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I have a problem making a circuit like this working Taken from the link below

I got it from this TI application note. For testing purposes I have set:

\$R4=R2=R3=R1=47 \Omega\$

\$ Rf = 10 k\Omega\$

\$Rg= 26.70\Omega\$

The op amp \$V_S=+5V\$. This should give a gain of 750 having \$V_{SIG}\$ as 0.8E-4V/kg.

I agreed that 80kg would be enough for full scale. The problem is that this setup doesn't show any variations when i load the strain gauge. It just stays at 2.40V which doesn't make much sense since it doesn't reach near \$V_S\$. I am using the TL064. I have tried a lot and think my wiring is correct. Do you have any suggestions?

Update: I came up with the 750 of gain by assuming that:

\$R4=R2 <=> \dfrac{R4}{R2}=1\$
\$Gain=\dfrac{2Rf}{Rg}+1\$
\$(Sig_+)-(Sig_-) = V_{IN} => V_O=V_{IN} \times 1 \times Gain <=>\$
\$Gain = \dfrac{V_O}{V_{IN}} <=> \dfrac{2Rf}{Rg}+1 = \dfrac{V_O}{V_{IN}}.\$
\$V_{IN}/kg = 0.8E-4V/kg\$

I also i assumed that my absolute maximum load would be 80kg. So \$V_{IN(MAX)} = 0.8E-4 \times 80 = 6.4E-3\$ V

So we can determine the needed gain for

\$V_{OUT}=5V. G=\dfrac{V_{OUT}}{V_{IN}}=\dfrac{4.8}{6.4E-3}=750\$

Taking the previous equations: \$ \dfrac{2Rf}{Rg}+1 = 750 <=> 2Rf = 749 Rg <=> Rg = \dfrac{Rf}{374.5}\$

Knowing this and the resistors I had available I chose to set Rg as a function of Rf because i have to arrange the resistors(series or parallels to approximately get the desired gain) I chose:

Edit:
Corrected Equation mistake but still no results \$Rf=10 k\Omega => Rg = 10E3/374.5 = 26.70 \Omega\$ Considering I don't have this precise value I came up with \$Rg = 47 || 47 = 23.5\$ which is close to \$26.70 \Omega\$

Hope everything is clear for everybody, otherwise please ask for clarifications Thank you

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    \$\begingroup\$ The TL064 is a low-power JFET op-amp. You're driving some very low resistances with R1 through R4 = \$ 47 \Omega \$. I would scale those values up by at least 100x (to \$ 4.7k \Omega \$) and see if anything changes. \$\endgroup\$ – Adam Lawrence Aug 22 '11 at 19:16
  • \$\begingroup\$ Why are you using an instrumentation amp? What kind of strain gauge? It generates a voltage? Are the voltages at SIG+ and SIG- between 0 and 5 V? \$\endgroup\$ – endolith Aug 22 '11 at 19:55
  • \$\begingroup\$ Yes the voltages at SIG+ and SIG- are 5V. The calculations i made were based on my expected full scale voltage through measurements of the voltage change by 1 kilo. I will update the questions so as to reflect the comments and the requests for clarifications. The circuit is composed of 4, 3 wire, strain gauges in full bridge(i think it is the correct term) \$\endgroup\$ – Paulo Neves Aug 22 '11 at 21:03
  • \$\begingroup\$ @aiwarrior: If SIG+ and SIG- are 5V, then you won't get any output. Do you mean that they are between 0 and 5 V? The strain gauges have an integrated amplifier that outputs 0 to 5 V signal? Why are you using an instrumentation amp? \$\endgroup\$ – endolith Aug 22 '11 at 21:17
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    \$\begingroup\$ @endolith - Strain Gauges are almost universally wheatstone bridges ( en.wikipedia.org/wiki/Wheatstone_bridge ). Basically the strain is reflected as the difference in two voltages, where both voltages are very high-impedance. It's pretty much what instrumentation amplifiers were invented for. \$\endgroup\$ – Connor Wolf Aug 23 '11 at 8:27
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You've made an error in the amplification.

\$G = \dfrac{2 \times R_F + R_G}{R_G}\$

With the values you're using this gives

\$G = \dfrac{2 \times 10k\Omega + 25.9k\Omega}{25.9k\Omega} = 1.77\$

If you want a \$G\$ of 750 you'll have to make \$R_G\$ = 26.7\$\Omega\$, so that

\$G = \dfrac{2 \times 10k\Omega + 26.7\Omega}{26.7\Omega} = 750\$

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  • \$\begingroup\$ I had already made a comment saying i noticed that. And it didn't solve the problem. Sorry if i didnt edit it earlier to save you some time \$\endgroup\$ – Paulo Neves Aug 23 '11 at 10:44
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I suspect your gain is too low - how do you get a gain of 750? I get a gain of ~1.7

Try replacing Rg with e.g, a 100 ohm resistor and see if that works a bit better. As mentioned, scale R1,2,3,4 up also - you could add some gain here too by making R3 and R4 larger.

I'm assuming you are using a wheatstone bridge?

EDIT - just saw your comment above. If the voltages at SIG+ and SIG- are both the same you will get no change. This is a differential amplifier, which means it amplifies the difference between the two inputs, not the inputs and ground. Can you confirm how you are connecting your strain gauge exactly? A schematic including the rest of the circuit would help greatly.

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  • \$\begingroup\$ The voltages at SIG+ and SIG- are not the same. I even notice the change with the multimeter. It is just that it is really small. I don't have the schematic of the 3 wire gauge and i kind of only guessed through the pcb which kind of formed a bridge with the 4 Strain gauges having an excitation and signal terminals. I will prepare a schematic of how i think it works internally. The sensor is really similar to this just the wires are of different colors. The circuit is really just 4 strain gauges of the 3 wire type in full bridge \$\endgroup\$ – Paulo Neves Aug 22 '11 at 23:23
  • \$\begingroup\$ Okay, try Rg 500, Rf 10k, R1/2 1k and R3/4 10k (or something similar - values don't have to be exact) Let us know how this performs. \$\endgroup\$ – Oli Glaser Aug 23 '11 at 0:14
  • \$\begingroup\$ I dont have 1k or 500 resistors with me and the closer i got is 680 Ohm. Can you explain your reasoning with the Gain of 410, so maybe i can adapt with the material i have at my disposal now? Thanks \$\endgroup\$ – Paulo Neves Aug 23 '11 at 0:59
  • \$\begingroup\$ My current reasoning is to just get it working - the exact gain is not important as long as it is high enough to see a change at the output. After you achieve this you can fine tune the gain. So I suggested a gain of >10 on the first stage, and >10 on the second, to give >100 overall - this should be more than enough to detect a change if the rest of the circuit is correct. 680 ohm Rg sounds okay, combined with the same values (or similar) \$\endgroup\$ – Oli Glaser Aug 23 '11 at 1:07
  • \$\begingroup\$ @aiwarrior: What are the actual voltages on SIG+ and SIG-? What is the difference? \$\endgroup\$ – endolith Aug 23 '11 at 14:03

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