Transformer parameters.

I have got a 12-0-12 transformer that is rated for 5A.

But when I rectified, filtered, and added an LM338 regulator, I get not more than 3A current at 5v.

When I filter the ripple waves, I get a higher voltage ( I have not measured this with a true RMS meter.)

I calculate the transformer to be rated 60W.

But if I get about 18V (on non true RMS meter), Should I see it as 18x3~60W

SO does this means that the current rating of the transformer is an rms value?

Is Wattage a better way to describe the transformer ?

And How is VA different from Watts?

lord loh. said:

I have got a 12-0-12 transformer that is rated for 5A.

But when I rectified, filtered, and added an LM338 regulator, I get not more than 3A current at 5v.

Click to expand...

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.

..
I have got a 12-0-12 transformer that is rated for 5A.
..

I calculate the transformer to be rated 60W

Click to expand...

it should be of 120W not 60W

we use VA along with powerfactor , for purely resistive loads VA and W are same.

akg said:

..
I have got a 12-0-12 transformer that is rated for 5A.
..

I calculate the transformer to be rated 60W

Click to expand...

it should be of 120W not 60W

Click to expand...

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.

JimB

Nigel Goodwin said:

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.

Click to expand...

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...?

lord loh. said:

Nigel Goodwin said:

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.

Click to expand...

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...?

Click to expand...

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?).

Nigel Goodwin said:

(I'm presuming you already have ayt least 10,000uF?).

Click to expand...

:shock: :shock: :shock: :shock: :shock: :shock:

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.

10mF is a lot.... :shock: :shock: :shock:

lord loh. said:

Nigel Goodwin said:

(I'm presuming you already have at least 10,000uF?).

Click to expand...

:shock: :shock: :shock: :shock: :shock: :shock:

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.

10mF is a lot.... :shock: :shock: :shock:

Click to expand...

And 4400uF isn't much at all 8)

I'm also a little concerned about 'jumper settings'?, at these sorts of current short thick direct wires are what you should be looking for!.