Help - Hall Effect Sensor

[John Burns]

Hi there,
I am new to this hall effect stuff - I am trying to build an inductive
pickup for a lap timer I am building for a friends gokart.

I am trying to put together a circuit which uses a hall effect sensor and
outputs a pulse whenever a magnet is detected....regardless of the magnets
pole.

I see that the Hall effect sensors put out a normal (no magnetic field
detected) voltage of 2.5V and this will change by about +/- 1.17 volts for
the extremes.

My plan was to supply the sensor with a regulated 5v supply, then use a
voltage divider to produce a voltage of 1.5V to compare the sensor ouput
with.
The output from the sensor is separated to two separate transistors which
one detects voltages >2.9V and the other voltages <2.1V - The two outputs
are combined and fed into an optoisolator to then be connected to a PIC.
What I want to know, is if I am making this too technical and there is a
simpler way to do it. I do not want to use the ADC built into the pic as I
want to have the chip do other things and get interrupted for the pulse.

Any ideas?

-John

 

[Phil Allison]

"John Burns" news:c65j96$5rn$1@lust.ihug.co.nz...

Hi there,

** Baaaaaahhhhhhhhhhhh.......

I am new to this hall effect stuff - I am trying to build an inductive
pickup for a lap timer I am building for a friends gokart.

** That reads like a self contradiction.

I am trying to put together a circuit which uses a hall effect sensor and
outputs a pulse whenever a magnet is detected....regardless of the magnets
pole.

** The range of a permanent magnet is a few inches, at best - a HE
sensor is tiny.

Just how is your Go Kart detecting idea supposed to work ?????







............ Phil

 

[John Tserkezis]

John Burns wrote:

The output from the sensor is separated to two separate transistors which
one detects voltages >2.9V and the other voltages <2.1V - The two outputs
are combined and fed into an optoisolator to then be connected to a PIC.
What I want to know, is if I am making this too technical and there is a
simpler way to do it. I do not want to use the ADC built into the pic as I
want to have the chip do other things and get interrupted for the pulse.

Any specific reason you're using a hall effect device?

You don't mention if it's used for speed/distance monitoring, or angular
detection/measurement.

If you're just counting pulses, a simple inductive coil pickup and associated
preamp circuitry would be a lot cheaper and simpler.

--
Linux Registered User # 302622 <http://counter.li.org>

 

[John Burns]

"John Tserkezis" <jt@techniciansyndrome.org.invalid> wrote in message
news:408666a4$0$6018$afc38c87@news.optusnet.com.au...

John Burns wrote:

The output from the sensor is separated to two separate transistors
which
one detects voltages >2.9V and the other voltages <2.1V - The two
outputs
are combined and fed into an optoisolator to then be connected to a PIC.
What I want to know, is if I am making this too technical and there is a
simpler way to do it. I do not want to use the ADC built into the pic
as I
want to have the chip do other things and get interrupted for the pulse.

Any specific reason you're using a hall effect device?

You don't mention if it's used for speed/distance monitoring, or angular
detection/measurement.

If you're just counting pulses, a simple inductive coil pickup and
associated
preamp circuitry would be a lot cheaper and simpler.

--
Linux Registered User # 302622
http://counter.li.org

A little more detail for you....
The tracks around new zealand have a magnet built into the track - its
either a permanent magnet or an electromagnet. It runs under the start
finish line so that racers can keep track of their lap times via on board
equipment.

The only reason I am using a hall effect device is that I have read the
specs for a commercial version of the product that says it uses 'high
sensitivity hall effect magnetic sensor is placed on the floor of the kart'.

All I want is a pulse that is high when a magnetic field is detected -
regardless of North or South - I believe the magnet is about 800Gauss and
will need to be detected through about 3 inches (1 inch of concrete/seal, 2
inches of air)

Am I barking up the wrong or right tree?

Cheers

-John

 

[John Tserkezis]

John Burns wrote:

A little more detail for you....
The tracks around new zealand have a magnet built into the track - its
either a permanent magnet or an electromagnet. It runs under the start
finish line so that racers can keep track of their lap times via on board
equipment.

The only reason I am using a hall effect device is that I have read the
specs for a commercial version of the product that says it uses 'high
sensitivity hall effect magnetic sensor is placed on the floor of the kart'.

