Reading the datesheet of the LF00 voltage regulator series leads to a basic question to myself.

The basic facts at the beginning of the sheet telling me, there is a logic with an


inside the component. I was very interested in this thermal limit and tried to find out the concrete behaviour of this function. Unfortunately I found only a two relevant information of some temperature behaviour in the whole datasheet.

  • Operating Junction Temperature Range -40 to 125 °C
  • enter image description here

None of them helping me to understand the thermal limit of the voltage regulator. I also expected a strong surge of the voltage dropout at higher temperatures - but figure 2 is telling a different story.

So how is this thermal limit defined and how is the expected behaviour? Am I allowed to use the component without any heatsink and without need to worry about overheating (If a accept a shutdown at overheating of the regulator)?


but figure 2 is telling a different story.

That is because figure 2 does not relate to the current limiting nor does it relate to the temperature shutdown.

This figure just shows the minimum dropout voltage you can expect at 500 mA over temperature.

Example 1:

At 80 degrees this dropout voltage is 0.5 V at 500 mA. So if the regulated output voltage of the regulator is 3.0 V then the input voltage needs to be at least 3.5 V when loaded with 500 mA and at 80 C.

Lower temperature will be OK, lower current will be OK as well (the minimum dropout voltage decreases with lower current).

Example 2:

If the regulated output voltage of the regulator is 3.0 V and the minimum input voltage to the regulator is 0.45 V then we can load this LDO with up to 500 mA if we keep it at a maximum temperature of 40 C. Less current will be OK, lower temperature will be OK as well.

Am I allowed to use the component without any heatsink and without need to worry about overheating

In principle yes but this is not a proper way of designing-in an LDO. What you need to do is determine the maximum power the LDO will drop. The maximum power is the largest current the LDO will need to supply times the voltage drop across the LDO.

Page 2 of the datasheet lists the thermal resistances for each package. Without a heatsink you need to take the junction-to-ambient value. For the TO220 case it is 50 C/W meaning that if the regulator dissipates 1 W it will get 50 degrees hotter than its surroundings. If the regulator would therefore become 80 C (the surrounding air is 30 C) that would be OK-ish.

If the regulator will heat up above 80 C I recommend a heatsink.

If you dissipate too much power without a heatsink the output voltage will drop and/or the regulator will shut down. In most designs this is unacceptable so why would your design be OK with that ?

Also if the regulator operates at a too high temperature for a long time you limit its lifetime and it might break sooner than expected.

In general it is not a good idea to not use a heatsink when one is needed. You will need to do the calculation. With a TO 220 case I'd only not use a heatsink when dissipation stays below 0.5 W or thereabout.

Even for 1 W you really should consider a small heatsink. If the product is in a metal case you might be able to use that case as a heatsink.

The thermal and current limiting function work as such:

The current limiting will lower the output voltage when the maximum current is reached. This is like a short-circuit protection.

The thermal limiting only comes into play when the chip gets too hot. Often if kicks in at around 125 degrees Celcius (that's very hot). It also lowers the output voltage which also lower the current and this will lower the dissipation of the regulator making it cool down.

I do not think this regulator shuts off completely when overheating, it is not that clear from the datasheet. What I describe above is typical regulator behavior, the regulator tries to deliver some power without getting hotter than it should. At the cost of a lower output voltage of course.

  • \$\begingroup\$ Great answer! Exactly I was looking for, many thanks! \$\endgroup\$ – Fruchtzwerg Mar 3 '17 at 9:12
  • \$\begingroup\$ My LF33CV has an input voltage of 12V and 75mA. This leads to Ploss = 8.7V * 0.075A = 0.625W. With 50C/W this leads to a higher temperature of 31.25K. With 40°C environmnent 71.25°C should be ok without using a heatsink. Would be great if you can verify my calculation @FakeMoustache. \$\endgroup\$ – Fruchtzwerg Mar 7 '17 at 9:17
  • \$\begingroup\$ Yes, your calculation is correct. With a dissipation of only 0.625 W it is indeed possible to skip the heatsink. \$\endgroup\$ – Bimpelrekkie Mar 7 '17 at 10:40

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