So me and my quadcopter project continues.
Now I would like to measure up to 40A using the ADC on a PIC.
So far:
1. I have some resistive wire, 4.2R per meter. (28SWG) I also looked at making a PCB trace, but that would have to be huge to cope with the heat dissipation/current.
2. I would like the shunt to be about 0.000625R, which would mean 1watt of heat dissipation at 40A and very little voltage drop(about 25mV).
3. Back to the resistive wire.
a) 4.2R per meter gives 0.0042R(I think R0042 is also correct) per mm.
b) Put 20 strands in parallel and it should be 0.00021 per mm, meaning I would need 3mm length to give 0.00063R. Double the length to make it mechanically more workable and that should give a voltage drop of about 50mV and wattage=2watt at 40A giving 1.25mV per amp.
c) To use the full range of the PIC ADC the 50mV must be amplified to 4V, so gain = 4 / 0.05 = 80 which should easily be achieved with a 741 opamp using 56K/680R resistors giving gain of 82.
d) Accuracy. This does not need to be 100% as it is high current, but 1024 steps for 40A should be around 40mA steps, which is pretty much good enough(exceeds requirements actually).
I looked at buying a high wattage low resistance resistor to use as a shunt, but the best I could find readily available here in the UK is R005 (0.005R) 5watt, which is not a high enough rating for 40A as that would need 8watt.
This is the resistor I looked at HERE and they are quite expensive to fry. I also looked at the 7watt version but they are all rated for maximum 22A.
Any advice appreciated.
Now I would like to measure up to 40A using the ADC on a PIC.
So far:
1. I have some resistive wire, 4.2R per meter. (28SWG) I also looked at making a PCB trace, but that would have to be huge to cope with the heat dissipation/current.
2. I would like the shunt to be about 0.000625R, which would mean 1watt of heat dissipation at 40A and very little voltage drop(about 25mV).
3. Back to the resistive wire.
a) 4.2R per meter gives 0.0042R(I think R0042 is also correct) per mm.
b) Put 20 strands in parallel and it should be 0.00021 per mm, meaning I would need 3mm length to give 0.00063R. Double the length to make it mechanically more workable and that should give a voltage drop of about 50mV and wattage=2watt at 40A giving 1.25mV per amp.
c) To use the full range of the PIC ADC the 50mV must be amplified to 4V, so gain = 4 / 0.05 = 80 which should easily be achieved with a 741 opamp using 56K/680R resistors giving gain of 82.
d) Accuracy. This does not need to be 100% as it is high current, but 1024 steps for 40A should be around 40mA steps, which is pretty much good enough(exceeds requirements actually).
I looked at buying a high wattage low resistance resistor to use as a shunt, but the best I could find readily available here in the UK is R005 (0.005R) 5watt, which is not a high enough rating for 40A as that would need 8watt.
This is the resistor I looked at HERE and they are quite expensive to fry. I also looked at the 7watt version but they are all rated for maximum 22A.
Any advice appreciated.