Current sensing chips - Page 1
Forum section: electronics / general

author: shebu18
author_rank: Frequent Contributor
timestamp: July 09, 2012, 04:12:45 am
content: Hello, so after the discussion in the shoutbox i wanted to write all toghether.I have the main input set at 12V, using a 1ohm resistor and a max 4080SASA(60gain) and i wnat a current of 4.096i get this:Code:Udrop=I*R=4.096*1=4.096VP=I*I*R=4.096*4.096*1=16.77WorP=I*V=16.77WOK, now to use 0.02ohm instead of 1 ohm, so i get:Code:U=4.096*0.02=0.08192VP=I*I*R=4.096*4.096*0.02=0.3355WorP=I*V=0.3355WSo my 10W 0.02ohm resistor should be good for this even at 4A current.Now the voltage output of the MAX is Vout=Rsens*Imax*AV(gain), Vou=0.02*4.096*60=4.9152V So i get 4.9152V for 4,096A.Using the INA138 i could get the same Rsens values BUT Vout could be in a ratio of 1:1 with the current drawn. So 4.096A mean 4.096V. 9or using a max with a 5 Gain and a 0.2Rsens)Is everything right?

author: shebu18
author_rank: Frequent Contributor
timestamp: July 09, 2012, 06:12:03 am
content: i can use 0.02ohm sense and the MAX4080SASA chip. i will use a voltage divider with a 1,2 ratio, and so i will get 4.096V from 4.1952V.I hope the problem is solved. I will need to see the error that is acumulating.

author: shebu18
author_rank: Frequent Contributor
timestamp: July 09, 2012, 06:52:53 pm
content: Anybody? Some remarks anything.

author: Gall
author_rank: Frequent Contributor
timestamp: July 09, 2012, 07:55:31 pm
content: Sounds right.Unfortunately it is hard to tell the exact resistance for such a small resistor even if using Kelvin connection unless the resistor itself has 4 wires. Solder joints have resistance too. So for relatively large currents measurement of magnetic field may be preferable. To do that, compensate the magnetic field of the coil by magnetic field of another coil (use a Hall effect sensor for feedback). Then the current ratio will be exactly the same as winding ratio. The small current in the compensation winding is easy to measure.

author: chrome
author_rank: Regular Contributor
timestamp: July 09, 2012, 08:13:14 pm
content: Don't forget that the resistance will change when it heats up.

author: hans
author_rank: Super Contributor
timestamp: July 09, 2012, 08:49:32 pm
content: I once made a 60W LED driver (2.8A output) with undersized current shunts. I took a factor of 2 with the power dissipated v.s. maximum. The PCB itself was getting very hot (up to 65C) because of a small form factor (believe 45x60mm) and many SMT cooled components. I believe the board had to 'cool' 10W - 15W from all the components. I turned it on. Initial current about 2.7A. After 1 minute it was 2.85A. Turned it off, grabbed the board, burned my fingers. Ouch!Great.. I think I need a second fan to cool the electronics10W sounds a lot, but it was a buck-boost converter which doesn't have ultra high efficiencies. (not the converter I was using, anyway).Good job on overspeccing your current shunt. A 5W resistor may be fine too, but anything that helps make it even cooler is good. Still, avoid it being warmed up by MOSFET's in the area, if you dissipate 5W in a MOSFET on the same PCB or heatsink, it will still drift..Your current sense:4.096A @ 20mohm = 81.92mV. To get 4.096V, amplify by 50x.According to the datasheet, the INA138 outputs 200uA/V. The input is 81.92mV, the output will be 16.384uA. To get 4.096V it's simply R=U/I=4.096/16.384u = 25k. Parallel up 4x 100k's or something.The tolerance depends on your resistor initial accuracy, it's thermal drift and temperature, the gain error of the current sense amp and your output resistors. They pretty much all add up, so a 1% shunt, +/-2% sense amp and 1% output resistors is a total of 4%. Excluding temperature drift, which can easily add 0.5% if you push the limits of a horrible resistor.For e.g. a 100ppm/C resistor heating up 40C is 40*100ppm = 4000ppm -> 0.4% driftA 300ppm/C resistor heating up 70C -> 2.1% driftIf you want to use the max chip, you can divide by 1.2 indeed. 15k & 75k should do that.

author: Unknown
author_rank: Unknown
timestamp: July 09, 2012, 08:56:32 pm
content: 0.02 Ohms seems doable with a PCB trace.Obviously with some sort of tempco compensation.

author: shebu18
author_rank: Frequent Contributor
timestamp: July 10, 2012, 04:53:41 am
content: Thanks for the answers. I also thought of using a max4080T, with 20 Gain, and a 0.1R resistor, or 10x1R in parallel. So if the Vcc is above 8.192V i will get 8.192 on the output pin(0.1x4.096x20). After this i use a voltage splitter with a ration 2:1 and feed it in the opamp. This way i get at 8.192V output of the max a 4.096V input at the opamp.

author: Gall
author_rank: Frequent Contributor
timestamp: July 10, 2012, 06:46:13 am
content: Quote from: chrome on July 09, 2012, 08:13:14 pmDoes not matter - too small change.Better think about thermoelectrical effect. It may and will appear in solder joints. Fortunately it is in range of microvolts.