theskyishigh
Newbie level 5
Hi,
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.
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.