Transformer with central tap output waveform?

[carbon dude oxide]

Hello, if i have a transformer that transforms 230v rms to 35-0-35 what would its output waveform be? I have drawn 2 up which would allow for the center tap to always be 0 but i am unsure thats all, i am leaning towards the top graph as it seems more logical

[David_AVD]

Top one.

[carbon dude oxide]

Ok thanks

So to get both of these to be dc i would use 2 bridge rectifyers except one would be in reverse? To its + would be the 0v and - would be -35

[David_AVD]

A single bridge connected to the outer wires is what you need.  The centre tap is the 0V line and where the filter caps return to as well.

[carbon dude oxide]

So its like the bottom one correct? And not the top

[David_AVD]

I was too lazy to draw it, so here's one off the interwebs.   

EDIT ** Shitty (incorrect) diagram I found on the web removed! **

[carbon dude oxide]

thanks alot

I remember it better if ive drawn it
thanks alot.

So this is actualy a transformer that outputs 70v ac which we then rectify into 70dc volts and then turn it into a split +-35 supply. is that correct? Well technicaly no due to it being rms not peak

[carbon dude oxide]

Oh i didnt see that te picture says split supply XD

[Dave]

The schematic you've drawn on your board is correct, the schematic David_AVD has pasted is absolutely wrong. Look at the orientation of the diodes. That thing would blow up!

[carbon dude oxide]

So it would XD i didnt really look at the orientation as i already knew that i just know they face towards the + and away from the - and both ac inputs have two diodes facing oposite ways.

[David_AVD]

LOL - you're right about that diagram (from the web) I posted.  I totally missed the incorrect direction on the diodes!  That web site has more incorrect ones too.   

Here's a better one:

[carbon dude oxide]

Thank you very much,

Also when you build your own power supply some have the earth pins in the front are these just connected to mains earth or are they isolated from that? If they are isolated where does it go?

[mariush]

The earth goes to earth. Sometimes a large (in the range of megaohms) resistor is used.

With center tap, if you only need positive voltages you can use just two diodes:



If you want two positive voltages you can use this (but generally those making linear power supplies use a relay to switch between taps, they don't use this "tehnique"):



Don't forget the diodes have to handle the current, and they should be sized about 25% more than the maximum current, and about 30% more forward voltage and reverse voltage breakdown values than the voltage you have.
On a bridge rectifier there's two diodes conducting always, so the current gets shared between the two, so you can use diodes rated for less current. On the design in the first picture, each diode pushes full current so they have to be rated for more than the peak current.

Also... don't forget about transformers having their voltage rated as RMS value.  If they're rms values, the peak dc voltage will be aprox 1.41* voltage, so about 1.414x35 = 50v. So if you don't use negative rail, you'll have a peak of about 100v.
Naturally, if your transformer is 35-0-35 Vrms at let's say 150VA , this would mean it can do about 2.1A on each tap (75VA /35v = 2.1A) ... but that's AC. When rectified your 35v rms becomes 50v so your current in DC would actually be about 0.62 x (75va/35v)= 1.32A

Don't forget that at low loads, the transformer will output more than the rated voltage, usually up to 10-15% more depending on its VA rating.So your 35v could be 40v rms. This matters when choosing the diodes/bridge and the storage capacitor (which should have a voltage rating about 20% higher than the peak voltage)

[C]

mariush, Would it not be better for the beginners if you added
" but the caps charge for a short time so PEAK Current is ____"
after this?

Naturally, if your transformer is 35-0-35 Vrms at let's say 150VA , this would mean it can do about 2.1A on each tap (75VA /35v = 2.1A) ... but that's AC. When rectified your 35v rms becomes 50v so your current in DC would actually be about 0.62 x (75va/35v)= 1.32A

C

[mariush]

I thought I posted quite enough technical details in a single post and didn't want to scare him with more details. I just didn't want him get exploding capacitors in his eyes or wonder why diodes smoke out.

I also didn't mention the possibility of using an inductor (choke) on output (and what parameters it should have), which can improve current but affects the output voltage (hence you may need larger capacitors to smooth out the output)...  this touches a bit on this : http://www.qsl.net/i0jx/supply.html

Please do contribute with that bit if you want (formulas etc), maybe explained in simpler terms so that he can understand. I'm a bit busy right now to go into math and all that.

[carbon dude oxide]

i dont mind learning all the techy bits now ive finished school so i ahve plenty of time to read up, and learn things

[C]

mariush, I liked your wording and details, sorry if I caused insult!
Looking at it again I should have just suggested adding "Average" to the currents in that paragraph as you talk about peak current another paragraph.

C
 

[carbon dude oxide]

im curious, i have heard of chokes before and i have heard about them but what do they actually do? i know there used somewhere in rectification but i am usure

[mariush]

See the following:




.. and the following parts.

