Transformer operating principle and emf equation

A transformer is a static device that transfers electrical energy from one ac circuit to another at the same frequency, usually changing the voltage level, by mutual induction between two windings on a common magnetic core.

Construction (in brief)

Two windings — primary ($N_1$ turns) connected to the supply and secondary ($N_2$ turns) connected to the load — are placed on a laminated silicon-steel core. They are electrically isolated but magnetically coupled.

           phi ---->
     +---------------------+
 o---|]                   [|---o
 V1  |]  N1           N2  [|   V2 --> load
 o---|]  I1           I2  [|---o
     +---------------------+

Operating principle

1. An alternating voltage $V_1$ applied to the primary drives an alternating current, which sets up an alternating flux φ in the core.
2. Almost all of this flux is confined to the core and links the secondary winding.
3. By Faraday's law the changing flux induces an emf $E_2$ in the secondary (mutually induced emf) and an emf $E_1$ in the primary (self-induced back emf, which nearly balances $V_1$).
4. If a load is connected, current $I_2$ flows and power is delivered. The secondary mmf tends to reduce φ, so the primary draws extra current to restore it; in this way power passes from primary to secondary through the magnetic field.
5. Emf per turn is the same in both windings, so the voltages are in the ratio of turns: step-up if $N_2 > N_1$, step-down if $N_2 < N_1$. A dc supply gives no changing flux, so a transformer does not work on dc.

Derivation of emf equation

Let the core flux be $\phi = \phi_m \sin\omega t$, where $\omega = 2\pi f$.

Instantaneous emf in the primary:

$$\begin{aligned} e_1 &= -N_1\frac{d\phi}{dt} = -N_1 \phi_m \omega \cos\omega t \\ &= 2\pi f N_1\phi_m \sin(\omega t - 90^\circ) \end{aligned}$$

Maximum value: $E_{1,max} = 2\pi f N_1 \phi_m$.

RMS value:

$$\begin{aligned} E_1 &= \frac{2\pi f N_1\phi_m}{\sqrt2} = 4.44\, f N_1 \phi_m \ \text{V} \end{aligned}$$

Similarly $E_2 = 4.44\, f N_2 \phi_m$ V. Since $\phi_m = B_m A$, $E = 4.44\, f N B_m A$.

Voltage transformation ratio

$$K = \frac{E_2}{E_1} = \frac{N_2}{N_1}$$

The emf lags the flux by 90°.

Example: a 230/115 V, 50 Hz transformer with 400 primary turns has $\phi_m = 230/(4.44\times50\times400) = 2.59$ mWb, and needs $400\times115/230 = 200$ secondary turns.