Amplifier circuit, need more sensitivity or gain?

Ron H

Joined Apr 14, 2005
7,014
You need to adjust the offset. As you can calculate, your bridge offset voltage is
Vos = Vout/gain = 18.8mV/5 = 8.81/2358 ≈ 3.7mV. You need to add an offset adjustment circuit, as in Fig. 2 in the datasheet (see attachment). This will compensate for the offset at any gain setting. The highest gain should give the best settability.
BTW, how did you get gain=2358 for Rg=80? The equation is
Gain=5+200k/Rg.
When Rg=80, Gain=2505. Did you mean Rg=85?

EDIT: Keep in mind that no rail-to-rail op amp will go all the way to either rail. I would set the gain high, adjust the offset down until it won't go any lower, then turn it up (or down, as required) until the output barely starts to move.
Note that, if your offset happened to be the opposite polarity, prior to adjustment you would get near zero output with low or high gain, and the output would only start to move when the strain caused the differential input to be greater than 0.
 

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Thread Starter

Nevalite

Joined Feb 20, 2012
45
Yeah I meant G=2505, not sure where I got that other number. As for the offset, it all makes sense now. I will give this a try tomorrow.

Thanks
 

Ron H

Joined Apr 14, 2005
7,014
Yeah I meant G=2505, not sure where I got that other number. As for the offset, it all makes sense now. I will give this a try tomorrow.

Thanks
What baffles me is that 8.81v divided by 2358 is very close to 18.6mV divided by 5.
calculated offset=8.81V/2358=3.74mV
calculated offset=18.6mV/5=3.72mV

However,
calculated offset=8.81/2505=3.52mV

What's up with that?:confused:
 

Thread Starter

Nevalite

Joined Feb 20, 2012
45
So I got a few other values to verify the offset needed.

Rg=0, G=5, Vo=18.4mV, offset=3.68mV
Rg=74.8, G=2673.8, Vo=9.26V, offset=3.46mV
Rg=81.2, G=2468, Vo=8.54V, offset=3.46mV
Rg=85, G=2357.9, Vo=8.15V, offset=3.46mV

It would seem that all values with an Rg are the same but the G=5 value is different. If you add the 5 to the Rg value you get the same as the G= 5. Example, 5+(200k/74.8) vs. (200k/74.8+5)...

Anyways whether I need 3.46 or 3.68 mV my question is the following. Since I am seeing a positive voltage does that mean my offset circuit needs to inject negative voltage into the ref pin? How am I going to do that with a single supply circuit?
 

Ron H

Joined Apr 14, 2005
7,014
You are correct about needing a negative supply.
Attached is an idea I had. It should give you about ±6mV of offset range. You should use low TC resistors and trim pot, and keep the leads as short as possible. A ten turn pot might help, but you could try a one turn unit first.
 

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Thread Starter

Nevalite

Joined Feb 20, 2012
45
I thought the offset had to be introduced at the ref pin? Isn't this just reducing the voltage coming from the strain gauge which would change my results?

Also why is there a +5v beside the 100k trimpot? Is that +5v going into the wiper?

Thanks
 

Thread Starter

Nevalite

Joined Feb 20, 2012
45
Nevermind I understand the trim pot now, I hooked leg 1 to +5, leg 3 to GND and the wiper leg as the output. I was able to get it down to 3.4mV without the 50k resistors but when I have the two 50ks I can't go lower than 1.296V...

First things first I need to get this single to dual supply circuit going because regardless if I use this trimpot idea or the offset op amp circuit they both need negative voltage to get this down. I'm gonna try the circuit Mr.Chips suggested on page 1.
 

Ron H

Joined Apr 14, 2005
7,014
Nevermind I understand the trim pot now, I hooked leg 1 to +5, leg 3 to GND and the wiper leg as the output. I was able to get it down to 3.4mV without the 50k resistors but when I have the two 50ks I can't go lower than 1.296V...

