Need Help Current Source for 5A and 0,1V,,,Is It Possible?

[oneitusatu]

I think you are underestimating the problem or not fully explaining it. Before we go further, can you tell us what exactly it is you are measuring that has such low resistance?
Give us some idea of the range of resistance you want to measure, it make a big difference to the way the amplifier is specified.

To show you worst case scenario, for a normal 10-bit ADC with 5V reference to show full scale while measuring 1 uOhm would need a current of 5 million Amps !

Brian.

i realize with effect of the noise,what i dont get it is how we could predict how high the noise is and how we can avoid it,

my contact that would be measure in around 100 uOhm , and i will use 16 bit adc for that , if i use 5v ref i got 0,00007692 v/bit i think its enough,
hows the noise?

 

[betwixt]

That makes 100 uOhm give a reading of 6! You want it to be nearer 60,000 !!
If you amplify the voltage by 1,000 times you will see why noise becomes a problem. Even thermal noise and thermoelectric noise from the contacts (look up thermocouples) will be significant.

Brian.

 

[oneitusatu]

That makes 100 uOhm give a reading of 6! You want it to be nearer 60,000 !!
If you amplify the voltage by 1,000 times you will see why noise becomes a problem. Even thermal noise and thermoelectric noise from the contacts (look up thermocouples) will be significant.

Brian.

how do you get that 6?and why would i want to be near 60,000?i dont get it,please explain it slowly

 

[betwixt]

A 16-bit ADC gives results from 0000000000000000 to 11111111111111111 or in decimal that's 0 to 65535. To give any accuracy of reading, you need to use as many of those digits as possible. As you correctly point out, each count of the ADC is equivalent to a voltage of the reference voltage/65535 which makes 0.000076295 volts. Now look at the voltage you get when you pass 5A through a 100uOHm resistor V=IxR so you get 5x(100^-6) or 0.0005V. that is equivalent to about 6.5 ADC units and as it only works in integer amounts, it will either round to 6 or 7. If I draw an analogy, it's like using an analog volt meter but all your readings would be crammed into the first small division of the meter scale.

The noise problem comes from several sources:
1. the reference voltage (possibly this is the supply line to the ADC) has to be very clean and very accurate at 5V, even 4.999 or 5.001 would make your reading wrong.
2. where two different metal come into contact with each other a small voltage is generated, this is likely to be more than the IxR voltage caused by the resistance you are measuring and it is temperature dependent.
3. noise picked up by induction or capacitance or motion of plastics in the measurement cables will likely exceed the voltage you are measuring.
4. the ADC and it's sample circuits introduce tiny amounts of electrical noise.
5. all ADC measurements are only accurate to at most +/- one count, this makes a difference of about 15% in your resistance reading.

The only practical solution is to amplify the voltage you measure so it uses more of the ADC range, then each count becomes proportionately less of the total and therefore causes less possible inaccuracy. The trouble is you need to amplify the voltage by ideally about 1,000 times and that in itself is a problem. It's a lot of gain and although it makes the ADC more accurate, it also amplifies the other noise sources at the measurement point.

I hope that makes sense!

Brian.