View Single Post
  #23   Report Post  
Posted to sci.electronics.repair,sci.electronics.design
Ron(UK) Ron(UK) is offline
external usenet poster
 
Posts: 501
Default Opinions on adding fuses to power amp

N Cook wrote:

I'm just trying to engineer a more respectable equivalent of fuses for trace
rupture.
Eventually one DC rail ruptured as a fuse. I've now decided to do some
calculation to determine what the rupture current for that trace was.
Then as the common return almost fused and is the same dimension but in the
even worse case then carries twice the current , halving this trace rupture
current to use as the fuse rupture current and reduce that for current
carying capacity , by half again ?
If all output trannies fused short circuit all round so loading both DC
rails the combined common return would have ruptured at whatever that
rupture current is for the trace that did burn up.
Anyone know the term for laying double traces of solder along copper traces
to increase current carrying , so I can research it ?


Beefing up!

I'm intrigued how you lay quite accurate molten solder runs and parallel
over and with existing traces.


When we used to build classic 'Hitachi' mosfet amps, we would lay a
thickness of solder along the DC rails to the outputs transistors and to
the output terminal. It`s a bit like welding, once you get the hang of
it, you can lay down a respectable run of solder that looks quite
professional. I spose you could beef up with tinned copper wire, but it
wouldn`t look so neat. Some amps have hard wired connections directly to
the output trannies, or, the DC feed from the PSU go directly to the pcb
lands upon which the outputs are soldered/bolted.


Anyway measurements:
Copper trace 3.5 x .02mm and as 2 half approximated elipses of solder then
area of 1 elipse of tin+lead which is Pi x a x b , a and b minor and major
axes of .15mm and .8mm.
From tables of fusing currents for copper, also tin and lead (NB in the form
of circular wires ) for different diameters.

Copper fusing current of the trace = 12 amps
Lead+Tin elipse then 6 amps (not as much as I would have intuitively
thought)
Total 18 amps so 18/4 = 4.5 amps conventional fuse rating.
Presumably the circular to sheet allowance would up this 4.5 amp figure ,
but by how much ?
What sort of correction factor for thin sheet/non-circular heating then
rupture allowance?


Did you ever tell us what make and model of amplifier this is?

Is it really worth all this effort to redesign? You're never going to be
able to protect against stray metal object clattering around inside, and
I`m sure the original designers never gave it a thought.


Ron(UK)