+/- 18VDC PS basic questions

You don't say if the 18VAC transformer is center-tapped, and it's not clear if you're describing a bipolar power supply, but if you are, then that's the circuit you're looking for. If the 18V transformer is not center-tapped, then you're out of luck.

Yes - a full-wave bridge rectifier _is_ four diodes.

Not unless you use two of them, or an AC wall wart and floating voltage doubler.

Good Luck! Rich

--
No, it doesn\'t, unless it being used in a bipolar supply.

--- Each wall-wart is a separate supply with a floating output, either terminal of which can be "grounded".

If you want a positive supply you ground the negative DC end of the supply and take your output from the positive end. OTOH, if you want a negative supply you ground the positve end and take your output from the negative end.

If you want both a negaitve and a positive supply you ground the negative end of one and the positive end of the other and take your outputs from the ungrounded ends; + from the positive end of the supply with its - terminal grounded, and - from the supply with its

• terminal grounded.

-- John Fields Professional Circuit Designer

--
Also be aware that by the time you get finished rectifying and
filtering the output of that transformer the DC across each of the
the filter caps is going to be somewhere in the neighborhood of 25
to 30V.

I need to build a power supply which will give minimum +/- 12VDC to power an audio device.

I've got a transformer which outputs 18VAC from an old CD player and a simple schematic which shows four 1A diodes arranged to output DC voltage, followed 2 capacitors (3300uF electrolytic and 1uF disc, one of each on the negative and positive) with analog ground being the other side of each pair of capacitors.

I have two questions:

1. Can I use a full-wave bridge rectifier unit instead of the four diodes? If the answer is no, why not?

1. Can I use a cheap "wall wart" DC power pack as my DC supply instead of my AC transformer and diodes/rectifier? I would then wire in the capacitors to create my analog DC ground.

Any replies greatly appreciated.

Check out

it should make things a little more clear. Brian

How does one tell if a transformer is "center-tapped" or not? What properties does center-tapping convey to a transformer? I will guess (bad, I know, bad) that it IS center-tapped given its' previous life as a (relatively decent) CD player PS. This particular x-former has more than one set of secondary windings... one seems to give about 36VAC, one about

18VAC, and another about 9VAC. Would this lead you to any logical conclusions regarding the construction?

of

capacitors

Yes, it is a bipolar power supply which I am in need of. Why would I have to use two? I thought that the wall wart was a just a transformer and rectifier in one. Ah, but perhaps not full-wave? It would only have one set of diodes to put out voltage in one direction?

tap

guess

If it's multiple taps, would there necessarily BE a center tap? If so, would it not be the highest voltage? Or is there just no way to tell.

Is it okay to use a transformer with the correct voltage rating but a larger current rating? For example, I see a 18VAC 1A center-tapped transformer for $18 at an online retailer, but I really only need 250mA.

The 4 diodes are already configured to make a full wave bridge. Why do you want to replace them ?

No because the wall wart doesn't have a centre tap that I expect the transformer you have does have although you don't mention it. 2 capacitors won't create a DC ! ground.

Graham

Yes. I expect the 9V secondary is independent and intended to supply the regulated 5V DC for logic etc.

The other winding is likely 18-0-18 V AC and used for the split ( bipolar ) supply. The centre tap connects to the 'ground' of the supply.

Note that 18V AC rectified will be about 24V DC. A +/- 24V DC supply will toast most op-amps.You need voltage regulators to reduce it to typically +/- 15V. The regulators also remove most of the supply ripple.

Graham

Because it's *NEEDED* to make a centre tapped ( i.e. bipolar - plus AND minus volts ) DC supply.

There is a way it can be done without a centre tap actually but it's only half wave rectification and generally not recommended.

You need to learn some basics about power supply configurations.

Graham

bipolar )

There are two windings, one gives 5VAC single-tapped or two leads, the second has four leads with varying voltages depending on which leads are measured. No center tap. Too bad.

toast

15V. The

Ah, now this is useful information. I have been poring over schematics for power supplies and nowhere is it evident (and I obviously lack the training to know) that you get higher DC voltage out of a rectifier than the AC voltage in. Is there a "rule-of-thumb"? I see that above you note about

1.5X multiplier for the DC out of a rectifier. The particular circuit I'm looking at shows a 32VAC CT transformer which is rectified, filtered, regulated down to 18VDC, and filtered again. For my particular app, I don't necessarily WANT a regulated 18VDC, I am okay with letting it float as the op-amps are protected by more downstream regulators. 32VAC would give ~48VDC? You'd need one helluva beefy regulator (most of them that I've seen can handle up to 30VDC) and heatsink to drop 30VDC!!! If the voltage is filtered between the rectifier and regulator, why would you need to filter it again after the rectifier? Could one assume that you'd need smaller filter caps as you work your way downstream?

In this regard, I have a 12.6VDC CT x-former (which I thought would not be enough voltage). I really only need 15VDC, so can I expect ~18VDC out of my rectifier if I use 12.6VAC in?

While I'm at it with this enlightened audience, I have another basic question: After the rectifier I have a 3300uF 50V electrolytic "filter cap" and a 1uF disc "bypass cap". It is my understanding (and, hey, I'm not batting a thousand here so bear with me) that the filter cap stores up charge and compensates for voltage drops. Is this right? What does the bypass cap do?

