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[SOLVED]transformer efficiency VA and W specs and other trafo questions

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d123

Advanced Member level 5
Hi,

Too much time in DC and not enough in AC...

This transformer:

1) Is there any need to put a fuse at the input to the primary as it says it is 'short-circuit proof'? Even if not necessary due to transformer design, is it still a good idea to put a fuse at the primary input side anyway in a circuit such as the following (it's only a rough/preliminary sketch)? A MOV, maybe, but the fuse? And if needed do I size it, as I saw somewhere else online, it would be 2.3 VA/230 VAC = 10 mA? - I find that hard to believe and almost ridiculous, I don't remember having seen a 10mA AC fuse in my life, more usually 3A, 10A, 13A, etc.

2b) Why is it 1.6 and not 1.414? What does 'app. x factor' mean in English/Human Being language?

2c) 'No-load loss (typ.)' of 1.2W with reference to which side of the transformer? Primary, I'm guessing: if I have no load whatsoever on the secondary, the wasted power at the input/primary side is 1.2W?

2a) and 2d) I'm a bit lost as to what current I can hope to draw from each secondary output between VA, W and trafo efficiency. I've had this conversation before with a member but it was a few years ago and do not remember their exact answer.
Going on the datasheet specifications and the 'Efficiency' section in this tutorial called Transformer Basics and Transformer Principles, below comes yet another image, this one is of the tutorial's formulas:

When I calculate 2.3 VA x 0.52 efficiency = 1.196 W. So, wearing my angry dunce's cap, I really must ask, is that what REAL POWER I can expect at the combined dual secondary side? Is that AC, DC, what exactly - do I then need to do 1.196 W x 0.707 = ~0.845572 W in DC as a rough idea or not at all?
If I do the simple - and I think incorrect - 2.3VA/15V = 153mA. If I then do the equally incorrect and misguided 153mA x 0.52 efficiency = ~80mA. So, then each secondary cannot output 15V and 76mA, but only 15V and 40mA?
Or, as in a thread I just read here, from what I understood, I need to assume that output power is 1.2W, and the 52% efficiency refers to 2.3 VA x 1.52 = 3.496 VA used at the primary side to produce the 2.3 VA at the secondary side. So, my above calculations are even more meaningless, and useless - correct?

Finding something so seemingly simple a bit hateful to grasp what I want to know, so thanks. My main question in this lengthy essay of hopelessness is how much AC average current can I actually hope to get out of the secondaries?

___________________________________________________________________

Different question(s):

Looking at the schematic, thinking of a very simple dual power supply just to test little things and no great voltage regulation precision required, and then considering the option of separating the secondaries and using two bridge rectifiers (one forward biased for positive supply, the other reversed for the negative supply, etc.) what configuration is less problematic with regard to varying loads and unbalanced loads between V+ and V- outputs? Does the transformer care how much power each secondary individually uses? This may seem a (very) stupid question but I'm curious, if I draw the whole secondary side 80mA from the positive output only, then surely the negative output will droop to death or fight the positive to get its share of the current? What do dual secondary transformers do in that respect?

What are D5 and D6 for, by the way? I basically copied the schematic from a very old ST power supplies app note.

Thanks.

Hi,

Fuse: in primary. While 2.3VA/230V = 10mA ... you did not consider phase shift.
You should expect a higher primary current.

In short circuit the (primary) current is limited. To something above the nominal current.
(This has so far nothing to do with efficiency).
Now the problem. A fuse rated for 10mA will not trip with 10mA or a bit higher even at higher ambient temperature (details in the fuse datasheet). And a 10mA fuse maybe "safely" trip above 50mA (details ..), but a short circuit prove transformer maybe will not draw 50mA.
So there is a good chance that the fuse does not trip... and a fuse with less holding current may trip (even unloaded tx) at power on.

Now you have to ask yourself: What is the idea to use a fuse on a short circuit prove transformer.
Please tell us, so we can discuss about it.

******
2b)
Has nothing to do with 1.414. It just tells that a 15V (RMS) transformer will output 24V (RMS) in unloaded condition.
15V apply at nominal current draw.
1.6 has to do with series resistance (impedance) and coupling factor

2c)
Can not apply to the secondary, since there is zero current.
And is not only the primary coil loss, but also the core loss.

2a)
best if yo do it this way: 2.3VA / 15V = 153mA. RMS! Since the current is non sinusoidal you must not draw 153mA after the rectifier. Maybe 70..80mA. In detail it depends on transformer parameter, rectifier, capacitor....
We can´t tell you the exact value.

You never calculate "power x 0.707 "... this only applies to pure sinusoidal current and voltage.

Efficiency just says:
The output power is 52% of the input power.
Output power is 2.3W, thus input power is 4.42W and thus the (thermal) dissipation is 2.12W

"efficiency, secondary watts, primary VA"
I call it nonsense... (at least without context).
Efficiency is always watts and never VA.

*****
On a 2.3VA rated transformer you may get 2.3W of output power. But ony in one case: Pure resistive load with the correct Ohms.
Phase shift (power factor), even if pure sine --> will drop output power
Non sinusoidal current will also drop output power.

