Linear regulated DC supply transformer current rating?

[HighFidelityGuy]

Just a quick question about how the current output of the transformer relates to the potential current output from the linear regulator....

Lets say I have one of THESE (https://uk.rs-online.com/web/p/products/5339347/) 15V 5A linear regulators and I want to ensure that I can get the full voltage and current range from it. So I could perhaps use a toroid with 2x 9V secondaries to give me 18V before rectification and about 23.45V after rectification. I calculated this using 18 x 1.414 - 2 to take into account a full bridge rectifier. I think that's the right way to do it but correct me if I'm wrong. Anyway, that gives me plenty of headroom above the minimum 17V the regulator will probably need to make 15V.

So I've fulfilled the requirement of being able to get full voltage out of the regulator. Now I need to know how to ensure I can get the full 5A. So will I need to choose a transformer that can produce at least 5A on it's secondaries? Or is there an equation for working this out?

Cheers. :cool:

[Welder]

I think something’s gone a bit wrong here Dave but I dare say one of the electronics chaps will put it right.

If you have 18volt from your secondary windings and a given current then doesn’t the full bridge rectifier pass the same current as near as, but give a voltage drop, 0.7 volt per diode pair, I think.
Ohms Law V=IxR
Vdc=Vac=2xpeak Voltage divided by 1.141
Then you have any smoothing caps and they will also give a voltage drop and finally the regulator. If anything don’t you get a very small current increase for each voltage drop in the circuit?
So 18volt at secondary won’t give the 23.45 after rectification surely? Won’t it be more like 15. or16. something volts depending on what type of rectification you use schottky diode or a full wave bridge type?
So as long as the transformer gives the same or more current than required you should get the same or slightly greater current after rectification and smoothing which the regulator will sort out to a given value.

[HighFidelityGuy]

Thanks John, I think you're partially right. My calculations were a bit wrong as they were based on cobbled together bits of info. I've since found THIS (http://www.freewebs.com/valvewizard/fullwave.html) website which explains things better. Here's a quote on the voltage side of things:

When not under load, the voltage after rectification will be close to the peak AC voltage, which is equal to the 1.4 times (root 2) the RMS voltage.
For example, with a transformer rated at 300-0-300Vrms, the DC voltage after rectification will be close to:
1.4 * 300 = 420Vdc
(There will be a couple of volts lost across the diodes, but we normally ignore this.)
We can expect this DC voltage to fall by 10% to 15% when we start drawing current, due to voltage being dropped across the transformer winding. So, we can realistically expect to achieve about 365Vdc of HT under load using this transformer.

They simplified the RMS voltage to 1.4V instead of using 1.414V which is fair enough.

So I was almost right but the voltage drop across the bridge rectifier will be dependant on the diode type. For schottky diodes it will be less than 2V, so in that instance I would actually get more than 23.45V. I don't think I need to take transformer regulation into account like they did in the above example as I'm basing my calculations on rated output voltage rather than un-loaded output. I don't believe the smoothing cap causes a voltage drop as it's connected in parallel. If anything it might increase the DC voltage as it's removing ripple. :scratch: That's just a guess though. :confused:

Anyway, this is all about voltage and it's the current I'm interested in right now. I've read through THIS (http://valvewizard2.webs.com/Transformers.pdf) PDF and it seems to suggest that the transformers current rating will need to be higher than the current demand of any load connected after the rectifier (page 13). Unfortunately there doesn't seem to be anything specific like an equation mentioned about this, so I'm still unsure just how much more current is required from the transformer. :confused: I'll do some more reading tomorrow, it's time for my beauty sleep now. I need it. ;)

[lurcher]

http://www.duncanamps.com/psud2/index.html

[Reid Malenfant]

John you are slightly wrong chap. Assuming 18V AC the peak voltage will be 1.414 x 18 = 25.42V DC. As Dave says there will be some voltage drop accross the bridge rectifier.

