safe operating area

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Ah yes, the forum of disinformation strikes again!

You must always derate! Always! Determine the worst-case steady-state operating temperature of the device. Suppose for this example it is 60C. Now determine the worst-case current and assume the highest possible rail voltage. Use the derating factor, and find if this current and voltage are within the DC (or 100ms if DC not specified) SOA of the device. If it is, you will have a very reliable design. If it's outside the SOA, you will have a very unreliable device.

Simple, really.
 
Re: Alternative approach....

mikeks said:
If a transistor rated at 200W, for Tj=150 deg. C, and Tc=25deg. C, is only allowed to dissipate Pd=40W max.

What will Tj=? for this 40W, assuming Tc=70 deg. C?


Of course that depends on the heat sink. For the IRFP240, junction-ambient is 40C/watt, so without a heatsink at 40 watts we'll have a junction trying to go to 1600C and magic smoke everywhere. with the above mentioned infinite heat sink that keeps the case at 25, you won't hit maximum junction temperature until dissipating 150 watts.

However, in a real world application, with finite heat sinks and insulators with thermal resistance you must derate to keep the junction temperature within its limit.

If you have a heat sink that allows the case temperature to rise 100C above 25C ambient IR says derate the IRF240 to 30 watts. (1.2 watts per degree above 25 and 150 watts max) What is the temperature of the junction with 30 watts and a 125C case? Right at 150C, its maximum junction temperature. Mikeks, you can call it derating or using enough heat sink, it really doesn't matter, you must keep the junction from melting.

Use enough heat sink to keep the junction temperature well within its rated limit and you don't have to worry about the derating factor.
 
jwb said:
Ah yes, the forum of disinformation strikes again!

You must always derate! Always! Determine the worst-case steady-state operating temperature of the device. Suppose for this example it is 60C. Now determine the worst-case current and assume the highest possible rail voltage. Use the derating factor, and find if this current and voltage are within the DC (or 100ms if DC not specified) SOA of the device. If it is, you will have a very reliable design. If it's outside the SOA, you will have a very unreliable device.

Simple, really.

Hi,

This is rapidly turning out to be one of those great audio red herrings.....It's all very elementary arithmetic really.....:
 
Re: Re: Alternative approach....

BobEllis said:



Of course that depends on the heat sink. For the IRFP240, junction-ambient is 40C/watt, so without a heatsink at 40 watts we'll have a junction trying to go to 1600C and magic smoke everywhere. with the above mentioned infinite heat sink that keeps the case at 25, you won't hit maximum junction temperature until dissipating 150 watts.

However, in a real world application, with finite heat sinks and insulators with thermal resistance you must derate to keep the junction temperature within its limit.

If you have a heat sink that allows the case temperature to rise 100C above 25C ambient IR says derate the IRF240 to 30 watts. (1.2 watts per degree above 25 and 150 watts max) What is the temperature of the junction with 30 watts and a 125C case? Right at 150C, its maximum junction temperature. Mikeks, you can call it derating or using enough heat sink, it really doesn't matter, you must keep the junction from melting.

Use enough heat sink to keep the junction temperature well within its rated limit and you don't have to worry about the derating factor.

Hi

i think you missed something...Note:

mikeks said:
If a transistor rated at 200W, for Tj=150 deg. C, and Tc=25deg. C, is only allowed to dissipate Pd=40W max.

What will Tj=? for this 40W, assuming Tc=70 deg. C?

From this info. it is straightfoward to establish R(j-c), without invoking the IRFP240:

R(j-c)=(150deg. C-25 deg. C)/200W=0.6deg. C/W

Additionally, there is no need to discuss heatsinking in the context of this example, as this is in fact implicitly taken as given in Tc=70 deg. C.....
 
Hi,
Mikeks you almost answered your qestion from post 22 but stopped short.
Using your figures Tc=70c Pd=200w Tjmax=150, then Rjc=.625Cdeg giving temp rise at junction =25Cdeg for 40w dissipation and resulting in a Tj of 95degC.
This is well below Tjmax and allows for a considerable peak dissipation in excess of 40w and still maintain a junction temp below 150degC and even below 125degC.
However our disagreement is about derating as used by the manufacturer for their devices.
Again using the same data, the derating factor at Tc=70degC is 0.64. This tells the user that the device must not continuously exceed 200w*0.64=128w and the same derating factor must also be applied to all short term ratings extracted from the SOA curves. This come about because the manufacturers have standardised their published data on Tc=25degC and given the users a method to arrive at practical operational conditions.
I repeat again, I believe that the derating must be applied based on the case temperature and a design using this method will meet the manufacturers and users expectations for "normal" reliability.
To exceed the manufacturers rating will seriously reduce reliability but by simply restricting Tj to a lower value than Tjmax the user can considerably extend the life of the devices.
Finally, I only claim to be an amateur/novice/willing learner and if someone can show by reasoned argument that I am wrong I will gladly reverse my stance.
regards Andrew T.
 
