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Low-Resistance Meter

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Hello Sir,

I am interested to one circuit that can measure resistance value.
Take a look the schematics, I need to review it and understand whole circuit.

resistor_meter.PNGresistor_meter.PNG


Let me explain it little bit,

1. Seems like a constant current source has been configured using U1A. Like other simple circuits, a voltage divider is using RV1, 90k and 10k parallel combination gives 0.1 V reference voltage. Feedback forces a voltage VCC-VRef across R3 , giving an emitter current (and therefore an output current) Ie= (VCC - VRef)/R? Dont we consider base to emitter DC offset voltage? What about U1A input offset ? Grounded load, with high power resistor !

2. Looks like the emitter current need to sense by placing a unknown resistor in P2 connector ?

3. This differential amplifier inputs are not same, means U1B inverting input getting the input from parallel of R5 and R3, while non inverting inputs using voltage divider, keeping also allowable input offsets.


4. Not sure, what U2A is doing here, U2B looks a voltage follower?

Kindly share your knowledge.
 

dr pepper

Well-Known Member
Most Helpful Member
1. your equation isnt quite right, the current from the current source will always be 100ma vref/r3, because the op amps feedback comes from r3, the voltage across r3 will always be 0.1v and therefor current through it and the unknown resistor will remain at 100ma(assuming theres enough voltage from the power supply).
Q1 is an emmitter follower, looks like they used this as the op amp cannot source 100ma itself, so Q1 is a current amplifier.
2. your correct
3.U1b is configured as an instrumentation amplifier, it simply measures the difference in voltage across the unknown resistance, with an offset zero control from Rv2, and R8 is there to match the impedances seen by U1b to minimise voltage offset on its output.
4.U2a is an amplifier set by Rv3 from x1 to x10, so that you can get 1v = 1 ohm output, U2b is a buffer or follower your correct.

Personally I think you could do this with just the 1 dual op amp Ic, U1b could be made to have x10 gain, and if your driving a high impedance load like a meter you dont need the follower.
 
Dear Sir Pepper,

Thank you very much to take participate here.
I wish to learn from you !

1. your equation isnt quite right, the current from the current source will always be 100ma vref/r3, because the op amps feedback comes from r3, the voltage across r3 will always be 0.1v and therefor current through it and the unknown resistor will remain at 100ma(assuming theres enough voltage from the power supply).
Could you kindly elaborate it more according to ideal current source? I might be wrong to interpret the equation, this transistor Q1 is NPN, so it will change. What does it means 100mA vref/r3? Input of the U1A is fed to base of Q1, explain why 0nly 0.1V is considerable.

Q1 is an emmitter follower, looks like they used this as the op amp cannot source 100ma itself, so Q1 is a current amplifier.
Please, explain how this transistor opamp connection acts as current amplifier.

3.U1b is configured as an instrumentation amplifier, it simply measures the difference in voltage across the unknown resistance, with an offset zero control from Rv2, and R8 is there to match the impedances seen by U1b to minimise voltage offset on its output.
Could you kindly explain the necessity of R7, feedback resistor in differential amplifier? If we wont control zero offset, what would be the problem?

4.U2a is an amplifier set by Rv3 from x1 to x10, so that you can get 1v = 1 ohm output, U2b is a buffer or follower your correct.
At this moment why we need buffer follower and amplifier ? Do you mean to keep the same input change or input to output impedance correction?

Personally I think you could do this with just the 1 dual op amp Ic, U1b could be made to have x10 gain, and if your driving a high impedance load like a meter you dont need the follower.
Do you want me to use LM324 or other ? Any recommendation ?
 
Last edited:

KeepItSimpleStupid

Well-Known Member
Most Helpful Member
Why we need it?
U2A: You have 1% resistors. If you want better than 1%, you have to be able to trim the output.
U2B: You have to "do something" with the second OP-AMP in the package. It doesn't cost you anything. It's a dual OP-amp package.

U1A: Maintains 0.1 V across R2. Q2 increases the OP-amp sourcing ability in a unidirectional way.
It maintains 0.1 V across R2. The 0.1 V reference is tied to Vcc (5V), thus it is proportional to the power supply. It's not an independent reference. Microprocessors and automotive circuits typically use Vcc as the sensor reference. A DVM may give the wrong answer under certain circumstances.

U2A: A lot of info here: http://www.analog.com/en/analog-dialogue/articles/deeper-look-into-difference-amplifiers.html
 
Last edited:
Dear Sir ,
Nice to get your feedback again.
Let me draw points those are confusing to me.

U2A: You have 1% resistors. If you want better than 1%, you have to be able to trim the output.
Are you talking about 1% resistor in P2 connector ? Which percentage calculation you are considering ? Output of U1B is showing 100mV, its being 10 times increased at the output of U2B, your "trimming" means amplified by U2A?

U2B: You have to "do something" with the second OP-AMP in the package. It doesn't cost you anything. It's a dual OP-amp package
.
LM358, has 2 opamp, does it means one single package you want me to use 4 opamps?
Can you suggest anything ?

