Full wave rectifier: AC RMS voltage vs DC voltage obtained

[Labaris]

Hello,

I'd like to know how to choose the right transformer for building a power supply.

DC value obtained from a full-wave rectifier is near 1.4*AC, isn't it?

So if I want 12V DC output, a 2x6V AC transformer (like this) is ok? Or what other things I must consider?

Thanks!

[MartyMart]

Correct, that's a very powerful transformer, 50VA and over 4 amps max per secondary winding !!

What are you going to power with this ?

Do you just need a single +12v dc output rail ?

MM

[Labaris]

Correct, that's a very powerful transformer, 50VA and over 4 amps max per secondary winding !!

What are you going to power with this ?

Do you just need a single +12v dc output rail ?

MM

Yes, 12V DC output (maybe 4 regulated independent outputs)

This PS is for pedal use. The goal with it is to never build another one again, so it has to be "big" enough.

[merlinb]

Yes, 12V DC output (maybe 4 regulated independent outputs)
This PS is for pedal use. The goal with it is to never build another one again, so it has to be "big" enough.

If you never want to build anther then you'll want floating outputs to kill ground loops (if they share the same ground then they're not independent outputs!). That will require one transformer winding for each and every regulator/output. Since each only has to supply one or maybe two pedals, each transformer won't need to be particularly high power. For example, a pair of dual-secondary 0-12, 0-12V 12VA transformers would eventually result in four 300mA DC outputs. If you go for 20VA transformers then you can ultimately get 500mADC from each output, which is plenty even for the hungriest digital multi-effects!

[Labaris]

Yes, 12V DC output (maybe 4 regulated independent outputs)
This PS is for pedal use. The goal with it is to never build another one again, so it has to be "big" enough.

If you never want to build anther then you'll want floating outputs to kill ground loops (if they share the same ground then they're not independent outputs!). That will require one transformer winding for each and every regulator/output. Since each only has to supply one or maybe two pedals, each transformer won't need to be particularly high power. For example, a pair of dual-secondary 0-12, 0-12V 12VA transformers would eventually result in four 300mA DC outputs. If you go for 20VA transformers then you can ultimately get 500mADC from each output, which is plenty even for the hungriest digital multi-effects!

Thanks for the tips! Separated transformers seem to be a great idea.

Anyway, I'm not sure about the voltage of the secondary windings. I want 12V DC output, so it seems I have to use a bigger transformer (in voltage).
The math:

Vs: voltage of sec. windings (RMS)
Vdc: DC output voltage

Vdc ~ 0.9*Vs - 1.4 [V]

Those 1.4V would be the loss of the two conducting diodes each cycle.

So, if I want Vdc=12V:

Vs ~ (Vdc + 1.4)/0.9
Vs ~ 13.4/0.9 = 14.9 [V]

Therefore, I need 15V AC secondaries...

Am I right?

[defaced]

Vdc ~ 0.9*Vs - 1.4 [V]

Should be 1.4 times for full wave rectification, 0.9 is for half wave.  So 15*1.4 - 1.4 = 19.6.  Too high.  You'd actually want a 10 VAC secondary to get close non-regulated.  If you want regulated, you then have to account for the regulator dropout voltage which is based on what regulator you use.  I'd say plan on 12VAC secondary for a regulated circuit, but check the numbers first. 

[R.G.]

Read "Power supplies basics" at GEOFEX.

[Labaris]

Vdc ~ 0.9*Vs - 1.4 [V]

Should be 1.4 times for full wave rectification, 0.9 is for half wave.  So 15*1.4 - 1.4 = 19.6.  Too high.  You'd actually want a 10 VAC secondary to get close non-regulated.  If you want regulated, you then have to account for the regulator dropout voltage which is based on what regulator you use.  I'd say plan on 12VAC secondary for a regulated circuit, but check the numbers first.  

That's what I thought first, as I remember from electronic classes at univ. But I read something different on a webpage. Well, maybe I missed something.
Thanks!

Read "Power supplies basics" at GEOFEX.

checking! thanks

EDIT: Not checking, 'cause I can't find it 

[merlinb]

The 1.4 multiplier only works for off-load calculations. Under load there are losses in the transformer, losses across the diodes and losses due to the finite reservoir capacitance. In practice you generally get a DC voltage that is about 1.2 to 1.3 times the AC voltage. The DC current is typically limited to about 0.6 to 0.7 times the AC rated current.