All I want is a pulse that is high when a magnetic field is detected -
regardless of North or South - I believe the magnet is about 800Gauss and
will need to be detected through about 3 inches (1 inch of concrete/seal, 2
inches of air)

Am I barking up the wrong or right tree?

No, a coil is just another option.

As it turns out, with a coil, you don't have to worry about the polarity of
the field, because if it doesn't trigger on the rising edge, it'll trigger on
the falling edge.
For your purpose, it won't matter, because I'll be the same every time you go
around the track (since you are only ever travelling in one direction).

With hall effect, you can use a capacitor in series with the output, which
will get you a similar result, but you're basically getting an expensive sensor
to do what a bit of wire can do. And it's heavier on the power consumption too.

I would advise to position the sensor as close to the ground as possible,
nearer to one of the front wheels, because if your tracks and driving habits
are anything like ours, you'll bottom out and tear the sensor off in the first
lap.
Also a good idea to say away from the rear to prevent false triggers from the
engine or magneto.


--
Linux Registered User # 302622 <http://counter.li.org>

 

[Steve]

All I want is a pulse that is high when a magnetic field is detected -
regardless of North or South - I believe the magnet is about 800Gauss and
will need to be detected through about 3 inches (1 inch of concrete/seal, 2
inches of air)

Well, stop asking us for a start.
Better to make up a mock up circuit (No PIC,
just the detector - actually, mostly just the hall
effect and a meter) and take it out to the track.

Then work out how to get the same effect (ie
meter reading) with a small magnet held closer
to the hall effect, - for your bench testing.

All my opinion, but worth atleast as much or little
as anything else you'll hear in this newsgroup

Regards
Steve
www.airborn.com.au

 

[Adrian Jansen]

Personally I would use Hall sensors in this application. Advantages are
that they sense at essentially zero speed, where an inductive pickup fails
at some minimum speed. But you DO need a fixed polarity magnet. The
commercial switching output sensors are biased to only detect one polarity.
Its usually pretty easy to ensure the polarity is correct when installing
the magnets. Cost difference between Hall sensors and inductive pickups
would be marginal, for any app like this. You will need sensors and magnets
suitable for the application. 3 inch sensing distance is fairly large. As
a very rough guide, you need a magnet of the same order of length as the
sensing distance.

You should not need any significant interface circuitry if you use sensors
with open collector outputs, just a pullup resistor, and possibly a spike
suppression capacitor, feeding the processor port inputs directly.

--
Regards,

Adrian Jansen
J & K MicroSystems
Microcomputer solutions for industrial control
"John Tserkezis" <jt@techniciansyndrome.org.invalid> wrote in message
news:4086ecf5$0$4547$afc38c87@news.optusnet.com.au...

John Burns wrote:

A little more detail for you....
The tracks around new zealand have a magnet built into the track - its
either a permanent magnet or an electromagnet. It runs under the start
finish line so that racers can keep track of their lap times via on
board
equipment.

The only reason I am using a hall effect device is that I have read the
specs for a commercial version of the product that says it uses 'high
sensitivity hall effect magnetic sensor is placed on the floor of the
kart'.

All I want is a pulse that is high when a magnetic field is detected -
regardless of North or South - I believe the magnet is about 800Gauss
and
will need to be detected through about 3 inches (1 inch of
concrete/seal, 2
inches of air)

Am I barking up the wrong or right tree?

No, a coil is just another option.

As it turns out, with a coil, you don't have to worry about the polarity
of
the field, because if it doesn't trigger on the rising edge, it'll trigger
on
the falling edge.
For your purpose, it won't matter, because I'll be the same every time
you go
around the track (since you are only ever travelling in one direction).

With hall effect, you can use a capacitor in series with the output,
which
will get you a similar result, but you're basically getting an expensive
sensor
to do what a bit of wire can do. And it's heavier on the power
consumption too.

I would advise to position the sensor as close to the ground as
possible,
nearer to one of the front wheels, because if your tracks and driving
habits
are anything like ours, you'll bottom out and tear the sensor off in the
first
lap.
Also a good idea to say away from the rear to prevent false triggers
from the
engine or magneto.


--
Linux Registered User # 302622
http://counter.li.org

 

[Phil Allison]

"John Tserkezis" <

John Burns wrote:

Am I barking up the wrong or right tree?

No, a coil is just another option.

