How do I modify the output of a Hall sensor ?

Thread Starter

vielle568

Joined Feb 3, 2012
13
Hello, Can someone please point me in the right direction ?

I am in the process of changing a Hall sensor that functions as a crankshaft position sensor in my 2CV engine. The existing sensor is about to be sawn in half by a shutter disc and a new set-up is becoming essential.

I now have a good industrial quality Hall sensor, but unfortunately there’s a difference in the output signals between the two sensors. The new sensor outputs 4.40 volts when open, and 0 volts when it detects. The original sensor output only 95 millivolts when open, and 12 millivolts when detecting.

I have tried using resistors to divide the output voltage, but sadly dropping down to 95 and 0 millivolts doesn’t work. The lower value of 12 millivolts seems to be essential as the ECU will not trigger the injectors at 0 volts. I have tried other circuits but without success.

My question to the forum how do I modify the output of the new Hall sensor to obtain the signal values from the original device ? Thanks for any suggestions.

vielle568
 

Alec_t

Joined Sep 17, 2013
12,065
I'm guessing how your sensor is constructed and powered, but the combination of resistors shown below may get you in the ball park.
AdapterShaftSensor.jpg
 

Thread Starter

vielle568

Joined Feb 3, 2012
13
Hello Alex,

Thanks for your rapid reply and for the schematic. I have all the components here on hand so I'll wire it up and start experimenting.
The original Hall sensor uses a shutter wheel to trigger the signal, but the replacement unit triggers on a ferrous insert in the plastic ventilation fan that's directly attached to the crankshaft.
I notice that both sensors are included in the schematic, so I assume that the original sensor (V1) is dormant. Both are three wire sensors; +5v , ground, and signal out.
I'll let you know how I get on :)
Vielle568
 

Alec_t

Joined Sep 17, 2013
12,065
I notice that both sensors are included in the schematic
No. Only one (in the left-hand rectangle). It's a functional simulation of the assumed new sensor. Your ECU is assumed to supply the 5V voltage source V1. You only need to add the 3 resistors in the right-hand rectangle.
 

Thread Starter

vielle568

Joined Feb 3, 2012
13
I'm sorry everyone. Please understand that my background is mechanical engineering and not electronics. I didn't recognise the two circular symbols in the schematics and I wrongly assumed that they were the Hall sensors. I wired up everything in this fashion and had an inverse output that just didn't conform to my expectations.
Yes, the Hall sensor is powered at 5v by the ECU. Looking at the right side of the schematic it's obvious that the bottom line must be ground, the top line that passes the diode must be Vcc, and I suppose that the sensor signal has to be at the centre.
I'll just try again.
Thanks for your patience !
Vielle568
 

Thread Starter

vielle568

Joined Feb 3, 2012
13
Yes, yes, yes ! Success !
This circuit has worked perfectly. With your combination of resistors I got exactly the 12 millivolts I was aiming for. The other voltage was not quite on target, but by replacing the 1K resistor with a pot I was able to adjust the output to the required level. I can simply replace the pot with an equivalent resistor and the problem is solved. I had never thought that such a simple solution would have been possible.
Thank you so much for your support and for sharing your knowledge.
vielle568
 
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