You don't HAVE to use an inductor at the output, but you do have to use some large capacitor(s) after the bridge rectifier to smooth out the DC voltage.

There are other things like Common mode chokes but some of these things with coils are mostly used for filtering "noise", they're not used as important parts to generate DC voltage.

Inductors are also used in filters to further smooth out the dc voltage after a DC voltage is produced, for example google "pi filter" which should give you some good info about it.

[carbon dude oxide]

Ok this is a quick schematic i drew up which in theory (well in my head anyway) should output a split rail +-1.2v -- +-30v at 1.2A aproximatly.

It uses the transformer circuit earlyer and the rectifyer and i have added in a lm317 and lm337 for the voltage regulation.

Have i drawn this out correctly and if so what value caps would i need to put there (a nice formula for me to learn and use might be helpfull

This uses a 230V to 24-0-24 52VA transformer and i am wondering if it needs some suppresion caps on the ac input.

[David_AVD]

Off the top of my head I'd say that I'd probably use 3300uF 50V for the main filter caps.

[carbon dude oxide]

that's a very beastly cap

is there a general formula that i can use to work out roughly what capacitance i would need for a required load ?

[mariush]

No, you don't need supression caps or varistor/thermistors .....

Depending on the size of those capacitors after the bridge rectifier, you will get a relatively large initial pulse of current, could be up to 5-10 A. So your 200mA better be slow blow/delay fuse or you'll go through a few fuses.

so 24v rms ... after it's rectified it's 33.9v ... but don't forget you have voltage drops on the diodes, about 0.4-1v on each, so take out 0.8-2v ...and you're left with about 31v.  On low loads like I said the transformer will output more than 24v rms but at high load, it's gonna be more accurate, close to like this 24v ac rms.

The capacitor after the bridge rectifier should be rated for about 50v because 31-33v is awfully close to 35v without even considering that 10-15% extra at low loads.
Capacitance wise I would guess about 820-1500 uF to have a low voltage ripple of a few volts .... approximate formula is :

capacitance = current (1.2a) x duty cycle (about 0.7)  / [2 x V ripple (how much you find acceptable to let the voltage sag) x ac frequency (50hz for eu, 60 hz for us) ]

assuming max 33.8v and you don't want it to drop below 30v then  =>  c = (1.2 x 0.7) / (2 *3.8* 60) = 0.84/456 = 0.0018 F  = ~ 1800 uF

^ that's at the peak of 1.2a, if you'll only pull 100-500mA naturally the voltage will not sag down to 30v

so realistically with about 1200-1500uF 50v (cheap capacitors) at 1.2a maximum output, you'll have about 28v before the linear regulators. Double or triple that (put capacitors in parallel or use 3300uF-4700uF 50 capacitor to get as close as possible to 33.8v you want.

Next, the LM317 and LM337 will have a voltage drop on them of about 2-2.5v at 1.2A so you'll get at most the input voltage - that 2-2.5v drop.

You WILL NOT get 1.2v at 1.2A, not even 5v at 1.2A with this regulator. 

The difference in voltage is dissipated by the regulators as heat - ( 30v input voltage - 1.2 output voltage ) x 1.2v = 28.8v x 1.2a = 34.5 watts.  You'd need a monster of a heatsink to dissipate that much (or a cpu cooler with a fan blowing on it to keep it cold).

If you check the lm317 and lm337 datasheets (should be the same) - you'll find that the maximum power dissipation is limited to about 15w for the classic 3 pin to-220 regulator and maybe about 25w for that to-3 (large round metal case) IC.
So if you want 5v with 30v input ... that's 15w / (30-5) = 15/25 = maximum 0.6A  on 5v with the lm317 in to-220 package... and the heatsink will be super hot. 

Even so, you'll need a quite expensive heatsink (with about 1.2-1.5c/w rating or less) if you want this to work without fans, and maybe 2-3c/w heatsink if you'll constantly blow air on it. \

See  http://www.ef-uk.net/data/heatsinking.htm  (the explanations section at the bottom explains how to do the math manually based on data from datasheet of the linear regulator)


Linear regulators are just not designed to work with such big voltage differences, traditionally the voltage differences are at most 5-10v, not up to 30v.  Linear power supplies use multiple taps, like 0-8-16-24-32v and have some circuitry to connect the next winding when needed so that at any point, only about 5-8v are dropped on regulator.

[David_AVD]

is there a general formula that i can use to work out roughly what capacitance i would need for a required load ?

One rule of thumb is 1000uF to 1500uF per Amp of output current required.  This assumes a full-wave or bridge rectifier (100Hz / 120Hz ripple).

[carbon dude oxide]

What kind of circuitry would be needed to select between several boltage taps? Would you use several comparators to determine it?