What does 25+25V mean?

[mossen]

Hi,

On this transformer page

Transformers and Power Converters From Avel Lindberg, Inc.

the secondary voltage is listed as something like 25+25. What does that mean? I want something that gives me +25, 0, -25V. Will this do it?

Thanks
mossen

 

[amc184]

Two separate secondary windings of 25VAC each. You can use one or two rectifiers to make a bipolar supply out of this, so yes, it will suit you.

However, if you're wanting a supply that provides +/- 25VDC, you'll want to choose a transformer with lower voltage secondaries than 25VAC.

 

[preiter]

To expand upon that answer. When you rectify a 25VAC input, you will end up with a 25 * sqrt(2) VDC output, which is about 35V.

For a 25VDC output, you are going to want closer to a 18+18 VAC transformer.

Whether your DC outputs are positive or negative depends on how you wire the power supply. The transformer is the same either way.

 

[nazaroo]

To expand upon that answer. When you rectify a 25VAC input, you will end up with a 25 * sqrt(2) VDC output, which is about 35V.

For a 25VDC output, you are going to want closer to a 18+18 VAC transformer.

Whether your DC outputs are positive or negative depends on how you wire the power supply. The transformer is the same either way.

You should also factor in a few extra volts on each rail if you want a REGULATED supply, because regulators need the extra volts to properly work.

The basic design would look like this:

An externally hosted image should be here but it was not working when we last tested it.

The 2200 uF here seems high (typical examples from regulator makers suggest .22 at input and .1 uF at output of chip).

Anyway, the top is a 7800 series (7812 = 12v +, 7809 = 9v +)
and the bottom should be a 7900 series (7912 = 12v -ve etc.)


The raw rectified DC in should be at least 2 volts higher than the regulated DC out you want, so a 12 volt plus and minus means about 15-20 volts going in.

Here's an example +- 15 volt supply:

An externally hosted image should be here but it was not working when we last tested it.

You need to get about 18 v from (c.t.) Ground on each input UNDER LOAD.

use Ohm's law to figure out what your load is.

If you know you need a 100 mA supply on each rail,
then your load will be Ohms = Volts / Amps = 15v / .1 = 150 ohm load.
Then calculate watts:
Watts = Volts X Amps = 15 x .1 = 1.5 watts.

Now hook up a 2 watt or larger 150 ohm resistor to the raw supply, and measure the D.C. voltage, under that load.
This will tell you what the transformer can put out at that current draw (do both sides like this at once, to load both sides of your supply to ground.

An externally hosted image should be here but it was not working when we last tested it.

In the above picture you'd hook up a resistor to both plus and minus supplies, and see what the voltage across each was.

For example, I needed a 9 volt supply (single sided) for my guitar pedal. I ended up using a 12 volt A.C. transformer, which actually delivered much more without a load. I ignored those voltages and hooked it up to a rectifier bridge, added a PS cap, and then put a 1 Amp load on it. I found it put out indeed about 13 volts D.C. under load, which was fine, so I added a 7909 regulator with a heatsink and some small caps, and got a pedal powersupply that delivered almost exactly 9 volts, and could power about 10 pedals at the same time, of various current draws.

 

[AndrewT]

Have you spotted the "message" yet?

You must avoid ambiguity between DC and AC and average and peak and peak to peak.
You must specify the units unambiguously.

If the +25Vdc, 0Vdc, -25Vdc had been seen you would not have to ask the question.
Similarly +-25Vdc is quite different from 25,0,25Vac

 

[nazaroo]

Have you spotted the "message" yet?

You must avoid ambiguity between DC and AC and average and peak and peak to peak.
You must specify the units unambiguously.

If the +25Vdc, 0Vdc, -25Vdc had been seen you would not have to ask the question.
Similarly +-25Vdc is quite different from 25,0,25Vac

Its a common standard to describe center-tap transformers as "300-0-300" or alternately as "600v c.t.", in each case meaning the same thing:

300 AC RMS from c.t. to either endpoint and/or 600 v AC RMS from the extreme ends of the secondary winding.

Sometimes this winding is really two 300 v.a.c. rms simply wired in series with a 3rd wire as the c.t., joined to the point where the two coils are connected internally.

