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MPC507: Input Voltage drop when power off

Shinichi Inoue
Guru 21791 points
Part Number: MPC507


Dear Specialists,

My customer is considering MPC507 and has a question.

I would be grateful if you could advise.

---

If voltage is applied to the input pin of this device when it is not powered, the voltage will drop.
I would like you to confirm this along with the estimation results.

■Situation (1)
Connect 10 in series with 10kΩ between 10V and ground when the power is OFF.

Then measure the voltage between each 10kΩ.

In this case, each voltage is 10V, 9V, 8V, etc.

However, the first 10V and 9V parts are not connected.

Power supply = 20kΩ, 10kΩ, 10kΩ...10kΩ = GND.

The same voltage was observed when the power was turned on.

■Situation (2)
Here, if this 10kΩ is changed to 100Ω,

When the power was turned off, each voltage became 5.8V, 4.7V, 4.2V, etc.

Each voltage becomes lower.

However, when the power is turned on, the voltage is divided normally.

Could you please see the attached file.

MPC507A　input resistance and voltage.pdf

■Considerations
According to the datasheet's FUNCTIONAL DIAGRAM, the input resistance of the device is approximately 1kΩ, and there is an overvoltage protection circuit at the subsequent stage.

When the power is OFF, there is no power to this protection circuit.

It becomes approximately ground potential through the power supply circuit, etc.

In the case of 10kΩ, it is larger than the built-in 1kΩ, so it is not affected.

However, since 100Ω is low, it is affected by the resistance inside the device, and the voltage at the measurement part drops significantly.

After that, it is divided into equal parts, but not in 1V increments.

---

I appreciate your great help in advance.

Best regards,

Shinichi

  • +1
    TI__Genius 10552 points

    Hi Shinichi,

    Thanks for reaching out here. Is the question here basically: why does the divided voltage vary between 100ohm case and 10kohm case? Or just to confirm that the observed behavior is correct?

    The observed behavior is as expected here. IN the 10kohms case, the voltage drop across the internal resistance is negligible so the voltage drop is determined by external resistors, therefore you see your input voltage divided into equal parts. In the 100ohm case, the voltage drop across the internal resistance is significant, therefore your input voltage is affected by both internal and external resistors which is why you see the voltage divided into unequal parts. 

    If the external resistance is much larger than the internal resistance, the voltage drop across the internal resistance is small. If the external resistance is comparable to or smaller than the internal resistance, the voltage drop across the internal resistance is larger.

    Does this help? Let me know if you need anything else here. 

    Regards,

    Alex

  • 0 in reply to Alex Vish
    Guru 21791 points

    Hi Alex,

    Thank you for your reply.

    This inquiry was a request to confirm the measurement results and considerations for the results.

    I'll share your answer with the customer.

    When the customer has an additional question, I consult you again.

    I appreciate your great help and cooperation.

    Best regards,

    Shinichi

  • 0 in reply to Shinichi Inoue
    TI__Genius 10552 points

    Thanks, Shinichi. I will go ahead and close this thread for now. 

  • 0 in reply to Alex Vish
    Guru 21791 points

    Hi Alex,

    The customer has additional questions.

    Could you please advise?

    ---

    (1) Voltage should be generated by connecting to GND internally. Where does the internal resistance (1kΩ) connect to GND?

    According to the block diagram in the data sheet,  the internal resistance (1kΩ) seems to be connected to GND via an internal protection circuit. Is this understanding correct?

    (2) In the attached circuit diagram, is there no problem even if signals are input to IN8A to IN1A pins when the power is OFF?

    The absolute maximum rating of the input pin(Vs) is 20V and input voltage operating range is +/-15V

    We think that there will be no problem as long as it is within this range. Is it correct?

     ---

    I appreciate your great help and cooperation.

    Best regards,

    Shinichi

  • 0 in reply to Shinichi Inoue
    TI__Genius 10552 points

    Hi Shinchi,

    You are correct that this internal 1k resistance is connected to ground through the internal protection circuit! 

    Additionally, your assumption of overvoltage conditions is correct as well. Input voltages can be +/-15V with no issue here!

    Regards,

    Alex

  • 0 in reply to Alex Vish
    Guru 21791 points

    Hi Alex,

    The customer has an additional question.

    Could you please advise?

    ---

    I understood that when the external resistance was set to 100Ω and the power was turned off, the voltage appearing at the input was different due to the effect of current flowing from the internal ESD protection circuit to GND.

    On the other hand, even if the external resistance is set to 100Ω, the input voltage is divided appropriately when the power is turned on. 

    Why is the phenomenon different when the power is turned on and off?
    Are the paths through which the current flows different?

    ---

    I appreciate your great help and cooperation.

    Best regards,

    Shinichi

  • 0 in reply to Shinichi Inoue
    TI__Genius 10552 points

    Hi Shinichi,

    That's correct and what you're seeing is powered off protection feature of this device. When the power is off, signal sources represent the 1Kohm resistance under this condition. This could cause a different voltage to appear at the input due to the current flowing from the internal ESD protection circuit to GND.

    When the power is turned on, the input voltage is divided appropriately because the multiplexer is active and the current paths are different. The current will flow through the selected channel of the multiplexer, and the external resistance will divide the voltage accordingly.

    So, yes, the paths through which the current flows are indeed different when the power is turned on and off. This difference in current paths is what causes the observed phenomenon. The multiplexer’s operation (on or off) changes the effective resistance at the input, which in turn affects the voltage division.

    Regards,

    Alex

  • 0 in reply to Alex Vish
    Guru 21791 points

    Hi Alex,

    Thank you for your reply.

    The customer has additional questions.

    Could you please advise?

    ---

    In the previous answer, it says that when the power is off, it is connected to GND via the protection circuit.

    Is it possible to understand that the resistance of the protection circuit part becomes almost 0Ω when the power is OFF?

    Also, is the protection circuit connected to GND only when the power is OFF?

    Does the resistance of the protection circuit have any effect when the power is turned on?

    ---

    I appreciate your great help and cooperation.

    Best regards,

    Shinichi

  • 0 in reply to Shinichi Inoue
    TI__Genius 10552 points

    Hi Shinichi,

    As mentioned before, when the power is off, the device presents a resistance of 1Kohms at the input. This means that the resistance of the protection circuit does not become almost 0Ω when the power is off, but rather it becomes 1kΩ.

    The protection circuit is indeed connected to GND when the power is off to prevent damage to the device.

    When the power is on, the protection circuit continues to function. It allows input voltages to exceed power supply voltage without damaging the device. Therefore, the resistance of the protection circuit does have an effect when the power is on, as it helps to maintain the integrity of the signal and protect the device from over-voltage conditions.

    Regards,

    Alex