Multi Voltage primary step down transformer

[capt.m]

Pls Help me for this, am confuse how to calculate the multi voltage of my project transformer. the voltage range are within 0-220Vac-230Vac-240Vac primary and 12Vac-0-12Vac secondary center tap at 100VA.

I want only the calculation for primary.

thanks

[W2PFY]

What do you want to know about the primary?

[N4LTA]

Do you want to know the primary current at full load?

If the transformer is 100VA  - divide 100 VV by the selected primary voltage. 

If 220 V   - then the current is 100VA / 220 V  =  .45 amps

The secondary is the same method    100 VA / 12 V  = 8.33 amps

[KE6DF]

I'm not sure what you are asking, but if you want 24v CT out then choose the input taps that are closest to your line voltage.

So, if your line voltage is 230v then if you connect to the 230v primary tap you will get 24.0 out.

Now if you connect to the 220v tap, you will get

230/220 * 24 = 25.1v out

If you connect to the 240v tap you will get

230/240 * 24 = 23v out

[N4LTA]

for 12-0-12 the rated voltage is 24 volts for the full winding

or 100 /24  = 4.16 amps

[capt.m]

hi,

What I mean is how to determine the correct AWG size for primary. what should be the basis for getting the total current is it by higher voltage or lower voltage.

the primary is selectable from 0-220Vac, 230Vac and 240Vac.

thanks

[Gito]

Hi

To determined the Correct AWG size is depended of how much current that will flow in the primary winding.
To know that we must know the secondary wattage of this Transformer ,in this case 12-0-12 vac with 100 watt power rating .
There are looses in the iron of this transformer,looses in the email cable and so...so on.
So the efficiency of the Transformer is around 80% .
Because of that we need a 125 watt trans former rating ,we need  a 125 : 220 = 0.57 A. email cable for the Primary winding.
The Email cable that can handle this current is about 0.6 mm or AWG 24.

To choose the "total" Current of the transformer,we must first determine the max load on the SECONDARY winding  we need ( WATTAGE ,not the voltage-high or low).
In this case 100 watt,so we must have a 100 watt + wattage looses in the Transformer itself(Iron and....)= 125 Watt transformer
 
So,Knowing  the Efficiency  Is about 80% ( depend of the quality of the transformer Iron),we can calculate the wattage of the transformer /.the current in the primary winding/looking at ARRL hand book finding the AWG of the email cable that can handle this current.

Gito

[capt.m]

hi,

how about 230 and 240 tap? if 220= 0.57A

[Gito]

Hi

Since the power /wattage needed is still the same, 125 watt than with 230 v is
125 : 230 = .54 A and with 240v is 125 : 240 = .52 A.

The primary winding is 0 -220 - 230 - 240 vac ,since the o - 220 v winding needs the largest Current,we choose email cable that can handle this current that's AWG 24 ,we don't need a different email cable for the 230 vac-240 vac ,since it's not practical .
and it needs lest current.

Gito

[capt.m]

actually in our area the main voltage is a little bit tricky because during night time the main voltage is within 240VAC. but in a day the voltage drops to about 220Vac, am wondering what should be the correct main voltage in our area.

[Gito]

Hi

Can you tell me whats the reason ,the purpose ,to design this multi voltage of your Transformer.
If its design using  as a power supply ?
It depends what you re you using it for.
If the highest main voltage is 240 vac,than of course you must use the 240 tap on the transformer,maybe with a 12.5 - 0 - 12.5  secondary ,so when the main voltage drops to 220 vac ,the secondary is 11.5 - 0 -11.5 . when the  main voltage drops to 230 vac the secondary is 11.9 - 0 -11.9 v.
So there's about 4%  tolerance against 12 - 0 -12 VAC

Of course it depends  if you don't  have  an AC automatic regulator transformer (AVR)

Gito

[W2PFY]

What does capt. m stand for??

[capt.m]

Hi

Can you tell me whats the reason ,the purpose ,to design this multi voltage of your Transformer.
If its design using  as a power supply ?
It depends what you re you using it for.
If the highest main voltage is 240 vac,than of course you must use the 240 tap on the transformer,maybe with a 12.5 - 0 - 12.5  secondary ,so when the main voltage drops to 220 vac ,the secondary is 11.5 - 0 -11.5 . when the  main voltage drops to 230 vac the secondary is 11.9 - 0 -11.9 v.
So there's about 4%  tolerance against 12 - 0 -12 VAC

Of course it depends  if you don't  have  an AC automatic regulator transformer (AVR)

Gito

Thanks Gito,
this is for my power supply transformer for audio amplifier low wattage only.

well done... the main voltage is 240Vac and drops to about 220Vac, so, i my calculation I rather get the 240Vac no matter if the main voltage fluctuates at 230 down to 220Vac.

