Parts list:

TR = 2 x 15 volt (30volt total) 6+- amps

D1...D4 = four MR750 (MR7510) diodes (MR750 = 6 Ampere diode) or 2 x 4 1N5401 (1N5408) diodes.

F1 = 1 Amp

F2 = 10 amp

R1 = 2k2 2,5 Watt

R2 = 240 ohm

R3,R4 = 0.1 ohm 10 watt

R7 = 6k8 ohm

R8 = 10k ohm

R9 = 47 0.5 watt

R10 = 8k2

C1,C7,C9 = 47nF

C11 = 22nF

C2 = 4700uF/50v - 6800uF/50v

C3,C5 = 10uF/50v

C4,C6 = 100nF

C8 = 330uF/50v

C10 = 1uF/16v

D5 = 1N4148, 1N4448, 1N4151

D6 = 1N4001

D10 = 1N5401

D11 = LED

D7, D8, D9 = 1N4001

IC1 = LM317

T1, T2 = 2N3055

P1 = 5k

P2 = 47 Ohm or 220 Ohm 1 watt

P3 = 10k trimmer

This is definitely an simple to create power supply which has reliable, clear and regulator 0 to 28 Volt 6/8 Ampere output voltage. By using two 2N3055 transistor, you'll get two times the amount of electric current.

Although the 7815 power regulator is going to kick in on brief circuit, overload and thermal overheating, the fuses within the primary section of the transformer and also the fuse F2 in the output will protected your power supply. The rectified voltage of: 30 volt x SQR2 = 30 x 1.41 = 42.30 volt measured on C1. So all capacitors ought to be rated at 50 volts. Caution: 42 volt is the voltage that might be around the output if one of the transistors ought to blow.

P1 allows you to 'regulate' the output voltage to anything between 0 and 28 volts. The LM317 lowest voltage is 1.2 volt. To have a zero voltage around the output I've put 3 diodes D7,D8 and D9 on the output of the LM317 to the base of the 2N3055 transistors. The LM317 optimum output voltage is 30 volts, but applying the diodes D7,D8 & D9 the output voltage is approx 30v - (3x 0.6v) = 28.2volt.

Calibrate your build-in voltmeter working with P3 and, of course, a good digital voltmeter is better solution.

P2 will certainly let you to set the limit of the optimum available electric current at the output +Vcc. When using a 100 Ohm / 1 watt variable resistor the current is limited to approx. 3 Amps @ 47 Ohm and +- 1 Amp @ 100 Ohms.

## Comments