I am following Figure 5 pg. 18 of this data sheet: Vp(HIGH), Vp(MID), Vp(HIGH), followed by 0-255 Vp(MID) pulses that increase the sensitivity each time Vp(MID) is applied.

However, when I bring the north and south poles of a strong fridge magnet close to the hall effect sensor I always get approximately 4.5668 volts for North and 0.3012 volts for South, no matter how many Vp(MID) pulses I send to VOUT pin.

The circuit I am using is here:


You may get a little confused about the 27V (Vp(HIGH)), 15V (Vp(MID)), and 3V (Vp(LOW)). Basically I have a single 30 Volts power supply that I am dialing up and down to get the high,mid, low voltage pulses needed to program the sensor.

For example, I set the power supply to 27 volts and connect it to hall-effect's VOUT pin. Then I unplug the jumper wire connecting to VOUT, dial down to 15 Volts, and plug the jumper wire back in. Unplug jumper wire, dial back up to 27 volts, plugin jumper wire into VOUT pin again. The voltmeter attached to VOUT pin gives me correct voltage readings when I plug the jumper wire into VOUT.

But this still does not explain why my north and south pole readings do not change significantly even though I have sent up to 30 Vp(MID) pulses after the initial Vp(HIGH)-Vp(MID)-Vp(HIGH) pulses.

I would greatly appreciate it if someone could point me in the right direction.

If you look at Figure 5 on pg. 18, you will see they send out a Vp(HIGH), then it drops down to Vp(LOW). Then it goes up to Vp(MID), then back down to Vp(LOW).

The way I am doing it, instead of dropping down to Vp(LOW), the voltmeter reads 2.0V when the jumper wire is unplugged, which is the "Pre-Programming Quiescent Voltage Output" as shown on pg.4 of the data sheet under "Pre-Programming Target".

So I don't specifically send out a Vp(LOW) pulse, but 2.0 volts is technically less than the 5.5 volts required for Vp(LOW) as shown on pg. 15 of the data sheet.

So I am quite confused about why the hall-effect sensor's sensitivity is not changing despite following Figure 5 exactly as shown.

  • \$\begingroup\$ Does contact bounce matter? If it does, how are you preventing it happening? How did you get to this stage from your prior stalled state? Do pins actually go all the way to the voltages that you specify. What resistor values are you using in the dividers ? Why? \$\endgroup\$
    – Russell McMahon
    Commented Feb 27, 2013 at 6:49
  • \$\begingroup\$ I don't think contact bounce is a problem because I have the voltmeter hooked up directly to VOUT and I can monitor the voltage levels rise and fall as I plug/unplug the jumper. Also, there is no mention in the data sheet regarding this. In my prior state I was using voltage dividers with low-impedeance resistors, however the VOUT pin of the hall-effect sensor never received the right voltage when a particular switch was closed. Therefore I decided to remove all the resisters out of the equation and to use the power supply directly (which appears to work better). \$\endgroup\$ Commented Feb 27, 2013 at 14:23
  • \$\begingroup\$ But now that VOUT pin is actually getting the right voltage reading, this does not explain why Try Mode Figure 5 pg. 18 of the data sheet is not working for me. \$\endgroup\$ Commented Feb 27, 2013 at 14:24
  • \$\begingroup\$ You can't see contact bounce using a normal volt meter. \$\endgroup\$ Commented Feb 27, 2013 at 15:04

1 Answer 1


You're probably suffering from bounce. If you were to connect an oscilloscope to the Vout pin (I suggest you try this) you might well see something like this:

Switch bounce

When you plug or unplug the jumper, it will be almost impossible to do this without causing several unintended pulses. I recommend not trying to program it by using jumper wires.

By far the best way to do it is to use a microcontroller and some transistors. However, if you really want to do it by hand, using this circuit might help:

Here, you can press S1 to pulse Vout to 27v, and S2 to pulse it to 3v. C1 will help to remove any switch bounce. I don't guarantee that this will work though.


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