Hall effect sensor
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Posted by lerameur on May 22, 2008, 10:25 am
Please log in for more thread options I hope I can clear up some confusion about reading specification sheet on hall sensor. First the UGN3503U chip. on the web page http://web-ee.com/schematics/instrumentation/build-gaussmeter/ the author calculates the Gauss with the following equation: The hall effect device has a sensitivity of 2.5mV/G which, when you invert it, is 1G/2.5mV or (multiplying top and bottom by 1000) 1000G/V. Now, you need to find out how many volts the output changed. It changed (2.48-4.56)V = -2.08V. So, we end up with B = -2.08V * 1000G/V = -2080G, and it is a South pole since it is negative. it makes sense so far because the spec sheet do not tell us anything about the maximum Gauss (voltage output) or could it be the maximum sensitivity ( 900Gauss). Now take a look at the ss495 chip , the spec sheet shows a graph from -640 to 640 Gauss, But from the calculation above we would get a swing of -1600 to +1600 Gauss. Do we choose the graph or the mV/Gauss approach ? which one is correct? K | ||||||||||||||||||||||||||||||||||
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Posted by Peter Bennett on May 22, 2008, 1:09 pm
Please log in for more thread options Math error - I make it 400 G/V, not 1000G/V. -- Peter Bennett, VE7CEI peterbb4 (at) interchange.ubc.ca GPS and NMEA info: http://vancouver-webpages.com/peter Vancouver Power Squadron: http://vancouver.powersquadron.ca | ||||||||||||||||||||||||||||||||||
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Posted by lerameur on May 22, 2008, 6:59 pm
Please log in for more thread options > On Thu, 22 May 2008 07:25:16 -0700 (PDT), lerameur
> > >Hello
>
> >I hope I can clear up some confusion about reading specification sheet
> >on hall sensor. > >First the UGN3503U chip. on the web page > >http://web-ee.com/schematics/instrumentation/build-gaussmeter/ > >the author calculates the Gauss with the following equation: >
> >The hall effect device has a sensitivity of 2.5mV/G which, when you
> > invert > >it, is 1G/2.5mV or (multiplying top and bottom by 1000) 1000G/V. >
> Math error - I make it 400 G/V, not 1000G/V. > > -- > Peter Bennett, VE7CEI > peterbb4 (at) interchange.ubc.ca > GPS and NMEA info:http://vancouver-webpages.com/peter > Vancouver Power Squadron:http://vancouver.powersquadron.ca thats nice, but if you do the same type of calculation on the ss495 chip, you get 1600G at 5v and they have a graph showing 640Gauss at 4.5 volt. That is where my confusion lies. ken | ||||||||||||||||||||||||||||||||||
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Posted by Mike on May 23, 2008, 9:06 am
Please log in for more thread options <snip>
>The hall effect device has a sensitivity of 2.5mV/G which, when you > invert >it, is 1G/2.5mV or (multiplying top and bottom by 1000) 1000G/V. >Now, you need to find out how many volts the output changed. >It changed (2.48-4.56)V = -2.08V. So, we end up with B = -2.08V * > 1000G/V = >-2080G, and it is a South pole since it is negative. > > >it makes sense so far because the spec sheet do not tell us anything >about the maximum Gauss (voltage output) or could it be the maximum >sensitivity ( 900Gauss). The UGN3503U datasheet I'm looking at specifies a sensitivity of 1.3mV/G over the range of +-900G If you exceed 900G the response will become nonlinear. -2.08V/1.3mV/G = 1600G. Which would be well above the linear range. >
The SS495 datasheet clearly specifies a range of +-670G with with a sensitivity
>Now take a look at the ss495 chip , the spec sheet shows a graph from >-640 to 640 Gauss, But from the calculation above we would get a swing >of -1600 to +1600 Gauss. >Do we choose the graph or the mV/Gauss approach ? which one is >correct? > of 3.125mV/G -2.08V/3.125mV/G = -665.6G Looking at the graph in the SS495 datasheet I see an output voltage of 4.5V at 640G, so (4.5-2.5)/640 = 3.125mV/G which is what the datasheet specifies for the sensitivity. The UGN3503U datasheet I saw doesn't have a graph like that. Neither of the sensors you mention has the required range to directly measure rare earth magnets. The field at the surface of those can easily exceed 900G. Take a look at the Honeywell SS94A2E if that's a problem. One of the Gaussmeters I used to make used a Honeywell sensor and a dual opamp to subtract the Vcc/2 offset and scale the output for direct readout on a panel meter. Another used an A/D with the math done in software. Either way works pretty well. Mike "In view of such harmony in the cosmos which I, with my limited human mind, am able to recognize, there are yet people who say there is no God. But what really makes me angry is that they quote me for the support of such views." Albert Einstein (theoretical Physicist) | ||||||||||||||||||||||||||||||||||
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Posted by lerameur on May 23, 2008, 11:23 am
Please log in for more thread options > <snip>
> > >The hall effect device has a sensitivity of 2.5mV/G which, when you
> > invert > >it, is 1G/2.5mV or (multiplying top and bottom by 1000) 1000G/V. > >Now, you need to find out how many volts the output changed. > >It changed (2.48-4.56)V = -2.08V. So, we end up with B = -2.08V * > > 1000G/V = > >-2080G, and it is a South pole since it is negative. >
> >it makes sense so far because the spec sheet do not tell us anything
> >about the maximum Gauss (voltage output) or could it be the maximum > >sensitivity ( 900Gauss). >
the range of +-900G
> The UGN3503U datasheet I'm looking at specifies a sensitivity of 1.3mV/G over > If you exceed 900G the response will become nonlinear.
> > -2.08V/1.3mV/G = 1600G. Which would be well above the linear range. > > > > >Now take a look at the ss495 chip , the spec sheet shows a graph from
> >-640 to 640 Gauss, But from the calculation above we would get a swing > >of -1600 to +1600 Gauss. > >Do we choose the graph or the mV/Gauss approach ? which one is > >correct? >
sensitivity of 3.125mV/G
> The SS495 datasheet clearly specifies a range of +-670G with with a > -2.08V/3.125mV/G = -665.6G
640G, so
> > Looking at the graph in the SS495 datasheet I see an output voltage of 4.5V at > (4.5-2.5)/640 = 3.125mV/G which is what the datasheet specifies for the
sensitivity.
>
rare earth magnets.
> The UGN3503U datasheet I saw doesn't have a graph like that. > > Neither of the sensors you mention has the required range to directly measure > The field at the surface of those can easily exceed 900G. Take a look at the
Honeywell SS94A2E
> if that's a problem.
to subtract
> > One of the Gaussmeters I used to make used a Honeywell sensor and a dual opamp > the Vcc/2 offset and scale the output for direct readout on a panel meter.
Another used
> an A/D with the math done in software. Either way works pretty well.
> > Mike > > "In view of such harmony in the cosmos which I, > with my limited human mind, am able to recognize, > there are yet people who say there is no God. > But what really makes me angry is that they quote > me for the support of such views." > Albert Einstein (theoretical Physicist) The Honeywell SS94A2E does not seem to be able to take more then 1000G Anybody know of a hall sensor that can read up to or about 10 000G ken | ||||||||||||||||||||||||||||||||||
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>
>I hope I can clear up some confusion about reading specification sheet
>on hall sensor.
>First the UGN3503U chip. on the web page
>http://web-ee.com/schematics/instrumentation/build-gaussmeter/
>the author calculates the Gauss with the following equation:
>
>The hall effect device has a sensitivity of 2.5mV/G which, when you
> invert
>it, is 1G/2.5mV or (multiplying top and bottom by 1000) 1000G/V.