Auto-transformer operating principal

Operating principal

Auto-Transformer

The operating principal and construction of auto-transformer is same of two winding transformers. Only difference is way in which the primary and secondary windings are inter-related. In two winding transformers both winding is electrically insulated but magnetically linked by common core while in auto-transformer primary and secondary are electrically and magnetically connected, in fact a part of the single continuous winding is common to primary and secondary. On same silicon steel core a single winding is wound so both primary and secondary sections of this one winding are on the same magnetic circuit.

There are two type of auto-transformer based on construction.

1. In this continuous winding with taps brought out at convenient points determined by the designed secondary voltage 
2. Here there are two or more distinct coils which are electrically connected to form a continuous winding.

As shown in fig the primary winding is AB having of whole turns in between terminals A and B. let it be N₁. And voltage between A and B is V₁. There is tapped at C to obtain secondary winding BC. It consists N₂ turns and terminal voltage of V₂.

Now consider a fig. voltage per turn is V₁/N₁ so total voltage across BC is N₂V₁/N₁.

From fig. auto-transformer is look like a potential divider but it’s operation is quite different.

Auto-transformer	Potential divider
Can step up voltage	Cannot step-up voltage
In step-down auto transformer load current is more than the input current	Input current is more than its output current
A part of power is conducted and rest is transferred to the load by transformer action.	Entire power to the load flows by conduction
Not in case of auto-transformer	Great loss of power in the different resistances

Now finding the ratio consider voltage V₁ is applied and voltage V₂ is available across the secondary terminals. Neglecting losses.

                                                          V₁ I₁ cosΦ₁ = V₂ I₂ cosΦ₂

If internal leakage impedance drop and losses are neglect, then

                                                                   cosΦ₁  = cosΦ₁

                                                                                             V₁ I₁ = V₁ I₁

                                                             V₂/V₁ = I₁/I₂ = N₂/N₁ = K                                    

Where k is transformation ratio

In auto-transformer two-way energy is transfer first primary to the secondary side by transformer action, second by directly from the primary to secondary side as mentioned above.

How much power transformed and power conductively transfer is depended on the transformer ratio. Shown below.

In ideal condition mmf due to both winding is same so I₂ will oppose I₁ in the common section. Current I₁ is flow from A to C and current supplied to load is I₂. So current is BC section is (I₂ - I₁) flowing from B to C.

Power delivered to load = V₂ I₂ volt-amperes

Power in winding AC = E_AC I₁ = (V₁ – V₂) I₁ volt-amperes

Power transformed = power in winding BC

                             = V₂ (I₂ - I₁) =  V₂ I₂ (1 - I₁ /I₂) = V₂ I₂ (1 - K)

Ratio of power transformed to total power delivered = V₂ I₂ (1 - K)/ V₂ I₂ = (1 - K)

Power conducted directly  = power delivered to load – power transferred by transformer action

                                       = V₂ I₂ - V₂ I₂ (1 - K) = K V₂ I₂ = K × power output

Phasor diagram

First consider the direction so positive direction of applied voltage V₂ be down wards i.e. from

A to B, so the positive direction of current I₁ round the primary circuit is clock-wise, and that of current I₂ round the load circuit is also clock-wise. The positive direction of the induced emf is upward i.e. from B to C and from C to A. The applied voltage V₁ is balanced by induced emf in there two sections, plus voltage drop in them. For drawing phasor diagram start from load side.

Phasor Diagram

Take a refence as flux in horizontal

Draw V₂ in downward of magnitude of OA

Now I₂ lag the V₂ so draw OB lag OA at Φ₂

Current in BC section is I₂ – I₁ represent by OC

Draw AD parallel to I₂ to show resistive drop

Draw DE perpendicular to I₂ to show reactive drop

Draw OE to represent E_BC which is sum of voltage of load terminal + drop

Draw E_CA in phase with E_BC but magnitude is E_CA = E_BC (1-K)/K

Draw OG representing -E_CA

Current I₂ is oppose to the I₁ and in magnitude I₁ = -KI₂

Now draw the impedance triangle GJK for the CA section where GJ=I₁R₁ and JK=I₁X₁

                                                V₁ = V_AC + V_CB = V_AC + V₂     

So draw -V₂ LK and connect OL which is V₁ 

Reference

Theory & Performance of electrical machine by J. B. Gupta