At that the LM338 would be dissipating something like 40W, and could well be thermal limiting - it's really too high a voltage transformer to feed a 5V linear regulator.
That depends on the intent of the original designer of the transformer.
A 12-0-12 winding infers that the designer intended the transformer to be used with a bi-phase rectifier. ie two diodes, one from each 12v output, then take the +ve rectified output from the junction of the two diodes and the -ve output from the 0v centre tap on the transformer.
Each half of the secondary winding is only conducting half of the time, so you can use a thinner wire.
At that the LM338 would be dissipating something like 40W, and could well be thermal limiting - it's really too high a voltage transformer to feed a 5V linear regulator.
So If I reduce my input voltage, could I use the linear regulator more efficiently? Say I use a halp wave rectifier instead of a full wave and filter it. Or use a 7.5-0-7.5 transformer...?
At that the LM338 would be dissipating something like 40W, and could well be thermal limiting - it's really too high a voltage transformer to feed a 5V linear regulator.
So If I reduce my input voltage, could I use the linear regulator more efficiently? Say I use a halp wave rectifier instead of a full wave and filter it. Or use a 7.5-0-7.5 transformer...?
I would suggest a different transformer would be best, 7.5-0-7.5 might be OK, or try a 9-0-9.
Using a half wave rectifier wouldn't help anything, and you would need far more massive reservoir capacitors (I'm presuming you already have ayt least 10,000uF?).
I have got a 2x2200uF=4400uf. And it drives two linear regulators. However, by a jumper setting, any one (or both :roll: ) of them can be disconnected.
I have got a 2x2200uF=4400uf. And it drives two linear regulators. However, by a jumper setting, any one (or both :roll: ) of them can be disconnected.