# Watt-squared-seconds: unit of safe operating area?

In the datasheet of Linear's LTC4364, page 15, the safe operating area (SOA) of a MOSFET is described as:

For short duration transients of less than 100ms, MOSFET survival is increasingly a matter of SOA, an intrinsic property of the MOSFET. SOA quantifies the time required at any given condition of VDS and ID to raise the junction temperature of the MOSFET to its rated maximum. MOSFET SOA is expressed in units of watt-squared-seconds (P2t), which is an integral of P(t)2dt over the duration of the transient. This figure is essentially constant for intervals of less than 100ms for any given device type, and rises to infinity under DC operating conditions. Destruction mechanisms other than bulk die temperature distort the lines of an accurately drawn SOA graph so that P2t is not the same for all combinations of ID and VDS. In particular P2t tends to degrade as VDS approaches the maximum rating, rendering some devices useless for absorbing energy above a certain voltage.

I've never encountered W^2s (watt-squared-seconds) before, and no datasheet that I've seen expresses SOA in W^2s -- rather in graphs of I_D vs. V_DS. Does anyone have any insight into how it's calculated based on datasheet graphs? Or how it's derived?

• I'd contact linear, I'll bet they mean watt seconds. Nov 3, 2016 at 20:59
• @laptop2d No. I just read the datasheet carefully. There is no possible way they are confused and made a typo in the document. Simple dimensional analysis of their later equations make this quite clear, too.
– jonk
Nov 3, 2016 at 22:03

When the pulse length is sufficiently short, the die temperature change is approximately constant for $P^2t$. So if you know the initial die temperature and the maximum die temperature you can estimate the maximum pulse the die can handle. See the below curve.

If you calculate $P^2t$ for the three points shown, it's around $300W^2 s$ for all three.

This will be derated to 0 for elevated die temperatures. See below from the LTC4233 datasheet:

You can see the cause of this effect in the transient thermal response curve of MOSFETs:

• Thanks. I knew it would be something like this, but hadn't been exposed to these particular units yet. It makes a lot of sense now that I see this nicely laid out.
– jonk
Nov 3, 2016 at 22:21
• @jonk I'd be happier if there was a bit more physics in this answer. Not all MOSFETs have quite as nice a response, as well. Nov 3, 2016 at 23:43
• I found a few white papers that I'll read later (I may also go back to basic physics and see if I can work it out myself.) These are: onsemi.com/pub_link/Collateral/AND8221-D.PDF and onsemi.com/pub_link/Collateral/AND8214-D.PDF and onsemi.com/pub_link/Collateral/AND8220-D.PDF
– jonk
Nov 4, 2016 at 0:40
• If the given formula held as pulse lengths approach zero, that would imply that the amount of temperature rise for a given amount of energy would be inversely proportional to the pulse length (cutting the pulse length by half would double the power, and thus quadruple the $P^2$ factor, which would be only partially offset by cutting the $t$ term in half). I could see that there would be a range of pulse lengths where heat rise would become more localized as pulse lengths get shorter, but the amount of thermal mass over which heat is distributed should approach a non-zero asymptotic limit. Nov 21, 2016 at 20:44