If you can include a resistor in series with the Hall sensor input, I think your best option would be a voltage clamp in parallel with the combined resistor and Hall sensor input. The simplest type of voltage clamp is a diode. The circuit looks like this:

The Hall sensor input has a certain resistance, R

_{HALL} (the data sheet says it is "less than 51 mΩ"). A fixed resistor, R1, is connected in series with the Hall sensor input.

These resistances add together to make a total shunt resistance that I will call

**R**_{S}. The voltage across R

_{S} is V

_{S} (marked on the diagram).

A high-current diode, D1, is connected across R

_{S}. As the current through R

_{S} (R1 plus R

_{HALL}) increases, V

_{S} increases in proportion, due to Ohm's Law. At a certain point, V

_{S} will reach a voltage where D1 will start to conduct, and it will start to shunt current away from the Hall sensor circuit.

For example, if D1 begins to conduct when V

_{S} reaches 0.5V, then you choose a value for R1 so that R

_{S} is 0.5Ω. Once the current through that path reaches 1A, V

_{S} will reach 0.5V and the diode will start to conduct.

I've had a brief look for a suitable diode and found a possible option - the Vishay FES16BT. It's available from Digi-Key for USD 2.17. See

http://www.digikey.com/product-detail/en/FES16BT-E3/45/FES16BT-E3/45GI-ND/2152949. Here is the V

_{F} vs. I

_{F} *typical* graph for that diode.

This graph shows the

*typical* relationship between V

_{F}, the

*voltage across* the diode in the forward direction (which is equal to V

_{S} in the circuit above) and I

_{F}, the

*current through* the diode in the forward direction.

EDIT ***** Previously (prior to 2015-02-10) I had estimated V

_{F} for I

_{F}=10A as 0.84V but I misread the graph; it should be about 0.88V, and the calculated value for R

_{S} is cleaner if I assume it's 0.885V. I have updated the following calculations to fix this error. *****

If you look at the right hand curve (for T

_{J}=25°C) you can see that an I

_{F} range of 1~10A corresponds to a V

_{F} range of about 0.7V to 0.88V (0.885V makes the calculations come out cleaner).

If we want to limit I

_{S}, the current through R1 and the Hall sensor, to a maximum of 1.5A when D1 is shunting 9A or 10A, we need to choose R1 so that when V

_{S} is 0.885V, I

_{S} will be 1.5A. That will work with a typical diode in the D1 position.

That works out to R

_{S} = 0.59Ω. If R

_{HALL} is 50 mΩ, that would mean R1 would be 0.54Ω. This would ensure that no more than 1.5A flows through the Hall sensor even when the total load current is 10A.

Edit: The maximum power dissipation in R1 will be quite low, because V

_{S} is limited to about 0.9V and I

_{S} is limited to about 1.5A, so a 2W resistor will be fine for R1. /edit

The problem is that we need the Hall sensor to measure up to 1A without interference from the diode. With R

_{S} = 0.59Ω, 1A current corresponds to 0.59V across the diode, and at V

_{F} = 0.59V, the diode will still conduct

*some* current. The graph doesn't go below V

_{F}=0.6V, but it also doesn't show what happens below I

_{F}=0.1A.

So with this solution, D1 will still pass some current and affect the Hall sensor's accuracy at currents of 1A and less. What is needed is a clamp with a V

_{F} vs. I

_{F} graph that's a lot closer to vertical.

I think it would be possible to make a clamp circuit using a MOSFET with very low V

_{GS} voltage and a few other components to provide the steep V-I curve. The MOSFET I'm looking at is the Vishay Siliconix SIA436DJ which is available from Digi-Key - product page at

www.digikey.com/product-detail/en/SIA436DJ-T1-GE3/SIA436DJ-T1-GE3CT-ND/3471119. This device is in a compact no-lead SMT package that would be hard to work with. Also, the clamp circuit would need about 2.0~2.5V across it before it could clamp; this would mean R1 would need to be around 2.2Ω.

Please read this post thoroughly until you understand it, then tell me what you think.