First things first I need to get this single to dual supply circuit going because regardless if I use this trimpot idea or the offset op amp circuit they both need negative voltage to get this down. I'm gonna try the circuit Mr.Chips suggested on page 1.
I don't think you connected it as I drew it. The wiper is NOT the output. Yes, the pot wiper is connected to +5V, but it could be connected to ground, with the same results. I think I would connect it to ground.
When the pot is centered, the offset circuit puts 100kΩ in parallel with each of the two 350Ω resistors on the top half of the bridge. The loss in sensitivity will be 0.18% when the pot is centered, and will vary from 0.12% to 0.35% when the pot is rotated fully to one side or the other.
 

MrChips

Joined Oct 2, 2009
25,928
No. Do what Ron H says. Connect the wiper to GROUND.
The purpose of the null circuit is to balance the currents in and out of the bridge.
 

Thread Starter

Nevalite

Joined Feb 20, 2012
45
Ok so that worked beautifully, I was able to turn the G=5 from 18.6mV down to whatever I wanted (above about 4mV). I set it at a value that allowed me to use G=8000 and still be in the 0-5V range on the output and the readings are now very sensitive while staying in the right range. I will post a picture of the completed circuit later.

Thank you very much Ron_H and Mr.Chips!
 

Thread Starter

Nevalite

Joined Feb 20, 2012
45
I've been thinking about this, and I think I may be opening a whole new can of worms with this one, but ultimately I need to make 5 of these. Originally my idea was that once I had 1 figured out it would be easy to string up 4 more in parallel with the battery and be done with it, that was before the offset voltage problem.

The thing about strain gauges is they are rated 350 ohms but they have a +/-15% tolerance on that value so they are rarely the same resistance. I would imagine that a different resistance in the bridge is going to require a different offset for every strain gauge?

I could probably put a null adjustment on each gauge and calibrate them all individually until they were all the same?

My other idea was that if I made the offset circuit to inject into the ref I could use the same circuit for all the gauges but since each gauge will require a different offset I would need a different circuit for each gauge...

Any thoughts on this?

Thanks
 

MrChips

Joined Oct 2, 2009
25,928
Generally speaking, each sensor is different and each one has to be calibrated.

I would suggest that you install two pots in the circuit to allow you to adjust the zero and gain. The zero pot will be the null pot as you are using now.

The gain pot will be a variable resistor placed in series with the gain resistor. Reduce the gain resistor by about 10% and select a trimpot resistance of 20% of your original gain resistor. This will provide +/- 10% adjustment.

I would choose a 10-turn trimpot for both adjustments. This will give you fine control.
 

Thread Starter

Nevalite

Joined Feb 20, 2012
45
That's what I figured. I currently have a trim pot for the null and I replaced that ugly one from the picture with another trim pot for the gain which is currently wired up as a variable resistor. I like your idea for the +/- 10% I think I will do that. I'm gonna complete this tomorrow and get picture of it.

Thanks again.
 

Thread Starter

Nevalite

Joined Feb 20, 2012
45
Here is the final version, the picture sucked so I added some lines fro clarity. How does it look?



I just noticed that 0.1uF off the regulator goes nowhere -_-, I'll fix that tomorrow lol.
 

Thread Starter

Nevalite

Joined Feb 20, 2012
45
I am trying to calibrate this amp so that I can get the best resolution while staying in the 0-5v range. I am a bit confused because G=Vo/Vi but with the offset the Vo/Vi won't give the right G, how do I factor the offset into this calculation?
 

Ron H

Joined Apr 14, 2005
7,014
I am trying to calibrate this amp so that I can get the best resolution while staying in the 0-5v range. I am a bit confused because G=Vo/Vi but with the offset the Vo/Vi won't give the right G, how do I factor the offset into this calculation?
The offset circuit has negligible effect on the sensitivity of the bridge, i.e., it doesn't affect the required gain.

Remember that, in your gain equation, Vi is the differential input voltage, i.e., the voltage between the inputs (across the bridge), not the voltage from the input(s) to ground.

Also, you have to set the output so that Vout=0 when the bridge is not under strain.
 
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