As it turns out, with a coil, you don't have to worry about the polarity
of
the field, because if it doesn't trigger on the rising edge, it'll trigger
on
the falling edge.

** If the Kart is moving slowly it may never trigger at all - unlike
with a Hall Effect device where it reacts to a stationary magnetic field.

With hall effect, you can use a capacitor in series with the output,
which
will get you a similar result, but you're basically getting an expensive
sensor
to do what a bit of wire can do.

** Try to get a bit of wire to output a steady voltage when placed near a
magnet.




............ Phil

 

[The Real Andy]

On Thu, 22 Apr 2004 08:32:07 +1000, "Adrian Jansen"
<qqv@noqqwhere.com> wrote:

Personally I would use Hall sensors in this application. Advantages are
that they sense at essentially zero speed, where an inductive pickup fails
at some minimum speed. But you DO need a fixed polarity magnet. The
commercial switching output sensors are biased to only detect one polarity.
Its usually pretty easy to ensure the polarity is correct when installing
the magnets. Cost difference between Hall sensors and inductive pickups
would be marginal, for any app like this. You will need sensors and magnets
suitable for the application. 3 inch sensing distance is fairly large. As
a very rough guide, you need a magnet of the same order of length as the
sensing distance.

You should not need any significant interface circuitry if you use sensors
with open collector outputs, just a pullup resistor, and possibly a spike
suppression capacitor, feeding the processor port inputs directly.

To get around pole orientation you could use more than one sensor?

 

[Franc Zabkar]

On Thu, 22 Apr 2004 08:32:07 +1000, "Adrian Jansen"
<qqv@noqqwhere.com> put finger to keyboard and composed:

Personally I would use Hall sensors in this application. Advantages are
that they sense at essentially zero speed, where an inductive pickup fails
at some minimum speed. But you DO need a fixed polarity magnet. The
commercial switching output sensors are biased to only detect one polarity.

Dick Smith sell the UGN3503U for $4.48. It's a linear device with a
sensitivity of ~1.3mV/G and is specified for 0G to +/- 900G. The
quiescent output at 0G is 2.50V.

Its usually pretty easy to ensure the polarity is correct when installing
the magnets. Cost difference between Hall sensors and inductive pickups
would be marginal, for any app like this. You will need sensors and magnets
suitable for the application. 3 inch sensing distance is fairly large. As
a very rough guide, you need a magnet of the same order of length as the
sensing distance.

You should not need any significant interface circuitry if you use sensors
with open collector outputs, just a pullup resistor, and possibly a spike
suppression capacitor, feeding the processor port inputs directly.

- Franc Zabkar
--
Please remove one 's' from my address when replying by email.

 

[John Burns]

"Adrian Jansen" <qqv@noqqwhere.com> wrote in message
news:40870f77$1@duster.adelaide.on.net...

Personally I would use Hall sensors in this application. Advantages are
that they sense at essentially zero speed, where an inductive pickup fails
at some minimum speed. But you DO need a fixed polarity magnet. The
commercial switching output sensors are biased to only detect one
polarity.
Its usually pretty easy to ensure the polarity is correct when installing
the magnets. Cost difference between Hall sensors and inductive pickups
would be marginal, for any app like this. You will need sensors and
magnets
suitable for the application. 3 inch sensing distance is fairly large.
As
a very rough guide, you need a magnet of the same order of length as the
sensing distance.

You should not need any significant interface circuitry if you use sensors
with open collector outputs, just a pullup resistor, and possibly a spike
suppression capacitor, feeding the processor port inputs directly.

I went out to the gokart track today with regulator, hall effect sensor and
multimeter in hand.
I have a quisient (spelling?) voltage of 2.55V. As the hall effect sensor
passes over the magenet in the track (at 3" distance away) the voltage
output from the sensor goes up to 2.69v and then down to 2.4v then up to
2.69v again before settling back to 2.55v - I assume the magnet in the track
has its North/South poles in the same direction as the karts travel on the
track.
The magnet runs the full width of the track and I have been told is about 2
inches deep (about half an inch wide).

I haven't had a lot to do with linear outputs and want to know the best way
of detecting the 2.69v (say 2.60v or higher) when 2.55v is the normal
output.
Do I simple use a voltage divider outputting around 2.2v with a transistor
across the 2.2v out and the hall effect sensor?

Any help would be greatly appreciated..

