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Original question:

INA212: Power Consumption with no V+ Applied

INA212: High Current Consumption through IN+ / IN- Pins

Andrew Kuikman
Prodigy 130 points
Part Number: INA212
Other Parts Discussed in Thread: INA214, INA240

Related to my former post here: 

https://e2e.ti.com/support/amplifiers/f/14/t/930415

The INA212 in our system is drawing 5mW - 6mW. I have confirmed this is not being drawn through VDD, but looks to be sunk through the IN+ / IN- Pins. Can you provide me with any input structures (caps, etc) that could cause this? The inputs are PWM signals with a typical carrier frequency around 300kHz - 400kHz. I suspect there is input capacitance in the INA212 that could be causing the current drain at these frequencies.

I have experimented with floating the device ground and VDD, but while the power draw can be improved slightly, it does not go away.

Even stranger, I have experimented with different PWM frequencies, and note that there seems to be a small window where power consumption is acceptable (400uA @ ~200kHz), but values above or below seem to increase back to the 6mW level.

Please also note that we are not using this opamp at these frequencies under normal circumstances. The INA212 in our system is used to take a periodic measurement (These 400kHz are disabled at this time), but when the periodic measurement is not taking place, the IN+ / IN- pins are exposed to these frequencies (with VDD grounded).

Any help you can offer would be greatly appreciated.

Thanks,

Andrew

  • 0
    TI__Mastermind 32395 points

    Hey Andrew,

    Thanks for the information.

    1.Could you please explain what the load is for the PWM bus?

    2. Is the PWM controlled by a MCU GPIO pin?

    3. Also, what does this input voltage (VCM) look like exactly? Is it a square wave with a 300kHz-400kHz frequency? I assume it goes from 0V to 3.3V.

    4. Which version of the INA214 are you using? A, B, or C?

    5. Can you check that the ground connections between host (MCU), INA214, ADC, and load are all low-impedance and have the same potential of 0V.

    Overall, the INA214 was not designed to be exposed to a rapidly switching input common-mode voltage (VCM) for extended periods of time regardless of whether the device is powered on or not. It could be that the fast switching is partially turning on input ESD cells and this is what is sinking the extra current. There could also be a discrepancy in the system ground between INA214 and the load. Any detectable difference in the ground voltage can create ground loops, which can also cause more current to sink through the device.

    We do not have capacitance information, but the INA214 was designed to operate on DC bus voltage. If you are using the A version of INA214, then I would try subbing in a B or C version as these have ESD cells that are less susceptible to fast switching VCM.

    If the problem is occurring with the B or C version, then you will want to use a device that was designed to withstand fast switching VCM (e.g., INA240).

    Sincerely,

    Peter

  • 0 in reply to Peter Iliya
    Prodigy 130 points

    Hi Peter,

    Thank you for your quick response!

    1) The PWM is driving a resistive load when we are experiencing the high power consumption. There is a 470nF capacitor connecting the PWM signal, and the inputs to the INA212.

    2) PWM is controlled by an MCU. Ive checked this, and the output is very clean with minimal ringing

    3) Correct, 0V - 3.3V

    4) The part number in quesiton is INA212 (rather than INA214) - full p/n is INA212BIRSW

    5) Ground connections are all very low impedance.

  • 0 in reply to Andrew Kuikman
    TI__Mastermind 32395 points

    Hey Andrew,

    I am confused with the answer #1. Could you please show a schematic a full schematic?

    Sincerely,

    Peter

  • 0 in reply to Peter Iliya
    Prodigy 130 points

    Hi Peter,

    In the state that draws excessive power, the schematic can be represented as follows:

  • 0 in reply to Andrew Kuikman
    TI__Mastermind 32395 points

    Hey Andrew,

    Are PWM1 and PWM2 different signals? Are they out of phase at all?

    Are the input pins really not connected to anything else? 

    I thought the V+ pin is grounded during the fault condition? Or does this not affect the behavior?

    This schematic alone shows that the INA212 input pins do not have a DC bias current return path since DC current can't flow through C3 and C2. This is not a good condition for many/most current sense amplifiers and instrumentation amplifiers.

