On 20/11/2022 21:03, Rob Jarratt via cctalk wrote:
The location of the diode is arrowed on this picture:
You can also see the heatsink where the transistor used to be.
The component you refer to as an inductor is actually a transformer,
providing isolation between the control PCB (riser card) and the
switching transistor. As such the control circuitry should not have been
harmed by the failure of the transistor.
You can test the control circuitry by applying about 15V from a bench
supply to the purple capacitor shown in the picture. This is the 2200uF
capacitor shown on "PSU sheet 1" of Tony Duell's schematics (available
on Bitsavers). You will then be able to see the switching drive signal
on the left hand pin of the control PCB.
On "PSU sheet 2" I'm assuming that the diode in question is the one in
parallel with the 2.7R resistor. My guess is that it is there to provide
fast turn-off of the switching transistor. If it were to fail then that
would lead to slower turn-off and probably overheating of the
transistor. That may explain the failure. Other components to check are
the ones in the snubber circuit. This is the two 500R resistors and
associated diode and capacitor connected to the collector.