Blog 14: How does a Transformer work?
Transformers are devices that transfers electrical energy from one circuit to another using electromagnetic induction. It is basically used to increase or decrease the voltage level between the circuits. The transformer in the simplest way can be described as a thing that steps up or steps down voltage. In a step-up transformer, the output voltage is increased and in a step-down transformer, the output voltage is decreased. The step-up transformer will decrease the output current and the step-down transformer will increase the output current for keeping the input and the output power of the system equal.
A transformer that increases voltage between the primary to secondary winding is defined as a step up transformer. Conversely, a transformer that decreases voltage between the primary to secondary winding is defined as a step down transformer.
Whether the transformer increases or decreases the voltage level depends on the relative number of turns between the primary and secondary side of the transformer.
If there are more turns on the primary coil than the secondary coil than the voltage will decrease (step down).
If there are less turns on the primary coil than the secondary coil than the voltage will increase (step up).
But how does transformer work?
As we have learned from the previous blog that the number of coil turns on the secondary winding compared to the primary winding, the turns ratio, affects the amount of voltage available from the secondary coil. But if the two windings are electrically isolated from each other, how is this secondary voltage produced?
A transformer basically consists of two coils wound around a common soft iron core. When an alternating voltage ( VP ) is applied to the primary coil, current flows through the coil which in turn sets up a magnetic field around itself. This effect is called mutual inductance according to Faraday’s Law of electromagnetic induction.
Say you have one winding (also known as a coil) which is supplied by an alternating electrical source. The alternating current through the winding produces a continually changing and alternating flux that surrounds the winding.
If another winding is brought close to this winding, some portion of this alternating flux will link with the second winding. As this flux is continually changing in its amplitude and direction, there must be a changing flux linkage in the second winding or coil.
According to Faraday’s Law of Electromagnetic Induction, there will be an EMF induced in the second winding. If the circuit of this secondary winding is closed, then a current will flow through it. This is the basic working principle of a transformer.
Let us use electrical symbols to help visualize this. The winding which receives electrical power from the source is known as the ‘primary winding’. In the diagram below this is the ‘First Coil’.
The winding which gives the desired output voltage due to mutual induction is commonly known as the ‘secondary winding’. This is the ‘Second Coil’ in the diagram above.
The strength of the magnetic field builds up as the current flow rises from zero to its maximum value which is given as dΦ/dt.
As the magnetic lines of force set up by this electromagnet expand outward from the coil the soft iron core forms a path for and concentrates the magnetic flux. This magnetic flux links the turns of both windings as it increases and decreases in opposite directions under the influence of the AC supply.
However, the strength of the magnetic field induced into the soft iron core depends upon the amount of current and the number of turns in the winding. When the current is reduced, the magnetic field strength reduces.
When the magnetic lines of flux flow around the core, they pass through the turns of the secondary winding, causing a voltage to be induced into the secondary coil. The amount of voltage induced will be determined by:
N*dΦ/dt (Faraday’s Law)
where N is the number of coil turns. Also, this induced voltage has the same frequency as the primary winding voltage.
Then we can see that the same voltage is induced in each coil turn of both windings because the same magnetic flux links the turns of both the windings together. As a result, the total induced voltage in each winding is directly proportional to the number of turns in that winding. However, the peak amplitude of the output voltage available on the secondary winding will be reduced if the magnetic losses of the core are high.
If we want the primary coil to produce a stronger magnetic field to overcome the core’s magnetic losses, we can either send a larger current through the coil, or keep the same current flowing, and instead increase the number of coil turns ( NP ) of the winding. The product of amperes times turns is called the “ampere-turns”, which determines the magnetizing force of the coil.
The rate of change of flux linkage depends upon the amount of linked flux with the second winding. So ideally almost all of the flux of primary winding should link to the secondary winding. This is effectively and efficiently done by using a core-type transformer. This provides a low reluctance path common to both of the windings.
The purpose of the transformer core is to provide a low reluctance path, through which the maximum amount of flux produced by the primary winding is passed and linked with the secondary winding.
The current that initially passes through the transformer when it is switched on is known as the transformer inrush current.
Transformer EMF Equation:
So assuming we have a transformer with a single turn in the primary, and only one turn in the secondary. If one volt is applied to the one turn of the primary coil, assuming no losses, enough current must flow and enough magnetic flux generated to induce one volt in the single turn of the secondary. That is, each winding supports the same number of volts per turn.
As the magnetic flux varies sinusoidally, Φ = Φmax sinωt, then the basic relationship between induced emf, ( E ) in a coil winding of N turns is given by:
emf = turns x rate of change
E= N dΦ/dt
E=N x ω x Φmax x cos(ωt)
Emax= N x ω x Φmax
Erms= (N x ω/√2) x Φmax = (2π/√2) x ƒ x N x Φmax
Erms= 4.44 x ƒ x N x Φmax
ƒ – is the flux frequency in Hertz, = ω/2π
Ν – is the number of coil windings.
Φ – is the amount of flux in webers
This is known as the Transformer EMF Equation. For the primary winding emf, N will be the number of primary turns, ( NP ) and for the secondary winding emf, N will be the number of secondary turns, ( NS ).
