Is this the right formula for transformer VA calculation?

[FriedMule]

After a lot or reading, learning Ohms law and abusing my paper block, I think that I have found the right formulars for calculating a transformers V and A

Calculate how many voltage you get out of a given VA: "V = sqr(VA * R)"
Calculate how many Amps the transformer can deliver: "A = VA / V"

I know that it properly are a lot more complicated then that, but are they a good thumb rule for a beginning beginner to know how many V and A you get out of a transformer with a certain VA?

[Benta]

The first one's wrong. The only thing defining the voltage from a transformer is N (winding ratio).
The second is good for a first estimation.

[Ian.M]

Nope. The first equation is *WTF* ? ? .  The second is valid, but only for pure linear (R, L or C - *NO* semiconductors, not even diodes) AC loads. (Edit: Benta got there before me  )

Neglecting losses, VA in = VA out.   That's quite simple - the Primary Voltage  x the Primary Current = the sum of all ( Secondary Voltage x Secondary Current ) for each secondary.  All voltages and currents are RMS.  In the absence of individual secondary current ratings you can assume that secondaries of all the same voltage split the available maximum VA rating evenly, and that FOR THE SAME WINDING WIRE SIZE, secondaries of unequal voltage split the VA rating in proportion to their voltage.

However as soon as you introduce a rectifier you'll get less current than the secondary's RMS rating as the waveform of the current into the rectifier will have a crest factor higher than unity, which will push up its RMS value with respect to the average DC current out of the rectifier.

Hammond Manufacturing Inc.'s Design Guide for Rectifier Use is a good place to start for the derating factors for various types of rectifier and load.

[FriedMule]

LOL wow I have created a wtf formula, that sounds like I am not spot on:-)

Why did I arrive to that totally wrong formula: V = sqr(VA * R)
I took some examples on the internet like "300VA did that delivered 50V and 6 amp, like my formula sayes:  sqr(300VA * 8Ohm) = 49V and 6 amp, an other was a 100VA that did deliver 28V and 3.5 amp.

My problem are that 100VA could deliver 100V 1amp or 1V and 100 amp

I want to calculate how much a 300VA vs a 600VA delivers to an amplifier, what are most important, V or A and so on.

I know that it is wary hard to answer, but are there not some crude generalisations that are good enough to dip the toes in the water?

[IanB]

My problem are that 100VA could deliver 100V 1amp or 1V and 100 amp

It could, but which one it is depends on the design of the transformer (primarily the turns ratio between primary and secondary, as someone else has said). So you can't just "work this out", you have to look at the specification and design of the specific transformer you have, and follow that. If you don't know (if it is a mystery transformer salvaged from something else), then you will have to make measurements to estimate what it can do.

[FriedMule]

My problem are that 100VA could deliver 100V 1amp or 1V and 100 amp

It could, but which one it is depends on the design of the transformer (primarily the turns ratio between primary and secondary, as someone else has said). So you can't just "work this out", you have to look at the specification and design of the specific transformer you have, and follow that. If you don't know (if it is a mystery transformer salvaged from something else), then you will have to make measurements to estimate what it can do.

So it is in some way "easy" If I have some old in a drawer, then maesure it and if it is a new, specify what I want it to do?

[Benta]

You specify a mains transformer by its output voltage = which voltage does you amplifier need.
Then you look at how much current your amplifier draws. From this you decide the physical size of the transformer (= VA).

[Brumby]

My problem are that 100VA could deliver 100V 1amp or 1V and 100 amp

It could, but which one it is depends on the design of the transformer (primarily the turns ratio between primary and secondary, as someone else has said). So you can't just "work this out", you have to look at the specification and design of the specific transformer you have, and follow that. If you don't know (if it is a mystery transformer salvaged from something else), then you will have to make measurements to estimate what it can do.

This ^^^

[FriedMule]

You specify a mains transformer by its output voltage = which voltage does you amplifier need.
Then you look at how much current your amplifier draws. From this you decide the physical size of the transformer (= VA).

That is the problem, I am "just for fun" and to learn, trying to build a small amplifier, nothing special, but to have some noise to listen to, while soldering or blowing my next project up:-)

I have trawled the internet an all talk about power: "how much power", "speaker vs power" nearly all af them talk about power as if it was one unit, not Amps x Voltage. The few places that does ends up in name calling and abuse.

Am I right in understanding that the amount of voltage determine when the amplifier are clipping, while the ampere delivers the "energy" that move the drivers in the speaker?

