12VDC Power Supply Help Needed

Discussion in 'The Projects Forum' started by Iggy, Feb 2, 2009.

1. Iggy Thread Starter New Member

Feb 2, 2009
2
0
I'm a newbie and want to make a 12VDC regulated power supply to experiment with various LED's combinations. I calculated that for 2.5% ripple, C1 should be a 4700uF 25V capacitor. However, I don't know how to determine what the value of C2 should be or if it is needed at all.

My calculations so far:
Transformer is rated 12vrms @ 500ma 60Hz
VP = (1.414 x 12) - 1.4 = (16.968) - 1.4 = 15.568
Vrip = 0.3892V X 2.828 = 1.10 V
C1 = [ ( 0.5A X 0.00833) / 1.10V] X 1000000 = 3786uF

Any help appreciated.

File size:
37.2 KB
Views:
119
2. SgtWookie Expert

Jul 17, 2007
22,182
1,728
4700uF should be fine for C1, but 25v is marginal. 12.6vac when rectified will become around 17v. The "rule of thumb" is to double the voltage rating for electrolytic caps, so you'd want a cap rated for 35v (standard value). This helps to minimize "leakage"; if leakage becomes excessive the capacitor will overheat and blow it's top.

Use a 0.1uF and 10uF on the output, just to suppress any oscillation tendencies. You won't have any real transients to speak of with LEDs.

Note that to ensure guaranteed regulation, a 7812 requires a minimum load of 5mA. You could provide that by adding a 2.2k resistor from the output terminal to ground, or simply ignore it if you have more than a 5mA load. Just don't be surprised if the output isn't close to 12v when you do not have a load.

Apr 5, 2008
15,649
2,348
Hello,

Just take a look at the datasheet of the 7812.
There are two smal capacitors on input and output to prevent the 7812 from oscillating.

Greetings,
Bertus

4. beenthere Retired Moderator

Apr 20, 2004
15,815
282
C2 is oversized. 1 uF would be plenty. In fact, you would do well with just a .1 uF ceramic close to the 7812's pins on the output. C1 is probably larger than necessary, but capacitors are inexpensive. The load will not exhibit dynamic characteristics.

5. Audioguru New Member

Dec 20, 2007
9,411
896
No.
An LM317 regulator has this problem because its idle current (current for its internal circuits) goes through the output so its ADJ pin's current is extremely low.
But the idle current of a 78xx regulator goes to ground through its ground pin so it does not have a minimum load current.

6. SgtWookie Expert

Jul 17, 2007
22,182
1,728
Please examine a datasheet from ST Microelectronics' L7800 Series or Fairchild's LM7800 Series positive regulators. You'll find that output voltage regulation is specified with a load of 5mA to 1A. To me, this indicates that a minimum load of 5mA is required to achieve guaranteed regulation. This is also consistent with other manufacturer's datasheets.

With an LM317, if a 120 Ohm resistor is used for R1, there is no additional minimum load required, as 10mA will flow from the OUT to ADJ terminal, and then to ground via R2.

With an LM317L, R1 can be 240 Ohms, as the minimum load requirement is 5mA.

7. Iggy Thread Starter New Member

Feb 2, 2009
2
0
Thanks! I'll have at least a 20mA load.

Thanks, I think I'll use a 2200uF 35V for C1 (5% ripple).

8. mik3 Senior Member

Feb 4, 2008
4,846
63
Iggy,

Have you considered that such a big capacitor (C1) will cause a large in rush current during power up which may blow the diodes?
This current must stay below the maximum current the diodes can conduct.

Also, diodes can withstand a surge current for an amount of time. If this surge current lasts more the diodes are gone.