Hello, is it possible to distinguish north and south pole of a magnet with the HCM5883L. I would be very happy about Ielfe. I'm not getting anywhere with my project. a reverb doesn't have enough distance, that's why I wanted to try the HMC5883l.
Hi loved the video! can you help me with something? I´m trying to add a gauss meter to other device and this was a lot of help. If I only need to measure gauss I could use one A1308 sensor right? but still I will need the ATMEGA328 and an arduino to burn the bootloader and program?
No, the A1308 outputs a voltage based on the field it's reading. 0 Gauss is 2.5 volts and min/max is 0.5/4.5. Can't remember off the top of my head what the mV/G is, but the datasheet has that info, but if you know the output voltage of the sensor you know the field strength. The arduino is just to make a nice display
@@HyperspacePirate oh thanks so much for your time. With that understanding I can display those volts then and in the near future update it with all the Arduino part.
The smallest readily available sensor IC's I can find at major suppliers seem to bottom out around +/- 5 mT, which is +/- 50 gauss. With a 16 bit ADC on the sensor, that comes out to a resolution of 0.0015 gauss (1.5 milliGauss). I'm sure there's scientific / laboratory grade sensors out there which can do much higher resolution, but at a certain point, the background field would completely override whatever you're trying to measure. Earth's magnetic field (which is very weak) averages around 0.5 gauss, which is 10 million times more than 50 nG.
@@HyperspacePirate thank you for the answer. I want to ask something; Is it possible to detect a magnetic flux coming from one direction only, parallel to the sensor? I mean a perception that has no effect of magnetic flux coming from the sides.
May I ask how you calculated the h value? Is it the sum of 3 axis? That's why I asked this. As you know, the size of the earth's magnetic field is known as 50 micro tesla on average. Especially in the compass application, which is one of the mobile phone applications, this value is shown in a corner. What I am wondering is: For example, what formula is used to reach this value, which is 47 micro tesla in my region?
The value is whatever magnetic flux intensity is normal to the face of the hall effect sensor. I guess in theory the same result could be gotten from the 3-axis mode with sqrt(h_x^2 + h_y^2 + h_z^2). I don't think the sensors im using are sensitive enough to detect the earth's field, though. Their sensitivity range is more suited to big permanent magnets (like >50 gauss)
@@HyperspacePirate Thank you for the answer. I did this experiment with hal effect sensors, but I couldn't measure the earth's magnetic field size. So I decided to buy a module. The LIS3MDL module is sensitive enough to do this job. However, the formula you gave above; Does sqrt(h_x^2 + h_y^2 + h_z^2) normally calculate earth magnetic field magnitude? Can you explain the formula a little bit?
Do you truly think an N52 magnet has a 1.4 Tesla magnetic field? Have you tested your "gauss meter" against the one built directly into you phone? Try a gauss meter app for free and compare.
Nominally, an N52 has an "internal" magnetic field of ~1.4 Tesla. But yeah, if you measured exactly on the surface, you'd get more like 0.5-0.7, and that would drop drastically as you moved a few mm away
Dear algorithm, this guy right here. I want more content like this. And so does everyone else.
Nicely job done
Phenomenal work
Can you tell me how strong the field is now, over canada?
Hello, is it possible to distinguish north and south pole of a magnet with the HCM5883L. I would be very happy about Ielfe. I'm not getting anywhere with my project. a reverb doesn't have enough distance, that's why I wanted to try the HMC5883l.
great video
Hi loved the video! can you help me with something? I´m trying to add a gauss meter to other device and this was a lot of help. If I only need to measure gauss I could use one A1308 sensor right? but still I will need the ATMEGA328 and an arduino to burn the bootloader and program?
No, the A1308 outputs a voltage based on the field it's reading. 0 Gauss is 2.5 volts and min/max is 0.5/4.5. Can't remember off the top of my head what the mV/G is, but the datasheet has that info, but if you know the output voltage of the sensor you know the field strength. The arduino is just to make a nice display
@@HyperspacePirate oh thanks so much for your time. With that understanding I can display those volts then and in the near future update it with all the Arduino part.
What is the range at which magnet will be detected, can we detect magnets upto 1 m away
What is the smallest detectable magnetic field size in gauss?
I've heard of a sensor that can detect 50 nanogauss.
The smallest readily available sensor IC's I can find at major suppliers seem to bottom out around +/- 5 mT, which is +/- 50 gauss. With a 16 bit ADC on the sensor, that comes out to a resolution of 0.0015 gauss (1.5 milliGauss).
I'm sure there's scientific / laboratory grade sensors out there which can do much higher resolution, but at a certain point, the background field would completely override whatever you're trying to measure. Earth's magnetic field (which is very weak) averages around 0.5 gauss, which is 10 million times more than 50 nG.
@@HyperspacePirate thank you for the answer. I want to ask something; Is it possible to detect a magnetic flux coming from one direction only, parallel to the sensor? I mean a perception that has no effect of magnetic flux coming from the sides.
@@ahmetozdemir7173 Yes. Almost all hall effect sensors only sense one axis. That's why magnetometers usually have 3 sensors.
Hello, it is very nice project, i like to make it, can i have the sketch code please? Thank you
May I ask how you calculated the h value? Is it the sum of 3 axis? That's why I asked this. As you know, the size of the earth's magnetic field is known as 50 micro tesla on average.
Especially in the compass application, which is one of the mobile phone applications, this value is shown in a corner. What I am wondering is: For example, what formula is used to reach this value, which is 47 micro tesla in my region?
The value is whatever magnetic flux intensity is normal to the face of the hall effect sensor. I guess in theory the same result could be gotten from the 3-axis mode with sqrt(h_x^2 + h_y^2 + h_z^2). I don't think the sensors im using are sensitive enough to detect the earth's field, though. Their sensitivity range is more suited to big permanent magnets (like >50 gauss)
@@HyperspacePirate Thank you for the answer. I did this experiment with hal effect sensors, but I couldn't measure the earth's magnetic field size. So I decided to buy a module. The LIS3MDL module is sensitive enough to do this job. However, the formula you gave above; Does sqrt(h_x^2 + h_y^2 + h_z^2) normally calculate earth magnetic field magnitude? Can you explain the formula a little bit?
Do you truly think an N52 magnet has a 1.4 Tesla magnetic field? Have you tested your "gauss meter" against the one built directly into you phone? Try a gauss meter app for free and compare.
Nominally, an N52 has an "internal" magnetic field of ~1.4 Tesla. But yeah, if you measured exactly on the surface, you'd get more like 0.5-0.7, and that would drop drastically as you moved a few mm away
@HyperspacePirate >>> 👍👍