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I have a quick noob question about current flow through a hall effect sensor.

Does the current flowing from one end of the sensor to the other purely depend on the load's draw from the power source? Or is it also affected by the voltage output?

In other words does the current draw stay static regardless of the output voltage changing? Tell me if I'm thinking about this the wrong way, I'm quite confused.

I'm specifically trying to figure out how to calculate what will happen to voltage and amperage when I put a resistor on the sensor output.

This is the diagram I have been looking at if it helps:

enter image description here

Thanks for your time!

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Your diagram shows a bare Hall element.

The current that flows along the element (excitation current) is dictated almost entirely by the resistance along the element, and its excitation voltage Vx.

The output or Hall voltage VH is the product of the B field, the excitation current, and the Hall sensitivity of the element. Any output current flowing is dictated by the magnitude of VH, the resistance of the element across the output terminals, and the load on them. It's good practice to have a load here that draws little or no current.

The Hall sensitivity tends to be so low that the excitation current varies little with output voltage.

However, you rarely see a bare element. What you buy as a 'Hall Sensor' has an amplifier following VH, to boost the mV output to a swing of volts that you can read sensibly on your Arduino. This further isolates any sensitivity of input current to changes in the output voltage.

If you load the sensor output amplifier output with a resistor, and the amplifier is capable of driving the current that resistor demands, then the amplifier will draw the extra supply current to drive the resistor load.

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  • \$\begingroup\$ One more question though... If I then hook the resistor up to some voltage sensing application will the voltage drop from the amplifier to the sensor? \$\endgroup\$ – MerpTheAwesome Dec 17 '16 at 7:26
  • \$\begingroup\$ If so how do I calculate that drop depending on which resistor I use? \$\endgroup\$ – MerpTheAwesome Dec 17 '16 at 7:27
  • \$\begingroup\$ Are you using a bare Hall element, or a packaged Hall Sensor? Provide a link to which ever one you are using. If you are going to integrate it into a system with an impedance in the supply driving it, and/or some load on the output, then provide a schematic diagram. I can't read your mind to figure out which 'drop' you mean, when you connect your resistor to which of the three or more positions that are possible. \$\endgroup\$ – Neil_UK Dec 17 '16 at 7:33
  • \$\begingroup\$ I will be using a packaged Hall Sensor. I don't know exactly which one I'm using, but it's inside a scooter throttle. I wanted to slightly limit the top voltage output of the Hall sensor in order to slow down the max speed of a motor by sticking a resistor on the output line. I was wondering how to calculate what the resistance will do to the voltage and current. The motor is a bldc motor plugged into a controller. I don't quite know how the controller reads the voltage and equates it to motor speed, nor do I have a diagram. I don't know if this is enough info, if not that's ok. \$\endgroup\$ – MerpTheAwesome Dec 17 '16 at 7:43
  • \$\begingroup\$ Ah! Use a potential divider between sensor and controller. This is more flexible, and can probably be adjusted to what you want. First, pick an impedance, 10k is usually a reasonable starting point. Put 10k across the controller input (signal to ground), and see if with that loading you can still get at least the top speed you want. If not, pick a bigger resistor until you can. Now put a smaller resistor between the sensor output and the controller input, for instance 1k will drop the gain of the speed control by about 10%, pick different resistors until it behaves like you want. \$\endgroup\$ – Neil_UK Dec 17 '16 at 7:52
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Hall cells have output impedance very close to the bulk resistance of the cell. treat it as a near ideal voltage source.

So, adding resistors of more than a few ohms is unlikely to have a signifcant effect on the output voltage.

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