Detecting drops in resistance in a low resistance circuit

Thread Starter

Jeffery Vahrenkamp

Joined Jun 16, 2016
42
I have a circuit where I am trying to detect when a systems resistance drops from ~5-20 mOhms to half or lower (~<1-10 mOhms) with a micro controller.
my microcontroller can read between 0-3.3v with 10 bit precision, or about 3.3 mV changes. I was hoping to get much larger changes out to give me confidence that I'm actually measuring the drop in resistance. My MC can also source 15-20mA of current at a pin.

My plan is to make a voltage divider with R1 much higher than R2, since making them even sort of equal will draw too much current from my pin. With V=IR being 3.3/.015= 220 ohms for the lowest resistor I can use for R1. At best this gives me ~ .3 mV of max voltage drop if my test circuit starts with 20 mOhms of resistance. I would like to build a voltage amplifier circuit to boost the voltage signal from the divider. I've drawn up a preliminary circuit ( you can replace "mosfet" with any transistor in your mind), but looking at diagrams of BJT and MOSFET transistor circuits that are used to amplify voltage, I am guessing I am missing something. I need around 1000x amplification.

I'm not sure the voltage that would be at the Gate of a MOSFET would be sufficient to turn it on at all, or that the voltage would be amplified. Maybe a darlington pair of BJTs? I'm worried that the power through the base of a BJT would interfere with my measurements.

Input would be most appreciated.
circuit.jpg
 

kaindub

Joined Oct 28, 2019
15
I have a circuit where I am trying to detect when a systems resistance drops from ~5-20 mOhms to half or lower (~<1-10 mOhms) with a micro controller.
my microcontroller can read between 0-3.3v with 10 bit precision, or about 3.3 mV changes. I was hoping to get much larger changes out to give me confidence that I'm actually measuring the drop in resistance. My MC can also source 15-20mA of current at a pin.

My plan is to make a voltage divider with R1 much higher than R2, since making them even sort of equal will draw too much current from my pin. With V=IR being 3.3/.015= 220 ohms for the lowest resistor I can use for R1. At best this gives me ~ .3 mV of max voltage drop if my test circuit starts with 20 mOhms of resistance. I would like to build a voltage amplifier circuit to boost the voltage signal from the divider. I've drawn up a preliminary circuit ( you can replace "mosfet" with any transistor in your mind), but looking at diagrams of BJT and MOSFET transistor circuits that are used to amplify voltage, I am guessing I am missing something. I need around 1000x amplification.

I'm not sure the voltage that would be at the Gate of a MOSFET would be sufficient to turn it on at all, or that the voltage would be amplified. Maybe a darlington pair of BJTs? I'm worried that the power through the base of a BJT would interfere with my measurements.

Input would be most appreciated.
View attachment 200128
You're sort of there
Firstly create a constant current source to put a known value of current through your "system" resistance
You then use an op amp to amplify the voltage that appears across your "system" resistance.
Calculate the amplifier gain so that the output voltage of the op amp circuit is near 3.3v
Then feed the opamp output to an analog pin of your micro
Does this make sense?
Robert
 

Thread Starter

Jeffery Vahrenkamp

Joined Jun 16, 2016
42
You're sort of there
Firstly create a constant current source to put a known value of current through your "system" resistance
You then use an op amp to amplify the voltage that appears across your "system" resistance.
Calculate the amplifier gain so that the output voltage of the op amp circuit is near 3.3v
Then feed the opamp output to an analog pin of your micro
Does this make sense?
Robert
I think so. I'm going to need about 10,000-100,000x amplification to get that, is that within the range of a reasonably priced opamp? Not sure where the max gain is on the op amp data sheets (or even which op amps to look at).
 

kaindub

Joined Oct 28, 2019
15
I think so. I'm going to need about 10,000-100,000x amplification to get that, is that within the range of a reasonably priced opamp? Not sure where the max gain is on the op amp data sheets (or even which op amps to look at).
If you put in 500mA, then at maximum resitance of 20mOhm the voltage across the resisitor will be 10mV
A gain of 330 on the op amp will give an opamp output of 3.3V
Since you are measuring such small resisitances you will need to use a 4 wire mearurement arrangement .
 

Reloadron

Joined Jan 15, 2015
5,370
If you look at DLRO (Digital Low Resistance Ohmmeter) circuits they normally source a current of 1 Amp up to 5 or 10 Amps. Placing two 20 mOhm (0.020 Ohm) resistors in parallel you have a Rtotal of 10 mOhm. With a standard applied current of 1.0 Amp you have 10 mV drop. If I remove one resistor my Rtotal is now 20 mOhm (.020 Ohm). With the same 1.0 Amp applied my voltage across the resistance is now 20 mV. Looking at a 10 bit ADC you have 1024 quantization levels so with your 3.3 volt reference you get 3.3 / 1024 = 0.00322 or 3.2 mV resolution on a good day and not taking into account any quantization errors. The numbers are also based on a 1.0 Amp constant current source.

