Lab 3 - ECE 420L 

Authored by Silvestre Solano,

Email: Solanos3@unlv.nevada.edu

2-20-2015

 

 

For the following questions and experiments assume VCC+ = +5V and VCC- = 0V. The LM324 Op-amp was used for the majority of this lab.

Knowing the non-inverting input, Vp, is at the same potential as the inverting input, Vm, (called the common-mode voltage, VCM) what are the maximum and minimum allowable common-mode voltages?

In an ideal op-amp, the output of the op amp is Vout = Aol(Vp-Vm), were Aol = infinity. In this case, the inputs can be any value as long as they are zero. However, a real op-amp has a finite power supply that limits its output, which means that the inputs can only be as high as the supply voltages. For this op amp, the supply voltages range from +16 to -16 volts, which means that the maximum input voltage is 16-(-16) = 32 Volts and the minimum is about 16-16 = 0 volts.
 
Support your answer with an entry from the electrical characteristics table in the datasheet.
 

 
As, shown in the above picture, the maximum input voltage is +32 volts and the minimum is -0.3 volts. This is very close to what I estimated previously.
 
What is a good estimate for the op-amp's open-loop gain?
Support your answer with a plot from the datasheet and an entry from the electrical characteristics table.
 
 I estimate that the open loop gain is about 100000. In the first page of the datasheet, it state that the gain is about 100dB
 


 
What is a good estimate for the offset voltage?
For worst case design what value would you use?
 
I would estimate a 100 mV offset, even though the datasheet says it has a max offset of 5mV. For the worst case scenario, I would use my value of 100 mV.
 
Build, and test, the following circuit. Note that a precise value for the 5k resistors isn't important. You can use 4.7k or a 5.1k resistors.
 

 
 
What is the common-mode voltage, VCM? Does VCM change? Why or why not?
 
The VCM is determined as follows:
 


The measured Vm, Vp and Vout are shown below.
 
Vm

 
Vp

 
Vout

 
The VCM in this circuit does not change because there is no current that flows through the Vp terminal, at least in the ideal model. However, there is apparantely some kind of bias input current of approximately 20 nA, but the current is very small and negligable. If the input changes,  Vm is affected as a result, but the Vp (VCM) should not change at all.
 
What is the ideal closed-loop gain?
 
From the experiment, the Vout appears to be about 2.5 volts, therefore the ideal closed-loop gain is just Vin/Vout = 2.5/2.5 = 1
 
What is the output swing and what is it centered around?
 
The output swing is limited by the power supply given to the op-amp chip. For this circuit, VCC+ = +5V and VCC- = 0V. This means that the op-amp cannot have a Vout exceeding 5 volts or going below 0 volts. The output swing is centered  around the midpoint between 0 and 5 volts, which is just 2.5
 
What happens if the input isn't centered around around VCM, that is, 2.5 V?
Provide a detailed discussion illustrating that you understand what is going on.

Using KCL, the Vout is derived as follows with Vm = VCM = 2.5 Volts
 

 

 

 
If the input is changed from 2.5 volts, then the Vout will be changed. For example, if Vin is changed to 3 volts, then the Vout should change to about 2 Volts. This was done in lab and the Vout is shown below. The Vout is not exactly at 2 Volts, but the op-amp is not perfect, so the slight inaccuracy is expected.
 
Vin=3  Volts, Vout = 2 Volts.

 
When the input is changed to 4 Volts, the output should decrease to 1 Volt. In the below picture, this appears to be the case. Like the previous picture, the output voltage vout is not exactly at 1 Volt, but intstead it appears as 808 mV.
 
Vin = 4 Volts, Vout = 1 Volt.

 
The above results seem to indicate that the ideal op-amp model seems fairly reliable in determining the characteristics of a real op-amp. The ideal op-amp model was primarily used to determine the Vp and Vm. Since Vm is held at a approximate 2.5 Volts, Vp must also be at about 2.5 volts. However, when the Vin is changed, the Vm remains constant at about 2.5 Volts. This means that the current through Rf must increase or decrease depending on the Vin. but Vm stays unchanged.
 

What is the maximum allowable input signal amplitude? Why?
 
The maximum allowable input is about 5 Volts. This is because Vout = 5-Vin and if Vin goes above 5 volts, say 5.5 volts, then the output Vout goes down to -0.5 volts. This is imposible since the power supply only allow a maximum output of 5 volts and a minimum output of 0 volts.
 
What is the maximum allowable input signal if the magnitude of the gain is increased to 10? Why?
 
The maximum allowable input remains the same as in the previous question, which is 5 volts. This is because the gain is increased, but the supply voltages remain the same and they limit the maximum output to 5 volts and minimum output to 0 volts. If the supply voltages are increased, then the input could go past 5 volts.
 
What is the point of the 0.01 uF capacitors from VCC and VCM to ground?
Are these values critical or could 0.1 uF, 1,000 pF, 1 uF, etc. capacitors be used?
 
The capacitors used in the circuit are not vital, but instead are used as so-called decoupling capacitors. They are there to hold the voltage steady by charging and staying at a given voltage. C1 charges to 2.5 volts and remains at that voltage. C2 charges and remains at 5 volts.
Their values choosen are not critical at could be any value listed in the question. I myself used 0.1 uF capacitors instead of the apparently very popular 0.01 uF.
 
The data sheet shows that this op-amp has an input bias current that flows out of the op-amp's inputs of typically 20 nA.
This current flows out of both the non-inverting and inverting inputs through the resistors connected to these inputs.
Show how the operation of the circuit can be effected if, for example, R1 and R2, are much larger. Explain what is going on.
 
If R1 and R2 are much larger, then they will reduce the current created by VCC and the small bias current will dominate, creating a small voltage with the large R2 resistor. For my experiment, the R2 and R1 resistors were at 2 Mega Ohms. The picture below shows this small voltage captured.
 

 
What is the input offset current? What does this term describe?
 
The input offset current is analogous to the offset voltage. It essentially describes imperfections in the current produced by the op-amp. This imperfection is modeled as an extra unwanted current.
 
Explain how the following circuit can be used to measure the op-amp's offset voltage.
Note that if the output voltage is precisely the same as VCM then the op-amp has no offset voltage (this is very possible).
To measure small offset voltages increase the gain by increasing RF to 100k or larger. Explain what is going on.
 

 
The circuit shown above can be used to measure offset voltage by realizing that ideally no current flows through Rf because (VCM-Vm)/1K = (VCM-VCM)/1K = 0 amps. With this information, the following relationship is derived assuming Vm = VCM+Voff, where Voff is the offset voltage, which could be either positive or negative.
 

 
Using  Vm = VCM+Voff arranging terms leads to the following equation.
 

 
The above equation shows that determining the offset voltage is fairly simple knowing VCM is about 2.5 volts. In order to get more accurate results, Rf must be increased so that it reduces any leakage current or the input bias current to prevent it from significantly affecting the results of the Vout measurement.
 
Measure the offset voltage of 4 different op-amps and compare them.
 
For the following measurements,  Rf was changed to 200K Ohms and Voff was calculated using Voff=Vout-VCM.
 
LM324
VCMVout
Voff calculated = 1.28 volts
 
 
 
LM741
VCMVout
Voff calculated = 0.44 Volts

 
 
TL082
VCMVout
Voff calculated = -1.08 Volts
 
 
 
LF351
VCMVout
Voff calculated  = 1 Volt




 As always, I will back up my stuff as shown below.

 


 



 


 

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