Why did you take apart the wall wart?
1. To find out what was inside.
2. To scrounge parts.
It was already in a useful configuration as a DC power supply in its original form. Now you have the same thing but in pieces without a housing to hold the transformer and related bits.
It's ok. I have a box of them. Way more than I need.
I don't see how this is an improvement.
To the wall wart, probably not.
To myself, possibly.
Parts scrounging is great but what parameters would you like for your power supply ? Random components may not be vary compatible with each other which can lead to early failure or not working at all . Also can eat up a budgets pretty quickly buying random components .
Constant voltage and constant current or adjustable voltage and current ? How many peak volts and current would you like ?
Once you set up the parameters then you can pick the components required.
Not all transformers are created equally and / or used in the same application or used at the same frequency.
EI core power transformers are around 75 - 85 % efficient where Toroid core power transformers are around 90 to 95% efficient . R core transformers roll in around 90% efficiency .
The RMS (Root Mean Square) is found by dividing the peak amplitude by the square root of 2 (approximately 1.414). This yields the actual, useable voltage .
Secondary AC voltage x 1.414 = rectified filtered DC voltage
So for example , after rectification and filter , a transformer with a 22VAC secondary winding would yield around 31V DC
The amps of the secondary can be calculated by The apparent power in volt-amps is equal to current in amps, times the voltage in volts .
VA = I x V Watts(P) = I x V VA = Watts(P)
Remember to double the VA to minimize voltage sag and and excessive heat. Heat is not your friend.
example:
So for a 2A transformer at 22V you would need 2 x (22 x 2) = 88VA
You probably wont find a transformer that's is exactly 88VA so go with 100VA .
Diodes or a full bridge rectifier must be able to handle the required current without producing excessive heat. So if you need 2 amps then the diodes or rectifier must have at least 2Amps forward current. Double that to reduce heat.
Rectifiers also need some load to properly measure with an oscilloscope.
An oscilloscope will only draw a very small amount of current in micro amps , And believe it it or not , diodes and rectifiers have a minimum forward current that can be greater than what the oscilloscope will draw . So some rectifiers will not function as expected with only the oscilloscope leds connected leading to incorrect measurements . This can happen more with larger rectifiers .
Filter capacitors must be Voltage rated higher than the voltage after rectification .
For example if your rectified voltage is 31 V then use at least 35V capacitors. 50V is better to cover potential surges . Using capacitors at voltages higher than the rated value will explode .
The size of capacitor is dependent on the exceptable ripple voltage .
Capacitance in Farads(C) is equal to load current in Amps(I) divided by 2 times the input frequency times exceptable ripple Voltage peak to peak(Vpp)
C = I / (2 * f * Vpp)
Generally Vpp is set at a reasonable 1V peak to peak
Frequency for North America is 60Hz and 50Hz for Europe
For example a 2 amp load current at at 60Hz
2/(2*60*1) = 0.0166F or 16600uF 1F=1000000uF
But be careful adding more and more capacitors to reduce the ripple voltage . After a point the capacitance needs to be exponentially higher just to reduce small amounts of ripple. There are a couple methods to reduce ripple without adding more and more capacitors . Chokes and capacitance multipliers are such methods .
This should get you started then we can discuss solutions for regulation. There are many . Some are complex and some are relatively simple . Each with advantages and disadvantages .