Mjl3281 Soar - diyAudio
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Old 2nd March 2007, 03:00 PM   #1
Pingrs is offline Pingrs  United Kingdom
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Default Mjl3281 Soar

I needed to plot the SOA curve for the MJL3281 to position a dual-slope protection locus prior to the resistance values being determined.
To do this requires converting the two SOA log-log graph slopes given in the manufacturerís data into a linear form.
This is easy for the first slope (power limit) because it normally follows the law Ic=P/Vce.
The secondary breakdown values are then tabulated from the law Ic=k(Vce)^n, with k and n determined from the SOA secondary breakdown slope.
However, I noticed that for the MJL3281, the first power slope does not conform to Ic=P/Vce, but needed to be calculated using Ic=k(Vce)^n.
Since I have seen articles written by respected authors which show this device as having a normal power curve, and also a differing break point to that given by the graph, Iím somewhat confused.

As an aside, if I perform the same analysis on the MJL 4281 it shows excellent correlation (as a check, P/Vce is equivalent to kVce^n, when, of course, k=P and n=-1).

Iíve attached the relevant sheet that shows my respective curves for the MJL3281.

Any comments, anyone??

Brian.
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File Type: pdf soar_mjl3281_3.pdf (36.0 KB, 98 views)
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Old 2nd March 2007, 03:10 PM   #2
AndrewT is offline AndrewT  Scotland
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Hi,
Email me if you would like to see Bensen's spreadsheet with protection locus added over the SOAR.
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regards Andrew T.
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Old 2nd March 2007, 07:49 PM   #3
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Hi Pingrs

I think you need to be careful about the latest SOA's from ON Semi. The curves are published for t=1s, not DC. There might be an error, but any time shorter than DC might also have a higher rating than the DC values.

I use the ratings from the data sheet to construct the DC curve. The two numbers are 4A at 50V; below which I assume that the slope is -1; and 1A @ 100V at which the slope I take to be -2.

You can construct the two lines from these assumptions to generate the DC SOA.

As usual, a caveat in that this does not guarantee that ON Semi will actually support such a curve!

cheers
John
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Old 8th March 2007, 02:14 PM   #4
Pingrs is offline Pingrs  United Kingdom
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John,

Thank you.

Sorry for the delay in my appreciation of your reply. I've been trying to get OnSemi to come up with something, but they say their data for this > 10yr old device is no longer available.

"1A @ 100V at which the slope I take to be -2."

Where does this slope come from? A kind of average of other devices, or does it have a natural physical relationship somehow?

The MJL4281, for example, is about -2.1, as far as my maths tells me.

Regards,

Brian.
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Old 8th March 2007, 02:17 PM   #5
Pingrs is offline Pingrs  United Kingdom
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Andrew T,

Not forgetting my thanks to you, Andrew.

Brian.
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Old 8th March 2007, 09:48 PM   #6
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Hi Brian

I only have this figure from past device characteristics. It is a guide not a calculation.

cheers
John
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Old 8th March 2007, 10:08 PM   #7
mikeks is offline mikeks  United Kingdom
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Brian

All you need in fact is to establish the nominal power rating of the device at 25 deg. cent. and then plot the secondary protection locus at the same temp. from the data sheet.

No need to indulge in the self-flagellation of exponentiated terms: the info. is in plain view on the datasheet.

You may then "derate" (adjust) the resulting linear-linear characteristic(s) according to the needs imposed by your thermal management (heatsinking).
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Old 9th March 2007, 12:47 PM   #8
Pingrs is offline Pingrs  United Kingdom
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Hi Mike,

Thank you.

If I understand you correctly, I find deducing plot points from a log-log graph difficult to judge, I've tried. I decided a bit of self flagellation is good for the soul, so sought the maths to convert just two relatively easy ones.

Your consideration for those non-English speakers on this forum whose vocabulary may not encompass "derate" and "thermal management" is admirable.

Regards,

Brian.
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