# Hall Effect Magnetometer-Amplifier?

Discussion in 'The Projects Forum' started by levangram, Oct 31, 2009.

1. ### levangram Thread Starter New Member

Oct 31, 2009
8
0
Hi everybody!
I have a llittle problem...I'm relatively new to these projects,so i would appreciate any help!
What i'm trying to do is to build a simple magnetometer with a hall effect sensor.I use a digital panel meter and i want it to show me not the voltage output of the sensor,but the present magnetc field's flux density.The equation is B=(V-2.5)/2.5
So that for example if the voltage output that i count is 2.5,my indication to be 0.if my voltage is 5,the indicator to display 1 and so on.
How exactly can i converty my voltage in this way?Do i have to use an amplifier or something like this and if so,what's the way to do it?
Many thanks in advance,i would appreciate any help given,as i am a total newbie to these electronic things!

2. ### hgmjr Moderator

Jan 28, 2005
9,030
214
Have you found any potential circuits from your research on the Internet?

hgmjr

3. ### levangram Thread Starter New Member

Oct 31, 2009
8
0
No,unfortunately i haven't...After all,i don't really know what i'm looking for,as i'm really new to these things..I'm trying for days to get into it and find a possible solution,but i haven't..

4. ### steveb Senior Member

Jul 3, 2008
2,433
469
The simplest way to do this is to use a panel meter powered by 5 VDC and allowing a reference point different from ground. You are simply trying to measure voltage off of a 2.5 VDC reference point. You can use a simple 2.5 VDC voltage reference IC chip as the reference point. Or, a Hall sensor that is not in a field can serve as the voltage reference during testing.

Then you can use an inverting OPAMP amplifier stage, also referenced to the 2.5 VDC reference voltage (apply it to the noninverting input on the opamp). Choose the two resistors on the amplifier to give you a gain of -0.4. The output of the opamp and the 2.5 VDC reference voltage then feed into your panel meter.

The panel meter, OPAMP and Hall sensor will operate off the same 5 VDC supply, if the parts are chosen accordingly. So, this is very simple. You need the 5 VDC power source, a few filter caps for the power supply, a 2.5000 V voltage reference (and recommended circuitry), a panel meter, opamp with 2 resistors (with a ratio of 0.4, e.g. 13.3K and 33.2 K) and the Hall sensor.

If you like this approach, start drawing up a schematic, and seek advice as you go (i.e. choosing parts, correct circuit schematic etc).

5. ### levangram Thread Starter New Member

Oct 31, 2009
8
0
I got 1/2 parts of your solution..I realised that i have to use an opamp with 2 resistors with a ratio of 0.4 to invert my voltage..What i didn't realise is how to invert Vout-2.5.
Look..I have a 9Vdc digital panel meter and my hall sensor with its sensitivity at 2.5mV/Gauss(that's why i need V/2.5).The hall sensor has 2 leads,one for the common and one for the Vout.That's where i get Vout from.How exactly can i allow a reference point different from ground with my hall sensor,as you said?What do i have to connect to my digital panel meter's Vin?
Sorry if my questions are stupid...

6. ### hgmjr Moderator

Jan 28, 2005
9,030
214
It looks like if you are interfacing a .0025 V/Gauss analog Hall-Effect sensor to a 9V full-scale digital panel meter, you are going to require a gain that could reach as high 3600 depending on the magnitude of the magnetic field you are investigating. I'm not sure where you got the 0.4 gain you quoted.

How large of a magnetic field are you going to need to measure.

hgmjr

7. ### steveb Senior Member

Jul 3, 2008
2,433
469
Not a stupid question at all. Can you post any details about the panel meter. Part number and/or details from a manual would help. In particular it's important to know if the voltage reading need to be taken off of ground, or whether there is a two wire connection allowed. If it's the latter, you can place the two leads between the 2.5 V reference and the output of the opamp. If it's the first case, then the opamp circuit has to be modified to include a shift from 2.5 VDC down to ground reference.

8. ### steveb Senior Member

Jul 3, 2008
2,433
469
The 0.4 gain comes from the fact that he wants +/- 2.5 V to be converted to +/- 1.000 V. Note that one Gauss is a very small magnetic field. The Earth's magnetic field at the surface is about half a Gauss. He is most likely interested in measuring the larger fields of permanent magnets, which can approach 1 Tesla=10000 Gauss. Of course, he will be limited to plus or minus 1000 Gauss = 0.1 Tesla.

9. ### hgmjr Moderator

Jan 28, 2005
9,030
214
That makes sense. I was thinking that the magnetometer was aimed at registering much smaller magnetic fields such as might be inherent in the earth's magnetic field.

hgmjr

Last edited: Nov 2, 2009
10. ### levangram Thread Starter New Member

Oct 31, 2009
8
0
This is my panel meter
http://www.vellemanusa.com/us/enu/product/view/?id=350726
and i also attach tha back view of it.I have to put my voltage leads to + and - of Vin.So,what do i have to put in them so that my indication will be zero when it's actually 2.5V??
As far as the other questions are concerned,i want to measure magnetic fields up to 3-4KG,but that's not the point,i can use any gain so that i can say it works with a scale.The important thing is to make it work after a zero adjust so that it will display zero in a zero magnetic field..

