The net magnitude of voltage induced across the primary winding of a transformer to its secondary winding is basically determined by the ratio of the number of turns wound over the primary and the secondary sections.

## TURNS AND VOLTAGE RATIOS

Let's take the example of the figure A below. Here we can see the primary winding consisting of around 10 turns, while the secondary with just a single turn winding. We know that in the process of electromagnetic induction, the lines of flux generated across the primary winding in response to the fed input AC, alternately expand and collapse, cutting through the ten turns of the winding. The same effect is also induced proportionately across the single turn present in the secondary section of the transformer.

However since the secondary has only a single turn, it experiences an identical magnitude of magnetic flux that may be equivalent across a single turn out of the 10 turns of the primary.

This implies that if the voltage applied across the primary is 12V, then each of it winding would be subjected with a counter EMF of 12/10 = 1.2V, and this is the magnitude of voltage that would be influencing the single turn present across the secondary section, because it has a single winding which is capable of extracting only the same equivalent amount of induction that may be available across the single turn over the primary.

Thus the secondary with a single turn would be able to extract 1.2V from the primary.

The above explanation also indicates that the number of turns over a transformer primary corresponds linearly with the supply voltage across it and the voltage simply gets divided by the number of turns. Thus in the above case since the voltage is 12V, and the number of turns are 12, the net counter EMF induced over each of the turns would be 12/10 = 1.2V

Now let's visualize the figure B, it shows a similar type of configuration as in figure A expect the secondary which now has 1 additional turn, that is 2 numbers of turns.

Needless to say, that now the secondary would be going through twice as many lines of flux compared to the figure A condition which had just a single turn.

So here the secondary winding would read around 12/10 x 2 = 2.4V because the two turns would be influenced by a magnitude of counter EMF that may be equivalent across the two winding on the primary side of the trafo.

Figure (1)

Therefore from the above discussion we can conclude that in a transformer the relation between the voltage and number of turns across the primary and the secondary are quite linear and also proportional.

## Universal Formula for Calculating Transformer Turn Numbers

Thus the derived formula for calculating the number of turns for any transformer can be expressed as:

**Es/Ep = Ns/Np**

where **Esec = Secondary Voltage,**

**Ep = Primary voltage,**

**Ns = Number of secondary turns,**

**Np = number of primary turns. **

### Primary Secondary Turn Ration

It would be interesting to note that the above formula indicates a straightforward relation between the ratio of the secondary to primary voltage and secondary to primary number of turns, which are indicated to be proportionate and equal.

Therefore the above equation can be also expressed as:

**Ep x Ns = Es x Np**

Further on, we can derive the above formula for solving the Esec and Eprim as shown below:

**Es = (Ep x Ns)/Np**

similarly,

**Eprim = (Es x Np)/Ns**

When some three of the proportions in the above formulas are identified, the fourth magnitude could be easily determined.

**Case in point.** *A transformer possesses 200 number of turns in the primary section, 50 number of turns in the secondary, and 120 volts connected across the primary (Ep). What could be the voltage across the secondary (E s)?*

Given:

**Np = 200 turns**

**Ns = 50 turns**

**Ep = 120 volts**

**Es = ? volts**

Answer:

**Es = EpNs/Np**

Substituting:

**Es = (120V x 50 turns)/200 turns**

**Es = 30 volts **

**Scenario**. *Suppose we have 400 number of turns of wire in an iron-core coil.*

Assuming coil is required to be employed as the primary winding of a transformer, Calculate the number of turns that needs to be wound on the coil to acquire the secondary winding of the transformer to ensure a secondary voltage of one volt with a situation where the primary voltage is 5 volts?

Given:

**Np = 400 turns**

**Ep = 5 volts**

**Es = 1 volts**

**Ns = ? turns**

Answer:

**EpNs = EsNp**

Transposing for Ns:

**Ns = EsNp/Ep**

Substituting:

**Ns = (1V x 400 turns)/ 5 volts**

**Ns = 80 turns **

Bear in mind: The ratio of the voltage (5:1) is equivalent to the winding ratio (400:80). Occasionally, as a substitute for particular values, you find yourself assigned with a turn or voltage ratio.