Regards

John Burns

 

[Phil Allison]

"John Burns"

I went out to the gokart track today with regulator, hall effect sensor
and
multimeter in hand.
I have a quisient (spelling?) voltage of 2.55V. As the hall effect sensor
passes over the magenet in the track (at 3" distance away) the voltage
output from the sensor goes up to 2.69v and then down to 2.4v then up to
2.69v again before settling back to 2.55v - I assume the magnet in the
track
has its North/South poles in the same direction as the karts travel on the
track.

** How about you amplify that +/- 0.14 volt level shift so it goes from 0
to 5 volts ??

A single supply rail op-amp (ie LM358 ) used as a differential amplifier
with gain of 22 times, say 10 k ohms each input and 220 kohms feedback
resistors. Create 2.5 volt voltage reference for the negative input using a
1 kohm multiturn trim pot off the 5 volt supply adjusted so the op-amp
output sits at 2.5 volts in the absence of the magnetic field.

As the buried coil is approached the voltage should swing up to 5 volts
from 2.5 and trip your timer circuit. A monostable may be added so the
pulse is always from 0 to +5 and ignores any subsequent signals for a few
seconds.




............. Phil

 

[John Burns]

"Phil Allison" <philallison@optusnet.com.au> wrote in message
news:4089dc43$0$20219$afc38c87@news.optusnet.com.au...

** How about you amplify that +/- 0.14 volt level shift so it goes from 0
to 5 volts ??

A single supply rail op-amp (ie LM358 ) used as a differential
amplifier
with gain of 22 times, say 10 k ohms each input and 220 kohms feedback
resistors. Create 2.5 volt voltage reference for the negative input using
a
1 kohm multiturn trim pot off the 5 volt supply adjusted so the op-amp
output sits at 2.5 volts in the absence of the magnetic field.

Phil,
Thanks for your reply..
I tried to reply directly to your email address but it came back with an
error, I was pretty sure its not munged to stop spam.

I have made up a circuit using an LM741 as my ciscuir simulation tool
wouldn't work with the LM358 unless the voltage was about 9v or higher,
while the 741 seems to work fine with the 5v supply that the pic and lcd is
using.
The signal is amplified to give about 2.5v at standby and swings from 0.8v
to 4.8v with the hall effect output near a similar strength magnet to the go
kart track.
I decided that I only want to detect the rising voltage as the hall effect
sensor ouput voltage rises and sinks when the magnet in the track is passed
over. Because of this, I have set up another LM741 as a schmitt trigger so
that the output is low, unless about 2.6v or higher is outputted from the
hall effect sensor.
The problem I have now, is that the schmitt trigger outputs about 0.7V when
low, and 4.7V when high. The high voltage is ok, but the pic still picks up
the 0.7 V as being high, it seems that anything above 0.5volts will take the
line high.
How should I go about reducing the output voltage to be as close to 0v as
possible when the schmitt is not triggered? Does a simple resistor do this,
or am I missing something. I did think about passing on the schmitt trigger
and using an opto coupler which does the same thing and doesn't seem to
switch until about 1.3V is going to it, but I am trying to build the circuit
for larger productions on a budget and the cost of the optocoupler is about
3 times that of a LM741.

Any ideas?

 

[Phil Allison]

"John Burns"

I have made up a circuit using an LM741 as my ciscuir simulation tool
wouldn't work with the LM358 unless the voltage was about 9v or higher,
while the 741 seems to work fine with the 5v supply that the pic and lcd
is
using.

** Get another circuit simulator - your one is wrong.

The LM 358 works perfectly from a single +5 volt supply and will swing
the output down to a few mV while the 741 will not work reliably from a
single + 5 supply.

Better still, protoype the thing with real parts.




............... Phil

 

[John Crighton]

On Mon, 26 Apr 2004 16:36:22 +1200, "John Burns"
<file[hyphen/dash]it@ihug.co.nz> wrote:

Snip>

I have made up a circuit using an LM741 as my ciscuir simulation tool
snip
Any ideas?

Hello John B,

sometimes it is possible for other peope to see your circuit if they
have the same circuit simulator. I have the free Switchercad III.
http://www.linear.com/software/
What are you using?
Regards,
John Crighton
Hornsby

 

[Adrian Jansen]

I see you have some advice already. And now I can see the arrangement a bit
better. Firstly you have to realise that the linear hall sensors you are
using will suffer from temperature drift, so the nominal zero voltage of
2.55 volts may drift, and of course different sensors, even from the same
batch, will have different offsets, so you might have to set the offset for
each sensor. That still leaves you with the temp drift problem.

