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Making a Hall effect sensor into a PCB
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This question is related to Is it possible to “trace†a hall effect sensor on a PCB?.
As far as I understand it, a Hall effect sensor consists of a metal plate, plus a current source and an amplifier.
Instead of buying a Hall effect sensing chip, would it be possible to implement this with a PCB and a microcontroller? A metal plate is easy to come by within a PCB, and current sources, amplifiers and ADCs are available in some microcontrollers (e.g. PSoC4)
Applications for this might include current sensing in a power supply, or position sensing of a magnet.
One drawback might be that the size of the plate would have a poor tolerance. Are there any other reasons this wouldn't be possible?
microcontroller analog current-measurement hall-effect
asked Aug 22 at 14:04
Rocketmagnet
20k1164130
add a comment |Â
up vote
6
down vote
favorite
This question is related to Is it possible to “trace†a hall effect sensor on a PCB?.
As far as I understand it, a Hall effect sensor consists of a metal plate, plus a current source and an amplifier.
Instead of buying a Hall effect sensing chip, would it be possible to implement this with a PCB and a microcontroller? A metal plate is easy to come by within a PCB, and current sources, amplifiers and ADCs are available in some microcontrollers (e.g. PSoC4)
Applications for this might include current sensing in a power supply, or position sensing of a magnet.
One drawback might be that the size of the plate would have a poor tolerance. Are there any other reasons this wouldn't be possible?
microcontroller analog current-measurement hall-effect
asked Aug 22 at 14:04
Rocketmagnet
20k1164130
Can you edit your question to include the current, magnetic field strength and the Hall voltage you would expect across your trace for your application?
– Transistor
Aug 22 at 14:09
@Transistor - I didn't have a specific range in mind for these values. The idea came to me a few minutes ago, while reading an article about low cost current sensing in power supplies. I tried to find anyone else who had done this, and found nobody. I just wondered why.
– Rocketmagnet
Aug 22 at 14:19
3
I suspect that if you plug some numbers in to the Hall formula you might come out with very small Hall voltages and this may give you a clue to why this is not practical - if that's the case. I don't know because I've never done it. I'd be interested to see your calculations.
– Transistor
Aug 22 at 14:21
1
I haven't done the calculations, but I have read about the discovery of the hall effect. The voltages you will be looking at are extremely small. The thickness of the "metal plate" also plays a role - to the point that the first experiments used gold leaf to get the "plate" thin enough for the currents and voltages to be large enough to measure.
– JRE
Aug 22 at 14:35
@JRE - Yup, I did assume they would be very small voltages. However, they aren't outside the possibility of amplification, since the Hall chips seem to manage. But the thickness is something I didn't know before.
– Rocketmagnet
Aug 22 at 14:39
add a comment |Â
up vote
6
down vote
favorite
up vote
6
down vote
favorite
This question is related to Is it possible to “trace†a hall effect sensor on a PCB?.
As far as I understand it, a Hall effect sensor consists of a metal plate, plus a current source and an amplifier.
Instead of buying a Hall effect sensing chip, would it be possible to implement this with a PCB and a microcontroller? A metal plate is easy to come by within a PCB, and current sources, amplifiers and ADCs are available in some microcontrollers (e.g. PSoC4)
Applications for this might include current sensing in a power supply, or position sensing of a magnet.
One drawback might be that the size of the plate would have a poor tolerance. Are there any other reasons this wouldn't be possible?
microcontroller analog current-measurement hall-effect
asked Aug 22 at 14:04
Rocketmagnet
20k1164130
This question is related to Is it possible to “trace†a hall effect sensor on a PCB?.
As far as I understand it, a Hall effect sensor consists of a metal plate, plus a current source and an amplifier.
Instead of buying a Hall effect sensing chip, would it be possible to implement this with a PCB and a microcontroller? A metal plate is easy to come by within a PCB, and current sources, amplifiers and ADCs are available in some microcontrollers (e.g. PSoC4)
Applications for this might include current sensing in a power supply, or position sensing of a magnet.
One drawback might be that the size of the plate would have a poor tolerance. Are there any other reasons this wouldn't be possible?
microcontroller analog current-measurement hall-effect
asked Aug 22 at 14:04
Rocketmagnet
20k1164130
asked Aug 22 at 14:04
Rocketmagnet
20k1164130
asked Aug 22 at 14:04
Rocketmagnet
20k1164130
asked Aug 22 at 14:04
Rocketmagnet
20k1164130
20k1164130
Can you edit your question to include the current, magnetic field strength and the Hall voltage you would expect across your trace for your application?
– Transistor
Aug 22 at 14:09
@Transistor - I didn't have a specific range in mind for these values. The idea came to me a few minutes ago, while reading an article about low cost current sensing in power supplies. I tried to find anyone else who had done this, and found nobody. I just wondered why.