Re: Re: Re: Alternative approach....

BobEllis said:

Use enough heat sink to keep the junction temperature well within its rated limit and you don't have to worry about the derating factor.

mikeks said:


Where I come from 25 deg C is an average day, designing for commercial use a hot day might be 37 deg C, thus the device must be derated even with an infinite passive heat sink. In the process of doing this: "Use enough heat sink to keep the junction temperature well within its rated limit" you will have derated the device.

The fact is that if we want to use a device rated for 200 W with passive cooling, in consumer applications where ambient will often be well over 25 deg C, even over 37 deg C, it will never be able to dissipate its rated power, 200 W. With a reasonably sized passive heat sink, on a hot day, dissipating anywhere near it's max (derated) power it will soon be at Tc = 60 - 70 deg C, in typical linear situations.

Most power devices at ambient on a hot day will be capable of 10% less than their rated power even assuming an infinite passive heat sink. In typical worst case music applications will be roughly, ballpark, 30-40% below rated power. Even more derating is necessary if the amp will be subjected to continuous burn in, or stressed applications. Thermal shutoff is always a good idea.

The situation is even worse when devices are used at high voltage due to secondary breakdown, the MJL4281As mentioned earlier are rated 230 W, but will only do 1 A at 100V or 100W, far below the spec. At half their rated voltage they're at less than half their rated power. Remember this is before derating for Tc > 25 deg C, to allow for a reasonable passive heat sink, and higher ambient temps.

Secondary breakdown is the interesting aspect of SOA, has there been a discussion of series versus parallel connection of output devices?

Power transistor specs are highly optimistic and must be derated even in consumer applications. This is why we see "safety factors" of 4, often much more, in the dissipation capability of output stages.

Then there is the thermal cycling consideration.

This is interesting, the author notes that efficiency is lower below max power but seem to miss the fact that max output stage dissipation is not at full power:
http://www.signaltransfer.freeuk.com/thermeqn.htm
 
AndrewT said:
...........This tells the user that the device must not continuously exceed 200w*0.64=128w ...............

Your definition of derating factor seems to differ from the conventional one...i.e DF=-1/R(j-C)

How did you arrive at 0.64?

Cheers.


AndrewT said:
Hi mikeks,
Can you give us a circuit to achieve "soa protection"?
regards Andrew T.


I thought i sent article out to you..? Check your mail....
 
Hi,
re post 33
Yes, Pd formula predicts Pmax at actual Tc as recommended by manufaturer.
But what do you do with the SOA curve?
You cannot apply the new Pd in any usefull way.
Work out the derating and then you can apply it to any values extracted from the SOA curve.
This applies particularly to voltages above the second knee in the curve where second breakdown becomes an issue.
eg. our hypothetical device has a Pdmax =200w and a derating =0.64. Suppose the second knee ocurrs at 50v and permissible power falls to 20w at 150v
Using the derating (0.64) times the power from the SOA will give the limiting power and current at any selected device voltage. At 150v the device can dissipate a continuous power of 12.8watts (85.3mA). This is way below 128w and significantly below 40w.
I believe this is the method the manufacturers intend.
BUT I have not yet seen an instruction from any of the data sheets suggesting that the derating and the SOA reduction must be applied together when voltage exceeds the second knee. That instruction is quite explicit at all powers and currents for voltages below the second knee. I have interpreted their recommended method as implying that the user should use the Tc derating for all regions of the SOA otherwise you can be in a situation just above the second knee where using SOA alone will give a higher power than just below the second knee with derating applied. This is contrary to the SOA purpose.
Finally, I asked a while back "are the manufacturers data sheets trying to achieve Tc control or Tj control or both". Since no one has answered I will propose that the manufaturers are giving the user a method that ensures both Tc and Tj are controlled in a manner for which their devices are designed.
regards Andrew T.
 
AndrewT said:
........But what do you do with the SOA curve?
You cannot apply the new Pd in any usefull way.


Work out the derating and then you can apply it to any values extracted from the SOA curve...........

But Pd(new max.)=128W IS the 'new' derated SOA characteristic at Tc=70deg. C...

In otherwords, 'derating' is implicit in post 33:


mikeks said:
post 33

However, I reckon this is more intuitive:

Pd(new max.)=(150deg. C-70deg. C)/R(j-c)=128W

Where, (as previously established), R(j-c)=0.625 deg. C/W

Moreover, note that in this hypothetical example i stated that Pd was not to exceed 40W.

Thus, in this context, Pd(new max.) is rather academic...
 
mikeks, I am glad to be able to point you in a good direction. That guy sure does manage to make thermal engineering interesting. Lots of rainy day reading there. It seems that lately I have been finding websites other people find useful... thank God for that!

When in doubt, be conservative eh!
 
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