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Hello George,
could you kindly tell me the electrical characteristics (amplitude and frequency) of the signal generated by the eddy current entering the first stage of the preamp?
Sorry ... I don't understand what you're asking for.
Please explain.
Sorry ... I don't understand what you're asking for.
Please explain.
I'd like to know the details of the signal present on pin 2 of U3 (NE5532) if possible...
everybody says we are working with microvolts but what kind of signal is it actually?
I'd like to know the details of the signal present on pin 2 of U3 (NE5532) if possible...
everybody says we are working with microvolts but what kind of signal is it actually?
Have a look at the attached image.
The main sample pulse captures a section of the preamp output on each TX cycle and puts it through an integrator. This produces a dc level that increases in the presence of a metal target. You will only see a signal change at the input of the preamp if the metal target is close to the coil, as the changes may only be a few microvolts for small targets or those far from the coil.
Try monitoring the preamp output on an oscilloscope using channel 2 (while triggering channel 1 on the falling edge of the TX pulse) and bring a relatively large metal target towards the coil. Note that the rising edge (where sampling takes place) moves towards the right. This will cause the sample pulse to see a signal that becomes more negative.
Then try the same procedure with a small coin at a distance that causes a weak audio response. I guarantee that you will not be able to see any movement in the preamp output, but the detector is still able to extract this miniscule change. Since the signal change at the preamp output cannot be detected visually on the scope, you will definitely not see any change at the input.
Thank you George, so the presence of the target either accelerates or slows down the curve we compose, right?
While the way it decays should tell us if it is a ferrous or not, obviously after a couple uSecs, and if we don't have a target the value shown is nothing more than the influence of the terrain that needs to be accounted. (SH_EF in your schematic)
Hello George,
could you kindly tell me the electrical characteristics (amplitude and frequency) of the signal generated by the eddy current entering the first stage of the preamp?
Scope pictures of a couple targets at different distances from the coil. Amplifier out, linear and log amplitude. Amplifier gain=500. Scope time base varied depending on target, 2us to 50us/div.
Scope pictures of a couple targets at different distances from the coil. Amplifier out, linear and log amplitude. Amplifier gain=500. Scope time base varied depending on target, 2us to 50us/div.
Question for those that may know: If using the Voodoo in highly mineralised ground, will the iron discrimination function still work if there is no ground balancing circuitry on the Rx channel?
Hi, I have a question to those who have more experience than me on this project, where I find information regarding PL1-13 contacts, power supply type or other. I searched the book but at the moment I couldn't. Thanks for your reply.
Hi, I have a question to those who have more experience than me on this project, where I find information regarding PL1-13 contacts, power supply type or other. I searched the book but at the moment I couldn't. Thanks for your reply.
PL1-13 in schematics, pages 26,18,16. Power supply on page 17.
Thanks Mr. Waltr, I have a problem with the power supply, at PL 15 I don't have +5 volts, instead on PL 14 I have -5, looking at the diagram on the book I noticed that the two ground of the two stabilizers should both be connected to the 0 volts, instead there is no continuity between them. Thanks for who can answer.
Thanks Mr. Waltr, I have a problem with the power supply, at PL 15 I don't have +5 volts, instead on PL 14 I have -5, looking at the diagram on the book I noticed that the two ground of the two stabilizers should both be connected to the 0 volts, instead there is no continuity between them. Thanks for who can answer.
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