File size:
106.9 KB
Views:
20
11. ### steveb Senior Member

Jul 3, 2008
2,433
469
OK, so if you put the negative on the output of a 2.500 VDC reference voltage IC and the positive on the output of the opamp, it should work. The 2.500 VDC reference voltage would be on the noninverting terminal of the OPAMP also.

The Hall effect sensor has an internal 2.500 voltage reference circuit in it. This is why it outputs 2.5 volts with no field. I'm suggesting that you use your own external 2.500 V reference IC chip as a reference point. Many different chips are available, but if you need help in selecting one, I'm sure someone here will recommend one.

12. ### levangram Thread Starter New Member

Oct 31, 2009
8
0
I don't have a 2.500 VDC reference voltage IC,so i managed to zero my field with the use of a pot.I connect the output of the sensor and the wiper of the pot to my indicator,and adjusting the pot,i actually zero my field.
What i'm trying now is to use the op amp so that my new output will be 0.4 of the original(original comes from the output of the sensor and the wiper of the pot).I have a 741,but i dont know which wire goes where!!
a)which wire goes to op amp -?
b)which wire goes to op amp +?
c)does an op amp needs to be supllied to work?And if so,with what voltage?

13. ### steveb Senior Member

Jul 3, 2008
2,433
469
I don't fully understand how you zeroed out your meter, but I'm concerned that you may not have done it in a way that will provide linearity and accuracy. The opamp does need power, and you can't just use the opamp with no additional circuitry. It really sounds like you are new at this; - no offense intended.

There is a limt to how much input I can provide right now, since I'm tied up with a few things. However, I slapped together a quick schematic and recommended parts for this. There are a number of ways to approach it and many different parts could be used. So, this is just a representation of what my first attempt would be, if it were my project.

Hopefully, other people can improve on this by correcting any mistakes, or recommending parts that are more commonly available. The schmatic should make it clear how the meter needs to be hooked up to the reference voltage and the output of the opamp, just as I recommended before.

File size:
30.2 KB
Views:
42
File size:
228.9 KB
Views:
25
File size:
234.6 KB
Views:
31
File size:
292.1 KB
Views:
25
File size:
229.8 KB
Views:
21
14. ### levangram Thread Starter New Member

Oct 31, 2009
8
0
This is my schematic.at your right is the digital panel meter.Adjusting the pot,gives me zero voltage at zero magnetic field.if i leave it this way,it will show me the voltage difference under the presence of a magnetic field.so i need to multiply it with 0.4 before getting into the panel meter..

File size:
15.6 KB
Views:
31
15. ### steveb Senior Member

Jul 3, 2008
2,433
469
OK, I think I understand what you are doing. It looks like you are powering your hall sensor with 4.5 V? This seems low to me, but it really depends on the sensor. I would think you want at least 5 V to have equal plus and minus range of 2.5V.

Another issue is that you are calibrating/zeroing off of the supply voltage which is not very accurate. Any change in voltage due to loading, or heating will throw the calibration off. This is why a voltage reference should be used. The Hall sensor has its own voltage reference internally set to 2.5 VDC for accuracy.

These are just recommendations. The complexity of implementation depends on your desired perfomance specificaitions and on the quality of the power supply you are depending on for calibration.

16. ### levangram Thread Starter New Member

Oct 31, 2009
8
0
The sensor characteristics refer to supply voltage from 4.5-12V.i use 4.5V because that's the battery that i have plenty of at home!
I zero off the suplly voltage with a pot,because actually null voltage at zero magnetic field is 2.38-2.4V,not 2.5 exactly.I thought i could to it with more accuracy and that it would be more flexible at the adjustment.
After all,i'm just working on hall effect theoretically.I build this magnetometer with the sensor just for my own use,thought it would be nice to try something like this!
So,to sum up..With my way if i put
-the hall output then R1 in Vin+ of the op amp
-the wiper of the pot in Vin- of the op amp
-the output of the op amp at V+ of the digital panel meter
-the wiper at V- of the digital panel meter
and supply the op amp with 9V....I think it's ok!(??)

17. ### steveb Senior Member

Jul 3, 2008
2,433
469
OK, good luck and have fun. The best way to learn is hands-on.

18. ### levangram Thread Starter New Member

Oct 31, 2009
8
0
Thanks a lot steveb,you've been of great help for me!
One last question if you don't mind.The op amp supply of 9V is ok or do i have to use a different suplly?

19. ### steveb Senior Member

Jul 3, 2008
2,433
469
It all depends on the OPAMP you choose. The one I recommeded works from 1.8V to 24 V. You need a rail to rail opamp in this application, and you should carefully review the datasheet of the part you choose.

There are potential issues running directly off the battery with no regulation, but that gets into higher level concepts. If you use a power supply filter capacitor near the chip, you should be ok given the simplicity of the circuit. If you end up seeing oscillations, then report back and someone will guide you further.