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Original question:

INA220: Shunt voltage random values (spikes) when measuring 0mA

INA220 - 9.2.4 Typical Application: 48-V Telecom Current/Voltage/Power Sense

Neel Desai
Prodigy 140 points
Part Number: INA220
Other Parts Discussed in Thread: OPA333, LM5056A, AMC1300, INA240, INA293, INA3221

Hello,

I am attempting to meaure bus voltages up to +48V with INA220 using the 9.2.4 typical application note in the datasheet, page 28, figure 32. http://www.ti.com/lit/ds/symlink/ina220.pdf?ts=1588366555281

I see there is a user guide for the -48V application note, but could not find anything for the +48V besides figure 32 in the DS. I am trying to understand the relationship between the RSHUNT resistor and the RL resistor (shown as 100 ohms in figure 32). I do not see any equations that show me the relationship between these to feed into the V+ and V- on INA220 to determine the resolution.

Am I missing something here? It seems like basically, they are using a 24V Zener to clamp the voltage "seen" by the INA220 on Vbus, correct? What would be the resistor tying the OPA333 and 5.1V Zener to GND? 

Any advice or another solution which I can use here for higher than 36V bus voltages?

Thank you.

  • 0
    TI__Mastermind 27715 points

    Hello Neel,

    Thank you for your question! Our engineers will analyze and respond when we return to work on Monday.

    Best regards,

    Ian Williams
    Applications Manager
    Current Sensing

  • 0
    TI__Mastermind 28485 points

    Hello Neel,

    Thanks for posting to the forum. Here is the application note for the circuit shown in Figure 32: http://www.ti.com/tool/TIDA-00528

    The Rshunt and RL set the gain of the circuit. The Rshunt is applied acros the RG resistor to create an output current that will in turn flow across RL and this voltage sensed by INA220.

    Resolution is determined by the INA220’s Vshunt LSB and Vbus LSB.

    The bus voltage is divided down to satisfy the 36V Abs Max rating of the device. The 24-V tranzorb is there to protect the pin from quick voltage spikes.

    The resistors tying the OPA333 and Zener is Rz and this is needed to bias the Zener diode into its reverse voltage, which is what is powering the ICs.

    The only current digital current amplifier/power monitor we have is the 80-V rated LM5056A.

    Sincerely,

    Peter

  • 0 in reply to Peter Iliya
    Prodigy 140 points

    Hello Peter,

    Thanks for your reply - This seems to be for 40-400V applications, the datasheet (9.2.4 typical application note in the datasheet, page 28, figure 32. http://www.ti.com/lit/ds/symlink/ina220.pdf?ts=1588366555281 seems to suggest up to 0-48V measurements? Am I reading this correctly?

    My measurement range would be from 0-50V, so I would need to go on the lower side of the 40V and not as high as the 400V. Do have any suggestions or reference designs for this?


    Thank you.

  • 0 in reply to Neel Desai
    TI__Mastermind 28485 points

    Hey Neel,

    Figure 32 of INA220 datasheet will not work at 0-V bus voltage for a couple reasons. One, the 5.1V Zener needs sufficient bias voltage and current to be a stable supply voltage. Secondly, the level-shifting MOSFET will need sufficient drain-to-source voltage to be biased properly. The advantage of this circuit is the level shifting of the shunt voltage from 48V (beyond the INA220 rating) to 0V (within INA220 operation) common-mode; however, it only works for high-side applications.

    Possible alternate solution for 0V to 48V capability is a device that has this rating. You could use an analog current sense amplifier such as the INA240 or INA293 and this would drive an ADC. Also, maybe and isolate amplifier such as AMC1300 or other located here: https://www.ti.com/isolation/isolated-amplifiers/overview.html

    Another solution is using the INA3221 which is a digital current monitor, but with 3 channels. So you could dedicate one channel & shunt to read the output current from OPA333 and another channel & shunt to measure the true, 0-V low side of the system.

    Hope this makes sense. Post back with more questions.

    Best,

    Peter