• State Resolved
  • Locked Locked
  • Replies 12 replies
  • Answers 3 answers
  • Subscribers 63 subscribers
  • Views 397 views
  • Users 0 members are here
Support feedback
Options
Options
Related

This thread has been locked.

If you have a related question, please click the "Ask a related question" button in the top right corner. The newly created question will be automatically linked to this question.

TPS2H160-Q1: Open load current consumption

Andrea D'Amaro
Prodigy 70 points
Part Number: TPS2H160-Q1

Tool/software:

Hello,

in the following conditions:

DIAG_EN -> pull-up to 5V

IN1,IN2 -> shorted, pull-down

THER -> pull-up to 5V

ST1, ST2 -> shorted, pull-up to 5V

OUT1, OUT2 -> shorted, open load

we're measuring at different supply voltages (12V, 18V, 24V, 30V) always the same power consumption: 27mA.

I see on datasheet that the maximum standby current with diagnostic enabled NOT in open load mode is 6mA.

However, since 27mA is quite high current consumption for standby (at 30V the device is dissipating 0.9W), is this value correct?

Thanks,

Regards,

Andrea

  • 0
    TI__Expert 8440 points

    Hi Andrea, 

    Can you provide a schematic? Are you implementing a pull up resistor on the output to handle leakage current? 

    Thank you,

    Sarah

  • 0 in reply to Sarah Raines
    Prodigy 70 points

    Hello Sarah,

    sorry I didn't get a notification for your reply.

    This is my circuit. I have 6.8kOhm pull-up resistor.

  • 0 in reply to Andrea D'Amaro
    TI__Expert 8440 points

    Hi Andrea, 

    A few things to note: 

    1.) I noticed you are using PSPICE. You will need to test this using the EVM to get "real" measurements. Simulation may not be perfect.

    2.) We recommend using a 20 kohm PU resistor not 6.8 kohm. 

    3.) I see you have a voltage source connected to your IN1/2 pins, you previously said you were pulling these down.  Just want to make sure it is not being overlooked. 

    Thank you,

    Sarah

  • 0 in reply to Sarah Raines
    Prodigy 70 points

    Dear Sarah,

    1) thanks, I will use EVM in future. However, I used it only to show you the schematic.

    2) I'm using 6.8kohm in order to guarantee a specific open load resistance detection range.

    3) I have a voltage source connected to IN1/IN2 just for simulation purpose. However, I'm talking about a real implementation in which the IN1/IN2 signals are pulled LOW by microcontroller.

    Do you think that the reason for this power consumption is the pull-up resistor value? By replacing the resistor the current consumption will be reduced?

    I do not see any current flowing in that resistor, all the current is flowing into VS pin. 

    Thanks,

    Andrea

  • +1 in reply to Andrea D'Amaro
    TI__Expert 8440 points

    Hi Andrea, 

    The PU resistor is intended to help with leakage current during an open load condition. The 20kohm could make a difference. i would recommend trying tis to verify. 

    Where are you probing current? Just at Vs? 

    Thank you,

    Sarah

  • 0 in reply to Sarah Raines
    Prodigy 70 points

    Hello Sarah,

    I've tested the circuit with 20kohm and the current consumption falls downto 4mA.

    Now my doubt is: which is the minimum load resistance that is always recognized as open load by the device with 20kohm pullup? Can you share with me the equation to compute it considering also the effect of leakage current I(ol,off)?

    Thanks,

    Regards,

    Andrea

  • 0 in reply to Andrea D'Amaro
    TI__Expert 8440 points

    Hi Andrea, 

    The data sheet lists a minimum and maximum V(ol,off) value that is the range for an open load condition in the electrical characteristics section. 

     

    Please consider that Vout = Vs* (R_PU/(R_PU+ R_Load))

    The data sheet indicates that an open load event occurs when Vs - Vout < V(ol,off). You can plug these two in to determine the minimum R load that would still trigger the open load threshold. 

    Thank you,

    Sarah

  • 0 in reply to Sarah Raines
    Prodigy 70 points

    Hi Sarah,

    are you sure about the Vout equation? That should be Vout = Vs* (R_Load/(R_PU+ R_Load))

    And what about I(ol,off)? It shouldn't be considered in this equation?

    Thanks,

    Regards,

    Andrea

  • 0 in reply to Andrea D'Amaro
    TI__Expert 8440 points

    Yes that's correct. I swapped the R load and R_PU incorrectly. 

    I(ol,off) is already accounted for in our recommendation for the R_PU value. I(ol, off) * R_PU = 1.5V < Vol,off.

  • 0 in reply to Sarah Raines
    Prodigy 70 points

    Ok thanks,

    last question: if output are connected in parallel should I use always the 20kohm pullup?

    Thanks,

    Regards,

    Andrea

  • 0 in reply to Andrea D'Amaro
    TI__Expert 8440 points

    Hi Andrea, 

    Each output channel requires a 20 kohm PU to Vs. If your channels are shorted together as shown in your above PSPICE model then you just need one, but still 20 kohm. 

    Is that what you meant by "in parallel"?

    Thank you,

    Sarah

  • 0 in reply to Sarah Raines
    Prodigy 70 points

    Yes, this is what I meant.

    Thanks,

    Regards,

    Andrea