If I understand the oscillator spec correctly, it will operate down to 2.5V, so you could put a resistor in series that would drop about a half volt or so (say 75Ω) at the typical operating current, and the circuit should still operate normally.
The short circuit power dissipated in the resistor would then be about 150mW.

Even a smaller 50Ω resistor, dropping about 330mW for normal operation, would only dissipate about 218mV for a short circuit.

Would either of those meet your requirements?

If not, you may have to go with a more elaborate current-limit circuit, which is likely not simple for such low voltage operation.
The simplest I know uses a matched, dual PNP transistor, plus a logic-level MOSFET.

You are correct, but there is the ATEX limitation of the power disspation from Uo=8V and Io=0.1A , we get about 1.36W ( atex Pd=Uo*Io*1.7. and so the resistor needs to be about 6.12W, since my resistor needs to qualify its rating at 2/3 its power and again 1/3 its power since it degrades at 120C.

 

I know what you mean but let me explain. its to do with the ATEX Qualification from standard 60079-11. The Uo=8V, Io=3.11A. But i have a fuse of about 0.1A. And then the qualification throws in all kinds of safety rules if the PWA is placed in zone 0. So even if the component is fed 3.3V. The atex qualifiers would take 8V an 0.1A for the power measurement. And since the RTHja of the component is too high at about 421C/W or so. te only way is to have a sufficient enough resistor to with high enough power working at 2/3 its rating and again by 1/3 since its be qualified at 120C. that would mean i would need an unusually large resistor capable of very high power dissipation. either ill need to skip this part and use the internal oscillator of my mcu instead.

If they are not paying attention to the actual voltage and current why would the actual RTHja make the slightest difference? Something about your interpretation of the qualification methodology makes no sense. Rather than rely on your intuition or our limited knowledge I'd suggest you consult an authoritative source before we waste more time helping you solve a problem that cannot be solved in any meaningful way by the methods you propose. Your proposed solution practically guarantees that the part won't work, so then where are you?

 

If they are not paying attention to the actual voltage and current why would the actual RTHja make the slightest difference? Something about your interpretation of the qualification methodology makes no sense. Rather than rely on your intuition or our limited knowledge I'd suggest you consult an authoritative source before we waste more time helping you solve a problem that cannot be solved in any meaningful way by the methods you propose. Your proposed solution practically guarantees that the part won't work, so then where are you?

Yes no worries. i understand. i am closing this case. maybe my interpretation is wrong. but from what i read on the qualification standards thats what it seems to indicate. Sorry to bother you, on this, its not really an engineering solution but more of a qualification related issue.

 

Yes no worries. i understand. i am closing this case. maybe my interpretation is wrong. but from what i read on the qualification standards thats what it seems to indicate. Sorry to bother you, on this, its not really an engineering solution but more of a qualification related issue.

No worries. We struggled with UL certification and CE mark when we tried the first time. Professional advice that you pay for (willingly) is the ONLY way to go.

 

If you really need a resistor to limit the fault current and still allow the oscillator module to function, that resistance must be low enough to not drop the Vdd below what is required. Long ago I did need to put fault current limiting resistors in tube circuits, and the solution was to have a very low power dissipation resistor with a low resistance value.
So for this instance, where the voltage is 3.3 volts and the normal current is only 0.0066 amps, you can get away with a resistor of perhaps 2 ohms and a power rating of 3.3 volts x 0.020 amps. That equals 6.6 x 10*-3 watts = 6.6 milliwatts.
In reality you need to know what the acceptable failed device current draw for that source connection would be OR you can install a fuse with a fast rating of 25 milliamps. I did need to do that for a vehicle crash data recorder system more recently. The fuse does work.
BUT since a failure of the clock oscillator will disable that portion of the system, a watch-dog time-out system to switch off the power if the clock signal stops is another option.