• State TI Thinks Resolved
  • Locked Locked
  • Replies 6 replies
  • Answers 1 answer
  • Subscribers 38 subscribers
  • Views 953 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.

How to calculate the effects of resistor self-heating

Vinod Kanoji
Prodigy 90 points

Thank you very much collin for your fruitful information. i have resistor which is 62Ω with 0.25W, +/-1% tolerance, 100ppm,  Tmax_pwr100% is 70°C and 1206 package as per the CRCW120662R0FKEA datasheet. with reference to above equations i have got the below values.

With reference to equation 5 i have got 340­°C/W
With reference to equation 3 i have got 85°C
With reference to equation 4 i have got +/-0.527Ω
With reference to equation 2 i have got +/- 0.62Ω

Now, I have two questions

1. Do i have to add the results of equation 4 and 2 in order to get overall resistance change.
2. What will be the exact state of resistance when goes to negative temperature will the resistance increase or decrease.

Thanks and Regards
Vinod Kanoji

  • 0
    TI__Guru 63965 points

    Hello,

    Can you please share your calculations and we'll respond back to verify.

    Thanks!

  • 0 in reply to Collin Wells
    Prodigy 90 points

    Hi

    Resistance at Negative_Temperature.docxThank you for you quick response. Please find the attached document.

    Regards

    Vinod Kanoji

    As per equation 5,
    = =340°C/W
    As per equation 3,
    Tsh =
    As per equation 1,
    Rtc = ±Ro = ±62 Ω ×  = ±0.651Ω        
    As per equation 4,
    Rtc_sh = ±62 Ω ×  = ±0.527 Ω
    So at 125°C does the total effective resistance becomes
    62 Ω+0.651Ω+0.527 Ω = 63.178 Ω ?
    My second question is in order to get the resistance change in negative temperature for example at -40°C
    Rtc = ±Ro = ±62 Ω ×  =
    -0.372Ω
    Hence at the -40°C does the total effective resistance becomes
    62Ω+0.527Ω-0.372Ω = 62.155 ?

  • 0 in reply to Vinod Kanoji
    TI__Guru 63965 points

    Hello,

    It looks like several of the images you tried to paste in didn't make it through.  If you type out the results instead of copy/pasting it may work better.

    It's unclear from me what the ambient temperature range of your measurement system operates in so it's hard to comment on some of the questions and results.

    A simpler approach for you may be to determine the maximum temperature swing of your 62Ohm resistor both from ambient temperature changes and also from the self-heating and then use the differences between those temperature ranges and 25C to calculate the resistance drift using Equations 1/2.

  • 0 in reply to Collin Wells
    Prodigy 90 points

    Hello,

       I have done what you have suggested but my question is does my calculations holds true for -40°C? Please go through the attached DOCX file in my previous post you will get better understanding of my problem. Ambient temperature of our measurement system is from -40°C to +125°C. 

    Regards

    Vinod Kanoji

  • 0 in reply to Vinod Kanoji
    TI__Guru 63965 points

    Hi Vinod,

    This is largely correct.  Note that you won't be able to operate at 125C ambient due to the expected 85C of self-heating from the 0.25W dissipation.

    You need to figure out the temperature swing of the resistor due to both ambient temperature range (-40C - 125C) and also the self-heating.  Then use those temperatures in Equation 1 (or 4, it's the same) to calculate the resistance change.

    At -40C ambient with 85C of self heating you'll have a resistor temperature of 45C.  So you'll use (45C-20C) * the thermal drift.

    If the resistor is not dissipating power then at -40C ambient, the resistor will also be at -40C and you would use (-40C - 20C) * the thermal drift.

  • 0 in reply to Collin Wells
    Prodigy 90 points

    Thanks a lot collins for helping me out..