I made a step-down power supply using the MC34063A switching regulator. The configuration was straight out of the datasheet (schematic attached). However, I neglected to connect the anode of a diode to ground on the output pin of the regulator. This diode was burning up. I tried replacing the 1N5819 with other diodes like the SS34, with no impact. As soon as I realized my mistake and connected the anode to ground, it came back down to room temperature.
My question is, how can a diode with only one side connected (cathode) heat up?
Thanks to all who can shed a little light on this.
It sounds awful that that happened though. Let's know where the anode lead of the diode was resting on. Seems like it was in contact with some higher voltage, higher current point and the IC driver was probably sinking a lot of current through it during the off-time.
It can't.
Power dissipation follows Ohms Law, W = Vf * I, and if I is zero so must be the power. If you connected both ends and had it reversed it would conduct continuous high current and I would expect the IC and diode to get very hot but you must have both ends connected for that to happen.