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.

  • Resolved

INA303: INA303 DELAY capacitor calculation (and a few other questions)

Prodigy 50 points

Replies: 8

Views: 140

Part Number: INA303

Hi I have been reviewing the INA303 and would like to use it with the delay. I followed the example you have on page 15 of the data sheet but the equation does not work out for me. I am using Vth = 1.22 V and Id = 5 uA per the Electrical Characteristics but that does not give me the same capacitor value. Do I need a pull-up resistor on the DELAY pin to set Id? What does Vdelay = 0.6 V in the Electrical Characteristics for Id imply - is that related?

I would also like to confirm with the DELAY, that if I am detecting an undercurrent condition and the input crosses under and then back over the threshold during the delay, I should not see any change in the ALERT2* output. This seems to be implied in the plot on page 17, and is stated more explicitly for the example on page 18, but I am using the configuration shown on page 17 and would just like to make sure this is the case.

A couple of other smaller questions, unrelated to the DELAY but specific to the INA303:

  • what is the current capacity of the ALERT* pins? The VOL of the comparators specs an IOL of 3 mA. I would like to use this output to sink 1-2 mA to toggle an LED.
  • the differential inputs of the amplifier are spec'd with a +/- range with Vref = Vs/2. For the INA303, it is spec'd at +/-25 mV. If my REF pin is grounded, does that mean the inout range is 0-50 mV? The example on page 19 seems to imply this.

Thank you!

Jeff

  • Hello Jeffrey,

    I agree with you calculation error that you see.  I will investigate what is the issue or if it is typo.    With regard to need a pull up you should not need it unless you are attempting to get a smaller delay.  The Vdelay in the conditions is basically a condition the part is in when the 5uA is measured.

    I will verify this but I assume you mean it returns above the hysteresis voltage point.

    The output is an open drain transistor that current will be limited and this will vary on many parameters.  You could get a variation of output voltage with a heavy load as you can see if the specification that the typical value at 3mA load is 70mV but can go as high as 400mV.    If you are using this to drive your LED take into account the possibility of voltage change.

    You are correct that that you will have that range but keep in mind you will be limited on the output by the Swing to Rail in the Electrical Specification Table.

    Best Regards,

    Javier Contreras, TI Sensing Products Applications Support

    Getting Started with Current Sensing Video Training Series

    All information in this correspondence and in any related correspondence is provided “as is” and “with all faults”, and it is subject to TI’s Important Notice (http://www.ti.com/corp/docs/legal/important-notice.shtml).

     

  • In reply to Javier Contreras4:

    Hi Javier, thanks so much for the reply. Yes, please let me know about the equation. I am trying to get a long delay (2-3 sec) and will follow the equation for now.

    And yes, returns above the hysteresis voltage point. What I am trying to do is allow enough time (via the DELAY feature) for the system to power on before the CSA starts detecting and reporting under/over currents. I don't want any ALERT outputs during power on. From the data sheet, it looks like during the DELAY, any over/undercurrents will not be reported so long as the sense voltage returns within range before the DELAY expires (plot on page 17, text on page 18). Is this correct? During normal operation, a 2-3 second delay is acceptable.

    Thank you for pointing out the voltage variability for VOL. I did notice that and have sized the currently limit resistor accordingly.

    And thank you also for pointing out the swing to Vs spec - noted.

    Thanks again!

    Jeff

  • In reply to Jeffrey Nguyen:

    Hello Jeffrey,

    I am confident that the equation is correct and you will need about 10µF to get a 2.4sec delay.  I can verify this but I will not be back in the lab until next week if you can wait that long.

    The Alert should also not activate unless the stays that long.  The one concern is that the capacitor could not discharge fast enough to start a new count cycle.  To verify this we would also need to wait till next week for verification.

    Please let me know if you can wait till early next week. 

    Best Regards,

    Javier Contreras, TI Sensing Products Applications Support

    Getting Started with Current Sensing Video Training Series

    All information in this correspondence and in any related correspondence is provided “as is” and “with all faults”, and it is subject to TI’s Important Notice (http://www.ti.com/corp/docs/legal/important-notice.shtml).

     

  • In reply to Javier Contreras4:

    Thanks for following up Javier. I also calculated 10 uF for 2.44 sec so that's good. I can wait for verification. We are proceeding with layout so there will be an opportunity to update this value if needed. In an 0603 package, I was able to get 10V, which I believe is enough.

    And what would happen if the capacitor does not discharge fast enoigh - does that mean that the next delay could be even longer?

    Thanks again. I look forward to your update.

    Jeff

  • In reply to Jeffrey Nguyen:

    Hello Jeff,

    Please look at the specifications of the capacitor.  Please note a few items. 

    • The capacitance will have a large range in accuracy and will most likely be your cause for variation from each variation you have. 
    • Be aware of parasitic values such as leakage.     
    • The higher the voltage the greater the chance of the capacitor having the same capacitance across voltage and other drift parameters.  If cost/size is more important please be aware of the difference from capacitor to capacitor variance through all your design specifications.  

    As for the capacitor not discharging it will make the next delay shorter.  I think there is a fast discharge path but I would like to verify this information.  

    Best Regards,

    Javier Contreras, TI Sensing Products Applications Support

    Getting Started with Current Sensing Video Training Series

    All information in this correspondence and in any related correspondence is provided “as is” and “with all faults”, and it is subject to TI’s Important Notice (http://www.ti.com/corp/docs/legal/important-notice.shtml).

     

  • In reply to Javier Contreras4:

    Thanks again Javier! The delay doesn't need to be very accurate so I think we will be OK - something like 2 sec +/- 1 sec is probably OK for us. But I agree with what you have said about the capacitance variance, and the derating across voltages. I tried to find the right balance between package, tempco, tolerance, and voltage and I think with an 0603 I can find some caps at 16V or even 25V.

    Good to know about the delay potentially being shorter too. Please provide a follow up on that if you can. Longer is better than shorter for us, but shorter may be OK too.

    Thanks!

    Jeff

  • In reply to Jeffrey Nguyen:

    Jeff,

    I verified the equation is correct.  And the example calculation is not correct.

    Once the output passes the hysteresis point to reset the alert condition the delay pin will reset to zero the delay timer will start again. This reset will discharge with a typical resistance of 70Ω to GND.  So the discharge of the capacitor will be quick.  I also verified this quick discharge.

    The delay will typically be the same for each instance of the alarm condition.  This is because of the quick discharge.

    Best Regards,

    Javier Contreras, TI Sensing Products Applications Support

    Getting Started with Current Sensing Video Training Series

    All information in this correspondence and in any related correspondence is provided “as is” and “with all faults”, and it is subject to TI’s Important Notice (http://www.ti.com/corp/docs/legal/important-notice.shtml).

     

  • In reply to Javier Contreras4:

    Great, thank you Javier for confirming and following up. Your support is much appreciated. This seems like a great chip.

    Jeff

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.