LM1084 5A/5V heating

Yes, the part will get very hot. No, this will not work for very long.

I don't recognize this heatsink, but I would guess that its thermal resistance might be in the area of  10°C/W to 15°C/W. Maybe.

The basic package power dissipation formula for a linear regulator is:

Pdiss = ((Vin - Vout) x Iout) = ((16V- 5V) x 4A) = 44W

Total thermal resistance needed, junction to ambient, to keep the internal junction temperature <= 125°C, while the part is dissipating 44W with an ambient temperature of 25°C:

   θja(max) = ((Tj(max) - Ta) / Pdiss) = ((125°C- 25°C) / 44W) = 2.3°C/W

The required heatsink-to-ambient (θha(max)) thermal resistance is determined by subtracting the known thermal resistances of the TO-220 junction-to-case and the case-to-heatsink interface (i.e. thermal grease):

    θha(max) = θja(max) − (θjc + θch)

Since the junction-to-case (θjc) of the TO-220 package is at 2.7°C/W, you will need either a heatsink with a negative thermal resistance, or a dramatic reduction in the ambient temperature, to make this work.

The excessive package dissipation of this application (16V in, 5V out, and 4A out) calls for SMPS solution, perhaps something in the Simple Switcher family.

The top recomendation from Webench Designer is LMZ12008, followed closely by a number of other equally suitable solutions.

http://www.ti.com/product/lmz12008

At the lower right of the LMZ12008 product folder page is the Webench tool to help optimize your design.

Alternately you can access Webench, and see the full list of possible solutions, from:

http://www.ti.com/lsds/ti/analog/webench/overview.page

Hey Donald Jones ,

Thanks a lot for your explanation about the LM1084's heating issue. As you recommended, I looked into LMZ12008.

It looks like a far better solution for my scenario. I have few concerns about the LMZ12008.  (Vin - 16V, Vout - 5V, Current - 5A)

• Is it necessary to use a fan and a heat sink?  
• is it necessary to use any BASIC type protection circuitry ?
• In the schematic, why does it say "qty=2" in Cin and Cout ??

Circuit attached.

" .... Is it necessary to use a fan and a heat sink ..."

At 5V and 5A out, the LMZ12008 efficiency would be near 90%. For 25W at the load, that would be a bit less than 3W dissipated in the LMZ12008. The expectation is that the PCB would be the only heatsink required, and no fan would be needed. The pcb should be multi-layer with several thermal vias under the package to the internal ground copper plane, and 2oz copper on top and bottom sides. The LMZ12008 datasheet has a section for this, see : "POWER DISSIPATION AND BOARD THERMAL REQUIREMENTS"..

" ... is it necessary to use any BASIC type protection circuitry ?"

Not sure what qualifies for BASIC protection circutry in your application. But I would say 'probably'.

" ... In the schematic, why does it say "qty=2" in Cin and Cout ??"

The suggestion is to use two in parallel to lower the effects of ESR in the input and output capacitance. This is not mandatory, but it is certainly encouraged.

An LMZ12008 eval board is available from TI now, and Digi-Key shows 1@ in stock this morning. 

     TI part #: LMZ12008EVAL/NOPB  (approx $25USD)

     Digi-Key part #: LMZ12008EVAL/NOPB-ND

http://www.ti.com/tool/lmz12008eval

and the Users Guide for this eval board ...

http://www.ti.com/lit/ug/snva478c/snva478c.pdf

Dear Donald,

Just a question with regards to your info above, I have a similar application and I'm using 22v supply. the output needs to be 12v, and my load is also 4A. I was initially considering the LM1084, however there another regulator from T.I. the LM338.

Pdiss = ((Vin - Vout) x Iout) = ((22V- 12V) x 4A) = 40W

The data sheet of LM338 (for the NDE package) gives the  θJA = 50°C/W, and the θJC = 4°C/W, θCH is not given but it is usually small using thermal compound (for the LM1084 it is 0.2°C/W). Also figure 4 (data sheet) shows that at the 10V differential the output current limit is at 8A. 

θha(max) = θja(max) − (θjc + θch) = 50 - (4 +0.2) = 45.8°C/W, so according to that a heat sink that is spec to provide 44°C/W will suffice. Is the above correct? will the LM338 work in this case?

Thanks,

George.

Looks like you skipped a step, or two. A Theta(ja) of 45.8°C/W won't cut it.

     Tj = Ta + (Theta(ja) x Pdiss) = 25C + (40W x 45.8C/W) = 1857°C. 

The needed Theta(ja) is calculated from:

    Theta(ja) = ((Tj(max) - Ta(max)) / Pdiss)

You're dissipating 40W, the maximum junction temperature is 125°C, and the ambient temperature is what ... 25°C, 50°C, 85°C? Using 25°C here ...

    Theta(ja) = ((125°C - 25°C) / 40W) = 2.5°C/W

That's going to be difficult to achieve.