TRANSFORMER - BASICS

CONSTRUCTION OF THE TRANSFORMER

Transformer has two wingdings, made of insulated copper, enameled is used as insulation. Windings are supported of core, which is made of iron. One winding is connected to supply, defined as primary winding and one winding connected to the load or not connected to the supply, defined as secondary winding. Primary winding has N₁ number of turns and secondary has N₂ number of turns.

Below is the sketch of a transformer.

According to FARADAY'S of Electromagnetism:

V = N dø/dt

E_p = N_p dø/dt

E_s = N_s dø/dt

E_S /E_{P =} N_S / N_P

Therefore,

E_S = (N_S / N_P) * E_p

E_PI_P = E_SI_S

E_PI_{P =} (N_S / N_P) * E_p * I_S

I_p = (N_s/N_p).I_s

AMPERES LAW:

If there is a current carrying conductor then it will produce a magnetic field around the conductor with some intensity (strength) ‘H’.

Converse: If there is a magnetic field intensity around a conductor, then there will exist a current, which will flow through conductor.

Where,

H is the field intensity of a conductor.

Unit - Ampere/meter

To find the current direction we used Right Hand Rule and Ampere’s Law:

Mathematical representation of Ampere’s Law:

Where,

N – Number of turns

H – Magnetic field intensity

L_m – Magnetic path length (conductor length)

mmf – Magnetomotive force

Hl_m = NI

H = NI / l_m     Amp/m

Flux Density:

Where,

1.      Magnetic Field intensity (H) = NI/L_m  = mmf/L_m

2.      Magnetic Flux Density (β) = ø/A_c

3.      Magnetic Flux Density (β) = µH

LOSSES IN TRANSFORMER:

• Hysteresis Loss
• Eddy current Loss

 

TOTAL LOSS IN TRANSFORMER:

E_sI_s = E_pI_p – P_h P_e P_cu

Where,

E_s – Secondary side voltage

I_s – Secondary side current

E_{p –} Primary side voltage

I_p  - Primary side current

P_h – Hysteresis Loss

P_e – Eddy current loss

P_cu – Winding or copper loss

 

HYSTERESIS LOSS:

β = µH = µNI/L_m

I = βL_m / µN -------------- Eq. 1 by Ampere’s Law

V = N dø/dt               ( Since ø = βA_c)

V = NA_c dβ/dt   by Faraday’s Law

INSTANTANEOUS ENERGY

dE = VI dt

= NA_c dβ/dt . βL_m / µN dt

V_c – Volume of core

V_c = A_c L_m

= V_c . (B / µ) dβ Joule

dE = (B / µ) dβ Energy per unit volume

Transverse:

“Higher the frequency a greater number of times this hysteresis loop is going to e traverse in a second therefore higher is going to be lost.”

Hysteresis energy (E_h) = 1/2 (BHV_C)

EDDY CURRENT LOSS:

“Eddy current flows in the high mass (solid core) of the core, if mass of the core is highly conductive then this current could be pretty high because resistance will be very low.

“To reduce the eddy current loss, what is generally done is to increase the resistance of the pathway (instead of solid core use laminated core) current flow.

What is lamination?

Construction of core done with the help of long strips of the shape of L, E and I to reduce the eddy current loss by narrowing the eddy current path.

ρ = RA/L

Where,

R is resistance of material (core)

A is cross section of material (core)

ρ is resistivity of a material (core)

“Ferrite material also can be used to reduce eddy current loss. Ferrite has very high resistivity,

Eddy current power loss:

P_e = (β_m)²  f²

B_m is the maximum density.

Core loss = Hysteresis loss + Eddy current loss

P_core = P_h + P_e

P_cu = i²R_cu

P_core + P_cu = Total loss in the transformer (P_loss)

In the next session to do with Transformer  Basics, we will look at Volt-Sec Balance, core material, polarity, Phasor  Diagram and power capacity of transformer and many more things... stay connected with us to learn in deep.