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amplifying a signal from a load cell with ad623

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Hi Eric etc
And thanks for the replies. Did not even consider effect of higher input voltage than supply rail for ad623. So thanks a lot for spotting that rather amateur error. I will make the necessary modifications and let you know how it goes.

I have one other question - what is the advantage of the input circuitry you have used for the input from the load cell to the ad623 as opposed to just directly connecting the sensor wires.

Thanks again,

Cheers.

hi,
That simplified filter on the inputs is in line with the AD623 d/s RF filtering.
 

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Hi,
Made the alterations and now it now works perfectly!

Thanks for all your help. No doubt i will be back periodically to ask more electronics related questions.

Cheers.
 
Sorry to dig this post, but I'm developing a circuit just like the posters', but I'm having some trouble.
The circuit is, actually, like the one attached.
Since I don't have a load cell with me right now, I am testing the circuit with a precision potentiometer. I have a 12V power supply to it and adjusted it to output around 40mV. With a 1k resistor, the amplification is ~100, so the output should be ~4V.
However, I'm getting around 180mV in the output. I also noticed that, when I measure the input of the AD623, the output voltage decreases slightly (to 160~170mV).
Anyone has any clues as to why this is happening? Thanks for the help, this has been slowing my work this past week...
 
hi and welcome,
I use the AD623 with load cells.
For a quick test set up, do it this way, connect a 100k resistor to one end of the 1Kpot and another 100K resistor to the other end of the 1K pot.

Connect the end of one of the 100K to +5V [ not 12v] and the end of the other 100K to 0V.
Join one end of the 1K pot to the +Inp of the AD623 and the wiper of the pot to the -Inp of the AD623

This will give you an adjustment range of approx 25mV

Depending how you connect your 1K pot, you may have to swap over the +Inp and -Inp connections.

Look at this image
 
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hi and welcome,
I use the AD623 with load cells.
For a quick test set up, do it this way, connect a 100k resistor to one end of the 1Kpot and another 100K resistor to the other end of the 1K pot.

Connect the end of one of the 100K to +5V [ not 12v] and the end of the other 100K to 0V.
Join one end of the 1K pot to the +Inp of the AD623 and the wiper of the pot to the -Inp of the AD623

This will give you an adjustment range of approx 25mV

Depending how you connect your 1K pot, you may have to swap over the +Inp and -Inp connections.

Look at this image


Should this +5V be the same one supplying the AD623?
But anyway, shouldn't my circuit also work as a test setup? Or maybe the problem is due since it has different ground references (one supply for the potentiometer and another for the amplifier)?
My test circuit is shown in the attachment, its actually a +6V power supply, not 12V.
 
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Hi,
I suspect in your set up , you have either the +Inp or -Inp connected to 0V.???is that correct.??

Try switching over the +Inp and -Inp connections to your pot.

What is the gain of your AD623 set for.??

EDIT:
Please post your full circuit diagram and I will check it for you
 
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Hi,
I suspect in your set up , you have either the +Inp or -Inp connected to 0V.???is that correct.??

Try switching over the +Inp and -Inp connections to your pot.

What is the gain of your AD623 set for.??

EDIT:
Please post your full circuit diagram and I will check it for you

The AD623 is set for a gain of 100.
Yes, the -Inp is connect to the 0V of the pot, which is actually different from the 0V of the -VS of the ADC623 (different power supplies).
The full circuit is attached.
 
hi,

You cannot ground an input on this type of Instrumentation amp, it will not work correctly.

Also with a 1K pot to 6V you cannot set the input voltage correctly, its far too coarse adjustment for an amp with a gain of 100. The amp will saturate at about + 3.5V outpur.

Look at this image, this is the way I do a quick test.
 
hi,

You cannot ground an input on this type of Instrumentation amp, it will not work correctly.

Also with a 1K pot to 6V you cannot set the input voltage correctly, its far too coarse adjustment for an amp with a gain of 100. The amp will saturate at about + 3.5V outpur.

Look at this image, this is the way I do a quick test.

