Power supply operation
This power supply uses LM317 for the positive regulation that can supply an output of 0V to 28V dc at a maximum current of 1.5A (with heatsink or cooling fan). The output voltage of the positive regulator is given by the formula:
Vout = 1.25(1 + 5K/240) + (100uA) (5K) - 1.4V
Where,
240 is the resistance from adjust and output terminal of LM317
5K is the resistance of the potentiometer connected to adjust terminal
-1.4V is the forward voltage of the two diode connection
Similar to LM317 operation, LM337 control the negative regulation of output voltage to ensure its output to 0V to -28V of a maximum current of -1.5A (of course wit heatsink). The variable negative output is given by the formula:
Vout = -1.25(1 + 5K/240) – (100uA) (5K) + 1.4V
Where,
240 is the resistance from adjust and output terminal of LM337
5K is the resistance of the potentiometer connected to adjust terminal
1.4V is the forward voltage of the two diode connection
Power supply materials
D is 1N4001 diode or any of the same maximum current capacity. Fuse must be 0.5A, 250V or much lower amperage rating. Transformer is 220VAC to 26V -0 -26V, 2A or 120V to 26V -0 – 26V rating of 60VA or higher, bridge diode must be rated 3A or higher for good circuit operation. The resistors used are rated 1/2W.The input capacitor must be 4700uF or larger for better regulation and stability. The output capacitor Cout is 10uF.
Note: This power supply project is powered through 220V main and can cause electrocution
I would like to know how to convert this power supply to 0 to 12v and 0 to -12v.
ReplyDeleteHpe you can help me soon,
ReplyDeleteRichard good day!
ReplyDeleteSince you ask about a +12V,0V,-12V version of the power supply, i design a very simple circuit for you.just follow the link.
dual 12V supply
Q: What are the 5.6K resistors from (+) (-) Diode Bridge to the four 1N4004 Diodes labeled D?
ReplyDeleteWhat is their function and why the four Diodes?
Q: What are the 5.6K resistors from (+) (-) Diode Bridge to the four 1N4004 Diodes labeled D?
ReplyDeleteWhat is their function and why the four Diodes?
LM317 and LM337 are has a minimum output voltage of 1.2V and -1.2V and not equal to zero.
By adding 5.6K with 2 diodes in series in each regulator, it ensures that the output of both regulator is zero since
new min output = (1.2V -2VD) where VD is the forward voltage of the diode and 1.2V is the original minimum voltage of the regulators.
As we know diode drop is around 0.6V to 0.7V, meaning the new minimum output voltage is zero and not 1.2V
I QUIT ELECTRONICS THIS THING IS DIFFICULT TO UNDERSTAND HEY
ReplyDeletewhere is the filter in this ckt?
ReplyDeletethe 10uf cap will act as the filter(voltage regulation).
ReplyDeleteYou can increase its value..The larger the value of capacitor the better. But larger means bigger and more expensive
how can filter capacitor act as regulator? for that we have reg ic right? filter should be before regulator..4470uf cap ?
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ReplyDeleteThe LM317T is the adjustable regulator IC, can adjust the output voltage of 1.25V to 37V.
ReplyDelete-terminal voltage regulator that is designed to supply in excess of 1.5 Amps of a DC output voltage range of (1.2-30) Volts.
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What is the purpose of dual power supply?
ReplyDeleteIs LM317 an IC?
ReplyDelete