125VA/240Vac = 0.52 A of current and #24AWg is within this current rating.

am I right Gito?

[capt.m]

What does capt. m stand for??

capt. m stands for captain marvel a comics superhero. but it doesn't exist anyway.

[Gito]

Hi

Of course for the safety reason we must used a 240 AC tap at the primary winding of this Transformer,to avoid transformer damaged,since at night times it  your main voltage soars  up to 240 AC,but if you can, make the secondary winding 12.5 - 0 - 12.5, so when the main voltage drops to 220 VAC,you still gets 11.5 - 0 - 11.5 .
As the Email cable ,still used AWG 24.

Usually audio amplifier has tolerances for their power supply units .

Gito

[N4LTA]

Transformers are rated with apparent power (VA) - Volt Amps - not by wattage - There is a big difference.

Transformers are among the most efficient energy transfer devices known - typically 96% for smaller transformers and up to 99% for large ones. The losses in  a typical transformer are the eddy current and magnetization losses which are fairly constant and the IR losses in the windings, which increase with load.

If you are feeding a resistive load such as a power amplifier, the VA and the Wattage are close to equal. Both the voltage and the current are in phase. Inductive and capacitive loads can have much higher VA than the actual power input (watts)

Apparent Power (VA) =  Power (Watts)/pf    where the pf (power factor) is the cosine of the phase difference between the voltage and current. The phase difference can be leading (capacitor) or lagging (inductor)

Feeding a capacitor or an inductor with AC can generate a very high current and have very close to zero power as the current and voltage are 90 degrees out of phase - practically zero watts (except for the capacitor  or inductor losses)  - the transformer can be loaded to full power or even overloaded and supplying near zero power (watts)


I find that a typical  240 volt ranged primary winding will easily withstand voltage variations of 20% or so (+- 24 volts). The only consequence is that the secondary will also vary +- 20%.  The windings and core are optimized for a designed volts/hz and going more than 20-25% out of the design range can lead to a drop in efficiency and heating.

[Gito]

Hi

You are right transformer are rated in VA.

And we are talking about a small 100 VA transformer to make a power Supply for An Audio  Amplifier ,so 100 VA is 100 Watt,right.

ARRL 1997  handbook page6.44 chapter 6 Power Ratio (transformer)  ..Po=n.Pi
Po =power output from secondary
Pi =power input to primary
n =efficiency factor
..........it stated .The full load efficiency of small power transformer used in radio recievers and transmitters usually lies between about 60 t0 90 % depending on the size and design.
So it's not wrong that I wrote the efficiency is around 80%

As Capt,M wrote that his main voltage at night is 240 VAC and at day is 220 VAC.
So of course we must used the 240 AC tap of this primary trafo.
If I' m not wrong  he means that all night it's 240 VAC and 220 VAC all day.

I think even if the transformer can withstand voltage variation ,its not wise to use the 220 VAC tap on the transformer at night since the main voltage is 240VAC all night ,sooner or later the transformer will burnt/damaged.

So used the 240 tap on the transformer,when the main voltage drops to 230 VAC or 220VAC ,only the secondary output will also drop .
But no harm  to the transformer if we use it all day and all night long.

Gito

[N4LTA]

100 VA transformer is not a 100 watt transformer - only if the load is 100% resistive - which is the common case for a small power transformer in a radio - switching mode transformers are not unity power factor, nor would be the transformer in question if it was supplying a motor or other reactive device.

Transformer are rated in VA - The transformer in question is rated at 24 volts  and  4.16 amps  into any load

It could supply 4.16 amps into a large capacitor and the total wattage would be just the losses in the capacitor (a few watts or less) and it would be fully loaded and would overload and  overheat if more load was added even though it was supplying just a watt or so..

220 volts on a 240 volt transformer is generally not a problem. It is not at all uncommon to see 208 volts on a 240 volt transformer used for distribution purposes.

