Lab3

Lab 3 - EE 420L

Dwayne K. Thomas

kendaleman@gmail.com

2/19/2015

Opamps I, basic topologies, finite gain, and offset

Experiment 1

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?

Common Mode Voltage is the sum of Vp and Vm divided by 2.

Therefore the Maximum Common Mode Voltage would ideally be (5 + 5) ÷ 2 = 5V

And our minimum Common Mode Voltage would be (0 + 0) ÷ 2 = 0V

But according to the datasheet for the Opamp, the maximum common mode voltage would be, VCC - 1.5V at ambient temperature.

So for our +VCC and –VCC chosen, the maximum would be 5V - 1.5 = 3.5V

What is a good estimate for the op-amp's open-loop gain?

According to the datasheet for the Opamp, the large signal Voltage Gain is 100V/mV which is 100,000.

What is a good estimate for the offset voltage?

For worst case design what value would you use?

The datasheet shows offset voltage varying with temperature, but normally sitting at 2mV. For worst case design, I would use the max offset voltage of 9mv.

We built the circuit below for Experiment 1

What is the common-mode voltage, VCM? Does VCM change? Why or why not?

Our common-mode voltage is 2.5V

It does not change for our circuit because it is the DC value we placed at VCM using a voltage divider.

In addition, the goal of the Opamp is to maintain the Vm and Vp voltages at the same value.

What is the ideal closed-loop gain?

The closed loop gain is –R₂/R₁ → = -1

What is the output swing and what is it centered around?

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.

The output swing is 200mv centered around 2.5V

If the input isn’t centered around VCM, the output swing would still be 200mv, it would instead swing around a different DC value inverted from the input. This is because the direction of current will always be from the positive input going toward the negative output in this circuit. Since both our feedback and input resistors are the same value, the voltage drop across each resistor will be the same. Therefore as the DC voltage at the input is increased positively, the DC voltage at the output increases negatively, or in other words goes down. But the AC voltage will remain the same amplitude.

If the offset voltage is increased or decreased too much, we will see clipping on our output. In the case below, our offset voltage was decreased which allowed our voltage swing on the negative cycle to reach our -VCC voltage of 0V. This causes clipping in the positive cycle of our output due to this limitation.

http://cmosedu.com/jbaker/courses/ee420L/s15/students/thomad1/Lab3/clipping_neg_input.png

What is the maximum allowable input signal amplitude? Why?

The Maximum allowable input signal amplitude is 1.25V, because if we increase the amplitude higher than this, our output will begin to clip in its negative cycle which coincides with positive cycle of our input. This is due to our limited Maximum Common mode voltage of 3.5V.

If we were just to calculate this, we would see the clipping at 1V, but this tells us that our actual max common mode voltage is measured at 3.75V

http://cmosedu.com/jbaker/courses/ee420L/s15/students/thomad1/Lab3/clipping_pos_input.png

What is the maximum allowable input signal if the magnitude of the gain is increased to 10? Why?

If the magnitude of the gain is increased by 10, the maximum allowable input signal would be 125mV due to our output being multiplied by the gain factor of 10. Vout = -10 × Vin
Once we go above this input, we would see clipping on our positive and negative cycles due to our VCC limitation as shown below

http://cmosedu.com/jbaker/courses/ee420L/s15/students/thomad1/Lab3/large_in_clipping_out.png

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 purpose of the capacitors from VCC and VCM to ground is to act as AC short to ground. We only want a constant DC voltage at our VCM, therefore these capacitors short out any noise and higher frequencies that might be coming from our source.

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.

What is the input offset current? What does this term describe?

Input offset current is the difference between the current flowing into and out of the V_plus and V_minusinputs of the Opamp which are the respective bias currents. This offset in current causes the difference in voltage at the Vp and Vm to increase as larger resistors are used. This is due to the larger voltage across the larger resistors which make this current difference mor visible.

This current can drastically affect the output of the circuit. When we built the circuit, we measured our output to swing around 900mv with the same AC amplitude of 100mv. And Vcm and Vm had a difference of 900mv. We simulated the circuit in spice simulation above by using a current source to simulate 20nA of offset current, and the spice simulation verified our real life results.

Experiment 2

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.

Measure the offset voltage of 4 different op-amps and compare them.

http://cmosedu.com/jbaker/courses/ee420L/s15/students/thomad1/Lab3/opamp2.PNG

The above circuit can be used to measure the offset voltage by placing this offset across the input resistor. This voltage can then be amplified by the output by a factor of 20 so that it is more noticeable due to Vout = (Rf/Ri)(Vcm-Vm) therefore (Vout – Vm)/20 = Vcm -Vm , since Vcm – Vm = Voffset , then

Voffset = (Vout – Vm)/20

file:///C:/Users/kendale/AppData/Local/Microsoft/Windows/Temporary%20Internet%20Files/Content.IE5/ZWL8NDKR/offset_voltage[1]. After measuring 4 different opamps, we found that three of them had an offset voltage of 2mV as per the datasheet. But one of them had an offset voltage of 4mV. This shows us that not all Opamps of the same model will the exact same test results. In real life situations, they can be slightly different.

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