– Rocketmagnet
Aug 22 at 14:19
3
I suspect that if you plug some numbers in to the Hall formula you might come out with very small Hall voltages and this may give you a clue to why this is not practical - if that's the case. I don't know because I've never done it. I'd be interested to see your calculations.
– Transistor
Aug 22 at 14:21
1
I haven't done the calculations, but I have read about the discovery of the hall effect. The voltages you will be looking at are extremely small. The thickness of the "metal plate" also plays a role - to the point that the first experiments used gold leaf to get the "plate" thin enough for the currents and voltages to be large enough to measure.
– JRE
Aug 22 at 14:35
@JRE - Yup, I did assume they would be very small voltages. However, they aren't outside the possibility of amplification, since the Hall chips seem to manage. But the thickness is something I didn't know before.
– Rocketmagnet
Aug 22 at 14:39
add a comment |Â
Can you edit your question to include the current, magnetic field strength and the Hall voltage you would expect across your trace for your application?
– Transistor
Aug 22 at 14:09
@Transistor - I didn't have a specific range in mind for these values. The idea came to me a few minutes ago, while reading an article about low cost current sensing in power supplies. I tried to find anyone else who had done this, and found nobody. I just wondered why.
– Rocketmagnet
Aug 22 at 14:19
3
I suspect that if you plug some numbers in to the Hall formula you might come out with very small Hall voltages and this may give you a clue to why this is not practical - if that's the case. I don't know because I've never done it. I'd be interested to see your calculations.
– Transistor
Aug 22 at 14:21
1
I haven't done the calculations, but I have read about the discovery of the hall effect. The voltages you will be looking at are extremely small. The thickness of the "metal plate" also plays a role - to the point that the first experiments used gold leaf to get the "plate" thin enough for the currents and voltages to be large enough to measure.
– JRE
Aug 22 at 14:35
@JRE - Yup, I did assume they would be very small voltages. However, they aren't outside the possibility of amplification, since the Hall chips seem to manage. But the thickness is something I didn't know before.
– Rocketmagnet
Aug 22 at 14:39
Can you edit your question to include the current, magnetic field strength and the Hall voltage you would expect across your trace for your application?
– Transistor
Aug 22 at 14:09
Can you edit your question to include the current, magnetic field strength and the Hall voltage you would expect across your trace for your application?
– Transistor
Aug 22 at 14:09
@Transistor - I didn't have a specific range in mind for these values. The idea came to me a few minutes ago, while reading an article about low cost current sensing in power supplies. I tried to find anyone else who had done this, and found nobody. I just wondered why.
– Rocketmagnet
Aug 22 at 14:19
@Transistor - I didn't have a specific range in mind for these values. The idea came to me a few minutes ago, while reading an article about low cost current sensing in power supplies. I tried to find anyone else who had done this, and found nobody. I just wondered why.
– Rocketmagnet
Aug 22 at 14:19
3
3
I suspect that if you plug some numbers in to the Hall formula you might come out with very small Hall voltages and this may give you a clue to why this is not practical - if that's the case. I don't know because I've never done it. I'd be interested to see your calculations.
– Transistor
Aug 22 at 14:21
I suspect that if you plug some numbers in to the Hall formula you might come out with very small Hall voltages and this may give you a clue to why this is not practical - if that's the case. I don't know because I've never done it. I'd be interested to see your calculations.
– Transistor
Aug 22 at 14:21
1
1
I haven't done the calculations, but I have read about the discovery of the hall effect. The voltages you will be looking at are extremely small. The thickness of the "metal plate" also plays a role - to the point that the first experiments used gold leaf to get the "plate" thin enough for the currents and voltages to be large enough to measure.
– JRE
Aug 22 at 14:35
I haven't done the calculations, but I have read about the discovery of the hall effect. The voltages you will be looking at are extremely small. The thickness of the "metal plate" also plays a role - to the point that the first experiments used gold leaf to get the "plate" thin enough for the currents and voltages to be large enough to measure.
– JRE
Aug 22 at 14:35
@JRE - Yup, I did assume they would be very small voltages. However, they aren't outside the possibility of amplification, since the Hall chips seem to manage. But the thickness is something I didn't know before.
– Rocketmagnet
Aug 22 at 14:39
@JRE - Yup, I did assume they would be very small voltages. However, they aren't outside the possibility of amplification, since the Hall chips seem to manage. But the thickness is something I didn't know before.
– Rocketmagnet
Aug 22 at 14:39
add a comment |Â
1 Answer
1
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up vote
11
down vote
I did this very experiment in physics class, upper school. We used the thinnest aluminium foil available, with a very significant current, in the 1A ballpark, and a very strong magnet. The actual voltage generated was very small, barely detectable.
Commercial Hall elements use semiconductors, which have carrier densities orders of magnitude lower than that of metals, so have a Hall coefficient orders of magnitude greater.
