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- Posted on
May 28, 2006, 11:26 am
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So I have a power supply that has a zener diode across the output to
trigger an SCR in the event of over voltage.......
The power supply's fuse is , say 10 amp. Does the SCR's rating have to be
at least 10 amp or will a lower rating one do given that the fuse in the
event of a short to ground through the SCR will fail pretty quickly
How is a request for information trolling?
The power supply in question is a Johnson Unit that I've repaired for a
The semiconductors only had in house type numbers on them.
My p was a genuine question. What seems elementary to you is not to me.
Is that Phil Allison you are having a discussion with?
If so, you are wasting your time trying to reason with him. Just killfile
the graceless prick. I did, six months ago and have not missed him.
His technical knowledge is quite good but he turns to psychotic abuse for no
reason. Or maybe his reason is the gratification of getting a shocked
reaction. Whatever. He is a troll and is not worth your trouble.
Ifsm is defined as "non-repetitive peak forward surge current 8.3ms
If you know something about your fuse, you *can* determine whether or
not the SCR will clear the fuse without failing, clear the fuse but
short-circuit in the process, or just plain old explode.
A fuse is characterised by the amount of energy it takes to melt it; the
term used is "I-squared-t" or I^2t. If you have an I^2t figure for your
fuse, you can compare it with the SCR's I^2t rating to see which one wins.
If SCR I^2t >> fuse I^2t, the SCR will happily blow the fuse, and keep
on ticking. This is what you really want to have happen.
If SCR I^2t << fuse I^2t, the SCR will blow itself to bits (or something
else will fail, eg a pcb track fry), without blowing the fuse. This is a
If SCR I^2t = fuse I^2t (or close to it) then the SCR will clear the
fuse, but will probably fail short-circuit in the process.
OK, so how do you calculate SCR I^2t? well, there are 2 ways:
1) read it off the datasheet. For really big diodes, this is an
incredibly important number - you really dont want to fuck up fuse
selection when directly connected to a 20MVA distribution transformer.
Look at the datasheet for a semikron SKKD15, I^2t is listed underneath Ifsm.
2) calculate it from Ifsm. You know its a sine wave, of say 30 Amps peak
(I'm looking at a UF400x which specs Ifsm for 8.3ms so f = 60Hz).
Square it, giving [30*sin(2*pi*f*t)]^2 = 900*sin^2[2*pi*f*t]
integrate it over one half period, giving:
I^2t = 900A^2*1/(4*60Hz) = 3.75A^2s
the general formula we have just worked out is:
I^2t = 0.5*Tfsm*Ifsm^2
because Tfsm = 0.5*1/f = 1/2f
lets check with the semikron SKKD15 diode:
Ifsm = 320A, 10ms (f = 50Hz, they are Germans) at 25C, so we calculate
I^2t_calc = 0.5*10ms*320^2 = 512A^2s.
the datasheet specs I^2t = 510A^2s at Tj = 25C
OK, lets get all brave and try it at Tj = 125C:
Ifsm = 280A
I^2t_calc = 392A^2s, the datasheet says 390 A^2s.
Just a word of caution though:
fuse I^2t is NOT even vaguely linear. take a look at this:
on page 2. You need to calculate I^2t rating from this graph, under the
same conditions as the SCR figure. 10ms = bottom of graph, so for a 30A
ABC fuse, 10ms I^2t = 450A^2*0.01s = 2000A^2s (this graph is RMS amps
not peak amps, and 0.5*peak^2 = rms^2).
the spec'd I^2t is 1429 A^2s, but thats specd at 1000A (ac interrupt
rating), which is off the sheet. if we run the calculation backwards,
1429 = (1000^2)*t so t = 1.4ms, which looks like a pretty good curve fit.
Semiconductors are well known for being great fuse protectors !
So in your sentence above you are speculating a lower rated SCR will
not fail (or presumably encounter any harm) whilst the fuse will, by
maybe restating the question and critiquing your own logic you may
well have answered it.
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