# Why does transformer winding need large numbers of turns?

In transformers there are large number of turns in winding for both primary and secondary sides around the core. So, my question is why can not we just put one or two turns in winding as we need to maintain the turns ratio only? It will be cheaper and less space consuming.

Is it that we have to create sufficiently strong magnetic field (B=uNI) and so we are to get N with a large value?

One or the main reasons for so many turns is to keep the winding current (and resulting energy loss) low.

Assuming a very low winding resistance, the no-load current flowing in the primary winding depends almost entirely on the applied voltage and the reactance of the winding ($I= \frac V{X_L}$). This current is 90 degrees out of phase with the applied voltage, so is not in itself drawing energy from the source, but does cause $I^2R$ loss in the winding.

The reactance ($X_L$) of the primary winding depends on it's inductance and the frequency of the applied voltage $(X_L = 2\pi fL)$ .

The inductance (all other things staying the same) is proportional to $N^2$ ($L=µN^2A/l$), where N = the number of turns, so as you reduce the number of turns the inductance drops exponentially (and the current therefore rises exponentially).

If the frequency is relatively low (e.g. 50 or 60 Hz) then the inductance must be kept high in order to keep the current from being too large (unless voltage is quite low). Higher current results in more heating ($I^2R$ loss ) in the winding, and the transformer then runs hotter. Thicker wire would help to minimise this but the heating is proportional to $I^2$ and inversely proportional to R so it is better to use more turns of thinner wire (higher R) than to run the circuit at higher current (much higher $I^2$).
e.g. halving the number of turns would result in $\frac 1 2$ of the winding resistance and $\frac 1 4$ of the winding inductance (= $\frac 1 4$ of the reactance). If the resistance is much less than the reactance then there will be close to 4 times the winding current. This will result in 8 times the $I^2R$ loss in the winding (assuming the same size wire).

Higher power transformers tend to have fewer turns (and thicker wire), to keep the winding resistance down, but are wound on larger cores (larger cross-sectional area, A, in the inductance formula above) to keep the inductance up.

At higher frequencies fewer turns and smaller lighter cores can be used, as seen in switch mode power supplies which typically operate at many kilohertz. These would rapidly overheat and burn out at 50Hz.

• Detailed and informed answer with a practical perspective on the problem too. Thank you very much Peter. Jan 27 at 10:07

$$v(t) = L{{di} \over {dt}}$$ The self-inductance of a coil is proportional to the square of the number of turns of wire forming the coil: $${{{L_{\sec }}} \over {{L_{pri}}}} = {{{{({N_{pri}})}^2}} \over {{{({N_{\sec }})}^2}}} = {a^2},$$ where a is turns ratio.

The mutual inductance also has the relationship: $${M_{21}} = {N_1}{N_2}{P_{21}}$$ where:

${M_{21}}$ is the mutual inductance, and the subscript specifies the relationship of the voltage induced in secondary coil due to the current in primary coil.

${N_1}$ is the number of turns in primary coil,

${N_2}$ is the number of turns in secondary coil,

${P_{21}}$ is the permeance of the space occupied by the flux. Above image is from https://en.wikipedia.org/wiki/Inductance

Refer to above circuit while doing so we can determine: $${v_1} = {L_1}{{d{i_1}} \over {dt}} - M{{d{i_2}} \over {dt}}$$ Thus induced voltage due to nearby coil depends mutual inductance which in turn depends on number of turns in the coil.

• Increasing number of turns also increases flux linkage. As flux linkage= NBA
– user78818
Nov 5 '15 at 11:56