Therefore, to get 12Vdc you can expect to use a 12/1.2 = 10Vac transformer. 12Vac is very common standard, and would allow you to get away with a smaller reservoir.
If the transformer winding is rated for 12Vac 10VA, then that's 10/12 = 0.83A ripple current, which translates to about 0.83*0.6 = 0.5A of available DC current (continuous).

[panterafanatic]

http://www.geofex.com/Article_Folders/Power-supplies/powersup.htm

First google link that wasn't sponsored. Always add Geofex to the add of a google search if you want to find something on it, simplifies it so much.

[Labaris]

http://www.geofex.com/Article_Folders/Power-supplies/powersup.htm

First google link that wasn't sponsored. Always add Geofex to the add of a google search if you want to find something on it, simplifies it so much.

Great, thanks!
I just tried by searching on the webpage...

[PRR]

> it has to be "big" enough.

But 48VA? Put a 48 Watt heater (an incandescent lamp) inside a pedal housing. It will run nearly 100 degrees C warmer than the room; it will boil spit.

For a low-volt regulated supply the "AC RMS" must (rule-of-thumb) be at least as large as the desired DC V. 12VAC times 1.414 is 17V DC, minus 2 diode drops (FWB) is 16V DC, maybe a Volt of ripple is 15V DC, which leaves 3V for regulator overhead. The 780x series needs most of 3V. So it "works", as long as you have ample capacitance and your utility company voltage does NOT sag. If it drops even 1%, the ripple-dips pass through the regulator which leaves large buzz on the "clean DC. If you don't want to do it again, I would start from 15VAC or even 18VAC. This also leaves the option to make 15VDC or 17V-18VDC for some odd uses.

You can often "cheat" without problem. Diode drop is often less than 1V each. 20VA transformers sag; if you only pull maybe 2 Watts DC then a "12VAC" winding may un-sag to 14V-15V, giving 20V DC to work with and a fine 12V regulated output.

> 1.4 times for full wave rectification, 0.9 is for half wave.

No.

Cap-input half-wave will approach 1.414 times Sine RMS, but you need more iron and capacitor.

The "0.9" factor is for CHOKE-input filters, which are very rare in audio. (Also for resistance-input when UN-loaded, a fairly useless mode.)

[Labaris]

Very interesting, thanks for the info/advices/tips

[defaced]

> 1.4 times for full wave rectification, 0.9 is for half wave.

No.

Cap-input half-wave will approach 1.414 times Sine RMS, but you need more iron and capacitor.

The "0.9" factor is for CHOKE-input filters, which are very rare in audio. (Also for resistance-input when UN-loaded, a fairly useless mode.)

Hey now, my statement was only half wrong.(pun intended)  But that does make sense now that I think it all the way through (regarding half wave rectifiers)

You can often "cheat" without problem. Diode drop is often less than 1V each. 20VA transformers sag; if you only pull maybe 2 Watts DC then a "12VAC" winding may un-sag to 14V-15V, giving 20V DC to work with and a fine 12V regulated output.

And add to that low dropout regulators.  For example, 0.5v before dropout and not too too much more expensive than standard regulators. http://www.fairchildsemi.com/ds/KA%2FKA78R12C.pdf

[merlinb]

I was guessing that Labaris wanted 12Vdc for the very purpose of feeding to a 9V regulator...

[R.G.]

That's a good guess, Merlin. At this point we don't know what he wanted other than 12v for four pedals. A reasonable guess would be for a local 9V regulator.

@Labaris: Unless you already have all the parts and pieces, and have experience with AC power wiring to do it safely, you might want to consider that there is an alternate path. There exist small plug in power adapters which will power over fifty normal pedals (or more) continuously with regulated 9Vdc power. The company I work for makes one (and I'm deliberately leaving the name and company out, as this is not intended as advertising, only technical options) as do other companies. These are on the order of $20 here in the USA. Import taxes an duties, shipping, etc. may make them more expensive other places, but not hugely so. They work on 100Vac to 240VAC 50 or 60Hz (at least one of them does).

It's a perfectly valid thing to do to make your own power supply, but there are options. If you want to make your own, great. If you think you have to make your own, there may be other options to consider.

