The output voltage may be rather higher unless the transformer is delivering its full load current. Since you have a 3A transformer, this may be the case even at full load using a 1A regulator.

You might find that adding something like a power resistor in series with the output from the reservoir capacitor helps to lower the regulator dissipation, but if you do that don't forget to add a fairly substantial capacitor between the regulator input and common.

Will the power resistor not affect the current output?

 

The first link isn't working.

It is fixed now. About two amperes DC from your transformer should be OK. Depending on your setup you may go somewhat higher. Here is a Design Guide for Rectifier Use, and selecting correct transformer specs
http://www.hammondmfg.com/pdf/5c007.pdf

 

Will the power resistor not affect the current output?

You need to calculate the resistor value, so that at full load the voltage at the regulator input at a ripple minimum is still not less than the minimum required to give full output.

Before doing this, you would need to check what the transformer and rectifier actually give you at full load. You would then know how much could safely be lost. Don't try to drop the voltage too far - the output will droop and develop ripple if you do. Remember that the resistor will have to be rated for the power which it will dissipate.

A resistor like this will not of course do much for the voltage with no load on the regulator output. If that is too high, some other method would be needed. One way would be a bleeding resistor in parallel with the regulator input to draw a minimum current, possible if you are desperate, but really a horrid inefficient kludge.

 

You need to find the how much load the transformer can handle first.

Here is a method a member suggested to find the safe current capacity of any given tx.

For this u need a couple of high power resistors.
This test is done using the transformer only. Disconnect the bridge and everything else first.

1. Note down the VAC of secondary without a load.

2. Calculate 5% of that Voltage.

3. Next connect a low Power ceramic like X* Ω 10W ones. But first keep the AC Voltage meter connected to the secondary prior to connecting the resistor.

4. You will notice the voltage decreasing when you connect the resistor to the secondary.

5. If the voltage is no less than 5% , add another X* Ω of the same power.

You should keep adding resistor until the AC voltage of the secondary drops to 5% of it's unloaded voltage..

Once the you reach the said voltage, You got the continues loading current of the your transformer.

Remember not to go lower than 5% of the no load voltage.

* Note

X ohms should be the numerical value of the unloaded AC voltage.
for ex: if the unloaded VAC is 30VAC, then X =30 Ω.

This way u will get 1 amps of current increase.
Easier to remember.

But if you have a rheostat then there is a much easier method.

 

The LM337 is complementary to the LM317 (it's a negative adjustable 1.5A regulator)

 

Hey thanks for all the reply. I tried the center-tapped 2 diode rectifier setup but I'm getting 28-29V instead of the supposed 22V. Any ideas?

A high quality 32V center-tapped transformer uses thick wire with a low resistance so without a load the rectified and filtered output is about 23VDC.
A cheap transformer uses thin wire with a high resistance so it is made with a higher voltage that drops when it is loaded.

Its high DC voltage when not loaded probably will not harm anything since it will drop with a load.

 

Thanks for the all the replies. I'll read each and try them out.

Happy holidays everyone!