**Single Phase Auto Transformer**

**1.**One-Winding Transformer in which a part of winding is common to both high-voltage and low-voltage sides.

**2.**There are two types of single phase auto transformers,Stepup auto-transformer and Stepdown auto-transformer but Generally Stepdown auto-transformer is used.

**Construction Of Stepdown Auto-transformer**

**1.**Consider a single winding

**abc**of below given fig.

**2.**

**a**and

**c**- High voltage terminals.

**3.**

**b**and

**c**- Low voltage terminals where

**b**is suitable tapping point.

**4.**Portion

**bc**of the full winding

**abc**is common to both high voltage and low voltage side.

**5.**Winding

**bc**is called common winding and smaller winding

**ab**is called Series winding(because it is connected in series with common winding).

**6.**Primary voltage is greater than Secondary voltage.

7. circuit diagram

Single Phase Auto Transformer,Construction Of Stepdown Auto-Transformer |

TH = Tac = number of turns of full winding

**abc**= number of turns of hv side
TL = Tbc = number of turns of common winding

**bc**= number of turns of lv side
Tab = TH - TL = number of turns of series winding

**ab**
VH = input voltage on hv side

VL = output voltage on lv side

IH = input current on hv side

IL = output current on lv side

IH = Iab = current in series winding

I = Icb = current in common winding

**bc**
8. In auto-transformer there are two voltage ratio namely circuit voltage ratio and winding voltage ratio.

current voltage ratio = VH/VL = TH/TL = aA

9. aA is called Transformation ratio of the autotransformer.

10. when the load is connected across the secondary terminals a current flows in the common winding

**bc.**It has tendency to reduce the main flux but the primary current IH increases to such a value that the mmf in winding**ab**neutralizes the mmf in winding**bc**. that is;**I**ab.

**T**ab =

**I**bc.

**T**bc

**I**H (TH - TL) =

**I**.TL

**I**/

**I**H = (TH - TL)/TL = TH/TL - 1 = aA - 1 (eq. 1)

The winding - voltage ratio is;

a = Vab/Vbc = Tab/Tbc = (TH - TL)/TL

a = aA - 1 (eq. 2)

since; RHS of eq.2 is a pure number

therefore; current in winding

**ab**and**bc**are in phase.
By KCL at point b;

**I**L =

**I**H +

**I**

**I**=

**I**L -

**I**H

Since

**I**H and**I**are in phase**I**/

**I**H = I/IH

therefore eq. 1 becomes

I/IH = aA - 1

(IL - IH)/IH = aA - 1

IH = IL/aA (eq.3)

From (eq.1) and (eq.3)

I/IL = (aA - 1)/aA