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TPS61161A: Looking for App Note or Help to calculate values, Info in data sheet appears to be missing several things

Jeff Sylvester55
Prodigy 70 points
Part Number: TPS61161A
Other Parts Discussed in Thread: TPS61196

Support Path: /Product/Development and troubleshooting/Search the forums for my problem/

From the data sheet on the led driver, it is confusing at best on how to Properly Calculate the part values for any given application.  so we can use a 10uH to 22uH inductor.  and control with PWM not analog dimming as the data sheet suggest.  if switching frequency is 600KHz, diode VF is 0.4v,  and V led is 30V ( 12 * 2.5V) and I am using 20mA  ILED,   and I will assume 85% efficiency,   how do i calculate the Cin, Cout, L, and ripple current, I peak,   so I can select the correct parts,  Because our application is space constrained, we can not use the parts in the example,  and I need to use an inductor that has higher DCR  

  • 0
    TI__Prodigy 600 points
    This Thread was assigned to the appropriate Engineer, it will be addressed in a short period of time.
  • 0
    TI__Expert 4550 points
    Hi Jeff,
    Please refer www.ti.com/.../snva763.pdf this application note, and also you could take a look at datasheet of TPS61196, which contains more calculation equations.
  • 0
    TI__Mastermind 22745 points

    Hi Jeff,

        As Caryss have answered, you can look the two guides to select the right parts. As your application is space constrained, I recommend you to use 10uH inductor. The calculate example is as follow:

        IL_DC=(VoutxIout)/(Vin_minxEfficiency)=30x0.02/(2.7x0.85)=260mA;

        IL_ripple=Vinx(Vout-Vin)/(LxFswxVout)=2.7x(30-2.7)/(10uHx600kHzx30)=410mA;

        IL_peak=IL_DC+1/2xIL_ripple=260+205=465mA;

        Add 20% tolerance, we recommend the inductor has more than 600mA saturation current. You can select the inductor with this parameter. And the DCR will affect the efficiency.

       About the Cin and Cout, we recommend you select 1uF-4.7uF ceramic capacitor, and we recommend 1uF to 10uF ceramic capacitor according to your output ripple as follow:

      Cout=(Vout-Vin)xIout/(VoutxFswxVripple)

  • 0 in reply to Sean Zhou
    Prodigy 70 points

    yes I understand this, my question is why are you using 2.7v  I understand this is the minimum of the chip,  but what if it is not the minimum of the application?  we are using 12v application, so

        IL_DC=(VoutxIout)/(Vin_minxEfficiency)=30.4x0.02/(12x0.85)= 60mA;  !!!  I would expect the current to go down if Vin Min gets higher

        IL_ripple=Vinx(Vout-Vin)/(LxFswxVout)=12x(30.4-12)/(10uHx600kHzx30.4)= 1.21 A???   this does not make sense to me;  how can the ripple current be 20 times higher than the DC current?

        IL_peak=IL_DC+1/2xIL_ripple=60+605=665mA; 

        Add 20% tolerance, we recommend the inductor has more than 800mA saturation current.     Is this calculation actually correct, or am I missing something, or did something wrong?

  • 0 in reply to Jeff Sylvester55
    TI__Mastermind 22745 points

    Hi Jeff,

        When the input voltage increase to 12V, the led driver will goes into DCM mode. So the equations is not as same as above, the IL_ripple and IL_peak may be is several times than IL_dc.

  • 0 in reply to Jeff Sylvester55
    TI__Expert 4550 points
    Hi Jeff,
    In your application, when input voltage goes up to 12V, the boost operates in DCM mode rather than CCM. The equation of IL_ripple used above is in CCM mode. So please refer to the application note I mentioned above to calculate again.