Lab 3 - ECE 420L 

Authored by Desi Battle, battled@unlv.nevada.edu

2/15/2016

  

VCC+ = 5             VCC- = 0

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.

-0.3 < Vcm < 3.5 V

lab3_img01_VCMrange.png

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.

100mV /V or 100K

lab3_img02_largegain.png

 

Typ 2 mV

Worst case 9 mV

lab3_img03_OffsetVoltage.png

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

VCM = 2.5 V It does not change because the

What is the ideal closed-loop gain?

1 V

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

+/- 50 mV centered around 625 mV

What happens if the input isn't centered around around VCM, that is, 2.5 V?

In this configuration the output and VCM will remain stable, but Vm will maintain its current relationship to the new input ( Vm = ½ Vin)

Provide a detailed discussion illustrating that you understand what is going on.

First consider the DC operating point.

·         The bottom half of the circuit is a DC power supply with a voltage divider forcing VCM to 2.5 volts at all times.

·         The input source at the top half of the circuit is also set to 2.5 Volts.  Ideally we can consider the Vm and VCM terminals to be shorted so Vm = 2.5 Volts as well.  Since 2.5 = 2.5 no current is drawn and there is no voltage drop across the feedback resistor and Vout = 2.5 V and everything is at equilibrium

Next we add the effects of the AC component

·         The 100mV Sine wave must drop evenly across the input and feedback resistors since they are equivalent. VCM is still tied to 2.5 Volts by the DC supply, but Vm now = ½ Vin or a Sine wave centered at 1.25 V. The op-amp internally has a dependent voltage source that measures the difference in Vm and VCM and it provides feedback at the value of the difference multiplied by a very large gain to make them equivalent.  As long as VCM is greater than Vm (or 1.25) all the op- amp can do is sink the current being driven by the AC input.

 

What is the maximum allowable input signal amplitude? Why?

2Vcc – 1.5 because you don’t want either of the input terminals to exceed Vcc – 1.5 and the op-input input is 1/2Vin

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

2Vcc – 1.5 because while the output does feedback to the input, it will lim

What is the point of the 0.01 uF capacitors from VCC and VCM to ground?

To prevent unstable start up behavior as Voltage across capacitors cannot change and instantaneously.  This is important for op amps since small variations in the input terminals are amplified by the large gain.

Are these values critical or could 0.1 uF, 1,000 pF, 1 uF, etc. capacitors be used?

I believe that different values can be used but within a certain range.  You want to be careful not to overload and change the fundamental behavior of the circuit.

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.

Higher Resistances would cause higher voltage drops at equivalent currents so the input voltage range to produce output within the operating range would be reduced.

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

The difference in the current flowing between the two op-amp input terminals.  In general this term would be 0 – 0 = 0 but in reality there is always some current flowing in/out of the nodes

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