If you arrange for your uC to use I2C I would think about using an ADC like the ADS1115 which has 16 bit ADC, can communicate over I2C with ease, runs on Vcc 3.3 volts and even has programmable gain. The chip is even available on a break out board ready for use as seen in the link. The chip offers 4 channel single ended but also 2 channel differential mode operation and considering what you are wanting a differential mode ADC is what you want. The chip also allows a programmable address so while you don't need it you could place four of them on the bus. With 16 bit and 16X gain you get 16x gain +/- 0.256V 1 bit = 0.125mV. That's 125 uV resolution. With the low resistance involved you really need a Kelvin type configuration.

I simply do not see what you have proposed as a viable solution and the above is merely my thinking on the subject.

Ron
 

wayne1956

Joined Sep 9, 2019
1
I have a circuit where I am trying to detect when a systems resistance drops from ~5-20 mOhms to half or lower (~<1-10 mOhms) with a micro controller.
my microcontroller can read between 0-3.3v with 10 bit precision, or about 3.3 mV changes. I was hoping to get much larger changes out to give me confidence that I'm actually measuring the drop in resistance. My MC can also source 15-20mA of current at a pin.

My plan is to make a voltage divider with R1 much higher than R2, since making them even sort of equal will draw too much current from my pin. With V=IR being 3.3/.015= 220 ohms for the lowest resistor I can use for R1. At best this gives me ~ .3 mV of max voltage drop if my test circuit starts with 20 mOhms of resistance. I would like to build a voltage amplifier circuit to boost the voltage signal from the divider. I've drawn up a preliminary circuit ( you can replace "mosfet" with any transistor in your mind), but looking at diagrams of BJT and MOSFET transistor circuits that are used to amplify voltage, I am guessing I am missing something. I need around 1000x amplification.

I'm not sure the voltage that would be at the Gate of a MOSFET would be sufficient to turn it on at all, or that the voltage would be amplified. Maybe a darlington pair of BJTs? I'm worried that the power through the base of a BJT would interfere with my measurements.

Input would be most appreciated.
View attachment 200128
Just a point of clarification... Do you need to measure these resistance or just be able to detect when that switch is open or closed?
 

Thread Starter

Jeffery Vahrenkamp

Joined Jun 16, 2016
42
Just a point of clarification... Do you need to measure these resistance or just be able to detect when that switch is open or closed?
Just need to be able to check when it is open or closed, but I only have access to the circuit where the first resistor is. To make this clear, I'm looking at a car and I want to be able to detect when the starter is being activated, and then activate a circuit at that time. The first small resistor is the battery and the second is the starter and it's solenoid.
 

Reloadron

Joined Jan 15, 2015
5,370
If what you want is to know when an automotive starter is cranking an engine I would consider using a shunt directly off the battery. Depending on engine and starter an unloaded starter draws about 60 to 150 amps and under a load can draw 250 amps or greater. If you know about what the starting current is on a nice sunny day just place a current shunt on the battery high side. Most current shunts have a 50 mV output at their full scale current. Another option is just monitor the battery voltage. A fully charged 12 volt lead acid battery has a voltage of 12.6 volts and when the starter engages that voltage drops. Can't you just use a voltage divider and look at system voltage? I have no idea how much of a drop there is but it has to drop noticeable on start even with a good healthy battery. Lastly why can't you just pick a voltage off the starter solenoid using a voltage divider network? Max voltage in an automotive 12 V system normally won't exceed 15 volts, that is calling it high. Using a simple two resistor 5:1 divider gets that down to 3.0 volts max.

If all you want to do is detect if a starter is cranking an engine there should be plenty of simple ways to get it done.
The first small resistor is the battery and the second is the starter and it's solenoid.
I don't see that as a good analogy?

Ron
 

Thread Starter

Jeffery Vahrenkamp

Joined Jun 16, 2016
42
If what you want is to know when an automotive starter is cranking an engine I would consider using a shunt directly off the battery. Depending on engine and starter an unloaded starter draws about 60 to 150 amps and under a load can draw 250 amps or greater. If you know about what the starting current is on a nice sunny day just place a current shunt on the battery high side. Most current shunts have a 50 mV output at their full scale current. Another option is just monitor the battery voltage. A fully charged 12 volt lead acid battery has a voltage of 12.6 volts and when the starter engages that voltage drops. Can't you just use a voltage divider and look at system voltage? I have no idea how much of a drop there is but it has to drop noticeable on start even with a good healthy battery. Lastly why can't you just pick a voltage off the starter solenoid using a voltage divider network? Max voltage in an automotive 12 V system normally won't exceed 15 volts, that is calling it high. Using a simple two resistor 5:1 divider gets that down to 3.0 volts max.

