Measuring negative voltage given by a coil using Electromagnetic induction

Hi, I am making a seismograph for a school project and ran into a problem while doing so. I am using a moveable magnet inside a coil that will produce a voltage proportional to the speed of a tremor when the magnet moves. To translate this analog data to digital data i am using an arduino uno. I am using a simple dc voltage meter as seen in the picture below. The problem is that the coil produces a voltage where the positive and negative terminal varies. This means half of the time i get negative voltage readings which the arduino cant read. I was told i coud use the 5v the arduino can give as my 0, so that when i get a -3 voltage read the arduino sees 2v. But how would i wire this? When i just plug in the 5v the arduino reads it but the coil doesnt give any voltage reading anymore when being exposed to a varying magnetic field. I have seen solutions with zenerdiodes but i have no access to these. Is there any way to do this without any more electronic components than the ones given in a standard arduino kit?

arduino-dc-voltmeter-schema.png

arduino-dc-voltmeter-schema.png

Not easy to work out from that! Perhaps give us a circuit diagram (schematic) instead of artwork?

In order to measure negative and positive Voltages, you will need to bias the analogue input to midway after dividing it down to a safe level. That way you will see readings of <512 for negative Voltages and > 512 for positive Voltages.

emielj:
clarify first, if the coil MUST be grounded.

Also, could you provide a link to your coil?

If its only 2-wire...

I guess that the peak voltages that your measuring didn't hurt the analog input, so the 100K you have in series with the signal and the input's internal protection diode gave adequate protection. So as per AJLElectronics suggestion, maybe you could try something like this (I'll let you do the circuit diagram and give us more details).

circuit.jpg

circuit.jpg

Here is how the biasing is usually done. Your coil is represented by "CT Output".

More at Learn | OpenEnergyMonitor

dlloyd:
Also, could you provide a link to your coil?

If its only 2-wire...

I guess that the peak voltages that your measuring didn't hurt the analog input, so the 100K you have in series with the signal and the input's internal protection diode gave adequate protection. So as per AJLElectronics suggestion, maybe you could try something like this (I'll let you do the circuit diagram and give us more details).

circuit.jpg

Hi, thank you for your answer. I cant find a link to my coil. It is a 12000 windings coil that is in fact 2-wired, what is the problem with this? I am also confused as to what extra info a circuit diagram would give. The wires going out of the breadboard are directly connected to the coil

Unless you are planning on having two or more of your devices all a known distance apart and will monitor the arrival time of the wave, you are only measuring the amplitude of the wave, not the speed.

Speed ALWAYS need a time and distance factor.

Paul

Just wondering what you actually have ... thanks for the info.

So, the coil is wound with 12,000 turns of wire. Lots of variables to consider, but for an example, if this info is used to calculate the voltage generated in the "Moving magnet toward coil" calculation here, we get 9.6V.

Do you know the maximum voltages you can generate?

I haven't tried this but I think it might work.

... and another example

dlloyd:
Just wondering what you actually have ... thanks for the info.

So, the coil is wound with 12,000 turns of wire. Lots of variables to consider, but for an example, if this info is used to calculate the voltage generated in the "Moving magnet toward coil" calculation here, we get 9.6V.

Do you know the maximum voltages you can generate?

The maximum voltages i can generate are reliant to the speed of the magnet. If i pull the magnet as hard as i can i get around 30v. But using the seismograph normally it probably wont succeed the arduino's 5v

"I am also confused as to what extra info a circuit diagram would give."

It shows how everything is connected and is the standard for all electronics. It is clear and unambiguous. Thos Fritzing artworks are useless to real electronics people and can cause those people to skip past your enquiry.

If i pull the magnet as hard as i can i get around 30v. But using the seismograph normally it probably wont succeed the arduino's 5v

Looks like you have some circuits to try and code to write and test. I haven't tested the circuit in my last reply, but it basically connects one of the coil wires to a 2.5V reference, and the other is your signal output.