U1A: Maintains 0.1 V across R2. Q2 increases the OP-amp sourcing ability in a unidirectional way.
It maintains 0.1 V across R2. The 0.1 V reference is tied to Vcc (5V), thus it is proportional to the power supply.
This little bit confusing, voltage divider rule has been applied on RV1 , portion of it also contributing in voltage distribution, parallel setting of R1 and R2 is taking part for keeping 0.1v in U1A input. First expart says "the voltage across r3 will always be 0.1v and therefor current through it and the unknown resistor will remain at 100ma(assuming theres enough voltage from the power supply)." Dont you agree it?

It's not an independent reference. Microprocessors and automotive circuits typically use Vcc as the sensor reference. A DVM may give the wrong answer under certain circumstances.
Yes, well said, I did see in many circuits.

Let me read this carefully.
 

KeepItSimpleStupid

Well-Known Member
Most Helpful Member
Are you talking about 1% resistor in P2 connector ? Which percentage calculation you are considering ? Output of U1B is showing 100mV, its being 10 times increased at the output of U2B, your "trimming" means amplified by U2A?
OK, spoke too soon. The 1% resistor is just one place. The differential amp does better with matched resistors Mostly in CMMR. There is going to be a gain error here too.

If the device is going to be manufactured on a large scale then one of the first things I would want to know is whether the trim would work for all resistors. Some being too high and some being too low. Yep, 0.1% resistors in the differential amplifier would give better CMMR, but is that necessary? Are these differential amp resistors 10%, 5%, 1%, 0.1%, 0.01%?

At low ohms, you really want a KELVIN type measurement. measure the current through the unknown and the voltage across the unknown. This meter won;t do that.
It also doesn't use a 4-terminal resistor. e.g. https://www.vishay.com/docs/30217/cpsl.pdf

Is the ZERO adjust a trim or is it a user adjustable part of the meter. Call it lead compensation?

How low of a resistance do you expect to measure? What are the limitations? What are the temperature effects?

Should some of these resistors be in a network package where the temperature coefficient is better?

There are 2, 4, 5 and 6 wire ohm measurements. See: https://www.tek.com/document/whitep...measurements-using-6-wire-ohms-measurement-te

What's the low end performance? With a single supply, it's difficult to null Vos completely and it won't go to zero, but it will get close.

it's not meant to be a 5 digit milli-ohmeter.

==

One OP-amp is used as a buffer. What can I say, They are dual OP-amp packages. You can elect not to use the buffer and configure it as one amplifying ground, just so it doesn't oscillate and warmup the IC package. The buffer can negative affect the temperature performance. but is the temperature performance necessary.

==

First expart says "the voltage across r3 will always be 0.1v and therefor current through it and the unknown resistor will remain at 100ma(assuming theres enough voltage from the power supply)." Dont you agree it?
First order approximation, yes. What if Vcc changes? Vcc can easily change with temperature. If the reference of the A/D is Vcc. then those effects go away, but a voltmeter can get the wrong answer.

Here http://cds.linear.com/docs/en/datasheet/1009ff.pdf is a reference. Here http://www.analog.com/en/design-cen...ge-reference-selection-evaluation-wizard.html are other references.
 
OK, spoke too soon. The 1% resistor is just one place. The differential amp does better with matched resistors Mostly in CMMR. There is going to be a gain error here too.
If you talk about U1B CMMR, datasheet says, at Vcm = 0 V to (V + −1.5 V), its about 65 to 85 dB. Not sure, why you are saying about gain error relating CMMR.
In practice, a CMRR in excess of 80dB to 100dB will be needed for high accuracy measuring system ,http://www.ntu.edu.sg/home/aschvun/FAQ/DiffAmp.html
This idea is telling that,
To eliminate the effect of the common mode component, we can either

(i) make the input common mode component equal to zero, i.e. make V2 = -V1
such that the average value of the two input signals equal to zero
or

(ii) choose the resistor values of R4 to R7 in such a way that ( Common Mode Gain ) Acm is zero.

If the device is going to be manufactured on a large scale then one of the first things I would want to know is whether the trim would work for all resistors. Some being too high and some being too low. Yep, 0.1% resistors in the differential amplifier would give better CMMR, but is that necessary? Are these differential amp resistors 10%, 5%, 1%, 0.1%, 0.01%?
What maximum resistor value do you think it can measure? Can you give me an idea how could I implement a wide range of differential amp resistor?

At low ohms, you really want a KELVIN type measurement. measure the current through the unknown and the voltage across the unknown. This meter won;t do that.
It also doesn't use a 4-terminal resistor. e.g. https://www.vishay.com/docs/30217/cpsl.pdf
Good point, but I dont think its a temperature dependent measurement. I saw other measurement system like "bio impedance " measurement.
People are using, smart MCU or DSP, generating high accurate constant current source. I started a PhD there, but did not continue.
LT,TI are offering many KIT to measure accurately, I have used those in 2015. Here, just I want to use simple circuit those components I have in my stock.

