Lab 3 – OP Amps I

EE 420L Analog IC Design

Lab Date: 2/06/19 Due: 2/20/19

By David Santiago – Email: santid4@unlv.nevada.edu

Last Edited on 2/19/19 at 9:45pm using Word

  

Lab description:

In this lab, we will be going over how Operational Amplifiers (Op Amps) work to make small signals become amplified.

 

Suppose we have a small signal such as a pulse or a sound and we would want to see it happen at a large scale. The Op Amp is here to help us see what the small signal looks like. The Op amp has a few important terminals; A VCC+ and VCC- to make sure that the Op amp is “powered,” two inputs that are connected to a very high impedance, and an output, which is equaled to the difference of the inputs multiplied by a very LARGE Open-Loop Gain (or AoL).

 

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Datasheet and Review:

-Given that VCC+ = +5 Volts and VCC- = 0 Volts; T

The Maximum allowable VCM Voltage is: VCC – 1.5V given that the Op amp is operating at T = 25˚C.

~So for our circuit, our max VCM will be 3.5V. We will choose to operate at VCM = 2.5V, so that we have an equal amplitude distance from VCC+ and VCC-

The Minimum allowable V­­CM Voltage is: 0V given that the Op amp is operating at T = 25˚C.

     ~So for our circuit, We will not be utilizing 0V because we will be using only ONE power supply.

The only reason why we could use VCM = 0V is if we use 2 Power Supplies and use VCC+ = 15V and VCC- = -15V so that we can have a LARGER Voltage swing but the drawback is that we will be using a dual power supply setup.

 

-A good estimate for a good Op Amp is usually AoL = 1MEG (106) V/V gain. For this Op Amp, the Open-Loop Gain is given by AVD (A.K.A. AoL) = 100 V/mV (or 105 V/V) gain.

     ~We want a higher Open-Loop Gain because we want to assume that theoretically Vp=Vm or:

              Vout

              -----     = AoL                And with the assumption, we will have a high Open-Loop Gain.

          (Vp – Vm)

 

          Here is a snip from the datasheet

 

-A good estimate for the offset voltage is usually 10mV offset, so that depending on our actual Op-Amp gain (Let’s assume Vout/Vin Gain = 100), we will have a Vout offset of around ±1V.

-In the worse case scenario (Lets say Voffset = 100mV), we would use a Vout/Vin gain of 5 or less, so that our Vout offset will be around ±500mV. It is a bad output, but if we used a larger gain, we will be amplifying the offset by a lot more. We generally want to use lower Gains to lower the effects of the offset voltage.

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Experiment 1: (Building an inverting Op-Amp Circuit)

 

Breadboard:

NOTE :All resistors are relatively near their respective values and precision is not important. We will be observing the basics of the Op-Amp and how to set it up under different scenarios.

 

Output:

Vin = 148mV, Vout = -121mV, Gain = Vout/Vin = -0.818 V/V

 

LTSpice Output:

Vin = 100mV, Vout = 100mV, Gain = -1 V/V

 

-In this Circuit, VCM = 2.5V, VCM does not change since it is a DC voltage. The decoupling capacitors are there to make sure we do not have any small variations of Voltage throughout our power circuit (The circuit providing the VCM).

The ideal closed-loop gain should theoretically be our Gain = -Rf/RI, so whatever resistor we place into our circuit would be theoretically seen by our gain factor. So if we use RF=RI=5k, then we theoretically should be getting Gain = -1 V/V

-The output swing for this circuit is about -121mV swing, and it swings around 2.5 Volts.

If we are not centered around 2.5 Volts, we will get clipping if our output swing goes above VCC+ = +5V or if our output swing goes below VCC- = 0V.

                   ~IF we increased the amplitude from 100mV to about 2.5-2.6Vpp, we will theoretically have a

                   2.5-2.6Vpp output swing. Below is what the Op-Amp outputs with these kinds of conditions:

                   From the picture, we can see that we are still at a Vout/Vin Gain of -1 V/V.

                   HOWEVER… If we increased the input voltage to something higher such as 3Vpp, we will get               clipping, as seen below:

This clipping occurs because we are going “Outside” the acceptable barriers of the Op-Amp. The VCC+ and VCC- of the Op-Amp act like a barrier or some kind of Max/Min output that can occur from the Output Swing.

Once the Output Swing is Greater than or equal to 2.5V(VCM) +2.5 Output Voltage Swing, our voltage then “clips” to the maximum value of 5 volts VCC+. This also can occur with the bottom, as if the Output Swing is Less than or equal to 2.5V(VCM) – 2.5 Output Swing, the bottom clips to the minimum value of 0 volts VCC-.

