Lab 3 - EE 421L 

Vallesm, Mario

2/20/2015

                    Opamps 1

        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?
• If Vp = 2.5 V and Vm= 2.5 V then Vcmmax = 5 and Vcmmin = 0
• datasheet confirms:
• What is a good estimate for the op-amp's open-loop gain?
• Vout = A(Vp-Vm) 
  
• 
• What is a good estimate for the offset voltage? 
• 2 mV
  
• For worst case design what value would you use?
• 9mV
  
• 
• 
  

The experimentation was performed in the following Circuit w

• What is the common-mode voltage, VCM? Does VCM change? Why or why not?
• VCM is 2.5 V
• No
• It doesn't change because is dc in a voltage divider
• What is the ideal closed-loop gain?
• -1 V/V
• What is the output swing and what is it centered around?
• The output is swinging in about 200mV
• What happens if the input isn't centered around around VCM, that is, 2.5 V? 
• The output will be negative and positive sinewave.
• Provide a detailed discussion illustrating that you understand what is going on.
• In our experiment we took the picture in AC so DC would be cancel
• What happens is that the output swings 200mV around 2.5 which is the bias voltage. This is done to have an amplified sinewave at the output, but only in the positive axis. Also as shown in the figure (b) if the DC offset is increase too much over the output will clip first at the most positive axis.

• What is the maximum allowable input signal amplitude? Why?
• The maximum would be 1.25V because if we increase the input voltage higher than this the output will clip. This is because we are providing the Op-amp to only 

• What is the maximum allowable input signal if the magnitude of the gain is increased to 10? Why?
• It is only .125 V since the input would clip at this voltage

• What is the point of the 0.01 uF capacitors from VCC and VCM to ground? 
• The point is short any AC voltage/current in the VCM circuit
• Are these values critical or could 0.1 uF, 1,000 pF, 1 uF, etc. capacitors be used?
• No any big enough capacitor would do a good shorting the current
• 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.
• Since the oscilloscope would no catch up the difference we simulate  the circuit to see what happens
• Show how the operation of the circuit can be effected if, for example, R1 and R2, are much larger. Explain what is going on.
• Since the Resistance is much bigger the voltage drop on these resistance increase and Vcm and Vp changes
• What is the input offset current? What does this term describe?
• Since the oscilloscope couldn't measure we assume 20nA

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.
• What happens is the Offset is amplified to factor of 20, this is becuase we use 20k resistance since the 100k did not work for us. In the following picture we show 4 different opamps. Some of them have only 2mV while one of them had 4mV

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