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
RS. 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.