I did not know about grounding the input.
So, since the amp will saturate at +3.5V output, I'll need an input up to +35mV to properly test it?
The pot i'm using is 2.2k.
Also, just to be sure, it doesn't matter that the test pot is wired to a power supply and the amp circuit is wired to a different one? Different grounds and stuff...
Since I'm at home I don't have many components around, but i'll see what I can do and try your test setup.

And what is this software you're using? Does it also simulates?

Thanks a whole lot man!!
 
I did not know about grounding the input.
So, since the amp will saturate at +3.5V output, I'll need an input up to +35mV to properly test it?
The pot i'm using is 2.2k.
Also, just to be sure, it doesn't matter that the test pot is wired to a power supply and the amp circuit is wired to a different one? Different grounds and stuff...
Since I'm at home I don't have many components around, but i'll see what I can do and try your test setup.

And what is this software you're using? Does it also simulates?

Thanks a whole lot man!!

hi,
Its VERY important that the test pot supply and the AD623 share the same 0V connection.

I have run another simulation for you with my test pot circuit, it shows the input Vin on one plot and the Vout plot on the other.

For a 2K test pot it will still work with the 100K's but it will a coarse adjustment

Its LTspice free simulator
 
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hi,
Its VERY important that the test pot supply and the AD623 share the same 0V connection.

I have run another simulation for you with my test pot circuit, it shows the input Vin on one plot and the Vout plot on the other.

For a 2K test pot it will still work with the 100K's but it will a coarse adjustment

Its LTspice free simulator

If they must share the same ground, then i'll supply the test pot with the 5V of the amp circuit (which will be the same for the load cell, anyway)
I have only found two 10k pots here, so i'll just solder them in series (the terminals with the full resistance) with that 2k2 pot.
 
If they must share the same ground, then i'll supply the test pot with the 5V of the amp circuit (which will be the same for the load cell, anyway)
I have only found two 10k pots here, so i'll just solder them in series (the terminals with the full resistance) with that 2k2 pot.

Take care when setting the pots, keep the common voltage around 2.5V.

I usually connect a 4k7 resistor in series with the +Inp and -Inp inputs, in order to protect the inputs from excessive voltage inputs, limits the input current.
 
Take care when setting the pots, keep the common voltage around 2.5V.

I usually connect a 4k7 resistor in series with the +Inp and -Inp inputs, in order to protect the inputs from excessive voltage inputs, limits the input current.

Hey man! thanks a lot! It is at least coeherent the amplification right now!!!
It's not exactly a gain of 100, but so close. I have 15.7mV on the test circuit pot, and that is giving me a 1.24V output.
This might be due to the 1k resistor's accuracy, right?
 
Hey man! thanks a lot! It is at least coeherent the amplification right now!!!
It's not exactly a gain of 100, but so close. I have 15.7mV on the test circuit pot, and that is giving me a 1.24V output.
This might be due to the 1k resistor's accuracy, right?

hi,
yes, I usually have a trim pot in series with the fixed resistor , it gives a 'fine' gain set adjustment
 
Do you think that a 100R resistor in series with an 1k pot is good enough for this gain of 100? or maybe a fixed resistor of 1k?

hi,
I would suggest a 820R fixed and a 500R multi turn trim pot if want precise setting of the gain
 
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View attachment 68515

Hello folk! I am having some trouble with this. I have set up the circuit. Thing is, when i put a 1K resistor between pins 1 and 8, i am supposed to get a gain of 100, instead i am getting a gain of 50. If i put 500ohm resistor, instead of getting a gain of 200, i am getting a gain of 100. What is going on here? My load cell is a bridge type. im getting an output in the mVs so i know my load cell is working. Just the amplification part. Note that the load cell and ad623 have the same ground and the reference is grounded too[0V]. Whats happening? any help appreciated. Diagram is attached.
 
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Sorry to dig up an old post . I am working with load cell and ad620A as amplifier. Using the above schematics i am getting very less amplification. So can anyone help ? if needed i can post my schematics
 
hi rd,
It would be better if you Created your own new Thread and added your circuit diagram
See if we can help you.

Eric
BTW: to get the same gain on a AD620 you have to half the value for Rg thats used on a AD623.
 
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