Beginner transformer question

Feb 26, 2006 16 Replies

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?



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?

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.

power is power, if the device can only deliver 100 Watts for example, that is what you get . when you scale down the voltage you simply get higher current at the lower voltage which can still work out to the same max watts. if you know the wattage of the supply source via the label on the unit? then you can calculate the amps you will expect to get . Watts/Volts = Amps. so if lets say your little step down supply is 100 watt handling and you have 9volts, the current should be a little less than 11 amps/. i say a little less is because of other factors that come in when using something like a xformer. you have losses and efficiency factors. that is about as basic as i can get it.

Real Programmers Do things like this. http://webpages.charter.net/jamie_5

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

So with the 2 sec> Ant_Magma wrote:

If so, it is a power supply, not a transformer.

Yes, you have choices.

How much DC current must this supply deliver?

Here you go:

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

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?

formatting link
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?

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

formatting link

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.

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

  • What's your major?

  • Frosh, Soph, Jr., Sr?

  • Is this the senior project?

  • When's it due?

Thanks Chris

No.

ehsjr's suggested circuit [WILL YOU GOOGLEGROUPIES ***PLEASE*** LEARN 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

Is the assignment "make a power supply", or is the assignment "these ICs need 3.3V @ 500 mA"?

If all you need is volts, then go here:

formatting link
[mind the word wrap] and look at: GFP051-A315: 3.3V/1.5A

They don't show prices, but it can't be as much as you'd spend trying to duplicate it.

If the assignment is "learn how to build a power supply", then any of the basics tutorials should help here, or "Art of Electronics", by Horowitz and Hill.

Good Luck! Rich

Join the Discussion

Have something to add? Share your thoughts — no account required.

Didn't find your answer?

Ask the community — no account required