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
12,734
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
12,734
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
 

PushToClose

Joined Aug 5, 2021
16
Hello,

Can I use this topic to ask a similar question? I have a electric wheelchair I’m trying to use for a project. The chair is controlled by a Hall effect joystick. The output voltages of this range from 1.1V to 3.9V, with 2.5V when the stick is neutral. I want to replace the joystick with a pedal, so I’ve purchased a Hall pedal. A brief test has shown that the output of this ranges from 0.8V with no deflection to 4.2V when depressed. Can I set a circuit up so that the output with no pedal deflection is 2.5V, and the max output is 3.9V?

Thanks for any suggestions.
 

LowQCab

Joined Nov 6, 2012
1,918
Now you're getting into a Mechanical-Actuation problem,
but it might be doable Electrically.
It depends on which way the Voltage goes in the original Joy-Stick setup when
pushing the Stick in the Forward direction.

Basically, it sounds like You need a Mechanical "Stop-Screw" that
holds the Pedal half-way down,
or however far down results in no forward motion.
.
.
.
 

MrSalts

Joined Apr 2, 2020
1,540
Hello,

Can I use this topic to ask a similar question? I have a electric wheelchair I’m trying to use for a project. The chair is controlled by a Hall effect joystick. The output voltages of this range from 1.1V to 3.9V, with 2.5V when the stick is neutral. I want to replace the joystick with a pedal, so I’ve purchased a Hall pedal. A brief test has shown that the output of this ranges from 0.8V with no deflection to 4.2V when depressed. Can I set a circuit up so that the output with no pedal deflection is 2.5V, and the max output is 3.9V?

Thanks for any suggestions.
without mechanical modifications, you can use one dual op amp (rail to rail input and rail to rail output (at least to 4.2v out with 5v supply. 12V supply is also fine without modifications.

A44B190B-31CA-427A-BB89-FDCF076CB729.jpeg
 

Ian0

Joined Aug 7, 2020
4,845
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.
Just curious - since when did a 2CV engine have hall effect sensors and fuel injection?
 
Thanks for the replies, that’s really useful. I’ve realised the problem is a bit more complicated than I thought, so may be worth its own thread.

Thanks again!
 
without mechanical modifications, you can use one dual op amp (rail to rail input and rail to rail output (at least to 4.2v out with 5v supply. 12V supply is also fine without modifications.

View attachment 245103
Thanks so much for this. I've just built this on a breadboard with an LMC 6484 op amp. However, the output I get starts at 2.9V, topping out at the desired 3.9V when the input is increased all the way to 4.2V. I'm really struggling to understand what values I need to change to bring the initial output to 2.5V. Any suggestions for guides or simulators that will help with this?

The circuit I need to build needs to output 2.5V to 3.9V in one mode, and then move to 2.5V to 1.1V in another mode (forward/reverse basically - these values are what a speed control module is expecting because they're the outputs given by a joystick. I'm replacing the joystick with a pedal). Measuring output to ground instead of output to Vcc, gives me a starting point of 2.1V, dropping to 1.1V with a fully depressed pedal (4.2V input from the Hall pedal). Again, the response to the high voltage input is spot on, but I need to adjust the low input response to be 2.5V.
 
Screenshot 2021-10-28 at 21.13.48.png

Sorry for the waffle. Does that ^^^ help explain what I'm trying to achieve? Circuit voltage is 5V.

(ETA: I missed a value off that diagram. The bottom of the 'pedal input' arrow should read 0.8V)
 
Last edited:
To update this, I managed to get the output range I was looking for by adjusting the first feedback resistor, and also the +ve voltage input to the second op amp. Thanks again for this circuit suggestion! Adjusted circuit diagram below:

Screenshot 2021-10-28 at 21.16.00.png

I now can't work out how to achieve going from 2.5V to 1.1V as the pedal is depressed, to give me what I need for a "reverse" signal. That's obviously what I get if put the voltmeter across the output and 0V, but that's not really something I can use as a signal I don't think...
 
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