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ELEC2130 - Circuit Design > CMRR and Differential Amplifiers > Flashcards

CMRR and Differential Amplifiers Flashcards

1
Q

Define “Common-mode-rejection-ratio”

A

The ability of an op-amp to not amplify common-mode signals.

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2
Q

What are common mode signals and what do op-amps do to them?

A

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

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3
Q

What are the equations for CMRR?

A

CMRR = 20Log(Adiff / Acom)

CMRR = ∆Vdiff / ∆Vcom

CMRR = -20Log(∆Vdiff / ∆Vcom)

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4
Q

What are the units of CMRR?

A

Units: μV/V or dB

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5
Q

What is the tropology of the Wheatstone bridge?

A

[Picture20]

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6
Q

What is the purpose of a Wheatstone bridge?

A

> 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

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7
Q

What are the advantages and disadvantages of the Wheatstone bridge?

A

Advantages

> Large DC offset can be accommodated >

Noise interference will the common for both inputs

Disadvantages

> Noise interference will definitely be present.

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8
Q

What is the circuit tropology of a differential amplifier?

A

[Picture21]

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9
Q

What is the equations for the output voltage of the differential amplifier?

A

Vout = V2 [R4 / (R3 + R4) × (R1 + R2 ) / R1] - V1[R2 / R1]

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10
Q

Derive the equations for the output voltage of the differential amplifier?

A

Ground V2 so: Vout = -V1 [R2 / R1]

Ground V1 so: Vout = V2 [R4 / (R3 + R4) × (R1 + R2) / R1]

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11
Q

How is the differential amplifier configured for no common-mode gain?

A

> We want R2 / R1 = R4 / R3

> Then the equation can be expressed as: Vout/(V2 - V1) = R2 / R1

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12
Q

What can be said about the limitations of the CMRR of a differential amplifier?

A

> 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

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13
Q

What is the equation to calculate the CMRR error?

A

CMRR err = 1/k × (1 + (R2 / R1))

> Where k is the fractional mismatch between resistor values.

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14
Q

How can a differential amplifier have equal input impedances?

A

> 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

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15
Q

What is the circuit for an instrumentation amplifier?

A

[Picture 22]

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16
Q

Explain the tropology of an instrumentation amplifier?

A

> On the left, two identical non-inverting amplifier

> On the right, a Standard differential amplifier

17
Q

What is the purpose of the left side of the instrumentation amplifier?

A

> 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.

18
Q

What are the equations for the current the left side?

A

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

19
Q

Why is common mode gain for the left side 0dB?

A

Because V1 = V2 and the current through RG = 0 which means we can treat it as if RG does not exist (for common signals)

20
Q

What is the CMRR of the right hand side?

A

> It is still dependent on precise resistor matching

> Dependent on the CMRR of the op-amp

21
Q

What is the CMRR of the instrumentation amplifier?

A

> Due to the pre-amplification of the left hand side, the overall CMRR is greatly improved

ELEC2130 - Circuit Design (15 decks)

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