Author Topic: INA125 for driving linear pot  (Read 767 times)

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Offline V_KingTopic starter

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INA125 for driving linear pot
« on: January 30, 2019, 07:47:59 am »
Hello,

I've been using  INA125 (http://www.ti.com/lit/ds/symlink/ina125.pdf) for various load cells and am very happy with the simple application. It has voltage reference built in which is very useful.
 
I am building a small board with a four channel ADC and want all four channels to be able to drive either a standard wheatstone bridge load cell or linear displacement pots (voltage divider).

Am I ok to tie the negative op-amp input to the ground when using to drive the linear pot or am I missing something there?

Thanks,
 

Offline capt bullshot

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Re: INA125 for driving linear pot
« Reply #1 on: January 30, 2019, 08:07:51 am »
Look at page 5 and page 11 of the datasheet.
So, you can't connect Vin- to GND if GND equals your negative supply for the INA125, but you can if you provide the INA125 with a negative supply and meet some conditions.
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Offline V_KingTopic starter

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Re: INA125 for driving linear pot
« Reply #2 on: January 30, 2019, 02:30:00 pm »
Thanks for the response.

That's where a confusion for me is coming in. I want to use IAref pin tied to 5v reference, so I could use tension and compression from a single 12V supply. So If I will connect V- to 2.5 Vref as a linear pot pseudo ground and win in to 10v Vref, I should have the 2.5-10V input range with a shifted output?  Also, can anybody tell me, why are they putting load resistor between pins 11 and 5 in all their reference circuits? 

Analog op-amps is an area I'm really struggling with  |O
 

Offline capt bullshot

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Re: INA125 for driving linear pot
« Reply #3 on: January 30, 2019, 03:10:42 pm »
Thanks for the response.

That's where a confusion for me is coming in. I want to use IAref pin tied to 5v reference, so I could use tension and compression from a single 12V supply.
So, you want to connect Pin 4 to Pin 15 to obtain a +5V reference voltage at Pin 4, to supply your load cell? Looks fine.
The load cell does "tension and compression" - so the voltage difference at its output might be negative or positive?
You want to feed the output of the INA125 into an ADC, having a full scale range of 0...5V? So you want to sit the INA125 output to sit at 2.5V when there is zero force applied to the load cell?
Now use RG to set the gain, and the voltage at IAref to set the offset, e.g. 2.5V at load cell zero, and an output range of maybe +/-2V related to IAref (0.5V ... 4.5V)?
Note, the 2.5V source to IAref must have near zero impedance - using Pin 14 (while configured for 5V ref) doesn't provide that. Usually one uses the buffered reference (or half the ADC reference) output from the connected ADC here. Or one might want to use the INA125 provided 5V reference as the ADC's reference? So you might have to add an additional 1:1 divider and buffer to provide the 2.5V to IAref.

To replace the load cell with a linear pot, I'd suggest now to connect a 1:1 divider from your Ref. voltage to GND as one half of the bridge, and the pot as the other. Effectively this sets Vin- to pot midscale. Turning the pot shall result in 0.5V ... 4.5V at the INA125 output? You'd want some other value for RG now, note you cannot create gains smaller than 1 here. Adding some resistors in series with both ends of the pot will help.

Quote
Also, can anybody tell me, why are they putting load resistor between pins 11 and 5 in all their reference circuits? 
That's symbolic, no real resistor required here. The ends of the resistor show the relevant voltage difference, say "where is my output voltage refererred to", this is always across the symbolic Load resistor.

Figure 6 shows a possible single supply configuration, note you don't have a 5V ref available here anymore. With the INA125 alone you can have either of the voltages, not two at the same time.

Posting a circuit diagram might help others (and me too) to follow your thoughts and be more precise with answers.

An instrumentation amplifier like this INA125 is rather simple to calculate:
Take the difference between Pin 6 and Pin 7, multiply by the gain set, add the voltage at Pin 5 - the result is the voltage at Pin 11 (within the limits set by supply voltage).
« Last Edit: January 30, 2019, 03:14:30 pm by capt bullshot »
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