[Brumby]

The terms "VA" and "Power" are very similar ... sort of.  VA is used to describe the capability of a power source.  Watts is a measure of the power used by the load.  Both are calculated using Voltage and Current, with a little extra something....

Let's start with the simple case:
If you have a purely resistive load then a 100VA rated transformer can deliver 100W into that load - assuming the resistance is such that the maths works out.  For example, if the secondary voltage is 20VRMS and the current capability is 5A (again, RMS) then the resistor would need to be 4 ohms.  If the secondary voltage is 100VRMS and the current capability is 1A (again, RMS) then the resistor would need to be 100 ohms.

In the case of the first example (20V/5A) the transformer is capable of deliver 20V at up to 5A - but the caution here is to pay attention to the phase angle between the voltage and the current drawn by the load.  For a purely resistive load, the phase angle is zero and the maths is easy (as shown above), but when you have an inductive or capacitive load, there is a real difference in that phase angle.

The end result is that for reactive loads, you may indeed have a transformer that is supplying 20V (as you measure it with a multimeter) at 5A (as you measure it with a multimeter), but the load may only be dissipating 80W - because of the phase angle between voltage and current is not zero.

This apparent discrepancy is called "Power factor".

The power factor for the resistive load examples above is 1, while the power factor for the reactive load is 0.8.

Aiming to keep the power factor as close to 1 as possible is a key objective in several situations - which is why you may have come across mentions of Power Factor Correction (PFC).

[FriedMule]

The terms "VA" and "Power" are very similar ... sort of.  VA is used to describe the capability of a power source.  Watts is a measure of the power used by the load.  Both are calculated using Voltage and Current, with a little extra something....

Let's start with the simple case:
If you have a purely resistive load then a 100VA rated transformer can deliver 100W into that load - assuming the resistance is such that the maths works out.  For example, if the secondary voltage is 20VRMS and the current capability is 5A (again, RMS) then the resistor would need to be 4 ohms.  If the secondary voltage is 100VRMS and the current capability is 1A (again, RMS) then the resistor would need to be 100 ohms.

In the case of the first example (20V/5A) the transformer is capable of deliver 20V at up to 5A - but the caution here is to pay attention to the phase angle between the voltage and the current drawn by the load.  For a purely resistive load, the phase angle is zero and the maths is easy (as shown above), but when you have an inductive or capacitive load, there is a real difference in that phase angle.

The end result is that for reactive loads, you may indeed have a transformer that is supplying 20V (as you measure it with a multimeter) at 5A (as you measure it with a multimeter), but the load may only be dissipating 80W - because of the phase angle between voltage and current is not zero.

This apparent discrepancy is called "Power factor".

The power factor for the resistive load examples above is 1, while the power factor for the reactive load is 0.8.

Aiming to keep the power factor as close to 1 as possible is a key objective in several situations - which is why you may have come across mentions of Power Factor Correction (PFC).

Great explanation on ohms law and PF, it is more clear now!

What I am missing now are what do current mean for the speaker and what do wattage mean?
Is it the amp or the wattages that influence the volumen?
Is it the amp or the wattages that influence the amplifiers control with the speaker?
What would low wattage + high current give, contra high voltage + low current?

(It have to be possible to know, Sony, Pioneer, Rotel and all other know it :-)

[T3sl4co1l]

FYI, use the symbol S for apparent power (unit: VA), P for real power (unit: W) and Q for reactive power (unit: VAR).

Note that Q is also used for "quality factor" (unit: dimensionless), so be sure to make the context clear when talking about the power flowing through resonant circuits!

Tim

[Brumby]

What I am missing now are what do current mean for the speaker and what do wattage mean?
Is it the amp or the wattages that influence the volumen?
Is it the amp or the wattages that influence the amplifiers control with the speaker?
What would low wattage + high current give, contra high voltage + low current?

OK - let's stick to a simple explanation.  (The actual numbers are a bit more involved for a real amplifier and speaker - but the following will start you of with the correct fundamental....)

If you have a 4 ohm speaker and the amplifier delivers 2Vrms to it, the RMS current will be 0.5A.  This is simple Ohms Law.  The power will be P = V.I which is 1 watt.  When the voltage from the amplifier is 4V rms to this same speaker, the current will be 1A and the power 4W.

Things start to look like this:
  2V => 0.5A => 1W
  4V => 1.0A => 4W
  6V => 1.5A => 9W
  8V => 2.0A => 16W
12V => 3.0A => 36W

You find the power by calculation - where the 3 possible parameters of voltage, current and resistance (impedance) are all related by Ohm's Law.