In cases like this, you could possibly anticipate any number for one of the voltages (or winding) and work out the alternative value from the ratio. As an illustration, suppose a winding ratio is assigned as 6:1, you could imagine a quantity of turn for the primary section and figure out the secondary section number of turns (60:10, 36:6, 30:5, etc.).

The transformer in all of the above examples carries a lesser number of turns in the secondary section compared to the primary section. For that reason, you can find a smaller amount of voltage across the secondary of the trafo rather than across the primary side.

A transformer wherein the voltage across the secondary side is lower than the voltage across the primary side is referred to as a STEP-DOWN transformer.

The ratio of a four-to-one step-down transformer is inscribed as 4:1. A transformer which includes lesser number of turns in the primary side compared to the secondary side will generate a higher voltage across the secondary side compared to the voltage connected across the primary side.

A transformer wherein the voltage across the secondary side is above the voltage present on the primary side is referred to as a STEP-UP transformer.

The ratio of a one-to-four step-up transformer needs to be inscribed as 1:4. As you can see in the two ratios that the magnitude of the primary side winding is consistently mentioned in the beginning.

### Impact Of Load

Whenever a load or an electrical device is hooked up across the secondary winding of a transformer, current or amps runs across the secondary side of the winding along with the load.

The magnetic flux generated by the current in the secondary winding interacts with the magnetic lines of flux generated by the amps in the primary side. This communication between the two lines of fluxes outputs from the shared inductance between the primary and secondary winding.

**Mutual Flux**

The absolute flux in the core material of the transformer is prevalent to both the primary and secondary windings. It is additionally the ways through which electrical power is conveyed from the primary winding to the secondary winding.

Given that this flux unites both the windings, the phenomenon generally known as MUTUAL FLUX. The inductance which generates this flux is furthermore prevalent to both windings and is termed mutual inductance.

Figure (2) exhibits the flux created by the currents in the primary and secondary windings of a transformer each time supply current is running in the primary winding.

Figure (2)

Whenever the load resistance is plugged into the secondary winding, the voltage stimulated into the secondary winding triggers current to circulate in the secondary winding.

This current produces a flux rings around the secondary winding (indicated as dotted lines) that may be as an alternative to the flux field around the primary (Lenz's law).

Consequently, the flux around the secondary winding cancels most of the flux around the primary winding.

With a smaller amount of flux encircling the primary winding, the reverse emf is cut down and more amp is sucked from the supply. The supplementary current in the primary winding releases additional lines of flux, pretty much reestablishing the initial amount of absolute flux lines.

### TURNS AND CURRENT RATIOS

The quantity of flux lines produced in a trafo core is proportional to the magnetizing force

**(IN AMPERE-TURNS)** of the primary and secondary windings.

The ampere-turn (I X N) is indicative of magneto motive force; it can be understood to be the magnetomotive force produced by one ampere of current running in a coil of 1 turn.

The flux which is available in the core of a transformer surrounds together the primary and secondary windings.

Given that the flux is identical for each windings, the ampere-turns in each the primary and secondary windings should always be the very same.

For that reason:

**IpNp = IsNs**

Where:

**IpNp** = ampere/turns in the primary winding

**IsNs** - ampere/turns in the secondary winding

By dividing both sides of the expression by

IpN s, we get:

**Np/Ns = Is/Ip**

since: **Es/Ep = Ns/Np**

Then: **Ep/Es = Np/Ns**

Also: **Ep/Es = Is/Ip**

where

Ep = voltage applied across primary in volts

Es = volatge across the secondary in volts

Ip = current in the primary in Amp

Is = current in the secondary in Amps

Observe that the equations indicate the ampere ratio to be the inverse of the winding or the turn ratio as well as the voltage ratio.

This implies, a transformer possessing fewer number of turns in the secondary side compared to the primary may likely step down the voltage, however it would step-up the current. For instance:

A transformer suppose has a 6:1 voltage ratio.

Try to find the current or amps in the secondary side if the current or amp in the primary side is 200 milliamperes.

Suppose

**Ep = 6V (as an example) **

**Es = 1V**

**Ip = 200mA or 0.2Amps**

**Is = ?**

Answer:

**Ep/Es = Is/Ip**

Transposing for Is:

**Is = EpIp/Es**

Substituting:

**Is = (6V x 0.2A)/1V**

**Is = 1.2A**

The above scenario addresses that despite the fact that the voltage across the secondary winding is one-sixth the voltage across the primary winding, the amps in the secondary winding is 6 times the amps in the primary winding.

