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Load Cell Amplifier AD620

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I know AD620 is an old part but it is available at my place. I intend to develop a Load Cell amplifier. Load cell is 2mV/V. I am planing to use AD620 for it. Following are my calculations:

Excitation Voltage for Load Cell = 5V
Full Scale for load of 140 kg = 2mV X 5V = 10mV
Gain = Vout/Vin = 5000mV/10mv = 500
From formula of Rg we calculate it to be 98.99 Ohm
I have worked on Following schematic:

What is the problem?

1.If I leave the pin 5 open then Min Voltage is 1.9 and max is 4.2 volts. If pin5 is grounded then min voltage is 0.6 and max is 1.9 (strange)

2.the max voltage (4V) is reached at 6 mv Input. even gain calculation is done. But we have added 100 Ohm resistance instead of 98.99 Ohm.

3.Is there any difference between AD620AN, AD620AR and AD620ANZ ?

4.In some circuits i see that at sense + and sense - sometimes before feeding these to operational amplifiers some resistance is added. what is the purpose of adding these keeping in mind the voltage is already in mV

5.Can you suggest load cell amplifier other than INA125 because that is not available
 

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Full Scale for load of 140 kg = 2mV X 5V = 10mV
Gain = Vout/Vin = 5000mV/10mv = 500
What is the load range that you want to measure (minimum to maximum)?
 
The load range is min 0.1 to 10 mv max

hi,
Are you using a single or dual supply, ? the AD620 is a dual.

If you need to work from a single supply the AD623 is OK.

Would you please post your circuit diagram.
 
hi,
Are you using a single or dual supply, ? the AD620 is a dual.

If you need to work from a single supply the AD623 is OK.

Would you please post your circuit diagram.
It's kinda hard to see, but he posted it in #1.:)
 
Updated with clear picture. I can use dual supply its no big deal. Its just one 7660. :)

Hi,
Ok, no problem.

A AD620 requires a dual supply, get an AD623.:D

Lets know how the AD620 works with a dual supply, pin 5 is used for output offset, if offset is not required, connect it to 0V.


Ron,
Thanks for the heads up.
 
Hi,
Ok, no problem.

A AD620 requires a dual supply, get an AD623.:D

Lets know how the AD620 works with a dual supply, pin 5 is used for output offset, if offset is not required, connect it to 0V.


Ron,
Thanks for the heads up.
If offset is required, use a ≈zero impedance source, such as a voltage follower.
 
I did an experiment with it. Gave AD620 Dual Supply of + and - 10 Volts. Placed a variable resistor at offset and managed ofset to 0. Now the thing is it works absloutly fine but according to me the weight extracted is approximately 1 kg less than it should be. The gain calculated is 98.99 OHM and we placed 100 Ohm. Is that the reason?

Secondly if you calculate Gain which is Vout/ Vin; does it depends on the VSS and VDD of the op amp? see if i have given +- 10 volts what if i give it + - 15 volts would the output swing be same ? if not we should include vcc and vdd somewhere.
 
I did an experiment with it. Gave AD620 Dual Supply of + and - 10 Volts. Placed a variable resistor at offset and managed ofset to 0. Now the thing is it works absloutly fine but according to me the weight extracted is approximately 1 kg less than it should be. The gain calculated is 98.99 OHM and we placed 100 Ohm. Is that the reason?

Secondly if you calculate Gain which is Vout/ Vin; does it depends on the VSS and VDD of the op amp? see if i have given +- 10 volts what if i give it + - 15 volts would the output swing be same ? if not we should include vcc and vdd somewhere.

Hi,
Its a bad idea to use a resistor for the Voffset control pin.

You need to use a low impedance drive, example: the output of a OPA .

If you use a resistor directly, its value is added in series with the internal resistor that connects to the Vref pin5. This has the effect of changing the overall gain of the AD620.

The Gain is not dependent upon the supply voltages to the AD620.

I would recommend also that you use a 150R 20 turn trimmer resistor if you require a precise gain in place of the 100R
 
Hi,
Its a bad idea to use a resistor for the Voffset control pin.

You need to use a low impedance drive, example: the output of a OPA .

If you use a resistor directly, its value is added in series with the internal resistor that connects to the Vref pin5. This has the effect of changing the overall gain of the AD620.

The Gain is not dependent upon the supply voltages to the AD620.

I would recommend also that you use a 150R 20 turn trimmer resistor if you require a precise gain in place of the 100R
I guess my post #8 went in one eye and out the other:
If offset is required, use a ≈zero impedance source, such as a voltage follower.
 
Just try an experiment by doing an A and B testing. First create a new set of equipment with a different voltage and analyse the behaviour of its output for "A” testing. With the testing of B, you must acquire a higher voltage with different resistors and monitor its behaviour. From this simple procedure you should be able to determine the difference and see which weight is more accurate. I personally use A B testing and it works on my electrical and mechanical experiment. Even professionals rely on this method.
 
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please upload straightforward circuit of loadcell with AD620 IC , fool proof working condition

It doesn't work that way, Uttam Dutta. If you want help, please start your own thread, and remember that members here generally don't just give out schematics. You need to figure it out for yourself, but we will be happy to help you out with any specific problems you may have.

Regards,
Matt
 
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