0
\$\begingroup\$

I was asking about the Al1676-30A LED Driver produced by Diodes Incorporated. (Product Overview)

The specified LED Driver is listed in Mouser Electronics website to provide 3 A of current to the connected load, however, when I was reading the datasheet of this product I have found in the recommended applications that the recommended maximum output current is only <=300 ma. So what is the difference between the recommended application value and the value specified in the product description. (Datasheet)

\$\endgroup\$
  • \$\begingroup\$ Because Mouser people misread the datasheet. They took "absolute maximum ratings" part only. Rdson =3ohm with 3A current cannot be good. \$\endgroup\$ – Todor Simeonov Jun 8 '17 at 16:49
  • \$\begingroup\$ Do you mean that it would burn the LED Module when connecting a low resistor value? \$\endgroup\$ – Ahmed M.Zahran Jun 8 '17 at 16:58
  • \$\begingroup\$ In the datasheet is stated that this chip has overtemperature protection. Most probalby it will work few seconds with 3A current, overheat and reduce output power or turn off. You can burn the LED module if current sense resistor provides current more than LEDs can handle. If you need 3A this is not your driver. Consider your LED module's voltage and if it is much lower than mains voltage, you should choose a flyback driver with transformer. \$\endgroup\$ – Todor Simeonov Jun 8 '17 at 17:05
  • \$\begingroup\$ @TodorSimeonov For the absolute maximum rating, they stated that AL1676-30A can provide a continuous drain current of 3A. \$\endgroup\$ – Ahmed M.Zahran Jun 8 '17 at 17:54
  • \$\begingroup\$ Continuous... but @ Tcase = 25C. Try to cool down couple of watts out of SO-8 to 25C ;) However such buck converters directly from mains are mostly intended to be used (and efficient) when LED array is significantly HV (like 70-80V USA or 150V Europe) and current is low - up to 150mA or 300mA. If your LED is 10V/3A this is not the right solution. \$\endgroup\$ – Todor Simeonov Jun 8 '17 at 18:19
0
\$\begingroup\$

The short answer is both specs are valid but your experience in reading and comprehending datasheets is weak. The Peak current is often > 10x the average current spec for switch efficiency reasons and small package thermal rise.

Duty Cycle, d (or Duty Factor) also plays a part in this ratio of Imax/Idc (for recommended <300mA rating.)

sage advice

for all questions on MOSFET current ratings used as switched current devices.

  • Keep this pk/avg ratio in mind when choosing MOSFETs and % losses in future. Also understand DC Motor Current starts with the ΔV/DCR=Istart ratio then reduces to 10% of this at full load at max Hp ( usually 80~90% max RPM). Also all capacitors including batteries have ESR which limits the pulse current to ΔV/ESR= ΔI. Thus peak currents for charging batteries or bridge caps can be much higher than average and driver RdsOn must be computed in charger losses thus that Pd * d/ESR=I² using duty cycle, d. This is simply Ohm's law, time averaged and re-arranged.

The first page and 1st line under Features on the datasheet says >90% Efficiency. You cannot achieve this unless your losses proportional to total loop current R (incl load) are <10% which means the RMS current must be <10% of the Imax and thus must be >10x to achieve the recommended Idc <300mA used and >90% efficiency.

Details

When you understand the use of Ohm's Law for power dissipation, efficiency , thermal resistance, heat sink design and temperature rise. It will all become clear. It is assumed you understand the following equations from Ohms Law and Thermodynamics.

  • \$P_D=I^2R_{DS(ON)}*d\$ .......(1)
    • for duty cycle d {0~1) and then junction temp rise
  • \$\Delta T_{JA}=P_D*(θ_{JC}+θ_{CA}) \text{ or } = Pd*(Rjc+Rca)\$ .........(2)
    • with no heatsink except 2"x 2" (or 26 cm and free air convection THERMAL RESISTANCE is defined as (depending on font format limits of text);
    • Datasheet specs are ALWAYS given with an infinite heatsink @ Tj=25'C but tested with a thermal junction thermal base regulator in a pulse mode (unless otherwise specified)

basil advice

(small pun)

  • Designers must understand this to DERATE the PD limits according to their Ohm's Law application of thermal resistance and limited thermodynamic understandings of heat velocity (slew rate °C/s), steady Thermal resistance °C/W and resulting junction rise, °C or °K ( same for ΔT)

