Hall Effect switch logic output

Thread Starter

mkenney

Joined Feb 14, 2009
24
The part is definitely out of spec, unless your controller board is loading the output. Have you tested it with the green wire disconnected?
Do you have the ground of the Hall effect device connected to the ground of the controller board (you should!)?
Hope I didn't mess up, I don't have it anywhere near or connected to the controller board, just on the bread board. The green lead is goiung to the positive side of my multimeter and nothing else. The negative side of the multimeter is going to the ground. Is this setup stopping it from working correctly somehow?

Thanks again for the help! Almost bedtime to sleep on it some like you suggested...

Mark [mkenney]
 

Ron H

Joined Apr 14, 2005
7,063
Hope I didn't mess up, I don't have it anywhere near or connected to the controller board, just on the bread board. The green lead is goiung to the positive side of my multimeter and nothing else. The negative side of the multimeter is going to the ground. Is this setup stopping it from working correctly somehow?

Thanks again for the help! Almost bedtime to sleep on it some like you suggested...

Mark [mkenney]
I thought you had the green lead connected to the controller board. Since you don't, that's good, as it can't possibly load your Hall effect device. Since you said you get 5V out when you remove the Hall effect device, I can't see how your multimeter could be causing a problem.
 

Thread Starter

mkenney

Joined Feb 14, 2009
24
Good morning all!

Alright I tried a couple things this AM. First I tried another sensor with doing my best to avoid any static discharge. Grounded myself and made sure to not touch the pins at all. The voltage outputs remainned the same and that makes three different ones that have the same output readings. I guess the whole lot of them could be burned but I would think that they would have more random outputs then but I don't have any experience with this so I'm just guessing.

I added a diode between the pullup resistor and the output and moved my multimeter probe to the junction between the resitor and the diode. The voltages here are 1.096V and .588V. This kind of sounds like it might work for my break out board but leaves the high side in the unspecified area of .8-2V.

I'm attaching a picture of one of the sensors to double check that they put the correct ones in to ship to me. (This is one of the ones I burnned up by removing the pullup resitor completely and feeding it 5volts directly, no point in being stupid if you can't show it!)

Thanks again for everyone's help and suggestions on this, its helped keep me out of the funny farm for another day or so;)

Mark [mkenney]
 

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Thread Starter

mkenney

Joined Feb 14, 2009
24
Hello,

Do you have a stronger magnet ?
At the melexis site the US5881 is stated as low sensetivity.
http://www.melexis.com/Sensor_ICs_Hall_effect/Hall_effect_Unipolar_Switches_/US5881_148.aspx

Greetings,
Bertus
Just tried a stronger magnet and the low output went to the same value. I tried using the north pole to see if I could turn it off more or something to boost the high side of the output but it stayed the same as before. What confuses me is that the sensor is supposed to be in the high state but seems to be taking ~4.5Volts from the circuit when sitting at rest.

Thanks!

Mark [mkenney]
 

Thread Starter

mkenney

Joined Feb 14, 2009
24
I found this on the melexis website and don't really understand what it is saying. Thought I would post it incase it is relavent somehow:

Hello.

I am using a US5881 Unipolar Hall Effect Sensor and a 10k pull up from Output to Vdd (+5V). This typically works well, however we have had a few that required a lower value resistor value. Is there a recommended pull up value that should be used?

Thanks in Advance!

Anthony
Dear Mr. Anthony,
The value of the pull-up resistor is limited only from the lower side - the output current should not to exceed the Absolute Maximum Rating (for US5881 it is 50mA ). The upper limit , if exists, comes from the application ( particularly the load resistance and capacitance). The pull-up together with the load resistance form a resistive divider which defines the HIGH logic level. This level should be appropriate to trigger the load.
Example: If Rload=Rpull-up , then HIGH logic level is equal to VDD/2 .
Be aware that the load capacitance together with the pull-up resistor influence the output rise and fall times. If you need a faster response decrease the value of the pull-up.
Best Regards,
Kolio
Mark [mkenney]
 

bertus

Joined Apr 5, 2008
22,270
Hello,

There is stated a maximum current of 50 mA.
You are running the circuit on 5 Volts.
So the minimum resistor is 5 Volts / 50 mA = 100 Ohm.
This value is the parallel value of the resistor connected to the sensor and the resistance of the controller you will use.

Greetings,
Bertus
 

Ron H

Joined Apr 14, 2005
7,063
It appears to be the correct part. Have you tested it with the 4.7nF cap on the output removed? I suspect it is oscillating when it is supposed to be off. Do you have an oscilloscope, which would show the oscillations?
 

Thread Starter

mkenney

Joined Feb 14, 2009
24
It appears to be the correct part. Have you tested it with the 4.7nF cap on the output removed? I suspect it is oscillating when it is supposed to be off. Do you have an oscilloscope, which would show the oscillations?
Seems to have the same outputs when the 4.7nF and/or the 100nF cap are removed. I don't have an oscilloscope, wish I did now.

