• State Resolved
  • Locked Locked
  • Replies 4 replies
  • Subscribers 48 subscribers
  • Views 160 views
  • Users 0 members are here
Support feedback
Options
Options
Similar topics

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.

ALM2403-Q1: temperature calculation problem

Paul Chang
Prodigy 20 points
Part Number: ALM2403-Q1

Tool/software:

Dear TI team

The temperature in front of me is ALM2403.
I have a small problem, the imported electricity is 20V, the calculation according to the standard is Vout for 20V/2, RL is 75 ohm, as shown below.

Calculated loss 1.37W.

Use RθJA  to calculation the temp is 149C, it is very high temp in IC.

but if i use RθJC(bot) to calculation the temp is 93C

1. Why should I use the individual calculations?

2. Do I have any problem with my calculations?

3. Or is it a calculation that I have in mind?

  • +1
    TI__Guru 55091 points

    Hi Paul,

    I can simulate it for you tomorrow, but I need additional information. 

    1. Lesl = 75ohm or impedance is 75ohm at 10kHz, What is winding resistance? Do you have interwinding capacitance of the resolver? what is inductance, sL = 75 ohm?

    2. What is Vs supply voltage? Assume that is single supply rail, say 18Vdc or 20Vdc or 24Vdc? Higher Vs, more heat dissipation. 

    3. Output swing is 7Vrms, total is 20Vpp, which is not problem. 

    I will use the following thermal information to calculate Tj. The ALM2403-Q1's thermal pad's heat  is dissipated via PCB board. 

    4. What is input signal type, SPWM, DAC or analog sinusoidal waveform? What is the gains?

    Let us assume power dissipation at ALM2403-Q1 is 1.37W, and the junction to PCB thermal resistance is 22.6C/W and Tamb =93C.

    This will generate 1.37W*22.6C/W + 93 = 116C at Tj, assume this is the worst case scenario. 

    Anyway, no problem in calculate these figures, but I can simulate the power dissipation, if I have the above information. 

    Best,

    Raymond

  • 0 in reply to Raymond Zhang1
    Prodigy 20 points

    Hi Raymond

    1. Lesl = 75ohm or impedance is 75ohm at 10kHz, What is winding resistance? Do you have interwinding capacitance of the resolver? what is inductance, sL = 75 ohm?

    impedance is 75ohm at 10kHz

    2. What is Vs supply voltage? Assume that is single supply rail, say 18Vdc or 20Vdc or 24Vdc? Higher Vs, more heat dissipation. 

    20Vdc

    4. What is input signal type, SPWM, DAC or analog sinusoidal waveform? What is the gains?

    analog sinusoidal waveform

    also Tamb=85C

  • 0 in reply to Paul Chang
    TI__Guru 55091 points

    Hi Paul,

    Since I do not have the necessary parameters to calculate the power dissipation vs. yours, here is an example. 

    The estimated power dissipation is < 1W per the configuration, please see the attached Tina Simulation. 

    If Tamb = 85C, Tj = 85C + 0.964W*22.6C/W = 107C per the configuration, which is acceptable for the worst case scenario.

    ALM2403-Q1 Power Dissipation 08302024.TSC

    If you have other questions, please let me know. 

    Best,

    Raymond

  • 0 in reply to Raymond Zhang1
    TI__Guru 55091 points

    Hi Paul, 

    I assumed that your issues are resolved. 

    If you have additional questions, please let me know. I am going to close this query. 

    Best,

    Raymond