The above equations could very well be viewed from an alternative perspective.

The winding ratio signifies the sum through which the transformer enhances or boosts or reduces the voltage connected to the primary side.

Just to illustrate, suppose if the secondary winding of a transformer has double as much number of turns as the primary winding, the voltage stimulated into the secondary side will probably be twice the voltage across the primary winding.

In case the secondary winding carries one-half the number of turns the primary side, the voltage across the secondary side is going to be one-half the voltage across the primary winding.

Having said that, the winding ratio along with the amp ratio of a transformer comprise an inverse association.

As a result, a 1:2 step-up transformer could have one-half the amp in the secondary side compared to the primary side. A 2:1 step-down transformer can have two times the amp in the secondary winding in relation to the primary side.

**Illustration:** *A transformer with a winding ratio of 1:12 possesses 3 amperes of current in the secondary side. Find out the magnitude of amps in the primary winding?*

Given:

**Np = 1 turn (for instance)**

**Ns = 12 turns**

**Is = 3Amp**

**Lp = ?**

Answer:

**Np/Ns = Is/Ip**

Substituting:

**Ip = (12 turns x 3 Amp)/1 turn**

**Ip = 36A **

Have questions? Please feel free to post them through comments! Comments will be moderated and solved ASAP.Basavesh BR says

Dear Sir

we devolved dc dc converter input 24v dc out put is 12v dc .and we need different outputs like 12v 15v 18v how to calculate the transformer winding.

Swagatam says

Dear Bhavesh, I am not an expert in designing transformers so won’t be ale to accurately explain you the details, I think the article already has all the info regarding the calculations, which can be used for calculating any trafo design…

Aabhishek Sharma says

hi swagatam sir,how much turns should be in secondary coil if in primary coil i have around 52 turns and transformer should be 230v to 12v.(ferite transformer)

Swagatam says

Hi Abhishek, It's explained in the above article, please use the given formulas to calculate it…by the way for a ferrite trafo the frequency must be also taken into consideration

Aziz Khalid says

Hi Swagatam,

I'm an Elec Eng senior and we were asked to build a single phase transformer. I already bought the core, I am trying to find the # of turns needed for primary and secondary windings. And I need to justify why I bought this specific core. That's why I need to figure out how to get turns per volt.

I'd really appreciate your insight,

Regards,

Aziz Khalid

California State University, Northridge

Swagatam says

Hi Aziz, If you are making an iron core transformer then you could probably refer to the following article for getting all the required details, It is explained with an example design:

http://www.homemade-circuits.com/2012/02/how-to-design-your-own-inverter.html

Taibani Imran says

Hello Swagatam,

Bro hope you're doing well.

Firstly thanks for all the great knowledge & information you have given here its really appreciated.

I am trying to do a project of home made low RPM VAWT generator which can generate enough power to run one small scale factory I need your help on winding section.

1) Correct copper winding design for low rpm.

2) Correct copper wire gauge.

3) Number of turns of winding.

4) What core material should be used for low drag ( Lenz effect ).

Please help me out & your readers with your great knowledge.

Thanks & Regards,

Taibani Imran.

Swagatam says

Hello Taibani, sorry I have no idea regarding the winding details of a vawt, but I think it would be a better idea to buy a readymade 10,000 watt dynamo and connect the VAWT with the dynamo through a calculated pulley ratio, where the VAWT pulley diameter could be 100 times bigger than the dynamo pulley.

here's an example dynamo which could be used for yur application

http://www.harborfreight.com/10000-watts-max-7200-watts-rated-belt-driven-generator-head-45416.html

Unknown says

Dear Swagatam,

I found a 100watt invertor circuit using 4047 ic and irfz44n and 12-0-12 transformer I made it and was working very well at 50hz thanks for the same,

then I tried the same with increased frequency above 2khz and using ferrite core transformer this time. I found that mosfets irfz44n heating up and ferrite core transformer making vibrating sound like squeeking sound

kindly guide me with help how to drive the mosfets above 1khz and what precautions to be taken

thanks

ananda

Swagatam says

Dear Ananda,

2kHz is too low for a ferrite based trafo, you must increase is to a minimum 30kHz to allow an optimal response from it.