To deal with the poor sensitivity, you would have to use a high current, a thin foil, and either a chopper/zero drift amplifier for DC excitation, or AC excitation and synchronous demodulation. It's not impossible, just very hard, and probably more expensive (to get the same sensitivity) than buying a commercial Hall IC and mounting that on the board.
answered Aug 22 at 14:42
Neil_UK
69.2k272152
add a comment |Â
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
11
down vote
I did this very experiment in physics class, upper school. We used the thinnest aluminium foil available, with a very significant current, in the 1A ballpark, and a very strong magnet. The actual voltage generated was very small, barely detectable.
Commercial Hall elements use semiconductors, which have carrier densities orders of magnitude lower than that of metals, so have a Hall coefficient orders of magnitude greater.
To deal with the poor sensitivity, you would have to use a high current, a thin foil, and either a chopper/zero drift amplifier for DC excitation, or AC excitation and synchronous demodulation. It's not impossible, just very hard, and probably more expensive (to get the same sensitivity) than buying a commercial Hall IC and mounting that on the board.
answered Aug 22 at 14:42
Neil_UK
69.2k272152
add a comment |Â
up vote
11
down vote
I did this very experiment in physics class, upper school. We used the thinnest aluminium foil available, with a very significant current, in the 1A ballpark, and a very strong magnet. The actual voltage generated was very small, barely detectable.
Commercial Hall elements use semiconductors, which have carrier densities orders of magnitude lower than that of metals, so have a Hall coefficient orders of magnitude greater.
To deal with the poor sensitivity, you would have to use a high current, a thin foil, and either a chopper/zero drift amplifier for DC excitation, or AC excitation and synchronous demodulation. It's not impossible, just very hard, and probably more expensive (to get the same sensitivity) than buying a commercial Hall IC and mounting that on the board.
answered Aug 22 at 14:42
Neil_UK
69.2k272152
add a comment |Â
up vote
11
down vote
up vote
11
down vote
I did this very experiment in physics class, upper school. We used the thinnest aluminium foil available, with a very significant current, in the 1A ballpark, and a very strong magnet. The actual voltage generated was very small, barely detectable.
Commercial Hall elements use semiconductors, which have carrier densities orders of magnitude lower than that of metals, so have a Hall coefficient orders of magnitude greater.
To deal with the poor sensitivity, you would have to use a high current, a thin foil, and either a chopper/zero drift amplifier for DC excitation, or AC excitation and synchronous demodulation. It's not impossible, just very hard, and probably more expensive (to get the same sensitivity) than buying a commercial Hall IC and mounting that on the board.
answered Aug 22 at 14:42
Neil_UK
69.2k272152
I did this very experiment in physics class, upper school. We used the thinnest aluminium foil available, with a very significant current, in the 1A ballpark, and a very strong magnet. The actual voltage generated was very small, barely detectable.
Commercial Hall elements use semiconductors, which have carrier densities orders of magnitude lower than that of metals, so have a Hall coefficient orders of magnitude greater.
To deal with the poor sensitivity, you would have to use a high current, a thin foil, and either a chopper/zero drift amplifier for DC excitation, or AC excitation and synchronous demodulation. It's not impossible, just very hard, and probably more expensive (to get the same sensitivity) than buying a commercial Hall IC and mounting that on the board.
answered Aug 22 at 14:42
Neil_UK
69.2k272152
answered Aug 22 at 14:42
Neil_UK
69.2k272152
answered Aug 22 at 14:42
Neil_UK
69.2k272152
answered Aug 22 at 14:42
Neil_UK
69.2k272152
69.2k272152
add a comment |Â
add a comment |Â
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Can you edit your question to include the current, magnetic field strength and the Hall voltage you would expect across your trace for your application?
– Transistor
Aug 22 at 14:09
@Transistor - I didn't have a specific range in mind for these values. The idea came to me a few minutes ago, while reading an article about low cost current sensing in power supplies. I tried to find anyone else who had done this, and found nobody. I just wondered why.
– Rocketmagnet
Aug 22 at 14:19
3
I suspect that if you plug some numbers in to the Hall formula you might come out with very small Hall voltages and this may give you a clue to why this is not practical - if that's the case. I don't know because I've never done it. I'd be interested to see your calculations.
– Transistor
Aug 22 at 14:21
1
I haven't done the calculations, but I have read about the discovery of the hall effect. The voltages you will be looking at are extremely small. The thickness of the "metal plate" also plays a role - to the point that the first experiments used gold leaf to get the "plate" thin enough for the currents and voltages to be large enough to measure.
– JRE
Aug 22 at 14:35
@JRE - Yup, I did assume they would be very small voltages. However, they aren't outside the possibility of amplification, since the Hall chips seem to manage. But the thickness is something I didn't know before.
– Rocketmagnet
Aug 22 at 14:39