Below is another example of if our VCM is lower than 2.5V and is clipping downward:

-Just as stated above, the Maximum Allowable Input Signal Amplitude is 2.5Vpp because if we

It is 2.5V since if we go above 5V (Our VCM plus 2.5V Swing), we start to clip to both the top and the bottom of the VCC barriers.

-Since we are multiplying our input by a Gain of 10, we should choose a gain that will not go pass the VCC barriers of 0V and +5V. For this circuit, the Maximum Allowable input signal is 0.25Vpp, so that our output swing will be theoretically 2.5Vpp. Passing the VCC barrier would then result in the clipping scenario as seen above.

Here below is our Gain of 10 V/V:

Vin = ~100mV, Vout = 1.06V

Here is a pic of our output swing being clipped if we increase the input voltage:

Vin = 332mV, Vout = ~2.56V

With the new higher gain, small input voltages will impact our output a lot more.

-The 0.01 uF capacitors in our power circuit are called Decoupling capacitors and are used to “clean up” the DC voltage from any small frequencies and to keep a constant clean DC current. The capacitor basically shunts any nasty noise that we don’t want in our Op-amp Circuit and will keep our circuit nice and happy.

A Great analogy for the Decoupling capacitor is to imagine the capacitor as a “Storage Containter” of water. If we have a river of water powering a Windmill, we want constant flow. Knowing that rivers are wild and random, there won’t be constant current throughout the river. To solve this, a “Container” is placed very close the Windmill so that we can always have a constant flow of water.

It is important to have decoupling capacitors so that noise is removed from any outside interference. It is also good practice to have them in your PCB designs. Here’s a pic for you :-)

If we use bigger capacitors, going back to our water container analogy, it would help out with some of the “bigger waves of water” or a high variation in frequency, however, at the cost of having more expensive pieces for your PCB. A small capacitor is used to clean up those very fast frequencies and ripples that are in the current.

-During all of our procedures, we have neglected a small input offset current that is coming out from the Op-Amp’s terminals. Here is what we are actually testing:

However, the current does little to no effect to our circuit, since the 20nA current through R2 of 10k will give us an added voltage drop of .2mV, which will not affect VCM at all (So VCM = 2.502V, which just rounds to 2.5V).

Experiment:

LTSPICE:

^Circuit doesn’t really change at all:

 

Suppose R1 = R2 = 10MEG, as seen below:

Now this is where things go crazy. Obviously by the Voltage divider, VCM = 2.5V,

HOWEVER… The 20nA offset current going through R2 will make an effect on our circuit by adding some unwanted voltage to our VCM circuit.

The new voltage drop will be V = 20nA(10MEG) = 0.2V, so our new VCM = 2.5+0.2 = 2.7V!

Now the voltage at the non-inverting and inverting terminals will be at a higher voltage (So higher VCM means Output Swing Offset will go up too), meaning that now the current through the Op-Amp circuit will and affect Vout, in which Vout’s DC Offset will be Greater Than Vin’s DC Offset, assuming that the Gain = 1 V/V.

Experiment:

NOTE: As you can see, Vout (Blue) is shifted up by about 0.2V, and the extra voltage is because of the small extra current that is running out from the Op-Amp into VCM.

LTSPICE:

LTSpice also proves this too.

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Experiment 2: (Measuring the Offset Voltage)

To measure the Offset Voltage, we will need to disconnect our Input sine voltage and replace it with our VCM voltage. Theoretically we can imagine VCM as our “Reference Ground” and assume that there is little to no current running through our circuit, so the only Theoretical voltage we should be seeing is the Offset voltage that is coming from the Non-Inverting Terminal. It works the same as any other circuit, calculate Vout and dividing it by your Gain, however, Vout in this case will be our Offset voltage that we want to find. To find it, we will be using a high gain of about 100k, and then dividing our DC voltage difference by a gain of 100 V/V.

Breadboard:

First Op-Amp:

 

 

VDifference = 20mV

 

After changing the RF = 100k, we got to see the differences between the input VCM and Vout, and the multiplier for this circuit is 100 V/V.

 

To Calculate this Offset, we get our VDifference and divide it by our Gain, which is 100 V/V, so for this first Op-Amp, our offset will be 0.2mV

 

Second Op-Amp:

VDifference = 60mV

Our offset will be .6mV

 

Third Op-Amp:

VDifference = 160mV

Offset = 1.6mV

 

Fourth Op-Amp:

VDifference = 120mV

Offset = 1.2mV

 

Chips tested: 4 Different LM324 Op-Amp Chips

 

Looking at all of the Op-Amp’s :

 

1st

2nd

3rd

4th

.2mV

.6mV

1.6mV

1.2mV

 

We can see that they are all not the same, even when they are from the same family of Op-Amps. The reason for this is that Manufacturing is not always perfect, and also the month and year each individual chip was made. For 2 chips to be identical, you’d have to make them both at the same time.

 

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