|
Hi Alan,
Of course when the reed switch closes, it must not (never, ever)
directly switch the drain current !
(To do that, one would need an "opener" which are quite exotic
devices for reed switches, with built-in permanent magnets).
What I would use the reed switch for is to 'remove the driving
signal', or kill the FET's drain voltage.
To break the drain current (if the FET had died), one would need
another -separate- MOSFET of course, which would then operate like
an electronic fuse. Use a flipflop with reset-input to enable the
supply voltage after the protection circuit has tripped. Or add
some more elaborate control circuitry which doesn't cut off the
drain supply voltage completely; instead reduce it to a level at
which the FET withstands even the worst output mismatch. This
would allow tuning (at low power), and then, after the SWR is ok,
reset the flipflop to switch back to "normal" (high) power.
All the best,
Wolf .
08.04.2013 23:01, schrieb Alan Melia:
Hi Wolf I cant remember the
ratings now but breaking 18A+ might be a bit cruel :-)) It
would probably need more than a tap with a pencil to unstick
it !
Alan
G3NYK
----- Original Message -----
Sent: Monday, April 08,
2013 9:18 PM
Subject: Re: LF: Re: Hall
Effect for Over-current shutdown?
.. or wind a few turns of fat
copper wire around a 'reed relay' contact tube. Virtually no
voltage loss then. I used this for a power supply, but it
would work for the DC current feed inside a PA as well. IIRC,
the product of number of turns * Amperes was about 50 for the
contact to close.
73,
Wolf .
Am 08.04.2013 21:39, schrieb Alan Melia:
Ah yes that is Class D and
can go that way if mis-matched......that is a lot of power
in a sensing resistor! I guess a Hall effect device or
even a temp sensor would do that job. I think in over
current condition a lot of power is dissipated in the
devices. High side sensors should be available at 30v The
sheets I have seen, suggest they require some assistance
above about 40v.
Alan
----- Original Message -----
Sent: Monday, April 08,
2013 8:07 PM
Subject: Re: LF: Re:
Hall Effect for Over-current shutdown?
Hi Alan,
I've modified the G0MRF amp with the M0BMU output
configuration, I'm using a 30V supply.
The G0MRF uses a current sense resistor to shut down
the amp quickly in case too much current is being draw.
I've modified the output transformer turns ratio to
produce more power at lower supply voltages (I'm using
500V, 55a FETs).
However with the new higher currents and with a new
appropriate value sense resistor its developing a LOT of
heat (12 W or so) and needs to be heat-sinked.
I'm thinking Hall-effect might be more efficient and
produce less heat - I'm pretty sure that others have
used Hall effect devices to protect MOSFET amplifier and
I'm looking for circuit tips.
73 & Tnx!
Warren
|
|
