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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=120dB

    2. You can use the approximation CMMR∼1/t, with t=tolerance. So you get CMRR=60dB for t=0.1%

    3. CMRR= 40dB, t=1%

    4. In an ideal opmap you see an infinite resistance at the non-inverting input (in fact it is typically used as a buffer configuration)

    5. In an ideal opamp you see R1 at the inverting input (in fact the non inverting input is a virtual ground)

    6. Not sure I understand 100% the question. But if you are to AC-couple both the inputs, you need to place capacitors at the inputs and the impedance will depend on frequency. 

Reply
  • 1. CMRR=120dB

    2. You can use the approximation CMMR∼1/t, with t=tolerance. So you get CMRR=60dB for t=0.1%

    3. CMRR= 40dB, t=1%

    4. In an ideal opmap you see an infinite resistance at the non-inverting input (in fact it is typically used as a buffer configuration)

    5. In an ideal opamp you see R1 at the inverting input (in fact the non inverting input is a virtual ground)

    6. Not sure I understand 100% the question. But if you are to AC-couple both the inputs, you need to place capacitors at the inputs and the impedance will depend on frequency. 

Children
  • Hi Gaetano,

    That was quick!

    The approximation 1/t is a little on the optimistic side for CMRR but is a good first approximation, taking into account the differential gain is 1, so congratulations on the CMRR questions.  If anyone else knows of a better estimate for CMRR, please share!

    You are correct that the op-amp has a close to infinite input impedance, and because of that IN+ sits at a voltage governed by the potential divider R3 and R4, in this case V2/2.  Therefore V2 'sees' a 2k input resistance to ground.

    The impedance that V1 'sees' is much more interesting.  You would think that it should be the value of R1, 1k, as you say, but the IN- has the 'virtual ground' voltage across it (V2/2) that is in anti-phase with V1, if V1 and V2 are equal and opposite in phase.  So, a bit like bootstrapping in reverse, it changes the effective resistance of R1, depending on the value of V2.

    The upshot of this for capacitively coupled inputs is that you need to modify the coupling capacitor values if you want to maintain the same cut-off frequency for the high-pass filter formed by the input R and C.