# Beginner transformer question

Discussion in 'Electronic Basics' started by Ant_Magma, Feb 26, 2006.

1. ### Ant_MagmaGuest

I'm looking for a transformer that is capable of stepping down 240V
from the wall outlet to maybe less than 20V depending on the type of
voltage regulator. My objective is to produce 3.3V from the wall outlet
to power my ICs.

1) Most of the times i see 2 same values for the secondary voltage, eg:
0-9,0-9 (9V). What does it mean? Does it mean i still get 9V output?

2) If i want to calculate the output current, do i take the power
rating lets say 3VA and divide with the output voltage, for example 3VA
/ 9V=0.33A?

2. ### John PopelishGuest

It means there are two secondaries of 9 volts, each. You can connect
them in parallel to act as a single secondary with the full VA rating
of the transformer and rectify them with a 4 diode bridge, or you can
connect them is series to produce 18 volts with the full VA rating,
and also rectify with a 4 diode bridge. Or you can connect them in
series and half wave rectify the each half of the total winding with a
single diode, and parallel the two diode outputs for a full wave 9
volt output. This uses each secondary only half of the time, so the
VA rating has to be derated a bit. But you get only one diode drop in
the rectification process, instead of two. Or you can use a 4 diode
bridge across the seriesed secondaries and have positive and negative
outputs with respect to the center tap node between the two
secondaries. The flexibility gained by having two equal secondaries,
instead of one is the reason this construction is so common.
That works if the transformer will be loaded with a resistor, but you
have to derate this if you are going to be charging a capacitor input
filter at the voltage peaks, only. You may get as little as half of
that (as filtered DC) for the same transformer temperature rise.

3. ### Ant_MagmaGuest

I thought the rectifiers and diodes comes together with the
transformer?

So with the 2 secondary voltages it's up to me to 'mix n match' to get
the desired output correct?

4. ### John PopelishGuest

If so, it is a power supply, not a transformer.
Yes, you have choices.

5. ### Ant_MagmaGuest

Is there reference circuit i can refer to in building a 3.3V power
supply from a 240V AC line voltage?

6. ### John PopelishGuest

How much DC current must this supply deliver?

7. ### Pooh BearGuest

www.powerint.com

8. ### ehsjrGuest

Here you go:

240AC to 12VDC
"Wall Wart"
--------- -------
| + |------+---Vin| LM317 |Vout---+-----+---> +3.3Volts
| | | ------- | |
240| |12 [C1].1uF Adj [220R] |
| | | | | |+
| - |------+ +-----------+ [C2] 1uF
-------- | | |
| [360R] |
| | |
+----------+-----------------+--- Gnd

9. ### Ant_MagmaGuest

Well, i'm guessing around 500mA?

Coz i have 2 main components, 1 is the Ethernet transceiver and d other
is the powerline module. The module states that it requires a current
of 370mA or a max of 530mA to operate.

I was wondering if this transformer will work?
http://www.rsmalaysia.com/cgi-bin/b...tscape&3318566216=3318566216&catoid=-83936776
The code is 3473518. It has a power rating of 3.2VA and a 6V output.
So 3.2/6= 0.533A

Is it correct?

10. ### Ant_MagmaGuest

In my case i should just replace the wall wart with a 240AC to 12V
transformer right?

11. ### John PopelishGuest

Sounds a bit light, to me. When you rectify the AC to DC and filter
that with a big capacitor, the current all bunches up around the
voltage peaks, so to get an average of 500mA, the pulses get up around
a couple amps. It doesn't hurt to have a transformer with extra
capability, and a larger one doesn't cost much more. I would go with
one rated between 1 and 2 amps, so you don't have to worry about it
getting hot. You'll have enough of that at the regulator, that wastes
all the extra voltage.

12. ### ChrisGuest

Hi, Ant. Just out of curiosity, could you answer a few questions?

* Frosh, Soph, Jr., Sr?

* Is this the senior project?

* When's it due?

Thanks
Chris

13. ### Rich GriseGuest

No.

TO QUOTE CONTEXT!!!!!]
uses a DC wall wart. It contains a transformer, rectifier, and filter
capacitor. To use just a transformer in its place, you would have to
provide these components yourself, as John Popelish has described.

And I'm way too lazy to draw that for you, but try googling for
"linear power supply basics" or "ac/dc tutorial" or any phrase with
"basics" or "tutorial" in it.

Good Luck!
Rich