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LM3404: Texas Instrument data sheet explanation

Federico Polidori
Prodigy 50 points
Part Number: LM3404

Good morning

I'm trying to design a driver for a CREE COB led and I'm wondering if LM3404 is suitable for this purpose or if you can suggest to me some more specific component.

I read attentively the LM3404 data sheet aveilable in your site and I repeated the calculation sample with my data but some resuts make me unsure.

In particular I don't understand why the inductor ripple current, in case of short circuit (ΔiL(led short)) in my calculation path comes smaller than Il (peak) while in the data sheet at the steps [29] and [30] page 24 is the opposite.

In addiction, calculating the output capacitor, I don't understand the difference  between ΔiL and ΔiF and if the target tolerance ripple current of 100 mA p-p indicated after step [31]  is suitable for the COB led I intend to drive.

Down here i put my calculation path 

Input voltage: 20 Vin [Volt] 1,34E-10 cost
LED forward voltage: 17,5 V0 [Volt] 0,4 inductor ripple curront increase
LED current:  0,75 IL [Amp]  
Switching frequency: 3,00E+05 Fsw [Hz]
RON =V0/(1.34 E-10*Fsw) 440298,5075 RON  [ohm] 470000 Ron comm
fSW =V0/1.34E-10*RON 281041,6005 Fsw comm  [Hz]
tON = 1.34E-10*(RON/VIN) 0,000003149 ton  [sec]    
ΔiL = 0.4 *If 0,3 ΔiL  [Amp]
LMIN = ((VIN - VO)*tON)/ΔiL  0,000028341
LMIN [H]
3,30E-05 uH L  comm
ΔiL= ((VIN - VO)*tON)/Ltip  0,257645455 [Amp] 3,00E-05 uH L  comm
ΔiL= ((VIN - VO)*tON)/LMax  2,83E-01 [Amp] 3,50E-05 uH L  comm
ΔiL= ((VIN - VO)*tON)/LMin  2,43E-01 [Amp]
IL(peak)=IL+0,5*ΔiLMax 8,92E-01 [Amp]
ΔiL(led short)= ((VIN - 0,2)*tON)/LMin  1,89E+00 [Amp]
IL(peak )=IL+0,5*ΔiLMax 8,92E-01
Rsns=(0,2*L)/((If*L)+(V0*tsns)-(((Vin-V0)/2))*ton) 0,270571089 ohm      

The mentioned COB LED is a CREE  XLamp CXA1507-0000-000F0YE435F with

Vf 17,5 V

If 750 mA

Thank you very much for the help

best regards

Federico Polidori

mail  kikopoli12@gmail.com

  • 0
    TI__Mastermind 28655 points
    Hello Frederico,

    In particular I don't understand why the inductor ripple current, in case of short circuit (ΔiL(led short)) in my calculation path comes smaller than Il (peak) while in the data sheet at the steps [29] and [30] page 24 is the opposite.

    You are comparing I delta to I peak, what you want to look at is I delta to Idelta, the short circuit current is higher in both cases. One a lot more because Vout is close to Vin and Vout is much higher than the other, (20V/17.5V versus 24V/6.9V).

    In addiction, calculating the output capacitor, I don't understand the difference between ΔiL and ΔiF and if the target tolerance ripple current of 100 mA p-p indicated after step [31] is suitable for the COB led I intend to drive.

    Delta iL is the peak to peak current ripple in the inductor. Delta iF is the desired peak to peak ripple current in the LED string WITH an added output capacitor. It's good you included the LED you are using. I come up with about 225 mV/(750 mA - 650 mA) = 2.25 ohms for rD, this is from the Vf versus Iled curve, I used the 55C but they are all close. Then I used the calculation setting delta iF to 100 mA (the desired peak to peak current ripple in the LED string) and come up with 0.349 uF so I would use 0.47 uF. It should be fine for the LED chosen, it just depends on how much ripple they want in the LED.

    Best Regards,
  • 0 in reply to Irwin Nederbragt
    Prodigy 50 points
    Hello Irwin
    Thanks for such a quick and exhaustive answer.
    In the next days I will go to the realization of the prototype.
    I will inform you about developments.
    For now, thank you again.
    best regards from Italy
    Federico Polidori