Lab

Lab 3 - EE 420L

Author: Victor Payumo

Date: February 19, 2020

Op-amps I, basic topologies, finite gain, and offset

Question 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?

Support your answer with an entry from the electrical characteristics table in the datasheet.

The minimum allowable common-mode voltage is 0 V and its maximum is 3.5 V.

Question 2:

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.

A good estimate for the op-amp's open-loop gain is 100 V/mV.

Question 3:

What is a good estimate for the offset voltage?

For worst case design what value would you use?

A good estimate for the offset voltage is 2 mV while 9 mV would be the worst case at higher temperatures.

Experiment 1:
LTSpice simulation:

Implementation & measurements:

What is the common-mode voltage, VCM? Does VCM change? Why or why not?
VCM here is 2.5 V which is being fed through a voltage divider. It is a reference and therefore does not change.

What is the ideal closed-loop gain?
Gain = - Rf/Ri = - 5 K/ 5 K = - 1 V/V

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

The output swing corresponds to the input signal and is centered around VCM @ 2.5 V.

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.

Clipping occurs when the input signal is excessively driven. Because of the capped open loop gain, when Vp rises Vm cannot match that potential.

What is the maximum allowable input signal amplitude? Why?
The maximum allowable input is 2.5 V which is half of Vp. Again, if it exceeds 2.5 V clipping occurs.

What is the maximum allowable input signal if the magnitude of the gain is increased to 10? Why?
If the gain is increased to 10 then the input signal must be divided by 10 resulting in 250mV.

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?

They filter out noise which could disrupt the AC signal and virtually any capacitance will suffice.

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?

Larger resistances would yield higher potential at Vm. Vp must match this potential in an ideal op amp thus resulting in a larger swing. Input offset current is the difference between the bias current in the inverting terminal and non-inverting terminal.
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.

Schematic and implementation

The circuit above shows that the amount of current that flows from Vout to Vm can be controlled by RF.

Measurements of LM348 chip, Vout (left) VCM (right)

The offset voltage is found by |Vout - VCM|/100
Here are the measurements including 3 other chips

Op-Amp	Vout	VCM	Offset
LM348	2.5451 V	2.5317 V	134 mV
NE556	2.6730 V	3.0059 V	3.33 V
LM324	2.5493 V	2.5324 V	169 mV
TP3054	2.5282 V	2.5286 V	4 mV

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