Thanks in advance for any and all replies.

Dave

will

for

training

don't

seen

my

I did some more research. Apparently my 12.6VAC transfermer (rated a 3A) UNLOADED should give me 12.6VAC x 120% x 1.414 =~ 21.4VDC. Perfect for my app. Max safe loading with 3300uF filter should be about 80% of that or

17VDC... that's unregulated. If I want regulated voltage I shouldn't count on more than 12.6VDC so I can safely regulate to 12VDC. No? I am using unoaded voltage because my load will be tiny ~150mA-250mA on a 3A rating.

cap"

Try looking further.

Remember that a common winding has DC continuity. You should be able to distinguish those windings that are entirely isolated from each other that way.

Independent windings way still however be combined to make a centre tapped supply. It's often more convenient for the transformer maker to avoid internal connections. For example an 18-0-18 supply can be made from 2 x 0-18 windings by series connecting them.

Yes.

It's the ratio of the rms voltage to the peak voltage. i.e. 1.414. Don't forget to subtract 1 V typically for the rectifier forward voltage drop though and be aware that the open circuit voltage is just that. It'll drop on load. The drop on load is greater for smaller transformers usually btw ( see 'regulation' on the specs )

About that figure certainly.

Well no.

A regulator only is 'beefy' if it has to dissipate lots of *power*. The voltage isn't the real problem ( although higher voltage regualtors are somewhat rare ).

Op-amps use mere milliamps and a regulator only dissipates ( Vin - Vout ) * I_load. Suppose Vin is 30V and Vout is 15V and I_load is 10mA. The regulator dissipation is 150mW. This is 'nothing'.

You misunderstand.

Once a voltage regulator is in play any further caps aren't there to provide extra 'filtering'. They are there to ensure circuit stability. See 'decoupling'.

I also think you are confusing filtering and regulating.

You mean the transformer is 6.3-0-6.3 ?

You *could* use it to power some op-amp circuitry at alowier than usual voltage. Depends on your performance criteria.

The 'bypass' cap is to do with RF stability. It should be located near to the active circuitry. The 3300 uF cap does the real work of 'filtering' the supply. It's more helpful to think of the 3300uF cap as a *storage* cap btw. I feel that 'filtering' is a misuse of the word in this respect.

Graham

(snip)

That's ~35 VDC out, from the most positive point ot the most negative point.

I should add, that each of the two storage capacitors connects between one output from the bridge rectifier and the center tap, to make two, stacked DC supplies.

Probably. But you should verify that the voltage from 1 to 3 is about

36 volts. It is possible (but not very likely) that there would be two windings with the same voltage but one of them is phased so that the total voltage between 1 and 3 adds up to approximately zero.

You would get 12.6 *1.414 - 1.2 = 16.6 volts between the positive supply output (+8.3 volts) and the negative supply output (-8.3 volts). The center tap just gives you a middle voltage between them to call zero, so the ends can be called a positive and a negative supply.

This is what you need, and each regulator will see half of the 25 *

1.414 = 35 or about 17 or 18 volts between one end of that 35 volts and the middle voltage of the center tap.

--
Like this, in Courier:


                   12.6VRMS
                  /
                 +--[CR1>]-----+---+--------> +17VDC
                 |             |   |
              +--+--[ -17VDC
           ||(So            |  |   |+    |
120AC>---+ ||(E             |  |  [C1]  [C1]
         P)||(C             |  |   |     |+
         R)|| +-CT----------|--|---+-----+--> GND/0V
         I)||(So            |  |          
120AC>---+ ||(E             |  |        
           ||(C             |  |
              +--+--[]-----+
                  \\
                   12.6VRMS

way.

internal

windings

Okay, I'll take a closer look... I know that there are two windings. I know that one of them has four leads. If I can identify three of those leads, call them leads (1-2-3) and two combinations give the same voltage, I can combine for a center-tapped winding? For example, if 1-2 and 2-3 each give

18VAC, that's what I'm looking for as a usable combination? And the common lead (2 in this case) would be my center tap, no?

be

of my

voltage.

Another light is flickering in the back of my brain. I am reading what you wrote, above, and I am thinking that if I use the center tap of a transformer, I would be effectively cutting the rated output voltage in half? In my example (12.6VDC CT, 6.3-0-6.3) I thought that if I used the outer leads to power my rectifier, I'd be getting an effective 12.6VAC input to the rectifier. Is this not the case?

My outputs from the regulator (~18VDC), then use the center tap as their ground. Does this cut the voltage in half? As you can tell I'm unclear on exactly what I need to get my +/-15-18VDC out.

Here's what I want:

120VAC --> transformer --> 12VAC --> rectifier --> 18VDC --> 2 sets of caps --> two 15-18VDC power supplies, one pos, one neg which use the CT lead for ground (common).

Do I need to start with a 12.6-0-12.6 transformer? This puts 25VAC into my rectifier, does it not? That's >35VAC out! It will cook any 30V regulator UNLESS by using the center tap for gorund you are cutting the voltage in half.

Geez, just when I was starting to think I was getting a handle on this...

Thanks again for all of your help.