How to connect this strain gauge?

[Micheal_Leo]

recently i bought this strain guage where vender said it is 4bridge, i am not sure how to connect this in wheatstone bridge , i have 120 ohms resistors, and 120ohm strain gauge.
please need guide thank you.

 

[sophiecentaur]

From the look of the image, I get the impression that those four areas are actually the four resistor elements of a bridge. Put volts across two terminals and measure the volts across the other pair, which should be around Zero with the gauge undistorted and balanced. Do you have a data sheet?

 

[Micheal_Leo]

From the look of the image, I get the impression that those four areas are actually the four resistor elements of a bridge. Put volts across two terminals and measure the volts across the other pair, which should be around Zero with the gauge undistorted and balanced. Do you have a data sheet?

thank you for reply, they did not have anything, from internet i found this

 

[sophiecentaur]

Looks ok to me. Wire it up and hit it with about 5V (arbitrary) and see what happens.

 

[Micheal_Leo]

Looks ok to me. Wire it up and hit it with about 5V (arbitrary) and see what happens.

i have AD620 FOR AMPLIFICATION, i connect strain guage with breadboard and arduino to get voltage output from arduino, arduino is not showing anything, please can you see the specimen principal stress direction and strain gauge orientation and selection

Looks ok to me. Wire it up and hit it with about 5V (arbitrary) and see what happens.

the strain gauge heat up when supply voltage but reading not coming out from arduino i think i made wrong in orientation starin guage

 

[Baluncore]

The voltage to the bridge power supply terminals need only be about 1 volt.
For a healthy symmetrical bridge, the resistance between the two signal output terminals, should be the same as the resistance between the two power terminals. If that is not the case, then the bridge has been damaged.

 

[Micheal_Leo]

The voltage to the bridge power supply terminals need only be about 1 volt.
For a healthy symmetrical bridge, the resistance between the two signal output terminals, should be the same as the resistance between the two power terminals. If that is not the case, then the bridge has been damaged.

but AD620 AMPLIFIER NEED 3-11VOLTAGE, I have supplied 3 voltage to both( strain guage and amplifier), does my strain gauge orientation and principal stress direction same?

 

[Baluncore]

Three volts should be OK, but heating the gauge can be a problem with higher voltages.

In an earlier picture, I see the strain gauge is glued onto a white structure. Maybe the connections you have used, or the orientation, is such that you get cancellation of the resistance changes due to the strain. Can you place the bridge gauge on a square of material for experimentation, until you have the electronics working?

 

[Micheal_Leo]

Three volts should be OK, but heating the gauge can be a problem with higher voltages.

In an earlier picture, I see the strain gauge is glued onto a white structure. Maybe the connections you have used, or the orientation, is such that you get cancellation of the resistance changes due to the strain. Can you place the bridge gauge on a square of material for experimentation, until you have the electronics working?

thank you for reply, you mean i print square shape 3D and put new bridge on it?

 

[Baluncore]

, you mean i print square shape 3D and put new bridge on it?

No printing required.
Just stick a bridge onto a scrap piece of flat material, something you can bend with your hand while you experiment with the electronics.

 

[Micheal_Leo]

No printing required.
Just stick a bridge onto a scrap piece of flat material, something you can bend with your hand while you experiment with the electronics.

yes i can do that

 

[sophiecentaur]

the strain gauge heat up when supply voltage but reading not coming out from arduino

I guess this could be a generational thing but did you consider using an independent supply and measuring the emerging volts with a DMM? An Arduino may be smart but you can't beat taking measurements in isolation if you don't know how the strain device will behave.

 

[Dullard]

You might want to take another? look at the AD620 datasheet.

 

[Baluncore]

You might want to take another? look at the AD620 datasheet.

... but AD620 AMPLIFIER NEED 3-11VOLTAGE, I have supplied 3 voltage to both( strain guage and amplifier), ...

The AD620 requires a minimum of ±2.3 V, that is 4.6 volts total. Depending on the circuit's range of output voltage, you will need more than a 5 volt supply. That will be too high for stable operation of the strain gauge.

 

[sophiecentaur]

The AD620 requires a minimum of ±2.3 V, that is 4.6 volts total. Depending on the circuit's range of output voltage, you will need more than a 5 volt supply. That will be too high for stable operation of the strain gauge.

