Selecting the Right Transformer Turns Ratio for a Given Voltage

[savvej]

Considering a practical transformer,
We can get a given voltage and corresponding current ratio by selecting a particular turns ratio.
ie. V1/V2=N1/N2=i2/i1
Say that I want to make a 240V/60V.
Now 240/60 volts ratio can be acheived by considering different no. of turns on primary(P) and secondary(S).
like:
1)P=4 turns ,S=1 turn
2)P=24 turns,S=6 turns
3)P=120 turns ,S=30 turns
4)P=240 turns,S=60 turns
5)P=4800 turns S=1200 turns
So we have various options for a given voltage ratio.So will each of the above turns ratio work?
Then why do we have 230-60V / or 230-12V transformers so bulky.How does current rating affect no. of turns?

 

[mdjensen22]

What do you know about transformers and how they alter the voltage & current?
If you want a specific secondary power (current @ a voltage), what do you think your primary power needs to be?

read http://sound.westhost.com/xfmr.htm as a starting point (there are 2 sections - read both). It may help address some of your questions.

 

[The Electrician]

Since you will be applying 240 VAC to the primary in all cases, the number of turns will be determined by the flux density you want in the core. The size of the core will be determined by the amount of power you want the transformer to handle.

Have a look at this: http://en.wikibooks.org/wiki/Electronics/Transformer_Design

There are many further references at the bottom of the page.

 

[savvej]

What I have come up with is this:
Consider a P=240 S=60 turn transformer.What I think is that if I wind such a transformer ,And if I check up Voltage at secondary,it should be indeed be 60V ,IF NO LOAD WILL BE CONNECTED.But if I connect a load,due to the current set up ,opposing flux will be set up in the core and to maintain flux,current will be drawn from the primary.But what will happen in this case is,though all flux of primary is linked with secondary,not all flux of secondary will be linked with primary(if the no. of turns are sufficiently small) and hence,the transformer will not be able to deliver any current.

Please correct me if I am wrong.

 

[I_am_learning]

You may be partly correct but that is not the main point.
The problem with too few turns is illustrated by this formula
E = N d(phi)/dt
Now considering the primary, if you put E = 220 V i.e. 220 Sin(wt), you can see that
phi = (220/N)Cos(wt)

So, the amplitute of flux density is inversely proportional* to the no. of turns. Put too few turns, and you have huge flux density.
Whats wrong with huge flux density ?
Well, if you recall the B-H curve for Iron, you will recognize that it has saturation characteristics. Which means, if your flux density substantially exceeds the saturation level, then huge magnetization current will be drawn.

What this practically means is that, Just when you connect the mains across your primary (with too few turns), huge current (magnetizing current) will flow, seen as a deadly spark. (don't try this).

So, you may be aware, that the tranformer primary (and the secondary) is just a continuous wire. Why don't it short (and spark) when we connect it to the mains?
Well, it does, as has been just shown, but when the no. of turns is large, due to inductive effect (i.e. due to magnetic effects), it don't happens.

I hope, I helped provide some intuitive feeling.

 

[I_am_learning]

Then why don't we wire up too many turns?
Who would? provided the cost of copper, if less can work. We generally, wind just as much wire as to limit the magnetizing current to a safe lower value (usually around 2.5 % of full load current). Also, there is an issue with space limitation too.

 

[savvej]

Thanks for explaining @I_am_learning.
Btw during no load,why is magnetization current drawn?
I didnt get what my textbook says when it says to set up flux in core.
Considering if I had air core transformer,what would happen?

 

[I_am_learning]

Its all kirchhoffs Law.
When you apply 220 to primary, (assume no load, i.e. secondary unloaded), it must be balanced by something (to satisfy kirchhoffs voltage rule)
So
E = I*r + E(back)
Where I*r is the voltage drop in the coil resistance and I is the current. (Since, there is no load, it must be the magnetization current). and E(back) is the back emf induced in the coil. Normally (for good designed transformers), E(back) is almost equal to E, so we generally have
E(back) ~= E. So, magnetization current is small.
How do we know if our assumption is good enough?
Here is how.
Suppose you apply 220V.
Assume E(back) is exactly equal to 220v.
The only way E(back) can be produced (be it iron core or air core) is
E(back) = N d(phi)/dt
So, we can solve for phi.
Given the cross sectional area of the core (or air if air core) we can find the magnetic flux density B.
Given, B and no. of turns we have and the permeability of the medium (air or iron), we can find the current required to set-up that amount of field.
Now, this is the same current I.
So, if I comes out to bee too small, we are fine.
If it comes out large enough that, I*r isn't Ignorable w.r.t. E, then we need to re-do our maths.
Generally, the redone maths would show that I would be slightly* less, becasue, initially, we had assumed whole of the applied voltage to be balanced by E(back) alone, but, after we realize that I is large, we need to take into account that the applied voltage is balanced by I*r as well. I hope you have realized what difference Iron core create Vs air core. If not ask. Also, much of the situation remains unchanged even at loading, except that the current now not only includes magnetizing current but also load current. And the I*r term will be larger, but practically still ignorable, as r is quite small in transformers. Note that, r is not only pure resistance, but also the inductive reactance.

 

[jim hardy]

It is very handy to measure the volts-per-turn of a core.
That way you don't have to do a lot of calculations.

What i do is this

before removing the windings from a transformer that i intend to use for a project,
i sneak a few turns of small wire around the core, ten if i can squeeze them in.
Then energize the core usually just plug it into wall
and measure the volts
usually it's in the tenths of a volt per turn
(i did have a huge one once gave 3v/turn
made a nice welder)

that tells me what is d(phi)/dt for that core at 60 hz.
If the core runs too warm i use more primary turns than the factory did
because it lowers the flux making the transformer run cooler and quieter.

Microwave Oven transformers are notorious for pushing flux high
if you're using one of those study it well.

old jim