Low ripple 12v fixed power supply... Good enough for you?

[expertmax1]

Hi fellow members

I am currently designing a little power supply for fun, but I have made this WONDERFUL schematic and I wanted to know if you like it or not (not my drawing skills tho).



ANY feedback is appreciated !

Max.

[c4757p]

How much current? That 7812 is going to be dissipating about 22 W per amp. Even at 50mA you'll need a heat sink.

[ecat]

I like the Max cad

But I think what you have designed is a one-shot asynchronous wide band audio pulse generator. Which is a long winded way of saying it will, one day, possibly when you are least expecting it, go BANG!

24V rms under load which gives 34V peak.
Off load you have, 28V (? I don't know but it will be more than 24V), so 40V peak.
When your mains voltage varies by 10%, heh do the maths.

The 7812 is good for a max input of 35V, even accounting for diode losses, it will one day go BANG!

Also what c4757p says

Perhaps a 15V transformer? 18V? Something like that?

[c4757p]

The 7812 is good for a max input of 35V, even accounting for diode losses, it will one day go BANG!

Ah crap, I forgot that! Yeah, they do not like excessive voltage one bit, and it's usually a gradual destruction. Worse, many of these regulators fail shorted input-to-output (not sure about 7812 specifically), so you'll stick the 30-40V right up whatever you are powering when it collapses.

[expertmax1]

What is the best thing I could do to improve my design ? Perhaps a new transformer ?

[mariush]

24v AC RMS is a bit much. 

You don't need that diode before the capacitor.

The capacitor is for smoothing the DC voltage before it goes to the 7812.

capacitance  =  ( Current  x Duty Capacitor [0.7 usually] ) / ( Vripple x frequency [2x50-60hz])

Let's say 60 Hz and 1A of current , you then have  0.0033F (3300uF ) = 1 x 0.7  / ( V ripple x 120)   => Vripple = 0.7 / 120 x 0.0033  = 1.76 v

So  at 1 A of current,  you have 24 V ac, that's rectified to 1.41 x 24v  = 33.8v DC, take out about 1.4v the voltage drop in the two diodes and you're left with about 31v.
At 1A of current, the voltage will be between 29 and 31v (1.76v ripple) ... your 7812 only needs 2 volts above the 12v to regulate properly, so you're giving it about 15v more than necessary.  Even with a 470uF capacitor, at 1A current you'll have only about 13v ripple, so you'd still be above 14v needed to get stable 12v.

But that's besides the point. The regulator will still have to dissipate the voltage difference as heat ... which will be ( V in - V out) x Current ... At 29v input and 1a current, you have 29-12  = 17 watts as heat.  I doubt the 7812 can even dissipate that much , with heatsink even.

If your transformer has a center tap, , use that to your advantage ...



Then calculate your capacitor size so that you'll have at least 14v all the time. 
V peak is about 17v, minus one diode drop (0.4-0.8v), so you'll afford about 2.5v ripple  ... so capacitance  = current x 0.7 / ( 2.5 x 120) ... for 1A of current that result in about 2300uF, so a 3300uF would do, but it wouldn't hurt to use 2 x 3300.

If it doesn't, your best bet is to get another transformer.

[expertmax1]

If I follow your schematics, I put the 7812 right after V D.C. ?

[mariush]

There's no resistor there where it says Vdc , it's just something to symbolize a load. After the capacitor you have + and -  where minus equals ground in your case.

So yes, you can put the regulator after the capacitor, Vin goes to + , GND goes to - ...  It won't hurt to add a 0.1uF / 35-50v  ceramic capacitor before the regulator, along with the large capacitor(s).

[croberts]

I think it would be helpful to increase the ripple of the input voltage to the regulator by lowering the value of the input capacitor (maybe 680uF for 12V ripple at 1A load). The ripple waveform approximates a sawtooth and the input voltage (to the 7812) will approximate Vripple min + 0.5 Vripple so the greater the input voltage ripple the less the input voltage to the 7812 and the less the power dissipation. Of course Vripple min should not fall below 15V  so the regulator doesn't drop out. Vpeak ripple will be approximately (24 x 1.414) - 1.9V (3 diode drops) as things are now.

