Electrical – Why is such amplifier configuration chosen for HB100 microwave sensor

amplifierconfigurationintegratorlinearoperational-amplifier

Continuous waveform amplifier

For HB100 microwave sensor module, I don't quite understand how this amplifier works and why it's been chosen instead of something simpler, such as two inverting or non-inverting op-amps. What can see for the diagram is that they are feeding integrator's output into an inverting differentiator's input. One friend suggested that a differentiator forms a band-pass filter and an integrator+differentiator combination cancels each other out, leaving you with a linear amplifier with a band-pass filter.

If that's that case, why would we want that?
How does that work, briefly described, and why has this circuit been chosen as an amplifier for this module (I am ignoring all the passive filters and coupling/bypass capacitors).

I and my co-worker are going to build it tomorrow and if anyone could shed some light I'd really appreciate it.

Datasheet (diagram at Annex 1): http://www.theorycircuit.com/wp-content/uploads/2016/09/HB100_Microwave_Sensor_datasheet.pdf

Best Answer

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Figure 1. The various filter componenents.

  • The 4.7u and 330k at (1) form a high-pass filter. Cut off is at \$ \frac {1}{2 \pi RC} = \frac {1}{2 \pi \cdot 4.7 \mu \cdot 330k} = 0.1 \ \mathrm {Hz} \$.

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Figure 2. An active high-pass filter. Source: Electronics Tutorials.

The RC networks (2-3) and (4-5) form active band-pass filters.

  • The low frequency roll-off is at \$ \frac {1}{2 \pi \cdot 4.7\mu \cdot 1M} = 0.034 \ \mathrm{Hz} \$.
  • The high-frequency roll-off is \$ \frac {1}{2 \pi \cdot 2n2 \cdot 10k} = 723 \ \mathrm {Hz} \$ for the 10k and you can work out the 8k2 version.

One friend suggested that a differentiator forms a band-pass filter and an integrator + differentiator combination cancels each other out, leaving you with a linear amplifier with a band-pass filter.

The integrator and differentiator would only cancel out if they had the same cut-off frequencies. They don't so you get a band-pass filter.

If that's that case, why would we want that?

The datasheet explains on page 3: The magnitude of the Doppler Shift is proportional to reflection of transmitted energy and is in the range of microvolts (µV). A high gain low frequency amplifier is usually connected to the IF terminal in order to amplify the Doppler shift to a processable level (see Annex 1). Frequency of Doppler shift is proportional to velocity of motion. Typical human walking generates Doppler shift below 100 Hz.