Continue to Site

Welcome to our site!

Electro Tech is an online community (with over 170,000 members) who enjoy talking about and building electronic circuits, projects and gadgets. To participate you need to register. Registration is free. Click here to register now.

  • Welcome to our site! Electro Tech is an online community (with over 170,000 members) who enjoy talking about and building electronic circuits, projects and gadgets. To participate you need to register. Registration is free. Click here to register now.

Transformer primary coil impedance And current questions

Status
Not open for further replies.

JohnMec

New Member
Hello there.

I'm a new to this forum as a member although i'm following topics periodically.

I have a question about transformers.

Well I feel weird asking this as i'm very experienced in electronic/electric circuits and i also know the transformer basics but i'm missing a lot of details as i haven't played much with these.

No one should be embarassed to ask. :D

To the question now...

Let's say we are about to wind up a transformer.

What should we becareful about the Impedance of the Primary Coil?



I will try to make this question as clear is possible for me:

Is the impedance of the primary the one which determines the maximum current that the secondary can get? (not counting flux current,copper losses,etc).

So for example if we want a 200v to 100v transformer that will output 10A at 100v should the primary coil have 40-Ohm impedance at e.g 50hz? 200v x 200v / 40Ohm = 1000w,as 5Ax 200 = 1000w?
(Again,counting with 100% efficiency for everything for ease)


I need to know this as any info about winding transformer primaries does not take into account the resistance but only the turns.


I've had some experiments in the past with "improvised transformers" based on existing cores for a 12-15V secondary...

But no matter WHAT magnet wire i was using (new,double checked for shortings) ,with hundreds of turns the primary coil always had 2-4-10-15Ohms,etc DC Resistance on the meter and as result when tested with thermal loads in series the load was operating just as if it had a 2,4,6R, resistor in series which means the primary would draw a nasty current if directly connected to the grid.

More turns and different ratio doesn't make sense as for an easy example 110:220v transformers that are on the market with 10A output use a bulky high ampacity wire on the primary yet are in reasonable size,if you try to get high resistance with like 2.5mm wires you will wind a few kilometers.

I'm definitely missing something here..



Keep wondering why my primary coil attempts are solely acting like resistors? Should i make out the ratio with higher resistance thinner wire and more turns? Like 40ohms for 5A/200v primary and e.g 10A/100v secondary when unplugged?


Thanks in advance,i will appreciate any reply as i'm looking forward to put Figure this out.
 
It's not the resistance that determines the primary impedance but the transformer inductance. The inductance is determined by the number of primary turns so that only a small amount of magnetizing current flows in the primary with the maximum input voltage and minimum input frequency, without magnetically saturating the core.

The wire size is then selected to give a resistance that doesn't overheat the transformer from the maximum load current. Also the wire size affects the regulation (change in voltage) of the transformer output from no-load to full-load that you desire.

All the above considerations then dictate the size magnetic core you need to accommodate the number of turns at the wire size you need. This can be somewhat of an iterative process.
 
Hello.

Thanks for your reply!

The truth is that i haven't think about the Inductance. :eek:

So if a transformer draws excessive current does that mean that is saturated?

My transformer attempts loaded aswell unloaded had the primary drawing so much current that it was actually bridging a 10A 230v heating element in series!

That's why i thought of it as a resistive matter.
 
Hi,

Wow, maybe there is a shorted turn then. A shorted turn in primary or secondary may cause this problem.

What is the normal primary rating, like 120v or 12v or what, and the frequency?

If you wound it yourself then you have to have a certain minimum number of turns for a give core size. If you dont have this min, the core enters saturation and that will draw a lot of current.
 
............................

So if a transformer draws excessive current does that mean that is saturated?

.................
It means it's saturated or it has a shorted primary or secondary turn.
 
Hello all.

The transformers i have modified with new primaries in the past most were small like a few amps at 12v or so,input was 230v 50hz,all with the necessary precautions of course like resistive load in series.

I don't remember the turns but they were a lot less than the original primaries,it's been a while,but i had checked the wires for short circuits,they were all good,insulation was intact,the coil was carefully winded,etc.

This was just for a quick experimentation.

I believe they were saturating due to less turns than it should have had.

Sure thing is that i need to study more on how to properly make one and use different size of cores,etc,as till now i have just experimented.
 
Hi,


You could use this formula:

B=E*10^8/(4.44*F*A*N)

E is RMS voltage in volts,
F is frequency in Hertz,
A is core cross sectional area in square centimeters,
N is the number of turns,
B is the flux density in G (Gauss)

Standard EI laminations can take 20kG (20000G) and with some safety margin design for B<=15kG.

Often a stacking factor is used with the above formula to account for some leakage, which results in a slight modification:
B=E*10^8/(4.44*F*A*N*K)

where K is the stacking factor which depends on the lamination thickness and how they are interleaved. This number ranges from about 0.80 to about 0.95 and affects the inductance by reducing permeability which in turn raises the flux density for a given number of turns.
 
Last edited:
There are two essential equations in transformer design. The first one is the EMF equation, which provide the number of turns (Np), given a primary RMS voltage (Ep), frequency (f), magnetic core effective cross-sectional area (A) and maximum flux density (B), which for common magnetic steels is 1.5 Tesla:

Np = E/(4.44*f*A*B)

Now the turns ratio is calculated:

Np/Ns = Ep/Es

Once that you finish those two steps, you go to a magnetic wire table, and determine the wire gage based on the load current. Since the primary and secondary currents are different, the wire gages will be different.
Finally, in the same wire table, obtain the wire's cross sectional area for the primary and secondary gauges, multiply them by the number of primary and secondary turns respectively, and see if they will fit inside the core's window area with sufficient margin. The window area is provided in the magnetic steel's datasheet.

Most likely your wire's area will be either too large or too small compared to the window area. You have three options: select a different steel size with a different window area, change the stack depth to adjust the core cross sectional area, or adjust your wire size and/or turns and have efficiency tradeoffs.

Transformer design is an iterative process. There are some software packages, like Col McLyman's, which will make some smart assumptions and give you a first-down solution. But you still have to check the suggested values and adjust them slightly.

Edit, Mr Al beat me to the response! And yes, you have to add a stacking factor because no steel lamination will be completely flat. He uses gauss in his formula, 1 Tesla = 10,000 gauss. Just be consistent with the units. And yes, you can run steel to 20,000 gauss, but 15,000 is more conservative and will yield lower losses.
 
Last edited:
Status
Not open for further replies.

Latest threads

New Articles From Microcontroller Tips

Back
Top