Hence my question about using a DMM as a starter, to see what the gauge actually does. There must be a few mV to measure.

 

[Micheal_Leo]

No printing required.
Just stick a bridge onto a scrap piece of flat material, something you can bend with your hand while you experiment with the electronics.

vide DC
unfortunately i can not give 1v to bridge , my adapter can not set on 1 v , please guide thank you

 

[sophiecentaur]

unfortunately i can not give 1v to bridge , my adapter can not set on 1 v ,

Do you have a soldering iron, by any chance - or perhaps you do your connections on a bread board? Make yourself a potential divider and 'manufacture' 1V that way from the same power supply that you are using for other things.

Edit: and do you have a DMM?

 

[Micheal_Leo]

No printing required.
Just stick a bridge onto a scrap piece of flat material, something you can bend with your hand while you experiment with the electronics.

vide DC
unfortunately i can not give 1v to bridge , my adapter can not set on 1 v , please guide thank you

Do you have a soldering iron, by any chance - or perhaps you do your connections on a bread board? Make yourself a potential divider and 'manufacture' 1V that way from the same power supply that you are using for other things.

Edit: and do you have a DMM?

yes i have these

how can i establish potential divider for 1V , thank you

 

[sophiecentaur]

how can i establish potential divider for 1V , thank you

You must know know the basics of a potential divider? This link will remind you about the basic potential divider idea. But, for a load of resistance as low as the 120Ω bridge, it may be better use either a low voltage regulator (off the shelf) to give you 1V from your amplifier power supply.
This is very basic circuit stuff and you really should try to learn about it for your future use. It's not really appropriate for me to design your circuits for you. There must be loads of information about interfacing the Arduino with simple devices - even your strain gauge bridge.

 

[Micheal_Leo]

You must know know the basics of a potential divider? This link will remind you about the basic potential divider idea. But, for a load of resistance as low as the 120Ω bridge, it may be better use either a low voltage regulator (off the shelf) to give you 1V from your amplifier power supply.
This is very basic circuit stuff and you really should try to learn about it for your future use. It's not really appropriate for me to design your circuits for you. There must be loads of information about interfacing the Arduino with simple devices - even your strain gauge bridge.

have tried to manage 1V using 2(1.5k) and 10k , i am thinking like this not sure i am right or wrong , appreciate your valuable time and guidance

the out from voltage divider circuit will go to breadboard down relay, and the strain guage + and - will connect to this relay

 

[sophiecentaur]

have tried to manage 1V using 2(1.5k) and 10k , i

Are you using the diagram in the link I quoted? Which is R2 and R1 and what is your supply voltage? Show your working how you calculated 1V out? Your output volts can be obtained by many different values for R2 but consider what happens when you connect the bridge across the output of the potential divider, the resistance ratio changes a lot (current will flow through the bridge). The bridge presents 120Ω and you need to consider this because it appears in parallel with R2 so you need a different R1. Google with "using potential divider to change supply volts" and you'll find loads of discussions which show you what I mean. If you have a box of resistors then your bread board will make this easy to experiment with.
A 1V voltage regulator may be better because the output volts do not depend on the value of the bridge resistance.

PS. I just found your earlier thread on the same topic. It may be just too hard for you and maybe you should try a simpler project first, to get familiar with the basics of electronic circuits. It's all good stuff though and you can progress to this project eventually but not if you are trying to sort out everything at once.

 

[Micheal_Leo]

Are you using the diagram in the link I quoted? Which is R2 and R1 and what is your supply voltage? Show your working how you calculated 1V out? Your output volts can be obtained by many different values for R2 but consider what happens when you connect the bridge across the output of the potential divider, the resistance ratio changes a lot (current will flow through the bridge). The bridge presents 120Ω and you need to consider this because it appears in parallel with R2 so you need a different R1. Google with "using potential divider to change supply volts" and you'll find loads of discussions which show you what I mean. If you have a box of resistors then your bread board will make this easy to experiment with.
A 1V voltage regulator may be better because the output volts do not depend on the value of the bridge resistance.