*****
Each output is unconditionally able to deliver 15V, 76mA RMS. The key is in "RMS". It takes care of wavefrom.

Again: 76mA RMS (on the AC side of the transformer) does not mean 76mA (average) from the capacitors.
Maybe it´s better to compare it with quare waves:
* 100mA DC (100% duty cycle) into a capacitor
--> you get 100mA average at the output of the capacitor.
--> input current also is 100mA RMS

* 100mA average (50% duty cycle: 0mA/200mA) into a capacitor
--> you get 100mA average at the output of the capacitor.
--> input current is 141mA RMS

* 100mA average (25% duty cycle: 0mA/400mA) into a capacitor
--> you get 100mA average at the output of the capacitor.
--> input current is 200mA RMS

Different questions:
"no great voltage regulation" ... just to get values:
In no load you get about 24V x 1.414 = about 33V
In full load condition you get less than 15V x 1.414. (due to high current during peak of sine) I rather guess 12V x 1.414 = 16V
It´s no "regulation" at all. It´s just a weak voltage control.

At the given schematic, when you only load the positive DC side, then sitll the AC current is symmetrical.
And for sure the primary current relates to the secondary current.

D5, D6 are useless in my opinion.

***
BTW:
I recommend not to ask that much questions with one post. Do it step by step. The answers are quite lengthy and it´s hard to find the correct answers to the according questions.

***
My opinion:
I personally don´t build those unregulated AC DC supplies anymore. I either use switch mode supplies or use the above AC DC supply with post buck converters.

Klaus

d123

d123

Points: 2
Short circuit proof means the output can be shorted without damaging the transformer. Without this feature, the transformer needs to be protected by output fuses to fulfill safety transformer requirements.

Although not strictly required, having an input fuse with e.g. 50 or 100 mA rating can be meaningful though. It reduces the impact of e.g. a primary winding short. If the 10 or 16 A mains fuse is the only protection, you can expect an electric arc making the transformer case burst.

d123

d123

Points: 2
Hi Klaus,

1) Exactly... I just wanted to be certain that no fuse is needed with this transformer.

2b) Okay, that's where 1.6 comes from.

2c) Oh yeah, good point, that was very thick of me.
Coil and core losses, okay.

2a) "2.3VA / 15V = 153mA. RMS! Since the current is non sinusoidal you must not draw 153mA after the rectifier. Maybe 70..80mA." and "Each output is unconditionally able to deliver 15V, 76mA RMS. The key is in "RMS". It takes care of wavefrom."

That answers my biggest doubt/confusion, great, now I have a reliable number to work from. It has also clarified some aspects I wasn't understanding.

Hate to ask, but in these two examples, and looking at the first of the two quoted, should the second be 282mA RMS, if not, why not?
"* 100mA average (50% duty cycle: 0mA/200mA) into a capacitor
--> you get 100mA average at the output of the capacitor.
--> input current is 141mA RMS

* 100mA average (25% duty cycle: 0mA/400mA) into a capacitor
--> you get 100mA average at the output of the capacitor.
--> input current is 200mA RMS"

Different questions:
"In no load you get about 24V x 1.414 = about 33V" Whoops, I'd thought 15V x 1.414 (before seeing the 1.6 no-load voltage bit in the datasheet), and "12V x 1.414 = 16V", so thanks.
There would be a voltage regulator on the positive output and (subject to change) a TL431 to regulate the negative output, they were not included because I just wanted to keep the schematic as simple as possible.

"At the given schematic, when you only load the positive DC side, then sitll the AC current is symmetrical.
And for sure the primary current relates to the secondary current."
I'm afraid I don't really grasp what that bit means, sorry, that's me, I'll have a look at transformer information again to save you the frustration of explaining that e.g. 2 + 2 = 4.

"D5, D6 are useless in my opinion." - Okayyyy!

***

Yes, sorry, I noticed that, too, after posting the questions, I agree with you, and will do in future.

***

Yeah, transformers in this day and age, what am I thinking (apart from wanting to use them up instead of just sitting in a parts bin doing nothing since the day I bought them)... I'm actually 'just starting' the laborious process of researching LLCs as - relatively speaking - they seem like a far less difficult SEPIC with regard to the control loop (RHPZs) and with other advantages (ZVS and ZCS). As I would like to do a 'baby' LLC (of low current load) just to understand this very useful and efficient DCDC converter (I see they are used a lot in EVs, for example, and seem to have advantages over other DCDC topologies), then the trafo-based supply would give me a sensible coward's low AC voltage - rather than working with the whole mains voltage and isolation devices I do not have and won't for some time - for me to make no doubt endless mistakes with. Not a great approach (Trafo + Vreg [mains AC > DC] >> LLC [DC > AC > DC]), admittedly, but then I'll also have a little dual power supply which I don't have at present.

- - - - - - - - - - -
Really, thank you very much, Klaus, I appreciate your answer and the time taken to write it, and the clear descriptions, very helpful, you've cleared up pretty much everything I wanted to know.

* 100mA average (25% duty cycle: 0mA/400mA) into a capacitor
--> you get 100mA average at the output of the capacitor.
--> input current is 200mA RMS"
Simply apply RMS formula
sqrt(0.4^2 * 0.25) = 0.2

d123

d123

Points: 2
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