It could never mean anything else, such as "D.C." or "Peak/p2p" and the RMS (root mean square) is because of convention, as it conveniently expresses the same amount of power that a D.C. voltage would, and so 10 volts RMS A.C. = 10 volts D.C..

Toroidal transformers always put out A.C. only (only a complete powersupply design could specify a D.C. voltage).
So for a transformer "25 + 25" means two output windings (not connected to each other) each which puts out 25 v A.C. RMS. hooking them end to end in proper phase would give 50 volts RMS A.C.

This A.C. (25 - 0 - 25 series c.t.) could produce either a 25 volt or a 50 volt supply depending upon how it is rectified and filtered. The actual D.C. voltage on the load would vary according to load and/or further regulating components.

In the case of a transformer, the "+" sign doesn't mean polarity or that a D.C. voltage is being measured.
Instead it simply means "and", that is "25 + 25" means 25 AND 25, that is two 25 volt windings (AC).

 

[AndrewT]

Nar,
if you want to be ambiguous just carry om omitting AC or DC from your stated units of measurement.

If you want to help beginners avoid being confused, then I strongly suggest you change your opinion in how to state voltages.

 

[nazaroo]

Nar,
if you want to be ambiguous just carry om omitting AC or DC from your stated units of measurement.

If you want to help beginners avoid being confused, then I strongly suggest you change your opinion in how to state voltages.

You're right of course Andrew.
I was merely trying to explain the practice of how the transformer seller
chose to list his transformers. (see OP)
Really the seller could have done better here,
but the convention used is very common.

 

[mossen]

One more question

Thanks for all the answers, everyone, it is making sense. About the power rating of the transformer: if it is listed as 160VA, how is that interpreted?

For example, if the transformer provides 18+18V, does this mean the transformer can deliver:

160VA / 36V = 4.44 amps?

or is the rating per secondary:

160VA / 18V = 8.88 amps?

or do I completely have the wrong idea here?

Thanks
mossen

 

[nazaroo]

Thanks for all the answers, everyone, it is making sense. About the power rating of the transformer: if it is listed as 160VA, how is that interpreted?

For example, if the transformer provides 18+18V, does this mean the transformer can deliver:

160VA / 36V = 4.44 amps?

or is the rating per secondary:

160VA / 18V = 8.88 amps?

or do I completely have the wrong idea here?

Thanks
mossen

I think the rated VA of a transformer is the total power drawn through it, when all secondaries (outputs) are loaded at their ratings.

Thus the total wattage (VA) is the MAINS current x MAINS voltage.

It might be possible (within sensible tolerances) to draw some more current through one winding,
while leaving another unused, keeping the total power within the VA specs.

Wire in the coils is rated by amperage only,
while insulation is rated by voltage.

A small increase in amperage may be well within specs, provided there is no heat buildup or breakdown of HV insulation, or any possibility of transient HV spikes (these are the typical cause of insulation breakdown.)

 

[AndrewT]

Generally we find that the VA rating for a transformer is arrived at using internal temperature as the critical limit.

If the 160VA transformer is an 18+18Vac then the maximum continuous output current is 160 / [18+18] = 4.44Aac. Note AC current rating. This can only happen when the transformer is driving a resistor load.

That 4.44Aac applies to both secondaries individually, unless the specification tells you otherwise.

Some transformers are rated by regulation, rather than temperature. Here one specifies the VA and the regulation, but again that regulation and VA apply only when the load is a resistor.

The input VA will always be higher than the output VA. This is due to the fact that all transformers must have an efficiency <100%.

Assuming the transformer is 95% efficient at maximum power then the input VA will be 160/0.95 = 168.4VA.
The input current when the rated input voltage is 230Vac will be 732mAac