If you are using linear, not switching type, regulators, it is best if the unregulated voltage you feed them to work from is as little above the regulated voltage as the regulators need to operate. This is because linear regulators have an input current that is a little larger than the output current. If one of them puts out 500ma, and uses 5ma for its internal operation, then it needs 505ma from the unregulated voltage feeding it. The power dissipated in the regulator is then the difference between the input and output voltages times the DC load current (approximately). Feed a 9V regulator supplying 100ma from an 11V supply, and it dissipates (11-9)*0.1A = 200mW. Feed the same regulator from 20V and it puts out (20-9)*0.1 = 1.1W. If the output current goes up to 600ma with more pedals added, then the power it dissipates goes up to (20-9)*0.6 = 6.6W and it definitely needs a heat sink now.

If where you're trying to get is 12Vdc regulated, then you need 12V plus the minimum voltage across the regulators (call it 2V) plus the amount the capacitor filters sag between pulses, plus two diode drops (in a full wave bridge) as a minimum when the AC voltage is sagging as low as you think it will; usually this is 10%. Doing the math:

12V(output) + 2V(regulator overhead) + 1V (capacitor ripple) + 1.4V (diode drops) = 16.4V minimum needed as raw DC.

The AC needed to make that is 16.4V/1.414 = 11.6Vac rms. That is voltage actually out of the transformer and into the rectifiers, and therefore this must include transformer sag.  The amount the transformer voltage sags under load gets worse as the transformer gets smaller. For fist-sized transformers it's often 5%-8%, and for smaller transformers it can run up to 15%. For a 48VA, call it 10% as a guess. This is also what the voltage drops at full load. You can guess that the sag will be proportional to the % load, and bear in mind that a full wave rectifier looks to the transformer as about 1.8 times the DC average current out because it's supplying big peaks of current. It gets complicated.

So let's just guess that the transformer will sag 5%. That's not going to be too far off, and we can correct it later if the numbers come out too critical. So instead of 11.6Vrms, we need 11.6/0.95 = 12.2Vac. This is at our guess of a 10% low AC line. So a "normal AC line voltage" rating for the transformer would be 13.42Vac. If the transformer sags 10% instead of 5%, we'd need 14Vac.

That's the minimum. How high you can go is limited by the regulator's ability to dissipate heat. If you are using a 7812 regulator, it can pass up to 1A, and is internally thermally protected. What that last sentence means is that it will shut down and provide no output voltage if you try to make it provide even slightly more current than it thinks it should, and also will shut down at an internal temperature which it thinks is safely below it's "meltdown" temperature of 150C inside.

The thermal resistance of the TO220 package to ambient air is about 50C/W; its internal temperature goes up 50C above the ambient air for each watt it dissipates. The makers do not tell us what the internal shutdown temperature is, but it's safely below 150C; many engineers guess at 130C.  So in 30C ambient air, the internal chip will be at 130C, 100C higher than ambient, when the power dissipated in the chip is 2W.

How much current you can get out of a linear regulator therefore depends on its peak current rating and also on its ability to get heat out of it without burning up or shutting down.

If you use the TO220 three terminal regulators, you can only get about 2W out of them if you don't put them on a heat sink. If you put it on a heat sink, the thermal resistance to a heat sink is about 4.5C/W, so even if you can somehow keep the heat sink down at 30C (which is impossible), the regulator could only dissipate (130C-30C)/4.5 = 22W. More realistic is perhaps 10W -15W because the heat sink itself will rise in temperature.

The reason I blather on about this is that heat dissipation rather than current rating is usually the limiting factor on linear regulators. The heat dissipated in a linear regulator goes up directly with the difference between input/unregulated voltage and the regulated output voltage, this multiplied by the load current.

And what voltage you feed it determines whether it will always be in regulation or drop out because the input voltage is too small, and also whether it will work continuously or shut down because of overheating.

[Labaris]

I was guessing that Labaris wanted 12Vdc for the very purpose of feeding to a 9V regulator...

I don't understand, sorry (vocabulary problems )

Thanks Mr. Keen for your explanation! It's very clear now

I'll try to explain what I from this project:

First of all, I've already build a 12V/300mA PS (LM317). It works fine with up to 6 pedals (I don't have more ).
So this project is in its "preliminary" stage of development, I just want to learn a lot so I can design it correctly. Someday I'll have lots of pedals to feed (hopefully it won't be the same with kids  ) and I'd like to have a "strong" PS to do that.

All your comments have been very helpful so far. That's exactly what I want with this post (as always): learn


Thanks again!


A possible resume could be:

Voltage: 12V DC
Current: 1A (2A maybe?)
Transformer: Picked to fit regulator's needs (minimum input-output voltages relation)
Regulator: choose it carefully! (and use heat sink)