0
\$\begingroup\$

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

\$\endgroup\$
  • \$\begingroup\$ I gave you a complete analysis of the circuit in your previous question of two days ago electronics.stackexchange.com/questions/337140/… and you accepted my answer and said you had "got it all". What have you forgotten? \$\endgroup\$ – Transistor Nov 1 '17 at 19:31
  • \$\begingroup\$ @Transistor, you explained the parts that I didn't understand at the time. Now when I look at the entire integrator/differentiator setup I don't understand why use something like that to amplify the signal, to be frank I have never seen something like that before. Any input is always appreciated. \$\endgroup\$ – Shibalicious Nov 1 '17 at 19:33
1
\$\begingroup\$

enter image description here

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} \$.

enter image description here

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.

\$\endgroup\$
  • \$\begingroup\$ really appreciate the explanation. But what puzzles me is that integrators and differentiators produce different waveforms at the outputs from the ones they had at the inputs. So when I said "cancel each other out" I meant, do you get the same waveform at the amplifiers output that you had at the input? \$\endgroup\$ – Shibalicious Nov 2 '17 at 15:54
  • \$\begingroup\$ (1) They are neither pure differentiators or integrators. For example, in Figure 2 above an integrator would have Cf but not Rf. The effect of this is to make it an integrator at high frequencies but a linear amplifier at low. (2) Meanwhile Rin and Cin are doing the opposite; it's a differentiator at low frequencies but becomes linear above its cut-off frequency. (3) The cut-off frequencies aren't the same so they don't cancel out. \$\endgroup\$ – Transistor Nov 2 '17 at 16:11
  • \$\begingroup\$ after re-reading this and the electronics-tutorials.ws page on this I finally get it. Thank you for your extreme patience! \$\endgroup\$ – Shibalicious Nov 2 '17 at 21:02
  • \$\begingroup\$ Good. Just check up that I've got my cut-off frequencies right. As you reverse engineer circuits like this you start to see patterns. My approach is usually to check the extremes. What happens if I put DC in - that gives a clue for low frequency response - and if I feed in a very high frequency. For DC all capacitors become open-circuit and for high frequency all capacitors become short-circuit (because their impedance drops). \$\endgroup\$ – Transistor Nov 2 '17 at 23:00
  • \$\begingroup\$ Yes, your calculations are correct and after the second stage the band-pass roll-offs are between ~3Hz to 72Hz, just like the datasheet suggests. Thank you again for your help! \$\endgroup\$ – Shibalicious Nov 3 '17 at 22:21

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.