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tlc5971 dissipation power

mahmut turk
Intellectual 300 points
Other Parts Discussed in Thread: TLC5971

Hello all,

I am designing a led screen using tlc5971. My design is same as the first application circuit in data sheet. I adjusted the Iref so that the current over leds is 30mA. And my vcc is 12v.

What is the difference in power consuption in TLC5971 if i use a single led per channel or three leds per channel? I mean if i use a single led, is most of the power dissipated as heat in tlc? because the current over leds is independent of number of leds per channel and the power over a single led is constant, where is the rest of the power going? Dissipated as heat in tlc or something else? is the choice of vcc dependent to number of leds? 

Thank you,

  • 0
    TI__Intellectual 955 points

     Hello Mahmut,

     TLC5971 dissipation power can be calculated by the following equation.

    PD = (VCC * ICC) +  (VOUT * IOLCMax * N * (BCn/127) * dPWM)

    Where:
    · PD = device power dissipation
    · VCC = device supply voltage
    · ICC = device supply current
    · VOUT = OUTn voltage when driving LED current
    · IOLCMax = LED current adjusted by RIREF resistor
    · BCn = BC value for OUTn
    · N = number of OUTn driving LED at the same time
    · dPWM = duty cycle defined by GS PWM value

     Also VOUT can be calculated by the following equation.

    VOUT = VLED - (Vf * N)

    Where:  VLED = Anode side voltage of LED lamp, Vf = LED lamp forward voltage, N = the number of LED lamp connected in series.

     Best regards,

     Kazuya Nakai

     

     

     

  • 0 in reply to Kazuya Nakai
    Intellectual 300 points

    Hello Kazuya,

    Thanks for your reply. From the equations it is clear that the number of leds per channel and the vcc should be adjusted accordingly in order not to dissipate much power on tlc5971. 

    Best regards,

  • 0 in reply to Kazuya Nakai
    Intellectual 670 points
    Hi Kazuya Nakai,

    thanks for your explanaition about the power dissipation of the TLC5971 -
    i have tried to follow it with these values:
    VCC 5,0 V
    ICC 22 mA
    VOUT 2,8 V
    IOLCMax 20 mA
    N 12
    Bcn 127
    dPWM 1
    then i get 672 mW.
    but i don't understand what the value range for dPWM is - can you explain this?!

    additional i have this chip in an hand-soldered prototype- and so the thermal pad is not soldered.
    the thermal vias are in the board - but for the prototype i have no oven available.
    i know from the data sheet that the chip has an over temperature shut down.
    so eventually this will trigger -
    but i want to know if this eventually will happen -
    so is my 672 mW my maximum? and is this fine for the chip without properly attached thermal pad?

    i have the calculations as excel sheet and iam willing to share it - but i have not found an option to attach a file here ....

    iam looking forward to read your answer and thanks for your help :-)
    sunny greetings
    stefan
  • 0 in reply to Stefan Krueger
    TI__Expert 3275 points
    Hi Stefan,dPWM in the above formula is duty cycle, = (GS data/65536).
    Whether it can trigger depends on the Temp inside the die. It will eventually trigger if exceed the limit.
    We don't have junction to ambient data f you didn't solder the thermal pad.
    Best regards,Feifei
  • 0 in reply to Feifei Shen
    Intellectual 670 points
    Hi Feifei,
    thanks for your reply and the explanation :-)
    so - just to add to my example:
    if i have GS = FFFF = 65535 = 100% light output that results to dPWM = 1;
    and if i have GS = 1 --> dPWM = 0,000015
    and GS=0 --> dPWM = 0
    so if i have 100% light output thats 'the worst case' for my power dissipation -

    PD = (VCC * ICC) + (VOUT * IOLCMax * N * (BCn/127) * (GS/65535) )

    ok - so for the maximal dissipation i can simplify the formular to

    PD = (VCC * ICC) + (VOUT * IOLCMax * N * 1 * 1) -->
    PD = (VCC * ICC) + (VOUT * IOLCMax * N)

    thanks for your help.


    sunny greetings
    stefan