Single Phase Auto Transformer|Construction Of Stepdown Auto-Transformer

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 = Tbcnumber 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
ILoutput current on lv side
IH = Iab = current in series winding
I = Icbcurrent 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;

Iab.Tab = Ibc.Tbc
IH (TH - TL) = I.TL
I/IH = (TH  - TL)/TL = TH/TL - 1 = a- 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;

IL = IH + I
I = IL - IH

Since IH and I are in phase


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