A small 100 VA transformer efficiency would approach 95% unless it is a very poor design. Think about your example - if the 100 VA transformer was passing 100 watts at full load at a PF of 1.0 - then 80% efficiency would cause the transformer to dissappate 20 watts in heat - a small 100 VA transformer would be smoking hot under those conditions.

I would suspect a decent 100 VA transformer would dissappate 4-6 watts with the proper line voltage to match the  volts/hertz design. Some bad quality "wall warts" are rated as low as 85% efficient.

It's an easy test to run - Take a small power transformer and hook the secondary to a resistor that loads it up to somewhere near full load. Measure the voltage across the resistor and the AC current through it. Multiply the numbers to get the VA and the watts since the load is resistive.

Measure the primary voltage and current and multiply them for the input VA and watts - if the secondary is resistive - the primary will also be per basic transformer theory.

The difference is the transformer loss and you can calculate efficiency.

[Gito]

Hi

First is a transformer itself  an resistive load,or inductive load? can a transformer have  a power factor of 1?

Yes you can supply a 4.16 A into a large capacitor  and looses in the capacitor is only a few watts,
Aren't  You forgetting that the transformer has A Resistance in the email cable winding, Say if it's only 2 Ohms
than there's power looses in the Transformer is 4.16 X 4.16 X 2 Ohm = around 32 watt?
There's a difference  between the power consumed by the load and the power dissipated in the Transformer it self .

Since a transformer  itself is an inductive load not a resistive load there's a power factor in this transformer.

So the real power is 100 watt ,but the apparent  power is higher than that
So if the transformer real current is 1 A for example ,the apparent current  is  1.2 A
Yes if there's no resistance in the email cable ,the power looses
 it's consume is The "same"
But since there's a resistance in the email cable there's difference power looses in this transformer.

Yes using a 220 VAC on a 240 VAC transformer is no problem,but if using a 240 Vac into 220 VAC transformer ? there' must be a problem.

Yes if the transformer is loaded full power all the time,it may smoked ,but the iron of the transformer is also a heat sink for the Transformer,but I think we don't load it all the time,since it's used in an audio amplifier

Thinking of it ,So the ARRL Handbook Handbook for radio amateurs is Wrong? because it stated ....the full load capacity of small power transformer is .......about 60% to 90 %.

Again there's difference between the power consumed by the load and the power needed of transformer
Like you wrote ,say we can load it with a resistive load to 4.16 A,yes it means it needs a 100 watt Transformer.
 
But the transformer itself  has a resistance in Cu cables, say it's only 1 Ohms so it dissipated 16 watt in the secondary winding,not including the looses in the primary winding.
So the transformer must consume 100 watt + 16 watt + looses in the primary winding.
Not mentioning the power looses in the iron ,that's 2.4 watt per Kg Iron

And don't forget the transformer it self is a Reacktanse load/Inductive load so the Power Factor  can not be 1


Gito

PS
 I have a 10-0-10 vac- 5A filament transformer,when I measured the email cable resistance it 's around 2 Ohm,so when we loaded full power ,there' is looses in this winding 5 x 5 x 2 = 50 watt only on the secondary winding?
Am I right?
Since A filament transformer is fuul  loaded all the time,Why doesn't it smoke? since it dissipated  50 watt as looses /heat?
I wonder?

[KE6DF]

 I have a 10-0-10 vac- 5A filament transformer,when I measured the email cable resistance it 's around 2 Ohm,so when we loaded full power ,there' is looses in this winding 5 x 5 x 2 = 50 watt only on the secondary winding?
Am I right?
Since A filament transformer is fuul  loaded all the time,Why doesn't it smoke? since it dissipated  50 watt as looses /heat?
I wonder?

A 2 Ohm resistance on the secondary doesn't sound right.

That would give you a voltage drop of 10 volts at 5 amps which doesn't make sense for a 10-0-10 v 5a transformer.

The resistance of a 10 volt 5a secondary should be more like 0.1 ohm (that would give you a 0.5v drop under full load.)

[capt.m]

hi to all,

I come up with this computation since our main voltage is 240Vac 60hz and drops/ fluctuates to about 220Vac. and I need a 12Vac rms at the secondary.

240/220*rms
1.0909091*12
=13.090909Vac

so my transformer should be

0-240Vac primary

13Vac-0-13 Vac secondary

there is a 10% increase allowance for the losses when the transformer is loaded.

[Gito]

Hi

Yes Capt.M ,if this 13 - 0 -13 vac is not dangerous /over voltage for your power supply,You can used  it Why not? .

