My project involves measuring the erosion rate in a pipe by monitoring the change in resistance of a small strip of carbon steel as it is eroded away (based on resistance being inversely proportional to its thickness).
Due to the small resistivity of carbon steel, and the small size of the sensor, I expect its resistance to be around 0.3 mOhms. Currently, I am planning to measure it using a simple potential divider circuit as shown in the attachments. I will have one 100 Ohms resistor, and a 50 Ohms resistor connected in series to my sensor, and I will supply 5V to the circuit. By measuring the potential drop across the 100 Ohms resistor, I should be able to calculate the resistance of my sensor.
I am using the 24-bit LTC2499 ADC to measure the voltage drop. I need a high precision measurement as I expect the change in resistance of the sensor due to erosion, to be very small; on the order of uOhms. As I am measuring the potential drop across the 100 Ohms resistor, I don't think I'll need any amplification.
Currently, I have only experimented with large resistances. I have tried to measure the resistance of a 100 Ohms +/- 1% resistor, and got an error of 5 3.5% with the LTC2499. The A0 port on the Arduino is much more accurate, with an error of only 0.6%. Also, the values obtained from the LTC2499 fluctuate a lot compared to the A0 port.
Is there any way I can obtain more accurate values. Also, is this a feasible approach to measure resistances of around 0.3 mOhms.
Can you enlighten as to exactly what and where the "pipe" is ?
Makes me wonder how you can measure such low resistances when its not in a controlled enviroment, surely changing humidity would have a strong effect on any such low resistance readings and also affect the rate of corrosion.
The sensor will be placed in a jet impingement rig. The only other factor that should affect the resistance is temperature, which will be accounted for. The idea is to replicate an ER erosion sensor (http://www.alspi.com/erintro.htm).
By passing a continuous current through the sensor, it should become passivated, and only affected by erosion, and not corrosion.
As I am measuring the potential drop across the 100 Ohms resistor, I don't think I'll need any amplification.
Have you made any calculations with 100 Ohms and a few milliohms? You'll have a very-tiny voltage drop across the smaller resistance and almost all of the voltage drop across the 100 Ohm resister whether the small resistance is a few milliohms or several milliohms, making your measurement useless.
I believe Ohmmeters (digital multimeters) normally use a constant current source. Then after measuring the voltage, resistance can be calculated using Ohm's Law. And for low-resistance measurements, the [u]4-wire[/u] method is used.
Higher current will help because that gives you a higher voltage to measure.
terryking228:
Hi,
I think you need more current in the sample to get meaningful results. Even 100 mA . Look up "Arduino Current Source"; many possibilities.
For very small resistances in one-ff situation I use a variable power supply to put 10 amps through the sample and read the voltage with a good DVM..
Yes, this is a good way to do this.
For testing purposes, power utility crews often use a CC of 10 amps through huge breaker contacts to measure resistance.
Note also that in the example in the link that you provided, the instrument uses AC excitation of the bridge. This is to prevent galvanic corrosion of the sensor.
DVDdoug:
Have you made any calculations with 100 Ohms and a few milliohms? You'll have a very-tiny voltage drop across the smaller resistance and almost all of the voltage drop across the 100 Ohm resister whether the small resistance is a few milliohms or several milliohms, making your measurement useless.
I believe Ohmmeters (digital multimeters) normally use a constant current source. Then after measuring the voltage, resistance can be calculated using Ohm's Law. And for low-resistance measurements, the [u]4-wire[/u] method is used.
Higher current will help because that gives you a higher voltage to measure.
I think a better way would be to use a differential amplifier with a very high gain (x10^5) to measure the potential drop across the sensor?
aali94:
I think a better way would be to use a differential amplifier with a very high gain (x10^5) to measure the potential drop across the sensor?
What he said plus you'll need to use a chopper stabilized amplifier to prevent drift with such low input levels. Analog Devices has many suitable chip level parts, light years ahead of what it took to do this with discreet components years ago.
aali94:
Is there any way I can obtain more accurate values. Also, is this a feasible approach to measure resistances of around 0.3 mOhms.
You need a 4-terminal measurement, constant current through the sample (your 100 ohm resistor
serves that just fine), and amplify the voltage across the sample using an instrumentation amplifier.
No need for a 24bit ADC, that's overkill.
aali94:
My project involves measuring the erosion rate in a pipe by monitoring the change in resistance of a small strip of carbon steel as it is eroded away (based on resistance being inversely proportional to its thickness).
Can you say more ?
Not connected to the pipe wall this may not give a comparable result compared to pipe erosion.
quote author=avr_fred link=msg=3061346 date=1483047318]
What he said plus you'll need to use a chopper stabilized amplifier to prevent drift with such low input levels. Analog Devices has many suitable chip level parts, light years ahead of what it took to do this with discreet components years ago.
[/quote]
I'll have 1.365338795 V at the non-inverting terminal and 1.365328895 V at the inverting terminal, giving a voltage difference of 9.9 uV. I'll have a feedback resistance of 20 MOhms, and an input resistance of 100 Ohms.
I'm not sure about the input offset voltage concept. If the voltage difference between the input terminals is 9.9 uV, does the input offset voltage need to be less than this? The LTC1150 has a maximum offset voltage of 10 uV.
