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?

__________________
Bharath Bhushan Lohray.
M.Sc. Electronics.

 

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.

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it should be of 120W not 60W

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

 

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

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Experience is directly proportional to the value of the equipment ruined.

 

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...?

__________________
Bharath Bhushan Lohray.
M.Sc. Electronics.

 

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?).

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: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:

__________________
Bharath Bhushan Lohray.
M.Sc. Electronics.

 

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!.

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PIC programmer software, and PIC Tutorials at:
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