On Mon, 22 Sep 2008 07:45:19 -0700 (PDT), poster
put finger to keyboard and composed:

On Sep 21, 2:21 am, Franc Zabkar wrote:
On Fri, 19 Sep 2008 20:05:27 -0700 (PDT), poster
put finger to keyboard and composed:

Microwaveburned the fast fuse and 'rectifier' 2x062 on this schematic
http://geocities.com/anglomont/mw.jpg
Replaced it with hvr 6x2 diode however once again it and fuse burned
though "high voltage rectifier" diode hvr1x stayed ok
(hvr 6x2 short, hvr 1x conducting 0.25 ma when tested with 4.5v)
Magnetron ok, so what could be the reason for this-perhaps capacitor
or some contact with oven interior along points A or B?
thanks in advance

I can't find anything in theMicrowaveRepair FAQ that discusses the
high voltage protector diode:

http://www.repairfaq.org/sam/micfaq.htm

I did find a datasheet for your HVR6x2 diode, though:http://www.dccomponents.com/products...igh_Voltage/HV...

This device consists of two back-to-back HV diodes, one drawn larger
than the other, the smaller one with a 1500V reverse breakdown
voltage, the larger with 6000V. AFAICS, if your HV capacitor were to
go open, then the smaller diode could see a reverse voltage close to
the peak secondary voltage of the power transformer (~2000V). This
would exceed its rating and presumably it would fail.

So IMO the capacitor is one possible suspect, although I would think
that anything which allowed a good capacitor to charge in the reverse
direction may also cause the same failure. Therefore it may be wise to
replace the HV diode as well, just in case it is leaky. Also check
that the magnetron's filament is not shorted to the case.

- Franc Zabkar

Thanks for the answer however
you are assuming the anode of the bigger diode is connected to the
transformer ...

No, I was thinking that the smaller diode connects to the transformer.
That's how it's drawn in the schematic diagram.

If the dual diode were reversed, then the smaller diode would always
see the voltage of a fully charged capacitor, which would presumably
exceed its breakdown voltage.

... but it was the other way around because this was the
original part
http://www.geocities.com/anglomont/hvr-d.jpg
On the other hand could it be that it would take 3seconds to charge
the capacitor to 1.5kv in reverse direction assuming 0.5ma leakage
current, however fuse blows up immediately after the oven is turned on
Checked filament -ok, and from schematic it seems that if grill-
heater contacts were problematic that would blow the main and not the
fast fuse

If the capacitor is open, then, during the negative half cycle of the
mains, you effectively have two reverse biased diodes (HVR and small).

10kV 6kV 1.5kV
Gnd o---||---|--||--||----o transformer secondary
HVR | big small
|
magnetron
|
Gnd

How this reverse voltage is shared between the two diodes would
probably depend on the reverse leakage characteristics of each. I'm
assuming that no reverse current flows through the magnetron (its
filament is cold). If current does flow via the magnetron, then the
small diode would see the full reverse voltage. Perhaps the diode
survives until just after the filament begins to warm up?

- Franc Zabkar
--
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