Mid voltage reference

Discussion in 'General Electronics Discussion' started by beedees, May 14, 2015.

1. beedees

17
1
Mar 30, 2015
Hi!
I'm working on a 120 ohm load cell. I decided to use an INA125 to power the bridge (3.4V circa) and to amplify output signal. The full-scale bridge output is +/- 1.06 mV and the initial offset can be in the +/- 1.94 mV range. In order to compensate this initial offset (due to tolerances and temperature) I decided to use a potentiometer in parallel to the bridge. Because the output of the bridge can be positive or negative, the output is referenced to midsupply (3.4/2=1.7 V). (My ADC range is 0-3.6V)
My issue is: how can use INA125 IAref terminal to obtain 1.7 V offset? First I have thought to operate with a voltage divider and a buffer which output is connected to IAref, but resistors must be very accurate i.e low tolerance and TCR. (In my application I have to pay attention to thermal effect).
Another solution I have thought is to use the potentiometer of the bridge in order to obtain an initial 1.7 V offset. In other words, instead of nulling the bridge I can move the wiper until the output is 1.7 V, but I don't know if this method produces a bigger non-linearity.
What do you think about? What is the best solution? Have you other ideas?
Thanks!

5,164
1,087
Dec 18, 2013
Can you please post a circuit diagram.

3. beedees

17
1
Mar 30, 2015
This is what I meant:

I use the potentiometer to create a virtual ground that is the mid reference voltage for the ADC. ADC works form 0V to 3.6V.
So instead of nulling the initial bridge offset I create an initial output of 3.3/2=1.65V and this is my mid reference voltage. I don't know if it could be a good idea or I'm neglecting some error.

5,164
1,087
Dec 18, 2013
Why don't you just balance the bridge by replacing one of the load cells with a variable resistor? The amplifier can run from 3.6 Volts why are you using 5 Volts?

5. beedees

17
1
Mar 30, 2015
Do you mean one of resistors of the bridge? I decided to use this configuration because it works in a ratiometric manner and so it is less temperature sensitive. In my application temperature can change also 5 degree and digipot are characterized by a high temperature coefficient in general. In my configuration temperature effect could be limited.

You're right. I didn't specified that INA125 provides a reference voltage for the bridge. So I use 5 V power supply for the INA and it powers the bridge with 3.3 V.

Initially I thought this configuration:

but I realized that in this way maybe I can't compensate a possible offset of the bridge. For example, if initial bridge offset (due to tolerances) is over 1.65V I can't add a negative voltage to obtain 1.65V! So I have decided to try the first configuration, but I don't know if it is viable alternative..

5,164
1,087
Dec 18, 2013
But this is what the Vref pin is for. This connects to the A/D ref pin to allow for ratiometric operation.

7. beedees

17
1
Mar 30, 2015
Sorry for my lack of experience, maybe I get confused...
As you can see at page 13 the pseudoground due to IAref=2.5 V increases the output voltage.So if [Vin(+) - Vin(-)] is positive then Vout>2.5V and if [Vin(+) - Vin(-)] is negative then Vout<2.5V. In this way I avoid a negative output because the mid reference is not zero but 2.5V.
I think I have to generate a mid reference value externally and use this to increase INA's ouput voltage. Are you according to me?
Can I set (digitally) Vref pin value of ADC and use it to connect IAref pin of INA?

5,164
1,087
Dec 18, 2013
How are you getting this negative voltage? You are using a single supply.

5,164
1,087
Dec 18, 2013
The 2.5 Volt reference on pin 5 is for biasing other opamp stages that operate on a single supply.

5,164
1,087
Dec 18, 2013
Also if you power the bridge from the amplifier you will maintain ratiometric measurements because the excitation voltage wont change much with variations in supply voltage.

11. beedees

17
1
Mar 30, 2015
I don't think so. Page 11, Offset trimming: "Most applications require no external offset adjustment. Figure 2 shows an optional circuit for trimming the output offset voltage. The voltage applied to the IAREF terminal is added to the output signal. The op amp buffer is used to provide low impedance at the IAREF terminal to preserve good common-mode rejection."
This provides to increase the output voltage. Suppose that the bridge is unbalanced and the output should be negative. This isn't possible because I use a single supply. So I need a mid reference voltage that is 3.3V/2=1.65V and this value represents the zero of the bridge i.e. when no load is applied.
Suppose now that a load is applied. The output of the bridge can be positive or negative, so INA's output can be greater then 1.65V or less then 1.65V, ok?
I can't use 0V for mid value, if you know what I meant... For this reason I want to create an offset.

Yes! It is so!. I decided to use INA125 for this reason.

(Sorry for my English)

5,164
1,087
Dec 18, 2013
Yes it will do both but most applications don't need this because the offset voltage is so low. Why do you think you need to adjust this, have you tried it in a circuit?

5,164
1,087
Dec 18, 2013
This is the whole point you balance the bridge first, you don't adjust out the bridge imbalance using the IAREF pin. Well I wouldn't.

14. beedees

17
1
Mar 30, 2015
I think you don't understand what I mean or I can't explain, maybe because I'm not mothertongue

Look at page 10 of INA125 datasheet: there is dual supply (V+ and V-). In this way the central voltage is zero, so IAref is connected to ground.

Look now at page 13: there is single supply (V+=5V and zero). In this way the central voltage is 2.5V, so IAref is connected to 2.5V and output is increased of 2.5V.

5,164
1,087
Dec 18, 2013
No your English is very good. Ok I think I now understand if the load cell goes the other way it needs to go below 2.5Volts. I thought you were only using it in one direction.

16. beedees

17
1
Mar 30, 2015
Exact! In post #5 I reported a way to create an output offset for the INA, but in this way if the initial offset of the bridge is over 1.65 (3.3/2), for example 1.90V, I can't achieve 1.65V because using single supply I can add only positive voltages.
So I decided to unbalance the bridge with an output of 1.65 V first, and then read the output.
Is this a good approach?