Hi,
I think this is a standard design for a linear power supply.
Not bad.
There's nothing wrong with it. It will work ... for most cases satisfactory
*****
I'd like to comment the voltage specifications. (From the view of an electronics engineer)
In the schematic the input capacitors are rated 25V.
So if one wants to chose a transformer...it's nominal output voltage should not be higher than (25V - 0.6V) / (1.2 × sqrt(2)) = about 14V.
(A transformer rated with 14V (@ nominal load) may have 1.2x higher voltage when not loaded, the capacitors will charge to peak of sine = sqrt(2) × RMS, minus the voltage drop of the diode of 0.6V @ low load).
In detail it depends on transformer specifications, but also in mains voltage stability.
One should also consider some headroom for the capacitor voltage rating.
So let's use this 14V transformer....but now at full load of 500mA output current.
The nominal output voltage now is 14V RMS. The capacitors now are pulse charged once per fullwave and maybe just in 20% of time.
This means every 20ms (@50Hz) there need to be a pulse current of 2.5A during 4ms. (Estimated).
This causes a higher voltage drop in the diode, let's assume 1.2V.
And additionally the sinewave gets a flattened top. It won't go up to 14V x sqrt(2). Let's assume a 2.5V reduced top (assumed 1 Ohms source impedance).
The peak value at the first capacitor will then be 14V x sqrt(2) - 1.2V - 2.5V = 16V.
The capacitor will be discharged (100% -20%) of 20ms with 500mA, this gives a ripple of 16ms x 500mA / 2200uF = 3.6Vpp
The second capacitor will reduce this peak-to-peak ripple, but let's assume it will be about 2.0Vpp at the second capacitor.
Additionally there will be an average voltage drop of 10 Ohms x 500mA = 5V at the resistor. (@2.5W power dissipation!!)
Thus the minimum voltage at the second capacitor will be 16V - 2.0V - 5V = 9V at full 500mA output current.
If one now considers a 2V voltage drop at the voltage regulators one can expect just 7V of clean, regulated output voltage at 500mA load.
Note: All the voltages are estimated, they may differ +/- several volts in the real circuit.
How to improve the situation:
* using input capacitors with higher voltage rating
* then you are able to use a transformer with higher output voltage
* using diodes with less voltage drop
* using full bridge rectifier
* using a transformer with more stable output voltage vs load current
* using larger capacitors
* reducing lower ohmic series resistor (now 10 Ohms), or replacing it with an inductor
* reducing specification for output voltage
(When you buy a power supply with rated 25V output voltage, then it should be able to supply this at full specified load current)
Klaus