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Practical IA characteristics; optical isolation

Short answer

Instrumentation amplifier features adjustable gain (1-1000) via one resistor, very high input impedance, high CMRR (100-130 dB), low offset and drift. The three-op-amp IA has buffered inputs for high equal impedance, single-resistor gain without CMRR degradation, and symmetric paths for better high-frequency CMRR than two-op-amp designs. An optically coupled isolation amplifier uses an LED and matched photodiodes with feedback to cancel non-linearity, providing Vo=(Rf2/R1)*Vin and kV isolation.

List out the practical characteristics of instrumentation amplifier. What are the advantages of 3 op-amp instrumentation amplifier over one op-amp and two op-amp instrumentation amplifiers? Explain the operation of optically coupled isolation amplifier.

Practical characteristics of an instrumentation amplifier

  • Differential gain from 1 to 1000, set accurately (about 0.1 %) by one external resistor RG
  • Very high input impedance, about 10⁹ Ω or more, on both inputs
  • Low output impedance, below 1 Ω
  • High CMRR of 100–130 dB, which falls at higher frequency
  • Low input offset voltage (µV range) and low drift (µV/°C)
  • Low input bias current and low noise
  • Moderate bandwidth (tens to hundreds of kHz, falling at high gain) and a good slew rate
  • Good gain linearity and stable gain with temperature

Advantages of the 3 op-amp IA over 1 and 2 op-amp IAs

  • Both inputs are buffered, so both have very high and equal input impedance. The 1 op-amp IA has only a few kΩ.
  • The gain is adjusted by a single resistor RG without affecting CMRR.
  • The input stage has unity common-mode gain but high differential gain, so CMRR is higher.
  • The signal paths are symmetric, which gives good CMRR at high frequency. In the 2 op-amp IA one input passes through two op-amps.

Optically coupled isolation amplifier

       INPUT SIDE (GND1)    |   OUTPUT SIDE (GND2)
                            |
 Vin -[R1]-> A1 --> LED ~~~~|~~~~> PD2 --> A2 --> Vo
             ^       ~      |      (I2)   (I to V,
             |       ~      |              Rf2)
             +----- PD1     |
           (feedback, I1)   |
                    isolation barrier
  1. A1 drives an LED with a current that depends on Vin.
  2. The light falls on two matched photodiodes, PD1 (input side) and PD2 (output side).
  3. PD1's current I1 is fed back to A1's inverting input. A1 sets the LED drive so that I1 = Vin/R1. This feedback cancels the LED's non-linearity and temperature drift.
  4. PD2 gets the same light, so I2 = I1. A2 converts it into voltage:
Vo = I2 · Rf2 = (Rf2/R1) · Vin

There is no conductive path across the barrier, so isolation of several kV is obtained. It is used in medical and high-voltage measurement.

✓ Answer checked by Claude Opus 5.5 (high effort)

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