    • \$Rja \text{ or } R_{JA} \text{ or } \theta_{JA}=123°C/W\$ for SO-7 case

    There are thermal limitations of instantaneous junction temperature rise and the peak current safely not given for Pd and this is what defines the part number and its corresponding current limit. e.g. AL1676-xxAS7-13 where xx means the Max x.x Amp peak limit for p.n.'s with xx= 10,20,30,40 \$ I_{DS} \$ and corresponds to the Max \$R_{DS(ON)} \$ of 12Ω, 6.5, 3, 2.5. Also note the xx is followed by a Letter A,B,C or D which must denote internal resistance (either electrical or thermal) for minimizing junction temp rise or fusing current margin of wirebonds.

Validation on assumptions

  • Using equation ...(2), for example, part number ratings, with Rca=0, and Rjc=19 °C/W (from spec Note 5, 1"x1" FR-4 2oz Cu with min. footprint), we get:

    • for 4A rated part Pd=(Ids)²•RdsOn = (4.0A)²•2.5Ω = 40 W
    • since we know Diodes Inc has a OTP shutdown at 150°C and from Rja we would expect it happen very fast or else you would use Ids (Max) with a duty cycle well below 20% or you have an excellent CPU style heatsink with Rca=0.2°C

Using the data in the datasheet there must be some sorting or process differences to separate each final part number. I suspect they are all made with same design but process controls and variations together with 100% testing with ICT test results will determine final part number.

AL1676-20AS7-13 = AL1676-xxyS7-13 with bins for xx and y.

p.n.               RdsOn    Ids Vds(max)    Pds     category
                    [Ω]     [A] =Rds*Ids    Max [W]  from p.n.
 AL1676-20AS7-13    6.5 Ω   2 A     13 V    26 W    A
 AL1676-30AS7-13    3.0 Ω   3 A      9 V    27 W    A
 AL1676-10BS7-13    12.0 Ω  1 A     12 V    12 W    B
 AL1676-20BS7-13    6.0 Ω   2 A     12 V    24 W    B
 AL1676-20CS7-13    5.5 Ω   2 A     11 V    22 W    C
 AL1676-40DS7-13    2.5 Ω   4 A     10 V    40 W    D
\$\endgroup\$
  • \$\begingroup\$ Does it means according to these specs, that the maximum power that the device can be allowed to dissipate would be equal to (Tjmax-Tamb)/Rjc which is equal to (150-25)/19 = 6.5W Considering that there is no heat sink is used in this case. Also the actual Pd of AL1676-30A would be (@ Ids = 3A) = 27W. \$\endgroup\$ – Ahmed M.Zahran Jun 9 '17 at 15:11
  • \$\begingroup\$ Not quite, that assumes infinite heatsink. Using the 2x2" copper substrate suggested with 300mA and RdsOn of say 12 Ω for the AL1676-10BS7-13 1.0A pk you can see that you get 0.3A^2*12=1.08W with 19'C rise to case and I know 4sqin of copper is about right. But for AL1676-30A 0.3A^2*3Ω =0.27W so you can determine how much more current and case rise depends on copper heatsink on board. ( other calc. needed by you for required Rca heatsink using copper substrate and proper pads) You do the same for power FETs, resistors and any SMT package design.... Using Ohm's Law for thermal resistance.... \$\endgroup\$ – Sunnyskyguy EE75 Jun 9 '17 at 21:26
  • \$\begingroup\$ Yes, so to be more precise; I have to consider the thermal resistance between all of : junction to case, case to heat sink (bond if used), and heat sink to ambient. \$\endgroup\$ – Ahmed M.Zahran Jun 10 '17 at 1:11
  • \$\begingroup\$ You got and now you ought to remember SMD parts this size are like prefer to run 0.25W If you need more current use a lower RdsOn so that I^2R=Pd suits the part with low T rise and always design a heatsink as much as possible with the copper layout. Learn the Rca for open copper topside vs bottom. \$\endgroup\$ – Sunnyskyguy EE75 Jun 10 '17 at 1:47
  • \$\begingroup\$ I Think that's why i found it to be too much cheaper than other 3 A ICs Produced by other manufacturers. Thanks for your great support provided, I really appreciate your effort and I intend to further communicate with you :) Your positive reputation is also appreciated. ;) \$\endgroup\$ – Ahmed M.Zahran Jun 10 '17 at 2:03

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.