indeed it is as what we expected from the calculations.

• In this experiment we had to design our own integrator that has an input of a square wave and the output is a triangle wave. For the design we had to use a frequency of 10kHz, and the output should be from 1V to 4V centered around 2.5V. To have the output centered around 2.5V we biased the circuit so that the DC offset is 2.5V. So the same circuit from experiment #1 was used but the values for the components were changed appropriately.
• Calculation is shown below of how the values for the resistor and capacitor were chosen. 

• Notice the phase is still ~90º in the wave form shown from the scope. And the change in the output signal is 3V just as calculation.

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