[ptricks]

Add some decoupling caps on the input and output of the 7812, not absolutely required , but a good idea.
Also multiple small caps are better than one large cap for filtering.

How much current do you need,  using a 7812 to produce 1A  from a 24+V supply will produce a lot of heat.

[expertmax1]

You know what? I'm thinking of making a 12V Postive and Negative DC power supply, so I will need to modify my schematic. I will then have +12V, GND, -12V. But I don't know where to start.

Another question, if I use the center tap on my 120V 12-0-12 1A transformer, I will get 12V to the 7812 but it will not be enough for a good 12V regulated output : the 7812 IC needs about 1.5V more than the source voltage.

Thanks.

[c4757p]

Another question, if I use the center tap on my 120V 12-0-12 1A transformer, I will get 12V to the 7812 but it will not be enough for a good 12V regulated output : the 7812 IC needs about 1.5V more than the source voltage.

12VAC is a sine wave oscillating between -16.97 and +16.97 (12 * sqrt(2)). The DC output of your rectifier will therefore peak around 16.3 (one diode drop less). All you
need is enough capacitance to keep the voltage above about 13.5 - 14V in between peaks (about 6800 uF for 1A).

[c4757p]

You know what? I'm thinking of making a 12V Postive and Negative DC power supply, so I will need to modify my schematic. I will then have +12V, GND, -12V. But I don't know where to start.

The negative circuit is pretty much the mirror image of the positive circuit. Like this.

Note that while I show just one 6800uF cap on each rail, it's better to use multiple capacitors that add up to near the total (like two 3300uF caps in parallel per rail), so that they share the ripple current.

Edit: 7812 and 7912 instead of 7805 and 7905, obviously. I'm too lazy to go back and fix it... 

Edit again: Holy crap I'm making a lot of typos and mistakes. 

[David_AVD]

6800uF seems like massive overkill for just 1A (or 1.5A) of output current.  Even 3300uF seems a tad large.  I'd probably use 2200uF (or 2x 1000uF in parallel) instead.

For a 12V regulated output I'd use a 15Vac transformer.  For a split supply (+/- 12V DC) that would be a 30Vac centre tapped unit.

[mariush]

If you want to make  -12v and +12v, you can use this :



On the + side, you can use 7812, on the - side, you can use 7912

Like c4757p says above, the transformer is 24v RMS , or 12v - 0v - 12v rms ... that means the peak voltage on both sides is  1.41 x 12v = 16.92v. Then the voltage drop of each diode working at one time for each AC pulse has to be taken off, so you're left with about 16.3-16.5v.

But that 17v is the peak voltage, the average is less, so you have to use those capacitors that get charged while the transformer gets close to the peak voltage and then as the sine wave goes down, the capacitor can provide energy and smooth out the DC output.

I've written the formula to compute the proper size of the capacitor a few posts above.  Capacitance =  [ Current x Duty Cycle Capacitor ] /  [Vripple x Frequency ]

Current is how much you want to pull from the transformer, let's say 1 A for this example.

Duty cycle is an estimated value, about how much time the capacitor has to provide power in contrast with how much time it can charge. a value of 70%  (0.7) is reasonable.

V ripple is how much you're willing to let that voltage go down, in volts... Since you want the voltage to go down as low as 12v + 2v  (you say 1.5v but 2v is safer) that means we can use 16.3v - 14.5 v = 1.8v here. But let's be on the safe side and say a maximum of 1.5v

Frequency is how many pulses of power come from the transformer and rectifier diodes. 
If you're in US, the AC frequency is 60 Hz. If you're in Europe, the frequency is 50 Hz. 

The way the diodes are layed out in the above circuit, it makes it that the cycles double, so if you're in US the frequency is 120 Hz, if you're in Europe the frequency is 100 Hz.

So now the formula becomes : Capacitance  =  1 x 0.7  /  (1.5 x 120)  = 0.7 / 180 = 0.003888 F    (I'm assuming you're in US, so I used 120Hz)

That's 3888 uF, so you'd have to use at least something above 3900 uF to have at most 1.5v ripple after the capacitors and keep the 7812 happy.   I would use 2x2200uF or 2x3300uF - the more the better but don't overdo it, it's not worth having more than 10000uF in this case.