PS. I just found your earlier thread on the same topic. It may be just too hard for you and maybe you should try a simpler project first, to get familiar with the basics of electronic circuits. It's all good stuff though and you can progress to this project eventually but not if you are trying to sort out everything at once.

the other possibility to have 1 V as shown by calculation and DMM

 

[Micheal_Leo]

Are you using the diagram in the link I quoted? Which is R2 and R1 and what is your supply voltage? Show your working how you calculated 1V out? Your output volts can be obtained by many different values for R2 but consider what happens when you connect the bridge across the output of the potential divider, the resistance ratio changes a lot (current will flow through the bridge). The bridge presents 120Ω and you need to consider this because it appears in parallel with R2 so you need a different R1. Google with "using potential divider to change supply volts" and you'll find loads of discussions which show you what I mean. If you have a box of resistors then your bread board will make this easy to experiment with.
A 1V voltage regulator may be better because the output volts do not depend on the value of the bridge resistance.

PS. I just found your earlier thread on the same topic. It may be just too hard for you and maybe you should try a simpler project first, to get familiar with the basics of electronic circuits. It's all good stuff though and you can progress to this project eventually but not if you are trying to sort out everything at once.

I have tried to get reading from wheatstone bridge , the beam not bend just in stationar position , although the reading from DMM not stable ,

Voltage output from wheatstone bridge
Voltage reading from AD620 output

 

[sophiecentaur]

the other possibility to have 1 V as shown by calculation and DMM

That 1.04V looks ok. It will change when you hang the bridge on the output? That's because you have changed the effective value of R2. Starting with that 120Ω, what R1 value is needed to get 1V? Insert new resistor and check that you get 1V.
Connect DMM across the other two terminals of the bridge. You should get low value (zero if perfectly balanced). Flex the bridge and see change in DMM volts

 

[Micheal_Leo]

That 1.04V looks ok. It will change when you hang the bridge on the output? That's because you have changed the effective value of R2. Starting with that 120Ω, what R1 value is needed to get 1V? Insert new resistor and check that you get 1V.
Connect DMM across the other two terminals of the bridge. You should get low value (zero if perfectly balanced). Flex the bridge and see change in DMM volts

several things i confirmed .
1) the voltage is receiving by bridge is 1V
2) the output from wheatstone bridge(DMM) is fluctuating a lot , never stable going to very high value and drop sudden.
3) the voltge in DMM is between 0.004-0.006 , mv fluctuates a lot

 

[sophiecentaur]

1) the voltage is receiving by bridge is 1V

Have you modified your potential divider resistors? If not then there is something wrong with your connections. The ratio of the 'pot' would be 120/10120.
Have you used your DMM to find the resistances between (all) the bridge terminal pairs? What do you get?

 

[Micheal_Leo]

Have you modified your potential divider resistors? If not then there is something wrong with your connections. The ratio of the 'pot' would be 120/10120.
Have you used your DMM to find the resistances between (all) the bridge terminal pairs? What do you get?

i am very much confused that previous i made voltage divider of 1V using fixed resistors( 10000 and 2k+510) and attach bridge to it, should i make voltage divider using pot (10k) and 120 resistor?

Edit: today the vender told me that
"Generally, the working current during static measurement is about 250mA. During dynamic measurement, the allowable working current can reach 75-100mA, while the allowable working current of the foil strain gauge can be greater."
so i want to go for 250mA

 

[Baluncore]

Edit: today the vender told me that
"Generally, the working current during static measurement is about 250mA. During dynamic measurement, the allowable working current can reach 75-100mA, while the allowable working current of the foil strain gauge can be greater."
so i want to go for 250mA

I would expect the bridge current to average between 5 mA and 20 mA.

If you put 250 mA through the bridge, it should not be for more than half a second in ten seconds.

 

[Micheal_Leo]

I would expect the bridge current to average between 5 mA and 20 mA.

If you put 250 mA through the bridge, it should not be for more than half a second in ten seconds.

yes i believe 250mA is still high , i will go for 5-20mA first

 

[Micheal_Leo]

I would expect the bridge current to average between 5 mA and 20 mA.

If you put 250 mA through the bridge, it should not be for more than half a second in ten seconds.

i just again confirm from vender , confirmation is 25mA

 

[sophiecentaur]

i am very much confused

That may be because you have not read more about potential dividers (and possibly not much about electrical circuits at all?). The idea of a potential divider is to arrange two resistors so that the voltage at their mid point is what you want. You have a formula, in the link I gave you and everywhere else that discusses potential dividers. Where did the idea of a 10k pot come from? What. voltage would you expect with that as R1? If you want to use a variable resistor for adjustment then you need to have a series fixed resistor to limit the maximum current to something safe.
Use the formula to make an equation with just one unknown (the required R1) and three knowns: Vsupply, Vbridge (=1V) and R2.