If all you want to do is detect if a starter is cranking an engine there should be plenty of simple ways to get it done.

I don't see that as a good analogy?

Ron
Just so the situation i'm talking about makes sense, this is for a super capacitor jump starter. So if I am working with a mostly depleted battery, I did think that I could just measure the voltage off of the battery and watch for it to drop to ~ 0 and then connect the super capacitors. However, I trying to figure out how to do it for a completely dead battery, where I would not see a voltage drop. So I am assuming an totally unloaded system for probing, except what I provide with maybe a small lithium battery pack that I use to power the micro controller and charge the super capacitors. I could probably do a pulsed 1A current to check the state, depending on the size of the lithium ion pack I use. I am probably going to use 2-4 18650s to keep the charge time for the super capacitor from being forever.

I consider the battery the first small resistor because it has an internal resistance of around 5-20 mOhm resistance when connected to the super capacitors, and who knows what the resistance is for the starter and solenoid coil.
 

Reloadron

Joined Jan 15, 2015
5,370
Wow, there's a new wrinkle. Normally when a 12 volt automotive battery gets down around 11.9 to 12.0 volts it is considered discharged or I guess we could say dead. While I have seen dead I can't say I ever saw one dead as in 0 volts. Actually and this is just a guess if the battery voltage is so low it can't even make a click on the starter solenoid we won't have an easy way to know that the starter is being tried. The existing system voltage would need to be enough to get the solenoid to engage. Short of that, using your drawing as an example, there will be no change in Rtotal. Now I know where you came up with 0.020 Ohm or 20 mOhm. That is a rough number for the internal resistance of a 12 volt lead acid battery in an automotive application. We can't really use that number unless we are concerned with maximum power transfer being when the external load resistance is equal to the internal battery resistance. That's what the power transfer theorem is about but I don't see it applying here.

Knowing what I know now I am not sure how to go about this because we never know what we have to start with making it hard to know to what level the voltage will drop when we turn a key and that assumes the dead battery can pull in a starter solenoid. Hopefully another member has some thoughts?

Ron
 

Thread Starter

Jeffery Vahrenkamp

Joined Jun 16, 2016
42
Yeah, as I was describing the situation to you, i did realize that the starter, it's solenoid, and even the relay to the solenoid are not going to engage without any power. I think it may just have to be a two man team thing where someone turns the key and then the jump starter is engaged. Or maybe a long cable so the person training the key can turn it over. And for the 99% of the time when the battery is not totally annihilated i can just measure the drop in voltage when the starter kicks in.

In terms of the low resistance from the battery, i think it's a reasonable concern since it's connected directly to the super caps which can deliver as much power as the battery can handle before the caps are discharged. This is the source of the problem as the battery is going to suck all the juice out of the caps before the starter gets any. Its fine if i can disconnect the battery cable, but I have a van where i can't really do that.
 

Reloadron

Joined Jan 15, 2015
5,370
When I was a kid I was a motor head. Not quite today's automotive world It was the 60s. :) Something I remember from those days was being able to start from under the hood, just a simple pendant we would clip on the solenoid and battery + terminal. With that in mind I would venture for a one man show using a pendant. Super capacitors were just coming into their own when I retired and they really never figured into my areas of work scope so never learned anything about them. I do see the gizmos like this one used to jump start automotive. I have no clue how well they work or just how much power for how long they can deliver. The linked unit just says 550 Amp? Obviously that is not AH so do we figure that as CCA? Even that CCA stuff is a bunch of crap. I liked the original Amp Hour rating system. With CCA the rating refers to the number of amps a 12-volt battery can deliver at 0°F for 30 seconds while maintaining a voltage of at least 7.2 volts. Right, 7.2 volts at which point a 12 volt battery is damaged. Anyway, yes, a remote pendant is something I would consider. :)

Ron
 

Thread Starter

Jeffery Vahrenkamp

Joined Jun 16, 2016
42
The super caps i have have enough juice to start the car, but just barely. I have two cable clamps with a resistance of .001ohm resistance each, and with both connected, i can't get the car to start, but with one connected and the post of the terminal capacitor on the + terminal can do it. I may have to stick another one on the end and push it up to 19v. Should be plenty of juice at that voltage, but I'm not sure if that might start damaging things in the car.
 
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