• 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.

primary winding current.

Status
Not open for further replies.

electricity86

New Member
Hello.
It is said in wiki that the primary current equals Is*Ns/Np.
If there's no load connected to the secondary winding, then there's still current flowing in the primary winding.
But according to the formula from Wiki (that is derived from Pin=Pout),
if Is=0 then Ip=0.

What is right then?
 

Sceadwian

Banned
Some current is always going to flow in the primary. It's connected to an AC source after all, the impedance of the primary is usually designed to be as high as possible so this current is as low as possible. Another example of where theory doesn't match the real world.
 
Last edited:

electricity86

New Member
Thanks.
I drew the next circuit:

By ohm's law, Ip = (V_amplitude/√2) / (ωL1).
On the other hand, by conservation law (Pin=Pout), Ip = (Ns/Np)*Vs/R1.
As you can see, the results are different, so what is Ip equal to?
 

Attachments

Sceadwian

Banned
Look at he schematic, the primary is nothing but an inductor dead shorting an AC source to ground. Calculate the inductive reactance add the DC resistance of the coil and that's how much current is going through it.

I popped the inductance of your coil at 50hz into the calculator bellow and got an impedance of 3.14159k there's no resistance listed in your circuit so the current should be around 110ma. I'm not sure if I'm missing anything.

http://www.cvs1.uklinux.net/cgi-bin/calculators/ind_imp.cgi
http://www.cvs1.uklinux.net/cgi-bin/calculators/ind_imp.cgi
 
Last edited:

Sceadwian

Banned
Pin=Pout ONLY in theoretically ideal situations, which never occur in the real world.
 

crutschow

Well-Known Member
Most Helpful Member
Pin and Pout (real power) are equal. The input current due to the transformer inductance is reactive power. It does not contribute to real power (power that generates heat or energy).
 

Sceadwian

Banned
The resistance of the coil will cause real power to be lost. EVERY single transformer based wallwart I have ever used always stays warm even if there's no load.
 

electricity86

New Member
Pin and Pout (real power) are equal. The input current due to the transformer inductance is reactive power. It does not contribute to real power (power that generates heat or energy).
Thanks.
I wonder, what real power is consumed by and transferred by the primary winding to the secondary winding?

Maybe you could please explain the equation Pin=Pout?
What is exactly Pin and what is Pout?
 

Roff

Well-Known Member
Look at he schematic, the primary is nothing but an inductor dead shorting an AC source to ground. Calculate the inductive reactance add the DC resistance of the coil and that's how much current is going through it.

I popped the inductance of your coil at 50hz into the calculator bellow and got an impedance of 3.14159k there's no resistance listed in your circuit so the current should be around 110ma. I'm not sure if I'm missing anything.

Inductor Impedance Calculator
You apparently calculated impedance at 50MHz, not 50Hz. The impedance of 10uH at 50Hz is 3.14mΩ.
 

colin55

Well-Known Member
The thing that amazes me is the fact someone with so much ability can make a mistake like saying a 10uH inductor can be in the order of 3k resistance.
I have been dealing with 10uH inductors for 20 years and the first thing I thought was: how can be it be 3k when a 10uH inductor is just a few turns of wire?
This is the very thing that brought down one of the editors of an Australian Electronics magazine. He said something about charging a battery that was incorrect and it was obvious he had never charged a battery in his life.
From that day on, he was TOAST.
 

electricity86

New Member
Well,
Could you explain please what Pin relates to, and what Pout relates too?
To my knowledge, the primary coil transfers magnetic energy to the secondary coil through the magnetic flux lines.
This magnetic energy is reactive energy, right? since the magnetizing current is the one who creates it.
So how do the primary coil transfers real power to the secondary coil?
 

ericgibbs

Well-Known Member
Most Helpful Member
Well,
Could you explain please what Pin relates to, and what Pout relates too?
To my knowledge, the primary coil transfers magnetic energy to the secondary coil through the magnetic flux lines.
This magnetic energy is reactive energy, right? since the magnetizing current is the one who creates it.
So how do the primary coil transfers real power to the secondary coil?
hi,
There is a lot of well explained information on the web.:)

Transformer

Transformer - HvWiki
 

Sceadwian

Banned
colin, it was a typo in a calculator, I don't do the math in my head. I apologize for the typo. You've made bigger gaffs and never fessed up so I wouldn't be one to sling mud =)
 

Roff

Well-Known Member
colin, it was a typo in a calculator, I don't do the math in my head. I apologize for the typo. You've made bigger gaffs and never fessed up so I wouldn't be one to sling mud =)
I originally had the same thought as Colin, but I realized that it takes experience to instill that automatic sanity check alarm in one's head. Colin and I both have that alarm. You obviously don't. This is not a negative thing. It's just a fact.
 

Sceadwian

Banned
Very true, I recant weekly if not daily =) This is what I get for being an armchair engineer => I don't mind it from your Roff, but I do mind if from Colin a bit, because I've pointed out some blatant, horrible errors he's made in a few forum posts that he never responded to. I really should check my math before I make posts though, those glaring errors make me look like such an ass.
 
Last edited:

electricity86

New Member
hi,
There is a lot of well explained information on the web.:)

Transformer

Transformer - HvWiki
Thank you very much :)
According to whats written there:
Its Sin=Sout, and nout Pin=Pout, where Sin is the apparent power dissipated in the primary coil, and Sout is the apparent power dissipated in the secondary coil.
So the question still remains:
- Does Ip equal (V_amplitude/√2) / (ωL1) (by ohm's law) or (Ns/Np)*Vs/R1 (by Sin=Sout)?

 

Roff

Well-Known Member
- Does Ip equal (V_amplitude/√2) / (ωL1) (by ohm's law) or (Ns/Np)*Vs/R1 (by Sin=Sout)?
The primary current will be (V_amplitude/√2) / (ωL1) when the secondary is unloaded. This is called the magnetizing current, and lags the primary voltage by 90 degrees.
When loaded, the primary current is approximately (Ns/Np)*Vs/R1when the secondary is loaded. This is an approximation for a real transformer, but is generally adequate for most applications. See the Wikipedia entry for Transformer. Pay special attention the the Equivalent circuit section.
 

audioguru

Well-Known Member
Most Helpful Member
Some transformers have a high loss because they are never used at low power, always at max power like the transformer in a microwave oven. Then the windings do not have enough expensive turns of wire to make the transformer low loss.
 

grim

New Member
The resistance of the coil will cause real power to be lost. EVERY single transformer based wallwart I have ever used always stays warm even if there's no load.
that's because the laminations are still energized and approx half the Watts loss comes from the laminations and half from the copper;)
 
Status
Not open for further replies.

Latest threads

EE World Online Articles

Loading
Top