I would pass the sensor output through an AC coupled amplifier first, to cut
out the DC levels entirely. This will fix both the sensor-sensor variation
problem, and the temp drift. Then use a comparator ( even just another
opamp section wired as a comparator - ie with a high gain ) to get the
nominal 0-5 volt signals your processor needs. Then use the processor to
sense the resulting high-low-high ( or low-high-low ) transitions you will
get from each sensor pass.

Get yourself a simple text on how to use opamps and so on to see how to do
the actual circuit design.

--
Regards,

Adrian Jansen
J & K MicroSystems
Microcomputer solutions for industrial control

 

[John Burns]

"John Crighton" <john_c@tpg.com.au> wrote in message
news:408cb8ed.5405677@News.individual.net...

On Mon, 26 Apr 2004 16:36:22 +1200, "John Burns"
file[hyphen/dash]it@ihug.co.nz> wrote:

Snip

I have made up a circuit using an LM741 as my ciscuir simulation tool
snip
Any ideas?

Hello John B,
sometimes it is possible for other peope to see your circuit if they
have the same circuit simulator. I have the free Switchercad III.
http://www.linear.com/software/
What are you using?
Regards,
John Crighton
Hornsby

Thanks for the link John. I have been using electronics workbench but will

try Switchercad shortly.
I have actually since made the circuit up physically, will change the 741 to
a 358 and see how it goes.

Cheers & thanks for the help
John

 

[Wayne]

"John Burns" <file[hyphen/dash]it@ihug.co.nz> wrote in message
news:c65j96$5rn$1@lust.ihug.co.nz...

Hi there,
I am new to this hall effect stuff - I am trying to build an inductive
pickup for a lap timer I am building for a friends gokart.

I am trying to put together a circuit which uses a hall effect sensor and
outputs a pulse whenever a magnet is detected....regardless of the magnets
pole.

I see that the Hall effect sensors put out a normal (no magnetic field
detected) voltage of 2.5V and this will change by about +/- 1.17 volts for
the extremes.

My plan was to supply the sensor with a regulated 5v supply, then use a
voltage divider to produce a voltage of 1.5V to compare the sensor ouput
with.
The output from the sensor is separated to two separate transistors which
one detects voltages >2.9V and the other voltages <2.1V - The two outputs
are combined and fed into an optoisolator to then be connected to a PIC.
What I want to know, is if I am making this too technical and there is a
simpler way to do it. I do not want to use the ADC built into the pic as
I
want to have the chip do other things and get interrupted for the pulse.

Any ideas?

-John

Hi,

Have you tried a simple reed switch?

OR, asked others what they are using.

Wayne.

 

[John Burns]

Hi,

Have you tried a simple reed switch?

OR, asked others what they are using.

Wayne.

Wayne,
the official 'alfano' product all use a hall effect sensor.

 

[John Crighton]

On Tue, 27 Apr 2004 22:46:20 +1200, "John Burns"
<file[hyphen/dash]it@ihug.co.nz> wrote:

"John Crighton" <john_c@tpg.com.au> wrote in message
news:408cb8ed.5405677@News.individual.net...
On Mon, 26 Apr 2004 16:36:22 +1200, "John Burns"
file[hyphen/dash]it@ihug.co.nz> wrote:

Snip

I have made up a circuit using an LM741 as my ciscuir simulation tool
snip
Any ideas?

Hello John B,
sometimes it is possible for other peope to see your circuit if they
have the same circuit simulator. I have the free Switchercad III.
http://www.linear.com/software/
What are you using?
Regards,
John Crighton
Hornsby

Thanks for the link John. I have been using electronics workbench but will
try Switchercad shortly.
I have actually since made the circuit up physically, will change the 741 to
a 358 and see how it goes.

Cheers & thanks for the help
John


Hello JohnB,

I am learning to use electronics workbench ver 4. myself.
Just for fun, can you send your circuit to me. The more
I play around, the more I learn about it.
The file suffix for my version has to end with .ca4
The file name, no longer than eight letters. Just mentioning
that as my copy of this program is old, maybe you have
a more up to date version.

No rush, whenever you get time.

Regards,
John Crighton
Hornsby