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Original question:

INA190: Part number typo in Space catalog?

INA190: INA190-SP

Rajan Bedi
Expert 1455 points
Part Number: INA190
Other Parts Discussed in Thread: INA901-SP

Hello,

What is the relationship between Vsense and Vout? Is it Vout = 20 x Vsense?

The typical value is Vsense is specified as 0.15V, but figures 17 and 18 only plot Vsense up to 24mV.

If Vsense = 150 mV, will Vout = 20 * 0.15 = 3V?

thank you

Rajan

  • 0
    Guru 140200 points

    Hi Rajan,

    do you have the latest datasheet of INA190?

    Kai

  • 0 in reply to kai klaas69
    Expert 1455 points

    The datasheet which I have is: SBOS938C –OCTOBER 2018–REVISED FEBRUARY 2020

    there is no mention of a Vref and the specified gain is 20 V/V.

  • 0 in reply to kai klaas69
    Expert 1455 points

    can the INA190-SP measure small negative currents?

  • 0 in reply to Rajan Bedi
    TI__Mastermind 26730 points

    Hello Rajan,

    I believe you are questioning the INA901-SP, which is not released yet for production currently so there is no official datasheet yet to go off of. Any datasheet online is outdated until it is public on ti.com.

    Anyway, I can make some comments about part functionality and address your questions. If the gain is 20 V/V, then the relationship between input (Vsense) and output (Vout) is simple as Vout = 20*Vsense given the device is operating within is linear region.

    So if Vsense = 150 mV, the Vout will equal 3 V only if the device’s output voltage is within the swing to rail specification. So if the Swing-to-Vs spec is 0.2 V, then the supply voltage must be greater than or equal to 3.2 V in order for Vout to drive to 3 V.

    I believe figures 17 and 18 represent the device’s performance when Vsense < 20 mV, thus the x-axis (Vsense) only goes to 24 mV.

    Also, I believe this device is unidirectional and thus cannot measure negative currents (Vsense < 0 mV). You can determine if a device is bidirectional (measure positive and negative currents) when the device has a reference pin.

    Hope this helps.

    Best,

    Peter