How to build a millivolt signal conditioning circuit with compensation?

Thread Starter

anoopak

Joined Nov 24, 2023
36
I want to build a miniature circuit which perform the following:
  • Measure a voltage signal between 0-20 millivolt DC signal
  • Measure a voltage signal between 0-24 V DC signal
  • Calculating a compensation factor as K = a Constant/voltage measured at step2
  • Multiply the compensation factor K with millivolt signal of step1
  • Transmit value got in step4 as standard 4-20Ma[500 Ohm] /RS485 signals
I want the circuit to be made on a single board with ICs [as tiny as possible].
Thank you.
 

crutschow

Joined Mar 14, 2008
33,962
The simplest solution would be to use a small microprocessor with a built-in A/D converter.
You would need to add an op amp amplifier to boost the 20mV signal to several volts for the A/D to read accurately.
Transmit value got in step4 as standard 4-20Ma[500 Ohm] /RS485 signals
4-20mA is an analog signal while RS485 is a digital signal, so which do you want?
 

Thread Starter

anoopak

Joined Nov 24, 2023
36
The simplest solution would be to use a small microprocessor with a built-in A/D converter.
You would need to add an op amp amplifier to boost the 20mV signal to several volts for the A/D to read accurately.
4-20mA is an analog signal while RS485 is a digital signal, so which do you want?
RS485 is preferred. Any suggestions for opamp and microprocessor
 

crutschow

Joined Mar 14, 2008
33,962
Any suggestions for opamp and microprocessor
Assuming you want to have just one power supply, a low offset voltage, rail-to-rail type op amp would likely be best, such as shown in this search.

I'm not familiar with what would be the best micro for your purpose, but some knowledgeable person on this forum will likely help you with that.

What resolution/accuracy do you want?
Please don't state more than you actually need if you want a simple, low-cost system.
 
Last edited:

Ian0

Joined Aug 7, 2020
9,428
For similar applications I use Microchip's MCP60xy series. where "x" represents the speed option and "y" represents the number of devices in the package.
The 6064 is a quad that is quick enough for most mains and DC monitoring applications. 6074 is a faster, quieter one.
 

Thread Starter

anoopak

Joined Nov 24, 2023
36
Assuming you want to have just one power supply, a low offset voltage, rail-to-rail type op amp would likely be best, such as shown in this search.

I'm not familiar with what would be the best micro for your purpose, but some knowledgeable person on this forum will likely help you with that.

What resolution/accuracy do you want?
Please don't state more than you actually need if you want a simple, low-cost system.
Resolution required is 1/2000 [i.e. 10 microvolt] [bare minimum]
 

Thread Starter

anoopak

Joined Nov 24, 2023
36
In that case, what is your source impedance, operating temperature and your measurement bandwidth, as component noise will start to become important.
Basically millivolt is generated from a Wheatstone bridge. Impedaence is 1.1K ohm. Operating temperature in 50 degree centigrade. I will confirm the measurement bandwidth later. [ I am a chemical guy so need to clarify about this parameter. I am assuming you are referring to RS485 bandwidth]
 

Ian0

Joined Aug 7, 2020
9,428
Basically millivolt is generated from a Wheatstone bridge. Impedaence is 1.1K ohm. Operating temperature in 50 degree centigrade. I will confirm the measurement bandwidth later. [ I am a chemical guy so need to clarify about this parameter. I am assuming you are referring to RS485 bandwidth]
No, I'm referring to the measurement bandwidth of the source, not exactly how fast the signal can change, but the speed of change that you wish to be able to measure.
Low impedance (1.1k) is good for noise, you should be able to achieve 10uV resolution with some care.
The resistors will generate noise according to Vn=√(4kTfR) where R is the resistance, f is the measurement bandwidth, T is absolute temperature and k is Boltzmann's constant. You will not be able to measure anything less than Vn.
Added to that is the amplifier noise.
 

Thread Starter

anoopak

Joined Nov 24, 2023
36
No, I'm referring to the measurement bandwidth of the source, not exactly how fast the signal can change, but the speed of change that you wish to be able to measure.
Low impedance (1.1k) is good for noise, you should be able to achieve 10uV resolution with some care.
The resistors will generate noise according to Vn=√(4kTfR) where R is the resistance, f is the measurement bandwidth, T is absolute temperature and k is Boltzmann's constant. You will not be able to measure anything less than Vn.
Added to that is the amplifier noise.
Ok. Thank you for the information.
 