Note: Arduino's serial plotter could be used to plot the signal waveform.

jremington:
Here is how the biasing is usually done. Your coil is represented by "CT Output".

More at Learn | OpenEnergyMonitor

Hi, i dont understand the capacitors use here. Could you elaborate? To my knowledge it blocks Dc and lets AC through

I'll jump in and say that this circuit is quite similar, except that the transformer is a CT (current transformer) and it has a burden resistor and reference capacitor.

The burden resistor will mainly limit (attenuate) the peak-peak voltages you get on the signal.
The capacitor filters any noise on the 2.5V reference (R1-R2 divider) and holds it stable.

emielj:
I am also confused as to what extra info a circuit diagram would give. The wires going out of the breadboard are directly connected to the coil

We are used to thinking of a circuit according to its function, not its physical layout, so the schematic diagram is the language in which we understand the circuit. There is little commonality between a proper schematic diagram and pretty Fritzing pictures, the use of the latter is just trying to dumb it down to "Lego" building blocks. Pretty Fritzing pictures are useless for engineering purposes.

Is the polarity important? If the intensity, I.E. movement of the magnet relative to the coil, is all that is needed, then just go through a full-wave bridge. (Use Schottky diodes).

Also, are you measuring P waves or S waves? The orientation of the coil/magnet makes a difference.

A detector at Parkfield, CA that I am familiar with measures S-waves. It is a coil buried about ten feet into the Earth with a magnet inside that is suspended by a spring. The S-waves will shake the coil up and down relative to the magnet, producing an output.

BTW, have you considered using an op-Amp to make a non-inverting summing amplifier. You can adjust the gain of the op-Amp to amplify the signal, and it adds the benefit that the output can't go past the positive rail.

SteveMann:
We are used to thinking of a circuit according to its function, not its physical layout, so the schematic diagram is the language in which we understand the circuit. There is little commonality between a proper schematic diagram and pretty Fritzing pictures, the use of the latter is just trying to dumb it down to "Lego" building blocks. Pretty Fritzing pictures are useless for engineering purposes.

Is the polarity important? If the intensity, I.E. movement of the magnet relative to the coil, is all that is needed, then just go through a full-wave bridge. (Use Schottky diodes).

Also, are you measuring P waves or S waves? The orientation of the coil/magnet makes a difference.

A detector at Parkfield, CA that I am familiar with measures S-waves. It is a coil buried about ten feet into the Earth with a magnet inside that is suspended by a spring. The S-waves will shake the coil up and down relative to the magnet, producing an output.

BTW, have you considered using an op-Amp to make a non-inverting summing amplifier. You can adjust the gain of the op-Amp to amplify the signal, and it adds the benefit that the output can't go past the positive rail.

I have thought about using an amplifier but the wiring seemed to hard for me to understand at the time.
What is the difference between measuring s and p waves?
What would schottky diodes do here, they are some type of rectifiers right?

The difference between S waves and P waves is 90 degrees. As in the quote you gave, S waves are vertical and P waves are horizontal movements. S waves bounce your house up in the air and then the P wave moves the foundation out from under it.

Paul

What would diodes do here, they are some type of rectifiers right?

Yes but there is a voltage drop across all diodes so that has to be exceeded before you can get any voltage through it. This limits the size of the smallest movement that can be detected. A schottky diode has the smallest volts drop you can have, but even so, it is in the order of about 0.25 V, which is why I decided not to use any diodes in my circuit.

I like dlloyd's solution, post number 9. Simple & effective

Is there any way to do this without any more electronic components than the ones given in a standard arduino kit?

every arduino "experiment kit" is different, but all you need is a couple of same value resistors (say 2* 1k - 2*10k) to provide a mid-point voltage; and I'd add a capacitor from mid-point to ground just to decouple;
and a 100k to protect the arduino ADC input.