This thread has been locked.

If you have a related question, please click the "Ask a related question" button in the top right corner. The newly created question will be automatically linked to this question.

INA240: Calculate the Vref error in midsupply divder circuit

Part Number: INA240

I would like to use INA240 in circuit shown below, here the REF1/2 are connected to split the

power supply by half.  See tab2, page 19, Vos_ref; what value of Vref shall I put into equation?

Vref seems to be nominal Vs/2 but I can guess REF2/1 resistor aren’t equal.

 

How such a connection affects error budget?

 Can Ti Support briefly explain and share the definitions of Reference divider accuracy, and RVRR?

 

In my situation Vref1=gnd/Vref2=Vs. I got roughly Half supply. Does it mean that Vos is near zero?

 Shall I put Vos ref as :

 Vosref = RVRR x [ Vs/2 -  Vs/2*(1+0.1%)]

 What is the Vref in the equation - ? (see drawing below)

Can I use 0.1% reference divider accuracy to calculate the Vos_ref?

 As You know we don’t use differential ADC, so the divider accuracy will put some offset to the INA240 internal connections.

The example from datasheet are not similiar to mine. The don’ use internal Ref divider.

---------------------------------------------

 

 

 

  • Hi,

    The ref divider accuracy directly affect the quiescent output level. As you found out it is near mid-supply but maybe some mV off. This deviation is bound by the 0.1% ref divider accuracy.

    RVRR defines how much additional input referred offset is caused when the reference voltage moves away from mid-supply. If you want to account for it, the way you plugged it into the equation is correct, however the effect should be negligible. Only when ref voltage is volts, not mV, away from mid-supply the effect becomes somewhat measurable.

    In your situation with single ended measurement, most of the error is going to come to ref divider accuracy spec.

    It is also important to have a quite supply to begin with.

    Regards, Guang

  • Dear Zhou,

    Glad You answered, please look at the picture, I just would like to confirm if the specified accuracy

    is R1up / (R1up+R1low) or using datasheet of INA240:

    Vout = |Vref1- Vref2|/2 at Vsense = 0 means in my situation

    Vout = (Vs-0) / 2 = Vs / 2

    and I shall put Vs/2 * 0.1 * 0.01  used 0.1%)

    Again the question arrises what is equation for RVRR ?

    if RVRR is input referred (see datasheet) than

    how to calculate voltage at output?

    if RVRR is for input refrenced, shall I multiply the RVRR * Gain?

    Best

    Krzysztof

  • Hi Krzysztof,

    Your understanding is correct – you should use RVRR*Gain to refer to the output. I’ll use a number example to illustrate how to account for RVRR.

    In this example, VS=5V; Gain=20; instead of setting Vref1=Vref2=2.5V (or equivalently splitting the supply) we set Vref1=Vref2=0.5V. To account for the RVRR effect, we calculate input referred offset to be RVRR*(2.5-0.5). To refer to the output, simply times the same term with gain, or RVRR*(2.5-0.5)*20.

    Regards, Guang

  • Dear Guang,

    Excus the lat answer.
    I can understand in my situation with splitted power supply Vref1=5.0V and Vref2 0V, I will amost get no error,
    but the didvider as I know has some tolerance. And using divider tolerance I will get 2,5V +/- tol
    so only the RVRR*(2,5 - 2,5 -/+ tol)*20 - true?

    Kind regards
    Krzysztof Ptak
  • Hi Krzysztof,

    I think the confusion is rooted in how to use these two parameters – divider accuracy and RVRR. I’ll try to explain.

    The reference divider can contribute to output error when splitting the supply. For example if VS=REF1=5V; REF2=0V, then the reference voltage that INA240 sees can have an error of up to 2.5mV (0.1% on 2.5V). This error will be directly reflected at the INA240 output. If both REF pins are tied together and driven with a single reference source, then the divider doesn’t contribute any error.

    The RVRR spec is used to define the effect of changing reference level, when it is different from mid-supply. It doesn’t matter where this reference comes from or how it is generated. If VS= 5V, and the reference INA240 sees is 0V(REF1=REF2=0V, for example), then there would be a 20uV*(2.5-0)*20=1mV additional offset at INA240A1 output.

    Regards, Guang

  • Hello Guang,

    Thank You provided more "inspect light" at this topic. As You know my circuit doesn't use diffrential ADC, so I have INA240 output connected to ADC. As You can see from the figure 29 (ina240 datasheet) which I use.

    Shall I assume 0,1% error of 2,5V ??

    Because I use splitted supply and not driven by common reference. Is it true?

    My best, Krzysztof
  • Hi Krzysztof,

    Yes in this case you’ll assume that the max error caused by the reference divider is 0.1% on 2.5V, or 2.5mV at the INA240 output. This is based on the assumption that the 5V supply is spot on. If VS=4.9V instead of 5V, then the quiescent output would be 2.45V±2.45mV, where the 2.45mV is due to reference divider tolerance.

    Regards, Guang

  • Dear Guang,

    Nice You helped to understand all the questions. Summarizing -
    1. Is there a equation for RVRR or method to measure?

    Found one of Your description in blog:
    "RVRR is “reference rejection ratio”, if there is a 1V ripple on REF pins (REF1=REF2=Vs/2 as spec table says, which equals 0V nominally in your configuration), it will cause a 20uV input referred ripple. When translated to output, it is Gain X 20uV. Hopefully you don't have such a big ripple to begin with, otherwise you'll need a better reference!"

    and Patrick Simmons answer at blog:
    "My test setup is illustrated in the diagram below. At each REF voltage, I measured both the ref voltage as well as the difference between the ref and output. Then I took the difference between measurements. Thereupon I divided the change in Vout-Vref by the gain and the change in Vref. So the ratio of change is plotted in the graph below."

    The setup circuit to measure RVRR (after Patrick Simmons)
    e2e.ti.com/.../BP101a.png

    Is the
    Delta(Vout - Vref) / (Delta Vref * Gain)
    a definiton of RVRR?

    2. Secondly, if I measure the outpuyt the error produced by divider is big (2,5mV nominal) - as I don't use differential ADC,
    can I try to use ratiometric supplied ADC reference to lower the error when INA240 Vsupply will change. (?)

    Kind regards, Krzysztof
  • Hi Krzysztof,

    Here are my thoughts on both questions -

    1 – The description and equation are equivalent. This is how RVRR is defined.

    2 – I’m not so sure on this proposal. It might lower the relative error if the output level is near full scale. However if output level is low, such as near 0V, there would be no error cancelling benefit. Besides, you probably want a known stable reference for the ADC.

    Regards, Guang