DESIGN AND CALCULATION OF AN LOW POWER ONE PHASE TRANSFORMATOR
Is very nessary to know the importance that this electric device it has, is an static electric machine, it was invented with the propouse to solve the big problem, concernent to transmition and distribution of the electric energy, for conduce easy to long distances, of economic and security manners. this type energy. Because is more easy and transportable the a.c. (alternating current) than d.c. (direct current).
The great problem is that direct current (d.c.)need great diameter solid electric wires, but is different in a.c.
Only A.S.C. (aluminium steel core) type electrics cables, wich permit it flowing electrons in external surface, in a.c. (Skin effect, refference) is no nessesary solid area.
the fundamental reasson is for transmition (d.c.), we need have low (voltage) and high current (I). but in a.c. is different it is possible high (voltage) and low electric current. These carcteristics permit us, to reduce area in all electric conductors (wires), to constant electric power in both cases.
The transformator it can to increase or decrease the voltage. we can to obtain increase o decrease the electric current, also.
Then, the electric power: in the primary winding and secundary winding is same.
When two coils of wire are inductively coupled, the flux that passes through one of them, also passes enterely or in part through the other.
This fact implies that the coils have a magnetic circuit that is common to both. Now then, if this flux created by varying current through one coil changes, the mutual flux will change; under this condition, there will be created an induced voltage in the second coil.
Note: that the secondary induced voltage results because the flux changes through the coil, although this flux ocasioned in the first place by the current in the first, or primary, coil. The induced voltage the mutual induction in the secondary, coil is called a transformer voltage, and the action that creates the emf (electrical motrice force) and the it known as transformer action.
In the transformator, one coil or set of coils is connected directly to an a.c. source so that the currentand the resulting flux automatically change periodically in both magnitude an direction.
Since de linking flux in the coupled coil or set of coils changes a voltage is induced in the latter by transformator action. Furthermore, if there is no relative motion between the coils, the frecuency of induced voltage in the second coil is exactly the same as of the frecuency of the primary impressed voltage.(because la alternating current it has a frecuency of 60 cicles per second)
If the second coil is now connected to an electrical load such as incandescent lamps, electric motor or heting devices, current will flow. Thus electrical energy can be transferred from ene electric circuit to another electric circuit by transformer actión (By electromagnetic inducction) even through there is absolutely no metallic and electric connection between the two.
The whole remarkable process of energy transfer takes place by THE PRINCIPLE OF ELECTROMAGNETIC INDUCTION: it was first discovery by Michael Faraday in 1831.
If we derivate this mathematical function
we will have:
ep=[[( - Np )(Fm)(cosine (2)(3.1416)(f)(t) ]x(()(3.1416)(f)(t) x(10exp8 ]] Eq. 4
Where:
ep= ( - NP ) (FM )(2)(3.1416)(f)( t )( 10 exp 8 ) Eq. 5
If previusly divide per square root 2 for obtain RMS values
eq. 5
We will have:
EP = eP = [(NP )(2)(3.1416/ 1.414)(f)(FM )(10 exp 8 )]
Eq. 6
Therefore:
EP =( 4.44)(NP)(f)(FM)(10 exp 8)
where:
(Np)=(Ep )(10 exp 8) / ( 4.44)(f)(FM) for calculate the turns of primary coil.
This right now the formule, with it you can determinate the number of turns of primary winding conductors.
Of similar manner we can consider
Es =( 4.44)(Ns)(f)(FM)(10 exp 8) for calculate # conductors of secondary winding.
where:
(Ns)=(Es )(10 exp 8) / ( 4.44)(f)(FM)
And (FM) is equal to (B)(Ac) in Gausses.
In the next week, I will be publishing more steps for to design the transformator.
for calculation the core area we use the next
mathematical function:
Sc= (0.054---0.065) square root [(kva tot)x(Gc/Gk)x10 exp 11) / ( total magnetic fluxe )x( (frecuency )x(density current)].
where: Gc/ Gk weihg ratio , is aproximaty 8.0, and density current, range from 1.5 to 2.6 (amp/ square milimeter).
The power Kva = (v) (I) we may consider in referency to the secondary winding.
Note. Is necesary multiplay Kva per (1.12) factor, because compense the eddy current, foucoult and copper losses, and magnetic core.
off course, for calculation we may take to total magnetic fluxe from 6300 to 16000 gauss for silicon- iron steel , wich the caracteristics and propierties specific, from 2% to 4 % silicon.
for calculations we can consider 8300 gauss. ok.
I will be very thankeful if you can comment about this topic, that I wrote for you. I hope to recibe some soon.
"Is good to know, but is better you can apply a little of the knowledge, or not ?"
This is my personnal thinking away.
Sincerily Eng. Miguel Ismerio Espinosa
Only A.S.C. (aluminium steel core) type electrics cables, wich permit it flowing electrons in external surface, in a.c. (Skin effect, refference) is no nessesary solid area.
the fundamental reasson is for transmition (d.c.), we need have low (voltage) and high current (I). but in a.c. is different it is possible high (voltage) and low electric current. These carcteristics permit us, to reduce area in all electric conductors (wires), to constant electric power in both cases.
The transformator it can to increase or decrease the voltage. we can to obtain increase o decrease the electric current, also.