    This schematic also begs the question of how current is even being sunk into the device when capacitors would theoretically block the current.

    How are you measuring this input current?

    Sincerely,

    Peter

  • 0 in reply to Peter Iliya
    Prodigy 130 points

    Hi Peter,

    Are PWM1 and PWM2 different signals? Are they out of phase at all?
    Yes, different signals. Same carrier freq, different duty cycle.

    Are the input pins really not connected to anything else?

    The PWM signals are also connected to a resistor, but the other side of the resistor is connected to high impedance through the MCU. I know the circuit doesnt look like it makes sense, but there are proprietary pieces that for all intents and purposes can be represented as the resistor to ground. When the INA212 is needed in our system, PWM is turned off, the resistors mentioned above are pulled high, and an AC component is passed into the INA212.

    I thought the V+ pin is grounded during the fault condition? Or does this not affect the behavior?

    When the INA212 is not being used, V+ is grounded, correct. From my testing, the V+ state doesnt impact the increase in power.

    This schematic alone shows that the INA212 input pins do not have a DC bias current return path since DC current can't flow through C3 and C2. This is not a good condition for many/most current sense amplifiers and instrumentation amplifiers.

    Thats a good point, I can try adding some high value resistors to ground to see if it makes any difference.

    This schematic also begs the question of how current is even being sunk into the device when capacitors would theoretically block the current.

    I agree, thats what seems so strange. Through those input caps, I would still expect to see small spikes of current if not for the high-impedance input.

    How are you measuring this input current?

    Input current is being measured as a function of our total system power. With the INA212 removed, our system power drops 5-6mW. When I re-add the INA212 (even with V+/GND/VREF supplied externally), the power increases. The only connections to the system are IN+ and IN- (and shared ground).

  • 0 in reply to Andrew Kuikman
    TI__Mastermind 32395 points

    Hey Andrew,

    Ok so PWM1 and PWM2 being different signals makes the INA212 see at some point large (3.3V) differential input voltages. For this device when the input voltage is sufficiently large (> couple hundred millivolts), the supply current can increase up to 1mA. However, you say that the state of V+ doesn't affect the power increase and the in-series capacitors should DC couple the voltages, but they will still allow the AC differential voltage to pass through.

    1. Is there a shunt resistor anywhere? Is it to the left or right of the in-series capacitors?

    I understand your system is proprietary, but in the end the INa212 (and all other current sense amplifiers) measure the voltage drop across a small shunt resistor which is the power path from source (MCU I assume) and load (the 800-Ω resistors?). I just need to understand where the shunt resistor is and how this relates to rest of system or else I cannot effectively help you. You can always send the full schematic to me privately with a direct message to my E2E account.

    2. How are you measuring total system power?

    3. When you re-add the INA212 and see power increase, is the power you are measuring including the V+ or VREF power? Or are you just measuring the PWM1 and PWM2 power?

    Once again based off the information you have provided, the INA212 is not in a recommended condition to generate its specified total input current (power).The input pins need to be tied together at all times with a small shunt resistor (Rshunt) so the input voltage (VIN = VSENSE = VIN+ - VIN-) is effectively a small voltage (VIN < 100mV). The only time VIN get large is when there is a large current flowing across Rshunt.

    Right now I can't tell if there is any Rshunt in your system so you could be significantly overdriving VIN with a 3.3V differential signal. Or you could just be floating the input pins separately causing the VIN to float to some unknown value to once again cause they VIN to become overdriven and put the INA212 in state where it will consume more current.

    I would short the input pins together with a small resistor (<1Ω) directly at the input pins and then pull-down the input pins with a large resistor (~10k- 100kΩ) to ground. The input pins should never be AC coupled with capacitors to a source voltage when trying to establish datasheet performance. I would also consider simplifying the circuit by grounding V+ and REF pins as well. Once you see the power drop, then begin adding things back like VREF = 1.5V and V+ = 3.3V and see if the power goes back up.

    Sincerely,

    Peter