Hall effect speed sensor question

[MICR0]

I have a Cherry GS1005 Hall effect sensor that I'm trying to use as a cam position sensor. It should be triggering off a tab on the cam sprocket but I'm getting no signal on my scope. I have verified the 5V Vcc is there, and I have a 1K pull up resister between Vcc and the signal out (as per Cherry's installation document). However on the outside of the bag the sensor came in was a sticker stating that current should be limited via a resistor to 25mA. I'm using a 20ohm resistor but I'm not sure I did that calculation right (I'm going on an electronics class I took a long time ago) I also don't know enough to know if this sensor is working properly or if the tab I'm trying to use as a trigger is triggering. If I pull the sensor out should I be able to trigger it or at least get some sort of out put by moving a piece of ferrous metal in front of if? Can someone give me some advise on testing this sensor?

 

[Skogsgurra]

First thing is that 20 ohm and 25 mA is 0.5 V. So I don't see where your 20 ohms came from.

Second, where did you put that 20 ohm resistor? Not loading the output - I hope.

Third, 1 kohm fom output to +5V sounds right.

Fourth, you didn't forget the 0 V connection? It happens sometimes.

Gunnar Englund

http://www.gke.org

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100 % recycled posting: Electrons, ideas, finger-tips have been used over and over again...

 

[MICR0]

The 20 ohm is on the input as a limiting resister. I got the 20 ohm value from the source voltage and from what I think I read somewhere is the voltage drop across the sensor. That's the way my very hazy memory told me to do it. The 0V is connected through the ECU sensor ground. The ECU is also supplying the 5Vcc.

 

[VEBill]

In addition to all the other points mentioned, did you confirm that the "tab" on the cam sprocket is already equipped with a magnet?

You might consider mocking up the circuit on a test bench and get it working there first.

 

[VEBill]

Correction: I see that the sensor has a built-in magnet so that it can detect ferrous metal directly.

So that leaves just my suggestion to get it working on the bench first. Good luck.

 

[MikeHalloran]

Sensors in this case type are normally adjusted so that they contact, e.g., the tip of a gear tooth, then backed off a bit, say 1/4 turn, to set the sensor gap. You should verify the range of acceptable sensor gaps with bench testing.

Note also that the operating temperature limit of 105C or 125C may not include your engine's normal oil temperature, assuming we are talking about an IC engine's cam.





Mike Halloran
Pembroke Pines, FL, USA

 

[Skogsgurra]

I really think that you need to brush up your Ohm's Law before trying to do any electrical work involving resistors (spelling!).

It would also be useful if you used the right terminology. Where, for instance, is the INPUT of a Hall sensor? In my world, the input is a change in magnetic field and no resistor can be connected there.

When you say that "The 20 ohm is on the input as a limiting resister" and that you calculated it from some data, I wonder what data? You have mentioned 5 V and 25 mA. If those are the data you used, you should arrive at 200 ohms. But still don't see where or why you should apply that. A diagram showing your hook-up could perhaps shed some light.


Gunnar Englund

http://www.gke.org

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100 % recycled posting: Electrons, ideas, finger-tips have been used over and over again...

 

[MICR0]

I guess "input" should be V+ or Vcc? Terminology across disciplines can cause confusion, and I'm certainly up on electronics terms as I should be. I should brush up.

I calculated the 20ohm value from the source voltage (5v) minus the the voltage drop across the sensor from memory of read a spec somewhere (4.5v), leaving a deference of .5v. divided by the max current (.025A) equals 20 (ohms). Is this not the way to size a current limiting resistor? Last night I was able to get a signal on the scope after placing a 51ohm resister at V+. the signal wasn't great and I was out of time, so it waits till Monday.

 

[Skogsgurra]

I like your attitude. You want to learn.

No, your calculation is based on wrong assumptions.

You do not need any current limiting resistor in series with your Vcc. Just connect to whatever voltage within specs that you have. Anyting between 4.5 and 24 V is OK.

The maximum current says how much you can pull out of or, in this case, sink with the output terminal.

Therefore, if you have +5V, you should never have less than 5/0.025 = 200 ohms pull-up from output to +5V. That's all there is to it. You have 1 kohm, so you are fine.


Gunnar Englund

http://www.gke.org

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100 % recycled posting: Electrons, ideas, finger-tips have been used over and over again...

 

[zappedagain]

5V / 20 ohms = 250 mA : that is well above the MAXIMUM current rating, so there is a possibility that you damaged your device. 5V / 1K = 5 mA, so that gives lots of safety margin. If the 1K resistor doesn't work you might want to try it on another device.

You want to hook the device up as shown on p. 15 of the data sheet. VCC (+5V) needs to be positive with respect to Ground (0V). 1K from the output to VCC.

John D