Lab 3 - EE 420L Engineering Electronics II 

Author: Matthew Meza

Email: mezam11@unlv.nevada.edu

February 6, 2015 

Op-Amps I, Basic Topoligies, Finite Gain, and Offset

• Watch the video op_amps_I, review op_amps_I.pdf (associated notes), and simulate the circuits in op_amps_I.zip.
• Read the write-up seen below before coming to lab.

• Learn how op-amps are used, their limitations, closed and open-loop gian, offset voltage, common-mode voltage (VCM),  
  max input/output, input bias current, offset current.

For the following questions and experiments assume VCC+ = +5V and VCC- = 0V.

• 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?
• 
• The maximum allowable common-mode voltage for our circuit is 3 [5-2] to 3.5 [5-1.5] volts.
• The minimum allowable common-mode voltage for our circuit is 0 volts.

• What is a good estimate for the op-amp's open-loop gain? 
• 
• At 0Hz we have around 110dB. AOL ~ 10^(110/20) = 316228 V/V
• The gain decreases by approximatly 20dB/Decade
  
  
• What is a good estimate for the offset voltage? 
• 
• A good estimate for the offset is 7mV.
• For worst case design, we would use the 9mV offset.
   
   

Build, and test, the following circuit.

 

DC coupled Offset Voltage Input and Output	AC Coupled AC Input and Output 180 degrees out of phase	Clipping when the input voltage is too high

• What is the common-mode voltage, VCM? Does VCM change? Why or why not?
• The common mode voltage is 2.5V DC. The VCM does not change because is used as a reference for the
  AC input signal to swing around. It is created by the 10k Voltage divider at the positive input.

• What is the ideal closed-loop gain?
• The ideal closed-loop gain is –Rf/Rin which is equal to 5k/5k = 1 or 0dB/Unity Gain.

• What is the output swing and what is it centered around?
• The output swing is approximately 200mV peak to peak (+/-100mV peak). The output is centered around VCM which is set at 2.5 V.

• What happens if the input isn't centered around VCM, that is, 2.5 V? 
• If the input is not centered around 2.5 V,  then one may encounter various problems such as clipping. A possible reason for an input to be
  off from VCM is that the op-amp’s open-loop gain is not sufficient to pull Vm up to Vp. If the input voltage at the Vp terminal increases,
  the output voltages will decrease, and vice versa. Vp must be appropriately within bound of the positive supply rail.
   
• Provide a detailed discussion illustrating that you understand what is going on.
• We have a 5V input at VCC which is voltage divided to 2.5 V at the positive input terminal, Vp. This voltage, then
  appears at Vm, due to the very high open-loop gain  of the nearly non-ideal Op-Amp. This voltage is then multiplied
  by the closed-loop gain of the topology which is designated by –Rf/Ri. In our case, this would be 1x (unity gain).
  For AC, Vp is at AC ground. Consequently, Vm is then at AC ground while the input AC voltage is multiplied by the same
  closed-loop gain. In conclusion, the output is centered around 2.5V with an AC swing of +/- 100mV (peak).
   
• What is the maximum allowable input signal amplitude? Why?
• Since our rails are set at 5V, and 0V. We could theoretically swing +/- 2.5 V. Anything larger, will result in clipping of
  the signal. The maximum allowable input signal would then be 2.5V. Since real op-amps are non-ideal, the maximum allowable
  input signal amplitude is less. In our case it may have an experimental max swing of +/- 500mV (worst case when max VCM is 3 volts)
  or a max swing of +/- 1V (most likely case when max VCM is 3.5 volts).
   
• What is the maximum allowable input signal if the magnitude of the gain is increased to 10? Why?
• By increasing the magnitude of the gain by 10, the largest input signal would be 250mV theoretically because
  the gain is ten times larger than before. (2.5/10).
   
• What is the point of the 0.01 uF capacitors from VCC and VCM to ground? 
• These capacitors are just to decouple the voltages, and detract noise from the power supply.

• Are these values critical or could 0.1 uF, 1,000 pF, 1 uF, etc. capacitors be used?
• These values could be anything. They are used to hold the voltage at a node to a specific value to detract from sudden
  swings. They are just charge buckets.
   
• 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.
• If we had a 20nA input bias current on each of the op-amp terminals, using a large resistor will produce a larger voltage at Vm, and 
  consequently, Vp. This mismatch of voltages would then change the output voltage of our op-amp.
  
  
• What is the offset current? What does this term describe?
• The input offset current is the difference between the input biase currents at VP and VM.
  It describes the mismatch between the two input currents of the op amp.

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

Op-Amp 1 ~ +20mV	Op-Amp 2 ~ -20mV
Op-Amp 3 ~ 0mV	Op-Amp 4 ~ +40mV

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