However as the transformer will only be conducting near voltage peaks & at about 25V you'd have to agree that power can't just be created from nothing :eyebrows:

What you need to do to balance things up is divide the transformer secondary current by 1.414 so you keep the VA at the transformer rating.

That'd be 3.53A DC. To get the full current from the regulator would require a transformer that can supply 18 (or two x 9V transformers) @ 7.07A :cool:

[Welder]

I wrote it didn’t look right when I posted ;)

I realised later I had used root mean square rather than peak voltage and completely ignored the fact that any smoothing caps would push the voltage after rectification back towards the peak voltage.

Good job I use a power supply calculator :doh:

[HighFidelityGuy]

John you are slightly wrong chap. Assuming 18V AC the peak voltage will be 1.414 x 18 = 25.42V DC. As Dave says there will be some voltage drop accross the bridge rectifier.

However as the transformer will only be conducting near voltage peaks & at about 25V you'd have to agree that power can't just be created from nothing :eyebrows:

What you need to do to balance things up is divide the transformer secondary current by 1.414 so you keep the VA at the transformer rating.

That'd be 3.53A DC. To get the full current from the regulator would require a transformer that can supply 18 (or two x 9V transformers) @ 7.07A :cool:

Awesome, thanks Mark! That clears that up. :cool:

There's just a couple of other things I'm a bit unsure about now. If I get transformer with 2x 9V secondaries, should I wire them in parallel to give me 18V, or as 9-0-9 connecting the centre tap to ground and the two 9V leads to the rectifier? I think you can do both but I think the 9-0-9 method is supposed to reject noise. Have I got that right? :scratch: Is there any difference between those methods in the amount of current the secondaries can deliver? :confused: Cheers. :cool:

[Reid Malenfant]

Awesome, thanks Mark! That clears that up. :cool:

There's just a couple of other things I'm a bit unsure about now. If I get transformer with 2x 9V secondaries, should I wire them in parallel to give me 18V, or as 9-0-9 connecting the centre tap to ground and the two 9V leads to the rectifier? I think you can do both but I think the 9-0-9 method is supposed to reject noise. Have I got that right? :scratch: Is there any difference between those methods in the amount of current the secondaries can deliver? :confused: Cheers. :cool:
If you wire the 9V secondaries in parallel then you'll still have 9V but at double the current ;)

If you happen to get a 9 0 9 V transformer simply connect to the 9V connections only & you'll have 18V AC.

As for the noise thing, no, not unless you are contemplating metres of cable between the transformer & rectifier... This isn't the same thing as balanced mains to reduce common mode interference ;)

[HighFidelityGuy]

If you wire the 9V secondaries in parallel then you'll still have 9V but at double the current ;)

If you happen to get a 9 0 9 V transformer simply connect to the 9V connections only & you'll have 18V AC.

As for the noise thing, no, not unless you are contemplating metres of cable between the transformer & rectifier... This isn't the same thing as balanced mains to reduce common mode interference ;)

Thanks Mark. I got my parallel and series mixed up. School boy error. :doh:
So it seems to me that the best way to get 18V and plenty of current is to get a 2x 18V transformer and wire it in parallel. I'll have a rummage on the RS website.

Cheers. :cool:

[HighFidelityGuy]

It looks like a 2x 18V 160VA jobbie will be the closest as that will give me 8.89A when wired in parallel. THIS (http://uk.rs-online.com/web/p/toroid-transformers/2575168/) is the cheapest one that RS have in stock. Is there any useful reason for me to spend an extra £10 to get THIS (http://uk.rs-online.com/web/p/toroid-transformers/2238774/) encapsulated version? :confused:

EDIT: I forgot to ask; can anyone recommend a good Schottky diode suitable for making a full wave bridge rectifier for this project? Cheers. :)
EDIT2: I have some spare 1N5822RLG (http://uk.rs-online.com/web/p/products/654-7139/) diodes, would they be suitable?