With 1V applied to the 120Ω bridge, you will have a sensible current. (Work it out)

Edit: today the vender told me that

The vendor can't be relied on to know any electronics at all; he may just be retailing meaningless (to him) components. The Power dissipated by a resistor value R, with I current flowing is P = I²R. What does that work out as for 250mA?

yes i believe 250mA is still high , i will go for 5-20mA first

How were you planning to do that?

Is your interest here in the results of a mechanical measurement exercise, rather than an electronics project? What is your level of expertise here? It may be better to go for another measurement method or even to buy a unit that works 'on its own' and will interface with your Arduino.

 

[Micheal_Leo]

That may be because you have not read more about potential dividers (and possibly not much about electrical circuits at all?). The idea of a potential divider is to arrange two resistors so that the voltage at their mid point is what you want. You have a formula, in the link I gave you and everywhere else that discusses potential dividers. Where did the idea of a 10k pot come from? What. voltage would you expect with that as R1? If you want to use a variable resistor for adjustment then you need to have a series fixed resistor to limit the maximum current to something safe.
Use the formula to make an equation with just one unknown (the required R1) and three knowns: Vsupply, Vbridge (=1V) and R2.

With 1V applied to the 120Ω bridge, you will have a sensible current. (Work it out)

The vendor can't be relied on to know any electronics at all; he may just be retailing meaningless (to him) components. The Power dissipated by a resistor value R, with I current flowing is P = I²R. What does that work out as for 250mA?

How were you planning to do that?

Is your interest here in the results of a mechanical measurement exercise, rather than an electronics project? What is your level of expertise here? It may be better to go for another measurement method or even to buy a unit that works 'on its own' and will interface with your Arduino.

• Vin=Vsupply= supply voltage= 5V
• Vout=Vbridge=1V
• R2 = 120 ohm ( Strain Gauge resistance)
• R1 = ( ( Vsupply - Vbridge ) R2 ) / Vbridge
• R1=(( 5 - 1) 120 ) / 1
• R1 = 480 ohm
• Current Flow through strain gauge Vbridge=IR ( Guage Resistance(R) =120 ohm , Vbridge= 1V)
• Vbridge / R(Gauge Resistance) = I
• 1 / 120=I
• 0.008A

The following calculation is to find Current through strain gauge and R1 needed ,
To get 1V , i need R1 = 480 ohm ,

i have interest in both areas( mechanical and electronics)

i have confusion about wiring

 

[Baluncore]

The bridge, 120R drops 1 volt;
1V / 120R = 8.333 mA;
If you have a 5 volt supply, you must drop four more volts.
Use a 120 ohm series resistor to drop each volt.
Four 120 ohm resistors in series with the bridge.
That may be two below and two above the bridge.

 

[Micheal_Leo]

The bridge, 120R drops 1 volt;
1V / 120R = 8.333 mA;
If you have a 5 volt supply, you must drop four more volts.
Use a 120 ohm series resistor to drop each volt.
Four 120 ohm resistors in series with the bridge.
That may be two below and two above the bridge.

do i need to connect 480 ohm with these 4(120) ohm series resistors

 

[Baluncore]

4 * 120 ohm is the 480 ohm series resistor.
You must study the common mode input voltage range of the amplifier, to work out if the 120 ohm resistors should be placed above or below the bridge.

 

[sophiecentaur]

4 * 120 ohm is the 480 ohm series resistor.
You must study the common mode input voltage range of the amplifier, to work out if the 120 ohm resistors should be placed above or below the bridge.

I was thinking that's the next stage in implementing. If the amp has + and - supplies then obvs the bridge should be hung between them so the common voltage would be around zero. But if single sided amp then the same thing applies but with half of the 480Ω above and half below. I was wondering if the OP will get this all in one go.

 

[sophiecentaur]

do i need to connect 480 ohm with these 4(120) ohm series resistors

They are not 'in series' on the bridge. They are in series / parallel - hence the 120 total.

 

[Micheal_Leo]

I was thinking that's the next stage in implementing. If the amp has + and - supplies then obvs the bridge should be hung between them so the common voltage would be around zero. But if single sided amp then the same thing applies but with half of the 480Ω above and half below. I was wondering if the OP will get this all in one go.

 

[Micheal_Leo]

They are not 'in series' on the bridge. They are in series / parallel - hence the 120 total.

please can you check like this , before attaching strain guage , the DMM show zero but when i attach strain gauge ( replacing one resistor in brdige with guage) it show 1.480V while beam is stationary

 

[sophiecentaur]

please can you check like this , before attaching strain guage , the DMM show zero but when i attach strain gauge ( replacing one resistor in brdige with guage) it show 1.480V while beam is stationary

what does that diagram mean? Look at other schematic diagrams. You will notice that connecting wires are actually joined together (often with a dot). Those 120Ω reasistors are not joined to anything so they don't "do" anything. You should try to stick to the conventions if you want to make sense. Have you seen a diagram of a bridge circuit? The power connects across one diagonal and the meter connects across the other diagonal. Your picture shows none of this.

PS the way you actually connect things together and where the wires go is best shown by a conventional circuit diagram. Your breadboard version means nothing to anyone else but you.

 

[sophiecentaur]

What should I understand from that picture? We don't seem to be getting anywhere with this at the moment.

 

[sophiecentaur]

We are talking in English and that's because we can rely (at least to some extent) on understanding what we both mean. I cannot converse with you is you don't use the equivalent convention when drawing diagrams. There is a universal standard for drawing circuits. You need to use it. It is not difficult. Just look at text books and serious web pages.

 

[Micheal_Leo]

what does that diagram mean? Look at other schematic diagrams. You will notice that connecting wires are actually joined together (often with a dot). Those 120Ω reasistors are not joined to anything so they don't "do" anything. You should try to stick to the conventions if you want to make sense. Have you seen a diagram of a bridge circuit? The power connects across one diagonal and the meter connects across the other diagonal. Your picture shows none of this.

PS the way you actually connect things together and where the wires go is best shown by a conventional circuit diagram. Your breadboard version means nothing to anyone else but you.

please fine the attached conventional diagram

 

[sophiecentaur]

Ahh!! I see what's wrong. The four strain gauge resistors ARE the bridge. No external 120Ω resistors are needed. (This is a great example of where one should start with a proper diagram; it can save a lot of wasted time.

 

[sophiecentaur]

The series resistor is used to reduce the 5V to 1V across the bridge supply.

 

[sophiecentaur]

I think you may have gottit!!!

 

[Micheal_Leo]

Ahh!! I see what's wrong. The four strain gauge resistors ARE the bridge. No external 120Ω resistors are needed. (This is a great example of where one should start with a proper diagram; it can save a lot of wasted time.

so should i make wheatstone bridge and connect this full strain bridge to it ? this is confusing that i should make a wheatstone bridge or not when these are full bridge strain gauge resistors

 

[sophiecentaur]

so should i make wheatstone bridge and connect this full strain bridge to it ? this is confusing that i should make a wheatstone bridge or not when these are full bridge strain gauge resistors

I thought you had got this. Why would you want another bridge when you have a strain gauge that IS a bridge? As the gauge flexes, resistors increase and decrease and the bridge becomes unbalanced proportionally with the strain. All you need is to reduce the supply volts to 1V by using the series resistor in the supply.

I suspect that you have not looked elsewhere to get an alternative view. there are dozens of web pages about the same problem. I cannot do the whole design job for. you.

 

[Micheal_Leo]

I thought you had got this. Why would you want another bridge when you have a strain gauge that IS a bridge? As the gauge flexes, resistors increase and decrease and the bridge becomes unbalanced proportionally with the strain. All you need is to reduce the supply volts to 1V by using the series resistor in the supply.

I suspect that you have not looked elsewhere to get an alternative view. there are dozens of web pages about the same problem. I cannot do the whole design job for. you.

yes sure , please can you check this reducing the voltage to 1V conventional diagram

 

[sophiecentaur]

yes sure , please can you check this reducing the voltage to 1V conventional diagram

You can check this yourself, actually. Show the workings of how you came to this circuit. What would the R1 be and what would the R2 be? The resistance of the bridge will affect how things work, remember.
Note: I will not just give you the answer because that won't help you for next time.