The input and output voltages of a power transformer allow us to derive the ratio of turns on the primary and secondary coils.

My question is how to find the exact number of turns of primary and secondary coils if input and output voltages and power in KVA are given?

For example a transformer with the characteristics Vp=11kV, Vs=400V and power=100KVA. Can we calculate the number of turns from this?

  • \$\begingroup\$ Can you not calculate it normally, just by forgetting power? \$\endgroup\$
    – Dzarda
    May 23 '14 at 20:39
  • 1
    \$\begingroup\$ No, you can find the ratio of the two. \$\endgroup\$ May 23 '14 at 20:52
  • 3
    \$\begingroup\$ You can make a fairly decent estimate of primary turns if you know the dimensions of the transformer core and assumed the primary magnetizing current was (say) 5% of full-load current. Calculate inductance required to take 5% F.L. current and note that L = N^2 / reluctance of core where N is number of turns. \$\endgroup\$
    – Andy aka
    May 23 '14 at 21:20
  • \$\begingroup\$ Is it a single phase or multi-phase transformer? What is the frequency? \$\endgroup\$
    – GR Tech
    May 23 '14 at 22:46
  • \$\begingroup\$ Will a Time Domain Reflectometer work to find the length of the wire in the primary and secondary? \$\endgroup\$ Jun 24 '14 at 15:07

I'm going to take a stab at this and maybe if I'm wrong about the material specifications of silicon steel laminates someone can point this out. There is almost enough information in the question to figure this out, given that you have to make some estimations about the size of the transformer....

I'd start from basics, B = \$\mu H\$ and, recognizing that the peak flux density of silicon steel\$^1\$ should be limited to about 2 teslas, H can be calculated from an estimation of the magnetizing current likely for a 100 kVA single phase power transformer.

Vp=11kV, Vs=400V and power=100KVA

Primary current will be about 10 A for full load (from above statement by OP) and, magnetization current will be about 5% of this (say) 0.5 A. Peak Mag Current will be 0.71 A. Noted. Also, using the classic B-H formula: -

B = \$4\pi\times 10^{-7}\times 40,000\times H\$ or

H = \$\dfrac{B}{4\pi\times 10^{-7}\times 40,000}\$ and for B = 2 teslas, H = 39.8 ampere-turns per metre.

What are ampere turns per metre - "amps" everyone knows, "turns" everyone knows and the "per metre" refers to the mean magnetic length of the core. For a toroid it is the average diameter of the ring multiplied by \$\pi\$. For a power transformer this might easily be 3 metres (a stab in the dark really because the kVA are out of my league!!)

So we have H = about 40 A.t/m and we know the amps (peak) at about 0.7 and I'm guessing the length is 3m. This means the number of turns is going to be: -

\$\dfrac{40\times 3}{0.7}\$ = 171 turns - can a sanity check be done on this I'm thinking?

I could estimate the cross section of the transformer core at 0.1m\$^2\$. Then I could calculate reluctance of the core based on this formula: -

Reluctance = \$\dfrac{effective\space length}{permeability\cdot effective\space area}\$. Plugging in real numbers we get: -

\$\dfrac{3m}{4\pi\times 10^{-7}\times 40,000\times 0.1m^2}\$ = 596.8

Knowing reluctance and turns I can calculate inductance: -

Inductance = turns-squared / reluctance = 171\$^2\$/597 = 49 henries.

Then I could estimate the reactance assuming 50 Hz and get 15.4kohms

Then I could calculate the magnetization current based on 11kV = \$\dfrac{11,000}{15,400}\$ = 0.71 amps.

This isn't far off from where I started when I assumed 0.5 A magnetization current so I guess the primary turns will be around about 171 turns and if the secondary voltage is 400 V then the secondary turns are about 6.218, so if I assume there are 7 secondary turns, the number of primary turns will be about 193.

I have no idea how far off this will be so don't go winding this at home and throwing a couple of wires over the nearest 11kV supply feed!

\$^1\$ Silicon steel is the likely candidate for the core of a big power transformer and this material has a relative permeability, \$\mu_R\$ of 40,000. See table halfway down this page. I'm no expert on power transformers so don't shoot me if it's wrong!!

EDIT - contrary to the site that states rel permeability is 40,000, wiki states \$\mu_R\$ for "electrical steel" is more like 4,000. This would make H, magnetic field strength calculated above more like 398 A.t/m. This would make the number of turns for the primary more like 1700. Anyone any idea what \$\mu_R\$ is going to be?

  • \$\begingroup\$ The number of turns seems low. Specs I've seen on transformers in this class usually have 50-100 turns on the secondary and over a thousand on the primary... \$\endgroup\$
    – Bitrex
    May 24 '14 at 6:52
  • \$\begingroup\$ @bitrex, maybe the quoted figure I got for relative permeability is wrong - it did seem rather high but I'm out of my comfort zone on this one! \$\endgroup\$
    – Andy aka
    May 24 '14 at 9:45

You'd need to count them.

You might be able to form a reasonable estimate by determining the gauge of the winding wire, then measuring the total resistance to find out how long the wire is. Then you figure out the length of the average loop (estimate).


Finding the exact number of turns on the primary and secondary of such a transformer, without further information, will be difficult. The turns ratio will be a rational approximation to the voltage ratio, and when there are thousands of turns on the primary there are many fractions primary turns/secodary turns which can approximate the voltage ratio with a great degree of accuracy.

In general, getting turns ratio from the input to output ratio alone becomed infeasible when the total number of turns on the transformer exceeds a few hundred or so.

Finding rational approximations to a given turns ratio is an application of a Stern-Brocot tree. If you could calculate the approximate number of turns on the primary and secondary using the method suggested by Daniel, a binary search through the tree would turn up candidate ratios that you might be able to narrow down with that additional information.


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