Hi Wb6
Yes after I re measured the resistance of my transformer it reads .5 ohm,maybe because it use " small" wire /the "economical "transformer,just enough for the Amperage it needed.
So there 's a 8 watt looses in the secondary winding + looses 8 watt in the primary winding + ( 4 x 2.4 watt) 4 kg iron = 24 watt.
I think the designer has calculated the drop in the voltage

Thanks for the correction

Gito

[N4LTA]

I would suggest that you read up on transformer theory.

If a transformer is supplying a purely resistive load on the secondary - then the primary sees a purely resistive load  at the primary - That is basic transformer treory.

Whatever the secondary is connected to - is reflected to the primary  - a transformer is not "inductive"

If the ARRL Handbook says a transformer is from 60 -90 % efficient - (I have not seen where it says that) then it is wrong - Badly wrong.

All you have to do is google transformer efficiency or read from any standard engineering text.


If we lived in a world where electrical transformers are from 60-90 % efficient - it would be a sad world for electrical distribution.

I do this for a living - 98% is the standard for a small power system transformer .

Pat
N4LTA

[Gito]

Hi Pad

In The ARRL 1997 hand book page 6.44 Chapter 6  it stated the full load efficiency of small  transformers such are used in radio reccievers an transmitter usually lies between about 60 and 90% depending on the size and design.

in another ARRL handbook  about rewinding transformer ,
Stated ..The primary volt-ampere rating of  the transformer to be rewound ,if known,can be used to determine its power handling capability. the secondary volt-ampere rating WILL BE TEN TO TWENTY percent lest than the primary rating.

But of course I believed that You can build more efficiency Transformer,Since You make it for living,
You must has more experience and skill in building , using high. Grade iron for the transformer ,using the best material for the enamel cable and so on.

But for me and most Amateurs can't build transformers like You,so it's natural that We believe books like ARRL, a world wide radio amateurs handbook and other books about transformer.( I read a lot of  books about it)

Here in Indonesia My friend also make transformer as a living and have small factory named "SEVTI Transfomer" ,but used Iron /enamel cable for the transformer imported from China.
And  as far as I know he  can't  build transformer as efficient as You.

Gito

[N4LTA]

It is a very simple concept (efficiency)

If a transformer used in a 1000 watt input amplifier was only 60% efficient - then the heat generated in the transformer would be 666 watts - that is absurd  - not remotely possible

1666 watts in x .6 =1000 watts  - losses are 1666 - 1000 = 666 watts

Even at 90 % the loss would be 111 watts -

You are saying that the loss would be nearly as much as the tubes disappate

Have you ever touched a large heat-sink disapating 111 watts? or 666 watts?

It doesn't happen.

I don't build transformers - I design systems and have for 35 years - typical distribution transformers are 98% efficient and over 99% for large transformers. The mails losses are copper IR losses which are very low in a well designed transformer.

BTW - My ARRL Handbooks - I looked in three - say absolutely nothing about transformer efficiency  - you are confusing power supply (rectifier, inductor and capacitor losses) with transformer losses

The DC winding resistance has little to do with this subject except in calculating losses - we are not dealing with DC - we are dealing with AC and mutual inductance.


BTW Below is a list of efficiencies for a standard distribution transformer copied right from the manufacturer.

While the larger size brings some efficiency of scale - typically a small well designed transformer would fall in the same ballpark - probably a couple of percent less.

Note that the cheaper aluminum winding one is 98.64 efficient???

Get a good engineering text on transformers and read about how they actually work. The calculations are fairly simple. This isn't rocket science.  I done on this topic.

 


   Standard (Aluminum)    High-Efficiency (Copper)    Standard (Aluminum)    High-Efficiency (Copper)
Load Factor**    65%    85%
Efficiency    98.64%    99.02%    98.47%    99.02%
Temp. Rise
(100% load)    150° C    80° C    150° C    80° C
Core Loss    4.3 kW    5.5 kW    4.3 kW    5.5 kW
Conductor
Loss    9.1 kW    4.1 kW    15.5 kW    7.1 kW
Total Loss    13.4 kW    9.6 kW    19.8 kW    12.6 kW
Power Saving    –    3.8 kW    –    7.2 kW
First Cost    $16,750    $22,650    $16,750    $22,650
Cost
Premium    –    $5,900    –    $5,900