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KCC' Quizzes AQQ270 about a difference amplifier challenge

1. Quote of the week: 

2. New challenge AQQ270 about a Difference Amplifier challenge (this is a proposal from our colleague

Martin Walker, Product Marketing Leader at ADI UK – warm thanks to him!):

Below is the classic difference amplifier circuit.

Questions:

 1. What is the Common Mode Rejection Ratio (CMRR) if we have perfectly matched resistors?  [Assume the amplifier’s CMRR is 120dB]

 2. What is the CMRR if the resistors have a tolerance of 0.1%?

 3. What is the CMRR if the resistors have a tolerance of 1%?

 4. What input resistance does V2 ‘see’?

 5. What input resistance does V1 ‘see’?

 6. If the difference amplifier was AC-coupled, what impact would this inequality in input impedance have?

Good luck!



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[edited by: GenevaCooper at 1:18 PM (GMT -4) on 20 Sep 2024]
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  • 1. CMRR is 120 dB (all resistors are exactly 1k, otherwise ideal opamp.).  120 = 20 log (Ad/Acm) implies Ad = 10^60 Acm;  Acm negligeable in front of Ad.

    3. With 1% resistor, I get a 1.5% max change for Ad  ( R1=1.01K, R2=R4=990 and R=1k), ( Acm still negligeable since it is opamp dependant).
    2. With  0.1% resistor, I get a 0.06% max change for Ad  (R1=999, R2=1001, R3=1000, R4=999).  I confess that  used a simulator to obtain those numbers. My simulation may be incorrect.

    4.  R3 + R4
    5.  R1+ R2   (again, confirmed with the simulator,  (V+in - Vout) / Ir1)

    6. Not sure about the question. I add a capacitor parallel to R2 or to R4 ?  And definitively, if the input is AC, the frequency would impact the amplification.

Reply
  • 1. CMRR is 120 dB (all resistors are exactly 1k, otherwise ideal opamp.).  120 = 20 log (Ad/Acm) implies Ad = 10^60 Acm;  Acm negligeable in front of Ad.

    3. With 1% resistor, I get a 1.5% max change for Ad  ( R1=1.01K, R2=R4=990 and R=1k), ( Acm still negligeable since it is opamp dependant).
    2. With  0.1% resistor, I get a 0.06% max change for Ad  (R1=999, R2=1001, R3=1000, R4=999).  I confess that  used a simulator to obtain those numbers. My simulation may be incorrect.

    4.  R3 + R4
    5.  R1+ R2   (again, confirmed with the simulator,  (V+in - Vout) / Ir1)

    6. Not sure about the question. I add a capacitor parallel to R2 or to R4 ?  And definitively, if the input is AC, the frequency would impact the amplification.

Children
  • Hi vanderghast,

    Yes, using a simulator with the resistance as a parameter you adjust to take into account the tolerance can help.  I make 1.5% about 36dB and 0.06% about -64dB, again matching closely with the theoretical work on this.

    Bingo on Q4.

    On Q5, try the simulator with equal and opposite voltage sources.  You'll find that the effective resistance that V1 'sees' is less than 1k!  Take a look at the current through R1.  Is it what you expect it to be?

    On question 6 I was more thinking about capacitively coupling the inputs and what effect the different input resistances would have on the cut-off frequency. 

  • About Q5, with opposite input voltages, V-in - Vout  will also increase. The current through R1 is then such it is still matching (V-in - Vout) / 2000.

    Here is a snapshot. I hope it is still readable.

  • So, if V1, 5V generates a current of 7.5mA in R1, what is the value of R1 by Ohm's Law?  Spooky, huh?

  • Remember that we have a virtual ground at a pin only if the other Vin pin is at ground level. It is not our case. The closed loop to consider is:
    ground, (V1=5), R1, R2, (Vout = -10V), ground.
    That loop is complet without leak if the same current passes through R1 and R2, which holds if the OpAmp doesn't capture any current (or a negligible current). So the current of 7.5 mA passes through R1 in series with R2 dropping a total of 15 Volt (even if V1 supplies only 5 volt).