Understanding how the sampling integrator works

[brumbarchris]

Great, thanks for clarifying.

[green]

Are these from a Geotech Baracuda or from something else? No, separate Tx and Rx inductance balanced. X signal is opposite R signal for non ferrous targets(quarter). If I start sampling when the amplifier is saturated the X signal subtracts from the R signal(less target signal). Was wondering if starting sample when amplifier is saturated with the Baracuda circuit does something similar. Don't have a Baracuda circuit to try.

[brumbarchris]

Some scope pictures with US quarter for target.

Hello!
Not sure if I can make sense of the scope pictures you posted. Are these from a Geotech Baracuda or from something else? My understanding of how the coil works (on the Baracuda) is that you should normally see a large negative spike at the "hot" end of the coil corresponding to each TX pulse; I cannot correlate this to CH1 in your plots, as it only shows a rather small amplitude positive spike.

[green]

You can sample into the static decay curve, but the more of the curve you sample the more offset you get at the integrator output. Usually this will reduce the dynamic range of target responses. One way to compensate is to slightly underdamp the coil. Another is to inject an offset current into the integrator. This could even be implemented with an autotune loop, which might have other interesting possibilities.

Some scope pictures with US quarter for target. If I started the sample while amplifier out was full scale positive the signal would go the wrong polarity in the beginning of the sample. Wondering if something similar happens with Baracuda circuit?

[Carl-NC]

Is it also true that circuit should be so tuned that when no metal target is under the coil, the Main Sample Pulse occurs after the complete decay of the RX preamp output? in such a case, the Main Sample Pulse and the Secondary Sample Pulse should eventually lead to no voltage build-up at the output of the integrator.

You can sample into the static decay curve, but the more of the curve you sample the more offset you get at the integrator output. Usually this will reduce the dynamic range of target responses. One way to compensate is to slightly underdamp the coil. Another is to inject an offset current into the integrator. This could even be implemented with an autotune loop, which might have other interesting possibilities.

[green]

Thank you for the explanation and also for finding the original post. I had searched for it also, but without success.

Is it also true that circuit should be so tuned that when no metal target is under the coil, the Main Sample Pulse occurs after the complete decay of the RX preamp output? in such a case, the Main Sample Pulse and the Secondary Sample Pulse should eventually lead to no voltage build-up at the output of the integrator.
On the other hand, as metal comes under the coil, the decay time of the RX pream output increases, and the Main Sample Pulse "catches" some of this decaying slope and feeds it to the input of the integrator, thus building up some voltage at the output of the integrator; subsequent samples under the same conditions gradually increase the voltage at the output of the integrator.

This is my understanding of how the circuit broadly operates, is it correct?

BR,
Cristian

Some scope traces I posted in another thread. Think sample could start after 4usec delay in top row, maybe 5us delay bottom row. No target signal doesn't need to be zero when sample starts(maybe less than 1/2 full scale), just needs to repeat. Ground could cause a signal which doesn't matter if it doesn't change. Probably not true about ground signal since it is going to change, would need to take a second sample to ground balance.

[brumbarchris]

If it were a true integrator, no feedback resistor. Feedback resistor limits gain, gain=Rfdbk/Rin*sample rate*sample time.

Thank you for the explanation and also for finding the original post. I had searched for it also, but without success.

Is it also true that circuit should be so tuned that when no metal target is under the coil, the Main Sample Pulse occurs after the complete decay of the RX preamp output? in such a case, the Main Sample Pulse and the Secondary Sample Pulse should eventually lead to no voltage build-up at the output of the integrator.
On the other hand, as metal comes under the coil, the decay time of the RX pream output increases, and the Main Sample Pulse "catches" some of this decaying slope and feeds it to the input of the integrator, thus building up some voltage at the output of the integrator; subsequent samples under the same conditions gradually increase the voltage at the output of the integrator.

This is my understanding of how the circuit broadly operates, is it correct?

BR,
Cristian

[waltr]

https://www.geotech1.com/forums/show...4582#post54582

Thanks Green.
This does help to put the post into context.

[green]

Sorry, I have tried looking for the original post but never found it.
When I was first reading the forum to learn about PI detectors I copy/pasted good posts/topics into Text files. Maybe some creative forum searching will find the post.

https://www.geotech1.com/forums/showthread.php?13113-Designing-an-Integrator-circuit&p=54582#post54582

[waltr]

That is great, thank you for sharing this. Do you know where I could find the original post? Carl mentions some component reference designators in his explanations, and I would like to match those to a schematic presumably found in the original post.

Sorry, I have tried looking for the original post but never found it.
When I was first reading the forum to learn about PI detectors I copy/pasted good posts/topics into Text files. Maybe some creative forum searching will find the post.

[green]

if a metal target is continuously kept under the sensor coil, the the integrator output will eventually saturate at the upper rail

If it were a true integrator, no feedback resistor. Feedback resistor limits gain, gain=Rfdbk/Rin*sample rate*sample time.

[brumbarchris]

Understanding how the sampling integrator works

Yeah, as per the forum rules, I cannot amend the first post, so I am adding here the pictures and attachments mentioned in the post. In addition to these, I have also added the zip file with the actual LTSpice simulation, if you would be so kind to check it. BR, Cristian

[brumbarchris]

It is better if you upload attachments directly to the forum instead of using an off-site server. Offsite attachments will eventually disappear.

Will do so, but at the time when I made the post I did not have the option to make attachments (possibly because it was my first post). I see that now I can add attachments and I will amend the first post to include pictures and simulation files.

Attached is a text file of Carl's explanation (copied from a forum post) of how a sampling integrator works.

That is great, thank you for sharing this. Do you know where I could find the original post? Carl mentions some component reference designators in his explanations, and I would like to match those to a schematic presumably found in the original post.
As a matter of fact, the explanations, tend to match my observations of the circuit in simulation.

I suppose the following hold true:
- circuit should be so tuned that when no metal target is under the coil, the Main Sample Pulse occurs after the complete decay of the RX preamp output; in such a case, the Main Sample Pulse and the Secondary Sample Pulse should eventually lead to no voltage build-up at the output of the integrator
- as metal comes under the coil, the decay time of the RX pream output increases, and the Main Sample Pulse "catches" some of this decaying slope and feeds it to the input of the integrator, thus building up some voltage at the output of the integrator; subsequent samples under the same conditions gradually increase the voltage at the output of the integrator
- if a metal target is continuously kept under the sensor coil, the the integrator output will eventually saturate at the upper rail

Are these understandings correct?

Best regards,
Cristian

[green]

From the documentation, however, I understand I should obtain a DC signal at the output of the sampling integrator. Any idea on what I am doing wrong?

Integrator time constant=about 125us. You need to sample over 5times longer to reach steady state.

Integrator gain over 10. Average volts during the sample needs to be .4V or less.

[Carl-NC]

Looks like you start sampling when the amplifier is still saturated, to soon.

Yes, you're sampling too early. You can do that but will need to also apply an offset.

It is better if you upload attachments directly to the forum instead of using an off-site server. Offsite attachments will eventually disappear.