[chrism]

It looks like a 2x 18V 160VA jobbie will be the closest as that will give me 8.89A when wired in parallel. THIS (http://uk.rs-online.com/web/p/toroid-transformers/2575168/) is the cheapest one that RS have in stock. Is there any useful reason for me to spend an extra £10 to get THIS (http://uk.rs-online.com/web/p/toroid-transformers/2238774/) encapsulated version? :confused:

EDIT: I forgot to ask; can anyone recommend a good Schottky diode suitable for making a full wave bridge rectifier for this project? Cheers. :)
EDIT2: I have some spare 1N5822RLG (http://uk.rs-online.com/web/p/products/654-7139/) diodes, would they be suitable?

Speak to Les at Avondale he has some Avel Lindberg encapsulated which do 2 x 17.5v @ 1.0A so you could get 2A. I have used them in my Caiman PSU projects and very impressed. They are quite cheap as well.

Les also sells a full wave bridge using Schottky diodes well worth looking at.

Sorry just noticed that you need 5A what are you powering up?



Regards

Chris

[Paul Hynes]

Hi Dave,

The diode rating needs consideration for reliable operation. The 1N5822RLG current and voltage ratings are not high enough for your application.

Running axial diodes at their maximum current rating can cause operating temperatures exceeding 100 degrees C depending on the diode construction, the forward voltage drop at the maximum continuous operating current and the chosen rectifier configuration. If you are planning on using the power supply at the full regulator output current of 5 amps it would be sensible to choose a diode package that can be bolted to the chassis or to a finned heat-sink to aid cool running.

Choose diode current ratings at least twice the maximum continuous load current drawn by the equipment. If you do not know what this maximum equipment load current is and/or you are likely to use the regulator at full current capability choose a diode with a rating twice the current delivery the regulator can provide. In this instance 10 amps.

For rectifying 18 vac I would also increase the voltage rating to at least 80 volts or preferably 100 volts to allow plenty of headroom for peak reverse voltages and for mains voltage surges.

Sorry to push your price expectations on these components but it is better to be safe than sorry with power supplies as quite a lot of damage can occur to the equipment being powered if a power supply component fails.

Regards
Paul

[HighFidelityGuy]

Hi Dave,

The diode rating needs consideration for reliable operation. The 1N5822RLG current and voltage ratings are not high enough for your application.

Running axial diodes at their maximum current rating can cause operating temperatures exceeding 100 degrees C depending on the diode construction, the forward voltage drop at the maximum continuous operating current and the chosen rectifier configuration. If you are planning on using the power supply at the full regulator output current of 5 amps it would be sensible to choose a diode package that can be bolted to the chassis or to a finned heat-sink to aid cool running.

Choose diode current ratings at least twice the maximum continuous load current drawn by the equipment. If you do not know what this maximum equipment load current is and/or you are likely to use the regulator at full current capability choose a diode with a rating twice the current delivery the regulator can provide. In this instance 10 amps.

For rectifying 18 vac I would also increase the voltage rating to at least 80 volts or preferably 100 volts to allow plenty of headroom for peak reverse voltages and for mains voltage surges.

Sorry to push your price expectations on these components but it is better to be safe than sorry with power supplies as quite a lot of damage can occur to the equipment being powered if a power supply component fails.

Regards
Paul

Hi Paul. Many thanks for your input! I couldn't agree more on the safety side of things, so I'll have a better look round for a much more chunky bridge.

Thanks again. :cool:

[Reid Malenfant]

I don't think there is too much point in going for the encapsulated version in all honesty. They are generally used in medical equipment (as a safety requirement), though if you want the last word in lack of transformer noise then it might be worth considering :scratch:

Personally i'd go with the standard transformer.

By the way, i think i have a 160VA 0-18 0-18V toroidal here doing nothing... Can i interest you in it - cheap :eyebrows: If i remember correctly it's on some form of PCB along with rectification & smoothing caps etc, mains input as well if i'm not mistaken...

I think it's wired for 36V out though...

Let me know & if you are interested i'll make sure it's a split secondary (99.9% sure) & test it etc..

:cool: