How to connect this strain gauge?

[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.

 

[Micheal_Leo]

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

THe best R1=10000ohm and R2=2500 in this way i can get 1V from 5V power supply

 

[sophiecentaur]

THe best R1=10000ohm and R2=2500 in this way i can get 1V from 5V power supply

This is ridiculous. We've been here before. Just try it and see that you get much less than 1V because. you are loading the potential divider with 120Ω in parallel with your 2500Ω. You clearly don't want to learn about this; you just want me to post a diagram with full instructions about how to set it up. PF doesn't work that way. (At least, I don't.)

Come back when you can show that you understand about Potential Dividers.

 

[Baluncore]

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

Not good.

THe best R1=10000ohm and R2=2500 in this way i can get 1V from 5V power supply

You are imagining a potentiometer, which will have a high output resistance = 2k, then shorting that with the 120R bridge.

If the bridge has elements of 120R, then the series resistance of each side is 240R, but there are two sides in parallel, which makes the whole bridge appear to be 120R.

Now you know that 120R will drop one volt.
If you need to eliminate 4 volts of the 5, you need four 120R resistors, in series with the bridge, carrying the same current.

 

[Micheal_Leo]

" If you need to eliminate 4 volts of the 5, you need four 120R resistors, in series with the bridge, carrying the same current"
this is very confusing , i have 120ohm fully strain guage resistors( 4 resistors in bridge) , i do not need external 120ohm resistors acting as wheatstone bridge till that point i understand , after that as you mentioned that " you need four 120R resistors, in series with the bridge" here bridge is fully strain gauge , than four 120ohm resistors where should i connect on breadboard

 

[Micheal_Leo]

This is ridiculous. We've been here before. Just try it and see that you get much less than 1V because. you are loading the potential divider with 120Ω in parallel with your 2500Ω. You clearly don't want to learn about this; you just want me to post a diagram with full instructions about how to set it up. PF doesn't work that way. (At least, I don't.)

Come back when you can show that you understand about Potential Dividers.

i am trying best to understand it

 

[Baluncore]

i am trying best to understand it

You are way out of your depth.
Study ohms law.

 

[sophiecentaur]

i do not need external 120ohm resistors acting as wheatstone bridge

What you need is 480Ω. It is a complete coincidence that you happen to have four 120Ω resistors and that they will give you 480Ω when connected in series. You should try to work out how you would get 1V if your supply volts were 6V - just one resistor is needed.

i am trying best to understand it

I think you are trying to get as much help as possible without actually learning anything. What is your level of Electronics / Science qualification? I assume that you want to do some mechanics measurements using electronics without actually learning the electronics. your only option, afaics, is to buy a working strain gauge unit and fix it to your mechanics experiment. If that would be too costly then choose a different mechanics project.

 

[Micheal_Leo]

What you need is 480Ω. It is a complete coincidence that you happen to have four 120Ω resistors and that they will give you 480Ω when connected in series. You should try to work out how you would get 1V if your supply volts were 6V - just one resistor is needed.

I think you are trying to get as much help as possible without actually learning anything. What is your level of Electronics / Science qualification? I assume that you want to do some mechanics measurements using electronics without actually learning the electronics. your only option, afaics, is to buy a working strain gauge unit and fix it to your mechanics experiment. If that would be too costly then choose a different mechanics project.

Voltage Source (VS) = 6V
Volts (V) = 1V( needed)
Resistance 1 (R1) = 480
Resistance 2 (R2) = 96
R2 should be 96 ohm to have 1V output

specifically voltage divider

i am learning from deep heart and thinking , today learn so many things , sometime get stuck

 

[sophiecentaur]

R2 should be 96 ohm to have 1V output

Alternatively, what R1 would you need so that you could just use the 120Ω for R2. Just re-arrange the equation to find the single resistor that can be used for R1. Certainly not 480Ω.

 

[Micheal_Leo]

Alternatively, what R1 would you need so that you could just use the 120Ω for R2. Just re-arrange the equation to find the single resistor that can be used for R1. Certainly not 480Ω.

after rearranging
R1= (Vin-Vout)R2 / Vout
IF Vin=5
Vin=5
R2=120
Vout=1
So R1=480 ohm
i always getting 480