CMRR and Differential Amplifiers Flashcards
Define “Common-mode-rejection-ratio”
The ability of an op-amp to not amplify common-mode signals.
What are common mode signals and what do op-amps do to them?
Due to op-amp imperfections, there will also be an output which is proportional to the average value of the two inputs. The amount of amplification to this average value is the common-mode-gain
What are the equations for CMRR?
CMRR = 20Log(Adiff / Acom)
CMRR = ∆Vdiff / ∆Vcom
CMRR = -20Log(∆Vdiff / ∆Vcom)
What are the units of CMRR?
Units: μV/V or dB
What is the tropology of the Wheatstone bridge?
[Picture20]
What is the purpose of a Wheatstone bridge?
> When used in conjunction with a differential amplifier of some sort, It allows a differential signal to be created.
> The unity point can be changed by adjusting the values of the resistors.
> Some sensors need a large DC offset to function and this is a circuit layout stops the op-amp from having a large DC offset on its input whilst still offsetting the sensor. It is also good for signal levels of 10s of millivolts
What are the advantages and disadvantages of the Wheatstone bridge?
Advantages
> Large DC offset can be accommodated >
Noise interference will the common for both inputs
Disadvantages
> Noise interference will definitely be present.
What is the circuit tropology of a differential amplifier?
[Picture21]
What is the equations for the output voltage of the differential amplifier?
Vout = V2 [R4 / (R3 + R4) × (R1 + R2 ) / R1] - V1[R2 / R1]
Derive the equations for the output voltage of the differential amplifier?
Ground V2 so: Vout = -V1 [R2 / R1]
Ground V1 so: Vout = V2 [R4 / (R3 + R4) × (R1 + R2) / R1]
How is the differential amplifier configured for no common-mode gain?
> We want R2 / R1 = R4 / R3
> Then the equation can be expressed as: Vout/(V2 - V1) = R2 / R1
What can be said about the limitations of the CMRR of a differential amplifier?
> The CMRR of the differential amplifier is still affected by the op-amps internal CMRR.
> Common-gain will only be zero if the resistor ratios are really equal which is hard to achieve in practice
What is the equation to calculate the CMRR error?
CMRR err = 1/k × (1 + (R2 / R1))
> Where k is the fractional mismatch between resistor values.
How can a differential amplifier have equal input impedances?
> If you follow R1 = R3 + R4 then this will be the case
> It is very hard to meet all these equations without ruining the CMRR
What is the circuit for an instrumentation amplifier?
[Picture 22]
Explain the tropology of an instrumentation amplifier?
> On the left, two identical non-inverting amplifier
> On the right, a Standard differential amplifier
What is the purpose of the left side of the instrumentation amplifier?
> To provide a higher (and symmetrical) input impedance.
> Designed so that differential signals are pre-amplified by a large amount (dependent on the resistors) but common mode signals are amplified by unity.
What are the equations for the current the left side?
Current through RG: IRG = (V2 - V1) / RG
Current through RF: IRF = (VO2 - VO1) / (2RF+ × RG)
Differential gain: (VO2 - VO1) / (V2 - V1) = (2RF+ RG) / RG = 1 + (2RF) / RG
Why is common mode gain for the left side 0dB?
Because V1 = V2 and the current through RG = 0 which means we can treat it as if RG does not exist (for common signals)
What is the CMRR of the right hand side?
> It is still dependent on precise resistor matching
> Dependent on the CMRR of the op-amp
What is the CMRR of the instrumentation amplifier?
> Due to the pre-amplification of the left hand side, the overall CMRR is greatly improved
