# +/- 18VDC PS basic questions

Discussion in 'Electronic Basics' started by Dave, Oct 28, 2005.

1. ### DaveGuest

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?

2. 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.

2. ### Rich GriseGuest

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

3. ### DaveGuest

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?
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?

4. ### DBLEXPOSUREGuest

If you can't look at the schematic you can use an Ohm Meter. The center tap
will have a resistance to each of the other two ends. This would not be
true of multiple secondary windings.

Center tapped xformers have multiple applications. A full wave bridge
rectifier requires a center tapped xformer. I believe old FM discrimators
use center tapped xformers. Bipolar Power supplies is another example.

Yours sounds more like either multiple taps or multiple windings. My guess
is multiple taps.

5. ### DaveGuest

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.

6. ### John FieldsGuest

---
No, it doesn't, unless it being used in a bipolar supply.
---
---
No, but if two of the windings were identical they could be
connected to become cetnter-tapped.
---
---
Not necessarily.
---
---
Yes, there's a very good way to tell. It involves using an ohmmeter
to determine which windings are connected to each other, which
aren't, and what the resistances of the windings are.

If you want to do that and post back with what you find, we'll be in
---

7. ### John FieldsGuest

---
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.

9. ### BrianGuest

Check out http://www.fncwired.com/XformerExamples/ it should make things a
little more clear.
Brian

10. ### Pooh BearGuest

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

11. ### Pooh BearGuest

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

12. ### Pooh BearGuest

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

13. ### DaveGuest

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.
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

14. ### DaveGuest

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
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.

15. ### Pooh BearGuest

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 )

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.
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

16. ### DaveGuest

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?
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.

17. ### John PopelishGuest

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.

18. ### Guest

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.

19. ### John FieldsGuest

---
Like this, in Courier:

12.6VRMS
/
+--[CR1>]-----+---+--------> +17VDC
| | |
+--+--[<CR2]--+--|---|-----+--> -17VDC
||(So | | |+ |
120AC>---+ ||(E | | [C1] [C1]
P)||(C | | | |+
R)|| +-CT----------|--|---+-----+--> GND/0V
I)||(So | |
120AC>---+ ||(E | |
||(C | |
+--+--[<CR3]--+ |
| |
+--[CR4>]-----+
\
12.6VRMS

20. ### cale

1
0
Oct 28, 2009
.....

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.

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.
>
> 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.
>
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.