On 20/11/2022 21:03, Rob Jarratt via cctalk wrote: > The location of the diode is arrowed on this picture: > https://rjarratt.files.wordpress.com/2022/11/img_20221120_205802-arrowed.jpg > > 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. Matt