MisterBill2

Joined Jan 23, 2018
17,717
OK, so there will be a 12 bit A/D converter. If the TS can afford two converters, one for the 0 to 20 millivolts and one for the 20 volts signal, then the op-amp with the high gain can be avoided. And an amplifier with that much gain is a challenge to make exact.
Is that "bridge derived signal" from a pressure sensor gage or a load cell? And what about that 20 volt signal. What is the range of the signal. (Min to max volts)
 

Thread Starter

anoopak

Joined Nov 24, 2023
36
OK, so there will be a 12 bit A/D converter. If the TS can afford two converters, one for the 0 to 20 millivolts and one for the 20 volts signal, then the op-amp with the high gain can be avoided. And an amplifier with that much gain is a challenge to make exact.
Is that "bridge derived signal" from a pressure sensor gage or a load cell? And what about that 20 volt signal. What is the range of the signal. (Min to max volts)
I will describe the problem in detail:
There is a calorimeter sensor which outputs a dc voltage in the range 0-20millivolt. This signal is generated from a Whetstone bridge working on an excitation current of 100milli Amperes, which is kept constant [Excitation voltage ~20V]. The sensor output is bound to change with ambient temperature [results in a change in excitation voltage also]. This excitation voltage is measured and a correction factor is derived from this excitation voltage as K. Then the corrected sensor output= indicated output X K.Presently, i am using two separate process indicators to transmit sensor output and Excitation voltage as Standard signal [4-20mA]to a PLC and correction is carried out in PLC.
To make this signal conditioning process simple and economic, I want to build a Custom signal conditioning board for the calorimetric sensor which can be mounted over the sensor.This board shall sense both millivolt signal from bridge and the excitation voltage and do the NECESSARY correction on sensor output and send this information as Standard signal [4-20mA or RS485] to the PLC. The resolution shall be 0.01millivolt.
For this purpose I searched for OP aMP, ADC, Microprocessor and RS485 converter chips.Being a Chemical Engineer [With limited Knowledge of electronics], I am not able to converge to specific models for the above ICs. So I want help to Choose the ICs and if possible the circuit to connect the ICs.
 

crutschow

Joined Mar 14, 2008
33,962
For this purpose I searched for OP aMP, ADC, Microprocessor and RS485 converter chips.Being a Chemical Engineer [With limited Knowledge of electronics], I am not able to converge to specific models for the above ICs.
Sounds like you may need more help than we can give on this forum.
Do you know anyone that can help you with the electronics?
Who will program the micro?
 

MisterBill2

Joined Jan 23, 2018
17,717
OK, now the application is clear, and there are two considerations immediately: First, both sides of the 20 millivolt signal are far off-ground, that is, floating exactly like a strain gauge. Second, that 20 volt signal is possibly referenced to a system common that may, or not, be common to the external instrumentation system.
Now for the interesting considerations: The bridge current is kept constant for a reason, which is to assure that the voltage drop across each bridge element is dependent on only the resistance. So immediately it is clear that the accuracy depends on the bridge current and not on the bridge voltage. (of course, this depends also on the change of resistance with ambient temperature is identical for all four bridge elements.) So really the second variable to be measured would be bridge current, NOT bridge voltage, because the voltage gets adjusted to hold the current constant. So you need to read and understand the bridge explanation more clearly.
The difference now is that a voltage developed from the bridge current can easily be in the same 20 millivolt range.
At this point iit becomes clear that what you need is "Instrument Amplifiers" ,possibly with digital outputs. These amplifiers must be able to work with up to a 15 volt common mode input voltage. They must also have isolated outputs, either digital or analog.
The good news is that this is the same as many bridge type force and pressure gauge transducers, meaning many standard products are available. The added expense will be a high accuracy shunt resistor to sample the bridge excitation current. That might cost up to $15 (USD) The second good news is that can connect directly to the analog inputs on the PLC, if it has them, using existing wiring, and probably existing code, with only coefficient changes.

This is exactly the sort of systems I have used in many test stands 20 years ago. The stability, resolution, and accuracy met factory requirements from startup to machine obsolescence years later.
 

BobTPH

Joined Jun 5, 2013
8,642
You need an instrumentation amplifier to handle that level of a differential signal. You will need to use 0.05% resistors at least. That exactly matches the resolution you want 1 part in 2000. To avoid losing precision, I would go with 0.01%.

You need any micro with a 12-bit ADC and a UART. If the sampling rate is more than about 1KHz, the speed of the micro might be an issue.

You need an RS485 line driver.

I don’t know anything about 4-20 mA. I assume there are chips that take digital input and output that signal. You need 12 bit minimum for that.
 

Ian0

Joined Aug 7, 2020
9,428
What is the equation for the temperature compensation? If it is simple then you may be able to get from input to output entirely in analogue, if you want 4-20mA output.
 

MisterBill2

Joined Jan 23, 2018
17,717
Keep in mind that actually you need to be aware of the variations in current because that has a linear effect on the output, as the fact that current is the regulated quantity, not the voltage. So that will change the calibration adjustment code a bit.
The one company that does sell ready to use packages is National Instruments, with a fairly broad product line. Analog Devices may also sell complete packages as well, it has been a lot of years since I purchased a few hundred instrument amplifiers from them (2B31J, as I recall) A small application specific package using an IA integrated circuit, coupled with a processor, is also quite possible.
 
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