Then, the electric power: in the primary winding and secundary winding is same.
 |
| figure a |
When two coils of wire are inductively coupled, the flux that passes through one of them, also passes enterely or in part through the other.
This fact implies that the coils have a magnetic circuit that is common to both. Now then, if this flux created by varying current through one coil changes, the mutual flux will change; under this condition, there will be created an induced voltage in the second coil.
Note: that the secondary induced voltage results because the flux changes through the coil, although this flux ocasioned in the first place by the current in the first, or primary, coil. The induced voltage the mutual induction in the secondary, coil is called a transformer voltage, and the action that creates the emf (electrical motrice force) and the it known as transformer action.
In the transformator, one coil or set of coils is connected directly to an a.c. source so that the currentand the resulting flux automatically change periodically in both magnitude an direction.
 |
| figure b |
Since de linking flux in the coupled coil or set of coils changes a voltage is induced in the latter by transformator action. Furthermore, if there is no relative motion between the coils, the frecuency of induced voltage in the second coil is exactly the same as of the frecuency of the primary impressed voltage.(because la alternating current it has a frecuency of 60 cicles per second)
If the second coil is now connected to an electrical load such as incandescent lamps, electric motor or heting devices, current will flow. Thus electrical energy can be transferred from ene electric circuit to another electric circuit by transformer actión (By electromagnetic inducction) even through there is absolutely no metallic and electric connection between the two.
 |
| figure c |
The whole remarkable process of energy transfer takes place by THE PRINCIPLE OF ELECTROMAGNETIC INDUCTION: it was first discovery by Michael Faraday in 1831.
 |
| figure d |
 |
| figure e
This device that most commonly utilizes the principle of transformator action is the static transformator..
The foregoing may be summarized as follows: a transformer is a device that transfer electrical energy from one electric circuit to another does so without one cahange of frecuency, does so by the principle of electromagnetic induction, and has electric circuits that are linked by a common magnetic circuit.
The figures a, b, c, d and e, ilustrates a very simple transformer, in wich two coils are linked by lamminated magnetic steel core.
Transformer Voltages and the General Transformer Equation:
We may consider the next variables Np # turns of magnetic wire in the primary winding Ns # turns of magnetic wire in the secondary winding Fm Flux magnetic in the transformator core, this is an instanius values. in Wb (webers). f frecuency en Hertz, 60 Hz. generaly. Now if we can consider idealy for calculation. Rprimary= 0, Then: the Vp values will be: ep=[[ - Np d(Fm) ]/(dt )]x( 10exp8)] ] ( Equation 1) 10exp 8 = Is the magnetic lines per square centimeter for induce 1 volt. Like the current that feed to transformator is alternating type this, is sinus function. Then Vp= (Vp maximum) x (Sine wt) Equation 2 Where wt = (2)x (3.1416)x(f) and (t) is tieme in seconds. I hope that you have could see the alternating wave in the osciloscope. Where 1 cicle is equal (2)(3.1416) The magnetic flux developed per the electrical energy of all conductors of primary winding. Then : ep=[[ - Np d(Fm)(sine (2)(3.1416)(f)(t) ]x( 10exp8)]/d(t) Equation 3 |
we will have:
ep=[[( - Np )(Fm)(cosine (2)(3.1416)(f)(t) ]x(()(3.1416)(f)(t) x(10exp8 ]] Eq. 4
Where:
ep= ( - NP ) (FM )(2)(3.1416)(f)( t )( 10 exp 8 ) Eq. 5
If previusly divide per square root 2 for obtain RMS values
eq. 5
We will have:
EP = eP = [(NP )(2)(3.1416/ 1.414)(f)(FM )(10 exp 8 )]
Eq. 6
Therefore:
EP =( 4.44)(NP)(f)(FM)(10 exp 8)
where:
(Np)=(Ep )(10 exp 8) / ( 4.44)(f)(FM) for calculate the turns of primary coil.
This right now the formule, with it you can determinate the number of turns of primary winding conductors.
Of similar manner we can consider
Es =( 4.44)(Ns)(f)(FM)(10 exp 8) for calculate # conductors of secondary winding.
where:
(Ns)=(Es )(10 exp 8) / ( 4.44)(f)(FM)
And (FM) is equal to (B)(Ac) in Gausses.
In the next week, I will be publishing more steps for to design the transformator.
for calculation the core area we use the next
mathematical function:
Sc= (0.054---0.065) square root [(kva tot)x(Gc/Gk)x10 exp 11) / ( total magnetic fluxe )x( (frecuency )x(density current)].
where: Gc/ Gk weihg ratio , is aproximaty 8.0, and density current, range from 1.5 to 2.6 (amp/ square milimeter).
The power Kva = (v) (I) we may consider in referency to the secondary winding.
Note. Is necesary multiplay Kva per (1.12) factor, because compense the eddy current, foucoult and copper losses, and magnetic core.
off course, for calculation we may take to total magnetic fluxe from 6300 to 16000 gauss for silicon- iron steel , wich the caracteristics and propierties specific, from 2% to 4 % silicon.
for calculations we can consider 8300 gauss. ok.
I will be very thankeful if you can comment about this topic, that I wrote for you. I hope to recibe some soon.
"Is good to know, but is better you can apply a little of the knowledge, or not ?"
This is my personnal thinking away.
Sincerily Eng. Miguel Ismerio Espinosa
