I found my multimeter and about five minutes of probing seems to imply the
diodes are there so that three DIFFERENT positive polarity strobes could be
triggered by the photosensor. Each of the two ports and the hot shoe itself
has a diode connected so that the photosensor could ground a positive
voltage.
Without the diodes if the ungrounded voltages of the strobes were different
current would flow from one strobe to the other with perhaps very unwanted
results.
I doubt (but don't know) that Olympus would change from what must be a
positive T32 to a negative F280. So Tim's guess that the voltage drop of the
diode prevents the F280 from tripping is still the best theory.
If you have a negative polarity strobe you could modify the shoe to short
out the diodes but you better be very careful hooking up more than one
strobe to the shoe adapter.
-jeff
> -----Original Message-----
> From: Tim Hughes
>
> Ok , I only just now caught my name being used in vain!
>
> The reference mentions soldering wires to short out the diodes inside.
> Since that is all that was said, and I don't have a
> schematic, I can only make an educated guess here:
>
> Ideally the slave needs to trigger either a positive or
> negative polarity strobe flash unit, so as to operate with
> different camera vendor's devices. (The OM flashes are all
> positive polarity) Conventional transistors (or scr's) only
> conduct well in one direction or have other limitations when
> operated inverted. It is likely, that the slave unit has a
> diode bridge or other series diode arrangements to allow it
> to trigger both positive or negative devices. The diodes add
> at least 0.6V and maybe 1.2V drop (4 diode bridge). Quite
> often a small low power SCR is used as output switch in the
> slave, to pull the flash trigger pin low. These drop about 1V
> when "on". So potentially the output only gets pulled down to
> 2.2V above ground, instead of ~0V. This may be above the
> threshold at which the strobe triggers correctly. This is
> particularly true if the slave-trigger steals it's power from
> the flash trigger pin. It then starts off lower than normal
> or even below the trigger level.
> Many
> newer flashes have extremely small currents required to
> trigger them, and stealing current from the trigger pins may
> stop them working altogether. Remember also, some new flash
> units have unloaded voltages as low as 4V.
>
> Regards,
> Tim Hughes
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