Looks like I'm contradicting David_AVD a bit, but I may also be wrong. Make the circuit on the breadboard and see!

later edit:  And I think it should be fairly obvious the capacitors should be rated for 25v or more, since your peak voltages will be over 16v... If you use 16v capacitors they may vent or pop, be careful!
(if you only have 16v caps at hand, put 1 or 2 diodes before the capacitor, in series... each diode will drop 0.4-0.7v so you'll have under 16v on the capacitor)

[c4757p]

6800uF seems like massive overkill for just 1A (or 1.5A) of output current.

Given I = C dv/dt, the voltage will drop 147V/s between rectified pulses. With half-wave rectification and a period of 16.67 ms, that's a drop of 2.45 V over one period, giving a minimum voltage into the 7812 of 14.5V.

With 2200uF, the minimum voltage will be 9.4V! Even with full wave (half the capacitance requirement), the voltage still drops to 13V.

[expertmax1]

Holly molly, that was very informative ! Thank you for your awesome reply. I will probably sink 0.7 A from the 7812 and the 7912 so a good heat-sink will be required of course just in case of a 1A momentary peak. I like where this post is going !

[c4757p]

mariush, follow the current in that circuit! You've got a bridge rectifier but only two diodes are active. You're not going to get full wave and bipolar output. You'll need to double the capacitance.

Edit: Or I'm being stupid. Um, how about we all just ignore me until I get some sleep?

[expertmax1]

Alright, so this is my final schematic, does it looks okay to you? I will buy the parts Monday.

[c4757p]

Looks good to me, but take that with a grain of salt:

Um, how about we all just ignore me until I get some sleep?

 

One point to make: The pins on the 7912 are in a different order, so check that before you connect everything.

[expertmax1]

One point to make: The pins on the 7912 are in a different order, so check that before you connect everything.

Will do, thanks ! Let's hope the 7912 is available on NTE.

EDIT: NTE Part # : NTE967. Yay.

[David_AVD]

6800uF seems like massive overkill for just 1A (or 1.5A) of output current.

Given I = C dv/dt, the voltage will drop 147V/s between rectified pulses. With half-wave rectification and a period of 16.67 ms, that's a drop of 2.45 V over one period, giving a minimum voltage into the 7812 of 14.5V.

With 2200uF, the minimum voltage will be 9.4V! Even with full wave (half the capacitance requirement), the voltage still drops to 13V.

Sorry, I was assuming full wave rectification, not half wave.  I also specified 15Vac, not 12Vac.

Practice over the years has seen 15Vac, full wave rectification and 2200uF used with a 7812 work well.

Using 12Vac to get 12Vdc regulated output is usually borderline and doesn't allow for any drop in line voltage without issues.

[jahonen]

I have sometimes experienced startup problems with this bipolar regulator configuration. Cure is to add anti-parallel diodes to the output. 1N400x diodes will work just fine. I think this issue is sometimes mentioned in regulator datasheets.

Regards,
Janne

[c4757p]

Using 12Vac to get 12Vdc regulated output is usually borderline and doesn't allow for any drop in line voltage without issues.

Good point. Still, he already has the 24V center-tapped transformer, so it ought to be at least worth trying for a little hobby project. I know at least at my home, the voltage is usually running pretty high (almost - occasionally over - 130V), so I myself wouldn't hesitate to use a borderline transformer like that for a quick thrown-together power supply if I already had the transformer. I hate buying transformers...

I have sometimes experienced startup problems with this bipolar regulator configuration. Cure is to add anti-parallel diodes to the output. 1N400x diodes will work just fine. I think this issue is sometimes mentioned in regulator datasheets.

Yes, the TI datasheet mentions this. In general, to fully protect the regulators during startup, shutdown and both input and output short circuits, you should have a diode from ground to out (adj to out for adjustable regs), and another from out to in. Obviously, reverse the diodes for the negative side.

[expertmax1]

It worked ! The device is running fine with 0.8A load on the +12V and -12V rails. Woohoo! No AC current left in the DC lines and I am almost certain there is absolutely no ripple on there ! Maybe it's time for a new scope eh!