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VA rating of mains transformer for 24W real power output

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cupoftea

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Hi,

We want a 50Hz, 240VAC mains transformer to supply 24V at 1A
We therefore selected the following 50VA transformer (504246)

https://docs.rs-online.com/c4ae/0900766b816919e8.pdf
its spec’d for 230VAC and 2*9V sec

However, the VA rating needed depends on the magnetising inductance, which depends on the primary inductance. However, the primary inductance isn’t stated for the above transformer.
So how can we work out if the VA rating is enough for the output power required?

I mean, if you observe the attached, then the transformer with a primary inductance of just 200mH ends up having a power factor of just 0.03. As such, it needs a VA rating of 860VA to supply the 24W output.

LTspice and PDF schem attached.
.....____
A mains transformer based solution has far better mains transient withstand than an offtheshelf SMPS. Also, it is far less EMC noisy. Also, it allows you to select your own electrolytic smoothing capacitors, and so you can select loads of really high quality 125degC ones, and so have really low ripple current in each cap, which means the product will last for decades.
 

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  • Mains transformer.pdf
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Sorry, that sounds really confused. You don't need to care about magnetizing inductance respectively magnetizing current because transformer VA rating refers to the secondary side. For technical mains transformers it's sufficiently low, e.g 5 to 10 % of the rated primary current.

The problem that actually matters for a rectifier power supply with filter capacitor is distortion power factor and respective reduction of available DC output power. A rule of thumb says that rated transformer power must be at least 1.6 times the DC power for a full bridge rectifier. The exact ratio depends on transformer series resistance and leakage inductance as well as filter capacitor size.

Having a the tranformer rating double the DC output power isn't bad.
 
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cupoftea said:
if you observe the attached, then the transformer with a primary inductance of just 200mH ends up having a power factor of just 0.03. As such, it needs a VA rating of 860VA to supply the 24W output.
Click to expand...
That makes little sense.
What makes you think that is true?

As noted, magnetizing inductance has nothing directly to do with the transformer VA rating.
Transformers with different power ratings can have the same inductance.
That inductance determines the magnetizing current, not the power rating.
 
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Hi,

I agree with the others.

If you want to supply 24V, 1A, then you basically just need a 24VA transformer.
But this is only true if the output current is purely resistive and of sinusoidal shape.

We neither know what's the load.
We don't even know whether you want a 24V AC output ... or something else.

I think it's a good idea to tell us all your requirements first.

Klaus
 
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When I place a transformer in a simulation, its default primary value is 4H. I think it approximates a typical power transformer. It admits sufficient current at 50 or 60 Hz for most purposes. Power factor is usually very good evidently as a benefit of transformer behavior (that is, properly designed transformer).

However lesser Henry values admit alarmingly high current (even with no load on the secondary).

I just tested several mains transformers on my meter. They read DC ohmic resistance (such as 30, 70, 200 ohms). Of course in addition to ohmic resistance there is inductive reactance, resulting in overall impedance which limits current.
 
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Hi,

BradtheRad said:
resulting in overall impedance which limits current.
Click to expand...

I'm sure it's meant the correct way. But I find some of the information a bit misleading.

Talking here about "limiting the current" ... is only true for "non loaded" transformer. The henry do not limit the loaded current.
Talking about "good power factor" ... is (in opposite to the above) only true for "loaded" transformer. Without load the power factor is very inductive.

I miss that "coupling" happens in a transformer.
Coupling happens in parallel to the henry .. and does not depend on henry (although a transformer needs some inductivity to operate)
Coupling means that the secondary power is feedbacked to the primary side. Including it's power factor (of the secondary load)

The total input current depends on the
* current that is caused by the henry (about independent of load)
* load current

... if one wants to step more deeper one has to consider
* stray inductance
* loss in core
* coupling factor
* loss in windings

Klaus
 

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That makes little sense.
What makes you think that is true?
Click to expand...
Thanks, by definition, power factor is [real power] / [apparent power.]

...so if you have loads of magnetising current in the primary, then your primary VA (apparent power) is very high

I must say that the failure of all 50Hz transformers (in their datasheets) to state L(pri) and R(pri) and r(sec) is very unfortunate.
They state efficiency, but dont state what the conditions under which that is measured.
So its not good.
 

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Hi,

The magnitizing current is rather low. And does not change with load.
The load current becomes added.