As such, this statement shows completely wrong thinking:

What would low wattage + high current give, contra high voltage + low current?

When it comes to AC (such as an audio signal from an amplifier) going into reactive loads (such as a speaker) the term resistance is inadequate to describe what is going on - so the term impedance is often used ... but as far as the basic idea of how they factor into basic formulae, impedance and resistance can (loosely) be considered equivalent.

[FriedMule]

FYI, use the symbol S for apparent power (unit: VA), P for real power (unit: W) and Q for reactive power (unit: VAR).

Note that Q is also used for "quality factor" (unit: dimensionless), so be sure to make the context clear when talking about the power flowing through resonant circuits!

Tim

Thanks a lot, I wil remember that! :-)

[FriedMule]

Brumby thanks a lot for trying to make such a great effort on explaining it!:-)

I feel that I am asking a question that shows that I do not understand what I am asking about :-)
But you all are learning me a lot!

My thought is that it have to make a differens if you select two transformers that both are 300VA, the one have 100V x 3A on the secondary and the other have 50V x 6A on the secondary?

[Brumby]

Just to make this simple, let's assume we have a power supply and amplifier that can deliver 40V rms to a speaker from the 50V transformer and 80V rms from the 100V transformer - at the rated current that you have given.

The important thing now is your speaker impedance.  Pick an impedance and use Ohms Law to work out the current - and check if this is actually achievable...

Note: all the current calculations here are simply using Ohms Law

Impedance	40V 6A	80V 3A
4 ohm	10A *EXCESSIVE*	20A *EXCESSIVE*
8 ohm	5A 200W	10A *EXCESSIVE*
16 ohm	2.5A 100W	5A *EXCESSIVE*
25 ohm	1.6A 64W	3.2 *EXCESSIVE*
32 ohm	1.25A 50W	2.5A 200W

This will give you an idea of how voltage, current, impedance and power are related.  They are fixed relationships and the key is centred on Ohms Law.  If you ever find yourself wondering about something here - go back to Ohms Law.  Always, go back to Ohms Law - and you will be able to work out your own numbers like I have above.

Armed with these figures, you can now look at finding an amplifier circuit that can fit your desired result.

[FriedMule]

Bromby I am so happy that you still try to help me!! :-)
You are just great!!
Yes Ohm's law, Ohm's law, Ohm's law... got it! :-)

Last question (I dare to hope:-)

As I understand you, when I choose the transformer, I also have to know the mean impedance of a speaker, like 4 or 8 ohm.

But if I order a 100VA transformer, would he not ask me how many voltages i want it to have on the secondary out?
And if I say 50V would it not give 50V 2A?

[IanB]

Actually, I think you should not choose the transformer first. That's the wrong end of the circuit to begin.

You should begin at the speaker. Decide what quality of speaker you want and what volume of sound you need. Then decide what amplifier circuit (or module) should be chosen to drive that speaker. Then find out what power supply that amplifier needs. Then, lastly, find out what transformer would be needed to create that power supply.

[Brumby]

^^^ What IanB said ^^^

I would lay out the steps like this:

1. Pick the power output you want to aim for and the impedance of the speaker that you can get to handle that power.  If you look at the table I produced above, you will see the speaker impedance has a huge effect on the voltage required to achieve a certain power level - and that having too high a voltage can actually make life difficult.  There is no mystery or magic here - it is simple maths.....

 2. Use the power equation and Ohm's Law to work out the voltage necessary to achieve that and the current it requires.

 3. Find an amplifier design that can deliver this and see what the power supply requirements are for it to do so.

 4. You will then need to consider the power supply design that can provide these requirements and work backwards to find the transformer specifications needed to achieve them.

As you go along this path, you are going to find some points where the numbers don't match up perfectly with real-life gear and you may have to revise your selections and/or expectations in order to do so.  One of the pesky realities that so often pokes its nose in is one of cost where you might ask yourself "I've got a solution for 168W and one that covers my target 200W - but is achieving my target worth tripling the cost of the amplifier board?"

"Perfect" solutions are rare and it is far more common to find a solution of "best fit".  This includes everything from amplifier output, to physical size, thermal design, component availability and cost (just to name a few).  Lowering expectations is common - but that isn't always the case and sometimes it may be more psychological than real.  For example, you aren't going to notice the volume difference of a 168W audio amplifier to that of a 200W one going through the same speakers.