# Single power supply with multiple linear regulated outputs

#### atferrari

Joined Jan 6, 2004
4,122
Another way to look at it is to just use 2 diodes for a start, like the 2 right hand ones. Have the center tap at 0V.
Now you will get around +12V, that is 9VAC x 1.414.
Next, connect it up using only the left hand pair of diodes, still CT at. 0V.
The output now is -12V.
Now, with all 4 diodes connected.......
So in effect, you have a dual power supply, +12V, 0V, -12V.
But if you take the CT off the 0V connection and put the -12V one there, you end up with +24V, +12V, 0V.
The voltages are the same, just the reference point has changed.

Now it is clear. So evident now...
I simulated to play a little. Values are made just to get some results.
Thanks again.

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#### Littlegee

Joined Mar 25, 2020
14
Dendad sorry i'm a bit confused here can you help me to understand how you worked this out please:

'So, 12V at 2.1A actually will be (18VAC*1.414) * 2.1A = 25*2.1 = 53VA.' 18*1.414 is 25 then * 2.1
Adding all the 5V regs together... = 5V @4.6A.
'So, (9VAC*1.414) * 4.6 = 21.5 * 4.6 = 28VA.' 9*1.414 is 12 but you have 21.5 then * 4.6 (2.1+1.5+1) where did you get the 21.5 from?

Joined Feb 20, 2016
3,617
When 18VAC is rectified, the actual voltage produced is 18 * 1.414.
The 18V is a sine wave that has a peak voltage 1.414 times. It has to do with the shape of the wave.
http://www.bristolwatch.com/ele/basic_ac_rectification.htm
The real world voltage will be lower, depending on the load current, capacitor size, and a few other things.

#### Littlegee

Joined Mar 25, 2020
14
I understand the peak value of 1.414 but I don't understand how you calculated 21.5 from 9*1.414 which equals 12 or 12.726 to be exact?

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Joined Feb 20, 2016
3,617
but I don't understand how you calculated 21.5 from 9*1.414 which equals 12 or 12.726 to be exact?
Neither do I

"21.5 * 4.6 = 28VA" does not work either!!!!

In my defense, that was just a bit before 4AM when I posted that so I was not at my best. I'd been awake for some time before that.
I think it should be 9*1.414 = 12.6V peak.
Then (12.6V * total current of the 5V supplies) = (12.6V* 4.6A) = 58VA

These numbers are not exact. There are the diode drops to take into account, and the capacitor charge dropping between cycles, but it is a good rough start.

I could say I put errors in there to see if you could pick them up, but I was not that clever. Just tired.

#### Littlegee

Joined Mar 25, 2020
14
No worries at all you are most helpful and patient, I thought maybe I missed something in the calculations as I am still learning. So given the new values:

53 + 58 = 111 - 48 = 63W to dissipate.

I will be using an aluminium chassis with huge aluminium side fins to attach the regulator to which act as a large heatsink something like this:

https://www.audiophonics.fr/en/alum...311x260x120mm-silver-front-panel-p-13647.html

111VA + 10% is 122VA so would you suggest a 130VA transformer or something? I will have a look at the 7.5-0-7.5 specs. I will purchase a variable linear bench power supply (Do you have any suggestions?), bread board and some components to start putting together and testing everything once I have worked out everything that I need in theory.

Joined Feb 20, 2016
3,617
Have you actually measured the currents each supply requires? It may be the numbers you gave are the max values, and working currents are a lot less.
But if the supply is 100VA, then a 130VA transformer will be better. It is always a good idea to not run components at their max ratings.
As for a variable supply, I have one similar to this..
https://www.ebay.com.au/itm/30V-5A-...997648&hash=item3b3f439254:g:bIAAAOSwiMBdqDOd

The only problem I have had with it over a few years is the hi/low voltage range relay died. That was a fairly easy repair.
But you can make a simple variable supply with an LM317 regulator. Many kits are available on Ebay.
For your breadboard, a fixed 5V supply , LM7805, is a good idea.
Or a small switch mode reg...

The trim pot can be removed and an external pot wired in place to make it variable. An old laptop 19V power brick is a good supply to run this reg.
There are many ways to go.

Ebay has lots of breadboard starter kits, usually associated with Arduinos. Have a look to see.
Enjoy your electronics adventure. It can be great fun, and a bit frustrating at times too

#### Littlegee

Joined Mar 25, 2020
14
Dendad, yes these values were taken from the output values found on the SMPS that come with the DAC and phono stage, the JCAT cards were minimum ratings found in the literature no other documentation is given on actual current or power values. I can check the DAC and phono but not the JCAT cards as they are installed on the motherboard and I don't have an LPSU for it yet, that will be the next project after building this LPSU. Maybe I can get this info direct from Chord and JCAT. First I should build a bench PSU then build this LPSU then my media server LPSU. Thanks again for your suggestions and Atferrari for chiming in, I'll update the thread as soon as I make progress.

#### Littlegee

Joined Mar 25, 2020
14
Current values from the manufacturer:

AC Qutest - 0.4A
AC Huei - 1.5A
USB Femto card - 1.5A
NET Femto card - 2A

Total power required 37.5VA
Total current required - 5.5A

Dendad I know we talked about the large dissipation with the 100VA 9-0-9 transformer and the 7.5-0-7.5 option. Please see the attached file from Jan Didden where he talks about the relation of input to output voltage for the super regulator boards. He states "
A word about the input voltage from your rectifier/reservoir caps: make sure is it at least 5V above the required output voltage. The regs will work with just a few volts headroom but performance will suffer. Ideally, you would use two completely separate windings, rectifiers and capacitors and connect each one to J1 and J5 respectively. If you have a center tap (CT) transformer connect the CT and the ground returns of J1 and J5 together at a star point with the reservoir capacitor returns."

If this is the case then:
5V above 12V would be 17V then 15VAC * 1.414 = 21VAC
5V above 5V would be 10V then 7.5VAC * 1.414 = 10VAC

So this in theory would still work but I am having difficulty finding a 108 VA transformer with 2 x 7.5VAC windings. Do you think it is better I just use 2 transformers closer to the voltages i need like one with a single 7.5VAC secondary and another transformer with a single 15VAC secondary winding or I get a custom 108VA transformer with the 2 x 7.5VAC secondaries made?

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