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On pg. 18 Figure 5 of this data sheet you will see a pulse train required to program the sensitivity of the A1362 Hall-Effect sensor.

I am using the following circuit:

my circuit

Using Figure 5 from the data sheet, I simply power on the Hall-Effect sensor by supplying the 5V VCC. Switches SW8,9,and 10 start out being in open state. Here are my steps:

  1. Close SW8 // high pulse
  2. Open SW8

  3. Close SW9 // mid pulse

  4. Open SW9

  5. Close SW8 // high pulse

  6. Open SW8

  7. Close SW9 // stream of mid pulses, each time incrementing sensitivity counter by 1

  8. Open SW9

repeat steps 7,8 0 - 255 times as necessary to set the sensitivity of hall-effect sensor.

When I bring the north / south pole sides of a magnet close to the hall-effect sensor, the voltage readings I get at the VOUT pin are almost identical. But if the sensitivity of the hall-effect sensor has changed I should get different voltage readings. This means I am not setting the sensitivity correctly. What am I missing?

From previous posts, others have said to watch out for my current readings too. So here is some more information:

At VpH the multimeter reads ~ 200mA At VpM the multimeter reads ~ 100mA At VpL the multimeter reads ~ 3.5mA

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Using VpH as an example the current will be limited to around 90mA by the resistor R7. The datasheet says the minimum current guaranteed to blow a fuse is 300mA and even recommends a 0.1uF capacitor to ensure enough current is available.

A simple solution more likely to work would be to scrap the resistor divider and use something like an LM317 adjustable regulator and switch the voltage set resistor shown as R2 in the typical application circuit instead. Using 1% resistors should keep you well within the tolerances given in the datasheet.

Also take note of markrages answer, switch bounce is likely to be a problem so ideally apart from a more robust supply you should be switching voltages from a microcontroller. The LM317 datasheet shows Iadj is limited to less than 100uA so you could achieve that using any form of transistor or analog switch that doesn't introduce too much resistance.

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  • \$\begingroup\$ @PeterJ- according to the datasheet of LM317, Figure 5 states the equation is Vo = 1.25 (1 + R2/R1)+Iadj*R2. Since I need Iadj=300mA=0.3A and Vo=27V for VpH, some simple math tells me R2=16.6129*R1. In order to get the VpM and VpL voltages I would need two more LM317s and perform similar calculations in order to determine correct resistance values. Is this correct or am I way off? \$\endgroup\$ – user1068636 Mar 9 '13 at 6:00
  • \$\begingroup\$ @user1068636, Iadj is guaranteed to be < 100uA, it doesn't directly follow the load current. There's no harm calculating at 0 and 100uA to see what difference it makes, but the datasheet mentions the error is negligible and can be ignored for most applications. \$\endgroup\$ – PeterJ Mar 9 '13 at 6:03
  • \$\begingroup\$ Are you claiming my problem to be the load current is not 300mA at VOUT pin? Your proposed solution is to replace the voltage dividers with LM317, but I am unable to understand what this achieves? Does this give me higher current values at VpH,VpM, VpL? \$\endgroup\$ – user1068636 Mar 9 '13 at 15:20
  • \$\begingroup\$ @user1068636, Yes that's right, the LM317 can deliver over 1.5A. The problem with voltage dividers is the impedance both limits the current and the voltage will vary depending on the load. You'd probably need a scope to measure it but I suspect with your circuit you'd see a voltage drop at the point it tried to blow the fuse. \$\endgroup\$ – PeterJ Mar 10 '13 at 0:00
  • \$\begingroup\$ Could you please verify this circuit is what you mean: circuitlab.com/circuit/b47yr5/… This is a visualization of what I think you are talking about. Please feel free to modify / correct it if you think something is wrong. \$\endgroup\$ – user1068636 Mar 10 '13 at 19:25
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If you are using regular mechanical switches, each switch closure is generating several pulses due to contact bounce.

You should use some switches or FETs controlled by a computer or microprocessor to do this I think.

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  • \$\begingroup\$ i am currently using regular switches from radio shack. Right now we are trying to stay way from using a microprocessor. we want to see if i can do it manually. Do I have any other options to reduce contact bounce? \$\endgroup\$ – user1068636 Mar 9 '13 at 15:33
  • \$\begingroup\$ The standard debounced switch is an SPDT driving an SR flip-flop. electronics-tutorials.ws/sequential/seq_1.html But you'll still need analog switches or FETs to switch those high (not logic) voltages. \$\endgroup\$ – markrages Mar 9 '13 at 18:30

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