Also please note that as transformers require an alternating magnetic flux to operate correctly, transformers cannot, therefore, be used to transform or supply DC voltages or currents, since the magnetic field must be changing to induce a voltage in the secondary winding. In other words, transformers DO NOT operate on steady-state DC voltages, only alternating or pulsating voltages.
If a transformer primary winding was connected to a DC supply, the inductive reactance of the winding would be zero as DC has no frequency, so the effective impedance of the winding will therefore be very low and equal only to the resistance of the copper used. Thus the winding will draw a very high current from the DC supply causing it to overheat and eventually burn out because as we know I = V/R.
Transformer Basics Example
A single-phase transformer has 480 turns on the primary winding and 90 turns on the secondary winding. The maximum value of the magnetic flux density is 1.1T when 2200 volts, 50Hz is applied to the transformer primary winding. Calculate:
a). The maximum flux in the core.
b). The cross-sectional area of the core.
c). The secondary induced emf.
Since the secondary voltage rating is equal to the secondary induced emf, another easier way to calulate the secondary voltage from the turns ratio is given as:
Electrical Power in Transformers
Another one of the transformer basics parameters is its power rating. The power rating of a transformer is obtained by simply multiplying the current by the voltage to obtain a rating in Volt-amperes, ( VA ). Small single-phase transformers may be rated in volt-amperes only, but much larger power transformers are rated in units of Kilo volt-amperes, ( kVA ) where 1-kilo volt-ampere is equal to 1,000 volt-amperes, and units of Mega volt-amperes, ( MVA ) where 1 mega volt-ampere is equal to 1 million volt-amperes.
In an ideal transformer (ignoring any losses), the power available in the secondary winding will be the same as the power in the primary winding, they are constant wattage devices and do not change the power only the voltage to current ratio. Thus, in an ideal transformer the Power Ratio is equal to one (unity) as the voltage, V multiplied by the current, I will remain constant.
That is the electric power at one voltage/current level on the primary is “transformed” into electric power, at the same frequency, to the same voltage/current level on the secondary side. Although the transformer can step-up (or step-down) voltage, it cannot step-up power. Thus, when a transformer steps up a voltage, it steps down the current and vice-versa, so that the output power is always at the same value as the input power. Then we can say that primary power equals secondary power, ( PP = PS ).
Where: ΦP is the primary phase angle and ΦS is the second phase angle.
Note that since power loss is proportional to the square of the current being transmitted, that is: I2R, increasing the voltage, let’s say doubling ( ×2 ) the voltage would decrease the current by the same amount, ( ÷2 ) while delivering the same amount of power to the load and therefore reducing losses by a factor of 4. If the voltage was increased by a factor of 10, the current would decrease by the same factor reducing overall losses by a factor of 100.
Transformers Basics – Efficiency
A transformer does not require any moving parts to transfer energy. This means no friction or windage losses are associated with other electrical machines. However, transformers do suffer from other types of losses called “copper losses” and “iron losses” but generally these are quite small.
Copper losses, also known as I2R loss is the electrical power that is lost in heat as a result of circulating the currents around the transformer’s copper windings, hence the name. Copper losses represent the greatest loss in the operation of a transformer. The actual watts of power loss can be determined (in each winding) by squaring the amperes and multiplying by the resistance in ohms of the winding (I2R).
Iron losses, also known as hysteresis are the lagging of the magnetic molecules within the core, in response to the alternating magnetic flux. This lagging (or out-of-phase) condition is due to the fact that it requires power to reverse magnetic molecules; they do not reverse until the flux has attained sufficient force to reverse them.
Their reversal results in friction, producing heat in the core which is a form of power loss. Hysteresis within the transformer can be reduced by making the core from special steel alloys.
The intensity of power loss in a transformer determines its efficiency. The efficiency of a transformer is reflected in power (wattage) loss between the primary (input) and secondary (output) windings. Then the resulting efficiency of a transformer is equal to the ratio of the power output of the secondary winding, PS to the power input of the primary winding, PP, and is therefore high.
An ideal transformer would be 100% efficient, passing all the electrical energy it receives from its primary side to its secondary side. But real transformers on the other hand are not 100% efficient. When operating at full load capacity their maximum efficiency is nearer 94% to 96%, which is still quite good for an electrical device. For a transformer operating at a constant AC voltage and frequency, its efficiency can be as high as 98%. The efficiency, η of a transformer is given as:
Where: Input, Output, and Losses are all expressed in units of power.
Generally, when dealing with transformers, the primary watts are called “volt-amps”, VA to differentiate them from the secondary watts. Then the efficiency equation above can be modified to:
It is sometimes easier to remember the relationship between the transformer’s input, output, and efficiency by using pictures. Here the three quantities of VA, W, and η have been superimposed into a triangle giving power in watts at the top with volt-amps and efficiency at the bottom. This arrangement represents the actual position of each quantity in the efficiency formulas.
Transformer Efficiency Triangle
and transposing the above triangle quantities give us the following combinations of the same equation:
Then, to find Watts (output) = VA x eff., or to find VA (input) = W/eff., or to find Efficiency, eff. = W/VA, etc.