Mark [mkenney]
 

Ron H

Joined Apr 14, 2005
7,063
Grasping at straws: Eliminate the loop of red wire and the 100 ohm resistor and replace with a short jumper directly to the +5V bus, similar to what you have on the ground pin of the 5881. Move the decoupling cap so that it is between the ground bus and the +5V bus.
 

Thread Starter

mkenney

Joined Feb 14, 2009
24
Grasping at straws: Eliminate the loop of red wire and the 100 ohm resistor and replace with a short jumper directly to the +5V bus, similar to what you have on the ground pin of the 5881. Move the decoupling cap so that it is between the ground bus and the +5V bus.
I gave both things a try and didn't see any change.

I'm going to order a couple more just to see if the ones in the mail got zapped or something. Probably order a different type if I can find one to give a try as well. Not giving up yet, just want to have some other options to try out.

Thanks again for the help!


Mark [mkenney]
 

Thread Starter

mkenney

Joined Feb 14, 2009
24
From another thread on cnczone where I originally found out about this idea. Here is the response that they had.

http://www.cnczone.com/forums/showthread.php?t=31601&page=5

With reference to ground or +VCC? I will bench test it today. OK, I just tested it. With an input voltage of 5.27V, between the output pin and ground I get .512V with no magent and .008V with the magnet. This switch is a current switch not like a regular mechanical switch. The voltage does not mean much. When the switch closes, it allows current to flow, so the reading of .022 looks correct for you. Think of it as reading across a mechanical switch, when it is closed you will not see any voltage. I still don't fully understand these solid state switches, but as long as they work, I don't try to. The change in voltage you are seeing tells me your switch is working.

Depending on the break out board you use and how their logic is wired, it could take from no resistor across the output to +VCC to 10K and just about anything in between. The resistor is to provide a voltage drop across it when the switch allows current to flow so the BOB can see a voltage at the pin.

Vince
Does this make sense?

Mark [mkenney]
 

Ron H

Joined Apr 14, 2005
7,063
From another thread on cnczone where I originally found out about this idea. Here is the response that they had.

http://www.cnczone.com/forums/showthread.php?t=31601&page=5



Does this make sense?

Mark [mkenney]
In a word, no. It doesn't work like the datasheet says, and the voltage change is not enough to operate your controller board.
Why don't you try +9V to the vcc pin on the 5881, while still maintaining the +5V to the pullup resistor. Perhaps the devices are out of spec.
 

Thread Starter

mkenney

Joined Feb 14, 2009
24
In a word, no. It doesn't work like the datasheet says, and the voltage change is not enough to operate your controller board.
Why don't you try +9V to the vcc pin on the 5881, while still maintaining the +5V to the pullup resistor. Perhaps the devices are out of spec.
I tried it with the 9 volts to the Vcc and got similar readings. Another greek to me response on the cnczone site:

That device you show is open collector with a current capable of 50ma, so at 5v, the Min. resistance that can be used from output to 5v would be 100ohms.
Normally with an open collector device, when it is off there should be close to 5v on the output (collector) and very low voltage, <.5 volts, when on.
You only need a pull up if the circuit it is connected to does not provide a resistive path to 5v.
This device is called a sinking device and the BOB would be termed source.
It is important, especially when using this type of device to know the exact nature of the BOB input.
The voltages I mention are WRT common, or ground as I believe they use on that BOB.
Al.
Your probably wishing I never found this site by now:) Feel guilty for taking up so much of your time but really appreciate the help and advice.

Mark [mkenney]
 

Ron H

Joined Apr 14, 2005
7,063
I tried it with the 9 volts to the Vcc and got similar readings. Another greek to me response on the cnczone site:



Your probably wishing I never found this site by now:) Feel guilty for taking up so much of your time but really appreciate the help and advice.

Mark [mkenney]
What he says is true, but you don't even have the BOB (what does that stand for?) connected. With a resistive pullup such as you have, the circuit is dead simple. It should switch between ≈0V and +5V. There is apparently a low resistance path, internal to the IC, from the output to ground. I would say that it might be in your circuit board, but you said you removed the IC and the voltage on the pullup resistor went to 5V, which eliminates the possibility of the problem being with the breadboard or the multimeter.
We'll cross the BOB interface bridge when and if we get the sensor working.:rolleyes:
 

Thread Starter

mkenney

Joined Feb 14, 2009
24
Sorry for the late reply. Busy day at work and switching cell phones and trying to get all my info back in.

Looking at the manual for my BOB, break out board, again. It was pointed out that they have a hook up description for open collector sensors which has been pointed out to be the case for the hall effect sensors. On my break out board it wants a 12 or 24 Volt external power supply to be used with these types of sensors. I'll give it a go as soon as I get a chance but wouldn't surprise me at all if the board then acts correctly??

Thanks again for the help and sorry for the late reply, jsut one of those days and its not over yet!!!

Mark [mkenney]
 
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