Voltage regulator power dissipation

I have a 5V fixed output voltage regulator (ZLDO1117G50TA, SOT-223 package) that will supply 0.8A to a circuit. The input voltage to the regulator is 13V which means 6.4W will be dissipated in the regulator. A chart in the datasheet claims that 0.8A with a 8V Vout-Vin is within its safe operating area. However that seems like a lot of heat that will be produced by this small SMD regulator. Am I worrying for nothing? I don't have a lot of space on the board but I had an idea of putting some SMD diodes in series to drop the 13V input voltage down a few volts. Would that be considered a sensible way to remove some of the power dissipation away from the regulator?
(BTW, I can not use a DCDC Converter).

You need additional heat dissipation, and lots of it, if you're pulling 800mA continuously! May still be really hard! See "Package Thermal Data" on page 3 of the datasheet.

If it's 800mA continuous, that really, really cries out for a DC-DC converter.

I used LM7809 to stepdown 12v to 9v but the regulator was producing a lot of heat and I feared it will melt my board. Later I used a switching regulator.

An SOT-223 can safely dissipate ~1watt on an average/small circuit board, but expect it to shutdown at ~1.5watt, depending on room temp and airflow. So Vin - Vout can only be ~1.5volt at 0.8A.

What are you trying to do with that 800mA. Why can't you use a DC/DC converter. RF interference?

Modern micro-power converters work on a very high frequency (2Mhz) that can be easilly filtered/shielded.

Pre-regulation with an LM317 or 7806 might be an option.

Thank you for the replies. I didn't know anything about package thermal data or Junction-to-Ambient or Junction-to-Case resistance so I spent about an hour googling and trying to research what these things mean, but I am still confused.

From the ZLDO1117 datasheet (SOT223 package):
Junction max temperature = 150C, recommended operating temperature is -40 to 125C
Junction-to-Ambient thermal resistance = 107 C/W

So does that mean if I'm dissipating 1W and ambient temperature is 25C then my Junction temperature will be 107+25 = 132C? Which is above the recommended junction temperature... so i should not even dissipate 1W at 25C??

And the datasheet also says Junction-to-Case thermal resistance is 16 C/W. I don't know what to do with that spec. The datasheet doesn't say anything about recommended case temperature, and how would I even know the case temperature?

And finally, on page 2 of the datasheet there is a chart showing a Safe Operation Area Curve which appears to show that i can operate the regulator at up to 1A at up to 10V Vin-Vout which would be 10Watts!! How can that be possible, 10W would result in a Junction temperature of 1070C??

Link to datasheet:

But getting back to my original problem, I think I will just add a big power resistor at the input of the voltage regulator and burn up heat in the resistor. I am making a board for a customer with sensitive analog electronics on it and they requested no DCDC Converters...

Thank you for any help & advice!

junction-to-Ambient thermal resistance = 107 C/W.
This means the package only. no heatsink.

Junction-to-Case thermal resistance is 16 C/W.
Mounted on an infinite heatsink.

There are lookup tables that tell you what e.g. one square inch of copper/circuit board is.
You will have to try/measure/estimate.

Have you read this one

And this


I am making a board for a customer with sensitive analog electronics on it and they requested no DCDC Converters...

Maybe but do they actually know it will be bad? I had engineeres with this attitude until I got them to try it and they found no interference at all on a sensitive system.

Just use enough filtering. And if it's really sensetive you could use a hybrid design. Use a DCDC to step down 13V to lets say 7V and use a linear regulator after that. But with 800mA you still burn 1,6W... And indeed, with a Junction-to-Ambient of 107 degree and 25 C ambient (which is low, especially in a case where 1,6W is burned...) the junction will be at 25C + 1,6W * 107C/W = 196C... So a propper heatsink is needed.

So, lets say we want a max junction temperature of 125C when ambient is 40C (because in a housing it will be hotter then outside...). That leaves 125C - 40C = 85C. 85C / 1,6W = 53C/W for cooling that we need. Junction-to-Case already takes 16C/W leaving us with 53C/W - 16C/W = 37C/W for the heatsink INCLUDING case-to-heatsink!

Do the sensors consume the 800mA? Or is it possible to split the power up into low current linear (or hybrid) for the sensors and DCDC for the power hungry part?