On 12 Apr, 17:51, "
wrote:

I have to replace a KBPC602 bridge rectifier. It is rated 200VRMS at
6.0 A. It is wired into the circuit using only the + and - terminals.
The AC connections are not used. If I were to build a bridge out of
four 6.0 amp 100V diodes, would I be able to directly replace this
device? Does the circuit basically require a 12.0A 200V device?
Conversely if I were to place four 3.0 amp 200V diodes in parallel
would that arrangement work also? Thanks. Lenny Stein, Barlen
Electronics.

Using devices in series parallel is an effective way to fail-safe
things. Any one of those diodes can short or go o/c and the rectifier
still works as a rectifier.

And you say its failed before - sounds like the designer was trying to
reduce the failure rate instead of getting to the bottom of the cause.

I dont know what type of bulb it uses, or why the BR is failing, but
if its a filament bulb, arc over will draw huge currents at fail time.
Fitting an mcb in the line to the bulb may save the diode sometimes,
maybe. This must be a magnetic one, not a thermal one.

Of course you could eliminate the problem entirely by using a mains
voltage bulb and losing the BR.

I'm not sure a CR across the bulb would help, as its current surge
thats killing the BR. But if you do that then 100 ohms and 0.1u in
series is typical.

Another solution would be to lose the BR and fit a capacitor dropper
with a fixed R to drop a bit of the power. Now the C&R will prevent
any monster currents when the bulb dies. Just a small R, not much
Pdiss and an X rated cap. The R helps limit peak i, though with the
lightbulb's R you should be fine with just the C alone.


NT