Klaus
 
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cupoftea said:
...so if you have loads of magnetising current in the primary, then your primary VA (apparent power) is very high
Click to expand...
Better say if you had, but you actually don't have. It's just a speculation due to lack of knowledge. Reading only this thread we won't expect that are you are working in electronic design since many years.
cupoftea said:
I must say that the failure of all 50Hz transformers (in their datasheets) to state L(pri) and R(pri) and r(sec) is very unfortunate.
They state efficiency, but dont state what the conditions under which that is measured.
So its not good.
Click to expand...
An average detailed datasheet of a 50 VA Talema toroid transformer specifies
copper losses 6 W (at rated load)
core losses 0.43 W
no-load current 5.0 mA

Using this numbers, you can calculate the the inductive component of magnetizing current as 4.6 mA, about 2 % of rated primary current. Also effective series resistance as sum of primary and transformed secondary winding resistance.

You may want to get also info about leakage inductance, but it's rarely specified for small mains transformers. You can either use typical values for the specific winding configuration or measure it yourself, if it matters.
 
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Thanks, yes, but its cheaper to make them with low magnetising current, and though quality ones have negligible magnetising current, there are those where its much higher than 2%.....its nice to be "on your gaurd" against this, and disconcerting when the datasheet fails to list such a simple-to-measure parameter. I think its worth remembering that the "middle men" who import this stuff over, will often do anything to save themselves money....not caring about the unfortunate consequences
 

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Hi
cupoftea said:
but its cheaper to make them with low magnetising current,
Click to expand...
Low magnetizing current means high impedance, means high inductivity, means high winding count.
But high winding cound is the opposite of cheap.

So how do I have to understand your statement?

Klaus
 

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So how do I have to understand your statement?
Thanks, thin, cheap winding , being cheaply done with not enough turns, would give low primary inductance. I am sure We've all seen small mains transformers which are sizzling hot even though on no load.
I must admit i dont see why its so difficult for these transformer datasheets to state Lpri and Rpri and Rsec...even if they give a wide tolerance
 

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Hi,

In the first statement you talk about low magnetizing current,
Now you talk about low inductance.

This is contradicting, because low inductance means high current.

Klaus
 
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Even a transformer with 10 % magnetizing current still has 0.99 power factor at rated load. I simply don't understand which problem you are talking about.
 
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Hi,
Tha attached mains transformer sim, shows very high magnetising current.
The power factor is very low.
cos(phi) is near zero
Its in LTspice
 

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  • mains transformer_high imag.zip
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for a typical cap input filter, 24Vac will give you the peak volts ( 32V assuming some flat topping of the mains & 2 diode drops ) of the sine at no load and near the average value at full load ( 19.8V ave - incl 2 diode drops ) with the ripple depending on the cap and load.

As to VA rating - the power factor for a cap input filter ( after the bridge rect ) is 06 - 0.65 typically

so for 24VDC x 1A = 24W divided by 0.6 = 40VA transformer needed, as the rms current is quite high due to high peaks near the mains voltage peak

hope this helps ...
 
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cupoftea, a factor you are failing to take into account in your simulations of extreme transformer magnetizing current is non-linearity of the core magnetic properties. The sim in post #1 assumes a primary inductance of 200 mH. Such an inductance would have so few turns that the core would pushed FAR into saturation. The core loss would be so high that the transformer would catch on fire.

The reason manufacturers don't specify the primary inductance is that it doesn't matter. If the manufacturer says the transformer can supply 50 VA, that means it can supply it at the rated temperature rise of the transformer. This translates into a maximum RMS secondary current.

So unless the manufacturer is outright lying, when they say the transformer can supply (safely is implied) 50 VA, they mean just that. The saturation characteristics of the core do not allow very low primary inductance, so you don't need to worry about exactly what that inductance is.
 
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cupoftea, a factor you are failing to take into account in your simulations of extreme transformer magnetizing current is non-linearity of the core magnetic properties. The sim in post #1 assumes a primary inductance of 200 mH. Such an inductance would have so few turns that the core would pushed FAR into saturation. The core loss would be so high that the transformer would catch on fire.
Click to expand...
Thankyou for that...great points, thats very interesting and worth looking in to....
If i have a 50Hz iron transformer with enough turns to give Lpri = 10H.....then say i reduce the turns (on the same core) such that Lpri now becomes 200mH. Then are we saying that putting it across the mains 240VAC again would now result in B(sat) being exceeded, and also "delta B" being pushed up to a point that the core loss grew much higher?

(Note, when the turns were reduced, lets pretend that the wire diameter was increased such that winding loss remained the same, as when Lpri was 10H)

Do we say that Ferrite is "linear" in that the B by H curve is a straight line up until almost saturation?
However, 50Hz iron core material is "non linear", and its B vs H curve is not ever straight?(unless of course looking over a vanishingly small section)
 

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