Is the ZERO adjust a trim or is it a user adjustable part of the meter. Call it lead compensation?
Yes, well said.

How low of a resistance do you expect to measure? What are the limitations? What are the temperature effects?
Its also arise in my mind, not sure how much I can measure in this design.

Should some of these resistors be in a network package where the temperature coefficient is better?

There are 2, 4, 5 and 6 wire ohm measurements. See: https://www.tek.com/document/whitep...measurements-using-6-wire-ohms-measurement-te

What's the low end performance? With a single supply, it's difficult to null Vos completely and it won't go to zero, but it will get close.

it's not meant to be a 5 digit milli-ohmeter.
Yes, these points are important, but as long as this design is poor we cant think all facilities.
==

One OP-amp is used as a buffer. What can I say, They are dual OP-amp packages. You can elect not to use the buffer and configure it as one amplifying ground, just so it doesn't oscillate and warmup the IC package. The buffer can negative affect the temperature performance. but is the temperature performance necessary.
Are you indicating not to use, U2B? Not need to set gain 1? not to shift from high output to low input impedance ?


First order approximation, yes. What if Vcc changes? Vcc can easily change with temperature. If the reference of the A/D is Vcc. then those effects go away, but a voltmeter can get the wrong answer.

Here http://cds.linear.com/docs/en/datasheet/1009ff.pdf is a reference. Here http://www.analog.com/en/design-cen...ge-reference-selection-evaluation-wizard.html are other references.
Any particular design?
 

Colin

Active Member
Explain exactly: what values of resistance you want to measure, are they resistors? Are they in a circuit? Is it wiring that you want to measure?
 
Explain exactly: what values of resistance you want to measure, are they resistors? Are they in a circuit? Is it wiring that you want to measure?
Thank you for your response, according to this circuit I have collected and posted here, the measurement is for 1Ω. We are discussing the circuit operation and for sure how to get better performance.
If you read this post, you would see some facts considering 1. CMRR 2. Temperature dependencies 3. Trimming output 4. buffer/voltage follower 5. Input resistance tolerance. 6. VCC dependencies etc.
 

KeepItSimpleStupid

Well-Known Member
Most Helpful Member
We are discussing the circuit operation and for sure how to get better performance.
First let's talk: How is the reading measured?
a) DVM
b) Microprocessor with it's A/D set as Vcc of the ohms converter as Vref?
(Is Vcc of the uP shared with the ohms converter?)
c) DPM (Digital Panel Meter) with external reference?
(the DPM can usually provide the gain trim)

Is this, a one of, for in house use? Should you be using it as a (Go)/(No Go) sort of design. Add comparators?

This is from the datasheet:
Output short circuits either to ground or to the positive power supply should be of short time duration. You can fix that if it's a problem. Put a current limiting resistor in the feedback loop of the buffer. i.e. Th e output of the OP amp goes to a resistor, and the negative feedback and output comes from the other end of the resistor

It has limitations at the low end of the range. From the datasheet: It allows allows Direct Sensing Near GND and V OUT
Also Goes to GND. Near GND mucks up the low end.

Lead resistance, I think is going to matter. Not sure if it can be adjusted completely out.

CMMR matters with lead length (AC line pickup) and lead dress (routing). IA's are best implemented in IC form because of the ability to trim the resistors to get the best CMMR.

There's all sorts of questions that should be answered to determine suitability of the design for the application. I just tried to tear it apart as though I was a "Devil's advocate".

What about ESD protection or touching something powered?
 
First let's talk: How is the reading measured?
a) DVM
b) Microprocessor with it's A/D set as Vcc of the ohms converter as Vref?
(Is Vcc of the uP shared with the ohms converter?)
c) DPM (Digital Panel Meter) with external reference?
(the DPM can usually provide the gain trim)

Is this, a one of, for in house use? Should you be using it as a (Go)/(No Go) sort of design. Add comparators?

This is from the datasheet:
Output short circuits either to ground or to the positive power supply should be of short time duration. You can fix that if it's a problem. Put a curre
nt limiting resistor in the feedback loop of the buffer. i.e. Th e output of the OP amp goes to a resistor, and the negative feedback and output comes from the other end of the resistor

It has limitations at the low end of the range. From the datasheet: It allows allows Direct Sensing Near GND and V OUT
Also Goes to GND. Near GND mucks up the low end.

Lead resistance, I think is going to matter. Not sure if it can be adjusted completely out.

CMMR matters with lead length (AC line pickup) and lead dress (routing). IA's are best implemented in IC form because of the ability to trim the resistors to get the best CMMR.

There's all sorts of questions that should be answered to determine suitability of the design for the application. I just tried to tear it apart as though I was a "Devil's advocate".

What about ESD protection or touching something powered?
Yes, seems like you want to make a new design.
Let me collect all component, I have LCD display, but to set Vcc , I dont have any MCU in mind.
Can I use AT89C2051, any suggestion to make A/D programming code?
Cant we replace a multi-meter by LCD?
 
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