Transformers formulas

$E_{rms}$ in V, $f$ in Hz, $N$ = turns of that winding, ${\Phi }_m$ = peak core flux in Wb (${\Phi }_m=B_m{A}_i$, with $A_i$ the net iron cross-section). The derivation assumes the core flux is sinusoidal. It applies to the primary and secondary separately, each with its own $N$.

An ideal transformer has no winding resistance, no leakage flux, no core loss and an infinitely permeable core, so it needs no magnetising current. In a practical transformer, $V_1/V_2=a$ holds only approximately (closest at no load), and $I_2/I_1=a$ holds only when the no-load current $I_0$ is neglected.

$P_h$ = hysteresis loss (the Steinmetz exponent $x$ is empirical, commonly taken as 1.6). $P_e$ = eddy-current loss, where $t$ is the lamination thickness. With $B_m\propto V/f$: at constant $V/f$, $P_h\propto f$ and $P_e\propto f^2$. At constant $f$, $P_e\propto V^2$. At constant $V$, $P_e$ does not depend on $f$, and $P_h\propto V^x{f}^{1-x}$.

Here $a=N_1/N_2$. To refer a primary quantity to the secondary, use the inverse: $V_1^{\prime }=V_1/a$, $I_1^{\prime }=a{I}_1$, $Z_1^{\prime }=Z_1/a^2$. An impedance always grows when it is referred to the side with more turns.

$Z_{01}=\sqrt{R_{01}^2+X_{01}^2}$ and $Z_{02}=\sqrt{R_{02}^2+X_{02}^2}=Z_{01}/a^2$. In the approximate equivalent circuit the shunt branch ($R_c$ in parallel with $X_m$) is moved to the supply terminals, ahead of the series impedance. This is justified because the no-load current $I_0$ is a small fraction of rated current.

These values hold at the voltage and frequency of the test. Core loss is not linear in flux, so $R_c$ and $X_m$ are not constants at other voltages.

$R_{eq}$ and $X_{eq}$ are referred to the energised winding. Multiply or divide by $a^2$ before combining them with quantities on the other side.

If the SC test was not run at rated current, scale its reading by the square of the current ratio to get the full-load copper loss. $\%Z$ is the SC-test voltage at rated current, expressed as a percentage of rated voltage. It has the same value whichever side it is calculated on, provided $I$, $Z$ and $V$ all belong to that side.

$V_{2,nl}$ is the no-load secondary voltage and $V_{2,fl}$ the full-load (rated) secondary voltage. Some texts divide by $V_{2,nl}$ instead ("regulation down"). These notes use the rated full-load voltage as the base.

Use $+$ for a lagging and $-$ for a leading power factor. $I$ and $V$ are the rated current and voltage on the same side as $R_{eq}$, $X_{eq}$ (that is, $R_{01}$, $X_{01}$ with primary quantities, or $R_{02}$, $X_{02}$ with secondary quantities). ${\epsilon }_r=I{R}_{eq}/V\times 100$ and ${\epsilon }_x=I{X}_{eq}/V\times 100$ are the percentage resistance and reactance. The formula neglects the quadrature component of the drop.

Zero regulation is possible only at a leading power factor, where the resistive drop in phase with $V$ is cancelled by the component of the reactive drop. At any leading power factor more leading than this, the regulation is negative: the terminal voltage rises on load.

Obtained by setting $\frac{d}{d\varphi }\left(R_{eq}\cos \varphi +X_{eq}\sin \varphi \right)=0$. Maximum regulation occurs when the load power-factor angle equals the impedance angle of the series impedance, and its value is then the percentage impedance.

$S$ is the rated VA. $xS\cos \varphi =V_2{I}_2\cos \varphi$ is the output. $P_i$ is the constant core loss (from the OC test at rated voltage). $P_{cu,fl}=I_{2,rated}^2{R}_{02}$ is the full-load copper loss (from the SC test at rated current). The formula assumes the terminal voltage stays at its rated value.

Efficiency is highest at the load where the variable copper loss equals the constant core loss. That load is $x_{\eta max}S$ VA, or $I_2=I_{2,rated}\sqrt{P_i/P_{cu,fl}}$. The load fraction does not depend on power factor. The value of ${\eta }_{max}$ does: it is highest at unity power factor.

This is the ratio of energy output to energy input over 24 hours, with $t_k$ in hours and $P_{out,k}=x_{k}S\cos {\varphi }_k$. The core loss is counted for all 24 h, including no-load periods, because it is present whenever the transformer is energised. Copper loss is counted only for the hours the transformer carries each load.

$Z_A$ and $Z_B$ are complex ohmic impedances referred to the same side. $z_A$ and $z_B$ are per-unit impedances, each on its own transformer's rating. Each unit's share is inversely proportional to its ohmic impedance. The share is proportional to its rating only when $z_A=z_B$. The unit with the lower per-unit impedance takes more than its proportional share and reaches full load first.

The common winding carries $I_{LV}-I_{HV}$ and the series winding carries $I_{HV}$. Only $S_{ind}$ passes through transformer action, so the windings and core need only be sized for $S_{ind}$.

This compares the same output and the same voltages, with the same current density and the same mean length of turn in both designs. The saving approaches 100% as $a\to 1$ and is small for large ratios. For example, $a=2$ saves 50% and $a=4$ saves only 25%.

Each winding keeps its rated voltage and current, so the inductively transformed power equals the two-winding rating $S_{2w}$. The auto-transformer's losses are the same as the two-winding transformer's, but its output is larger, so its efficiency is higher. Its per-unit series impedance on its own rating is $(1-1/a_{auto})$ times that of the two-winding transformer. This gives lower regulation and a higher short-circuit current.