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INA149: Differential Input Voltage Range

Karl Morant
Prodigy 20 points
Part Number: INA149


Tool/software:

Hello Ti,

We are using a TI INA149 in a circuit, see snipped below, and I need advice on the maximum differential input range. We are using this circuit with differential input voltages from 3V to around 6V just fine, but we would like to use it with a differential input voltage of 22.5V. Is this possible
 
The datasheet indicates that the differential input voltage is 1.5V from the power rails, but then I see a note on PG-13 about keeping the inputs of the internal applifier within 1.5V of the supply voltage.

Using the equation VOUT = (+IN) – (–IN) + 20 × REFA – 19 × REFB, I fill in the values from the schematic below, and with a 22.5V input I get 7.3V, and the inputs of the internal amplifier stay well under 1.5V.

Is this a workable solution in this application?

Thank you

  • 0
    TI__Mastermind 33151 points

    Hey Karl,

    So there are two gains here that need to be taken into account. There is the common-mode gain, and the differential gain. The common-mode will be attenuated by 20, which keeps the inputs in a linear operating range, however, the differential gain is 1. So if you want to input a voltage differential of 22.5V on the input, you will be expecting a 22.5V output.

    However, you now run into the output limitations of the amplifier. The output cannot exceed the supply rails, (in fact, it must remain within 1.5V of the supply rails as well). In your design case, you must remain within -1.5V < Vout < 8.5V, or increase your supply voltage to be able to support 22.5V output voltage.

    Best,
    Gerasimos

  • 0 in reply to Gerasimos Madalvanos
    Prodigy 20 points

    Hello Gerasimos, 

    Thank you for the reply. I think I understand your reply, however I am using the RefB input to offset the 22.5V differential, look carefully and I have 0.8V there. Using the formula from the datasheet I get:

    VOUT = (+IN) – (–IN) + 20 × REFA – 19 × REFB
    VOUT = (22.5V) – (0V) + 20 × 0 – 19 × 0.8V
    VOUT = 7.3V

    Am I misunderstanding something?

    Thank you

    Kind Regards

    Karl

  • 0 in reply to Karl Morant
    TI__Mastermind 33151 points

    Hey Karl,

    My apologies, I missed the 0.8V refb. You are correct, this will act like a weighted summer to your voltage, and will keep your input CM in a valid range, as well as your output.

    The Vout equation helps to look at the amplifier as a weighted summing circuit. The only addition to this would be to check the input CM to be within the valid CM range, which will be defined by IN+ of the diff amp and RefA. IN+ CM will be set by the input voltage divider, and will subsequently be tried to forced to this voltage by the weighted summer circuit on IN-. If this is already apparent, feel free to disregard this.

    Another quick way to check the validity of the input common mode voltages and output voltages is with the input/output checker tool. This is available for download on the product page, and can be simulated in TINA-TI.

    Best,
    Gerasimos

  • 0 in reply to Gerasimos Madalvanos
    Prodigy 20 points

    Thank you Gerasimos for the explanation and the example.

    I would also like to see what happens when we have 45V at IN- and 22.5 at IN+, but I am not understanding how to set the VCM value. Can you explain that to me?

    Regards

    Karl

  • +1 in reply to Karl Morant
    TI__Mastermind 33151 points

    Hey Karl,

    The VCM would be the midpoint between the two input voltages (33.75V), and the Vdiff would be a -22.5V voltage differential, since IN- is above IN+.

    Even though the circuit says "Do Not Touch" it's just a text box, not a law :). You may delete the voltage differential circuit (VCVS5, 6, 7 and 8), and then rename and connect VCM and VDIFF to different nodes as you see fit (see below for an example)

    22.5V at IN+ will give you an input common-mode voltage of 22.5*19k/399k (IN+ voltage divider) 1.071V. Now, the INA149 will try to force 1.071V at the node labeled VM. To do so, the currents through each resistor connected to IN- of the amplifier must sum to 0 when VM is 1.071V.

    The current across the 380k resistor on IN- will be (45-1.071)/380k = 115.6uA, the current from RefB will be (0.8 - 1.07)/19k = -13.57uA, so the amplifier output will need to sink 115uA - 13.57uA (102.03uA)  from the VM node. Translating this to a voltage, Vout will need to be (Vout - 1.07)/380k = -102.03uA. Solving for Vout you get -37.7V of output voltage needed at Vout. This is well below your V- voltage, and is an invalid output. Swapping the inputs so that IN+ is 45V and IN- is 22.5V

    Once these inputs are swapped, you get a valid output, your input common-mode is now 2.14V, and the Vout must be 7.3V to force 2.14V on the VM node.

    Please let me know if this explanation makes sense, or if there are any other ways I can help!

    Best,
    Gerasimos

  • 0 in reply to Gerasimos Madalvanos
    Prodigy 20 points

    Hello again Gerasimos and thank you so much!

    It appears that you proactively realised that I asked the question wrong -> "Swapping the inputs so that IN+ is 45V and IN- is 22.5V"

    You're second picture is exactly what I meant to ask.

    You have answered all of my questions.

    Kind Regards