1. Autotransformer

2. Autotransformer connected for step-down operation
NHS = # of turns on the High Side
NLS = # of turns “embraced by the Low Side

3. Autotransformer Example
Turns ratio = a = NHS / NLs = NA / NB = 80 / 20 = 4
VLS = VHS / a = 120 V / 4 = 30 V
ILS = VLS / ZLOAD = 30/0.5 = 60A >> IHS = ILS / a = 60/4 = 15A

4. Autotransformer Example continued
How did the load current become 60A?
15A provided directly to the load by VHS
45A provided to the load by “transformer action”

5. Example 3.1
A 400-turn autotransformer, operating in the step-down mode with a 25% tap, supplies a kVA, 0.85 Fp lagging load. The input to the transformer is 2400-V, 60-Hz. Neglecting the small losses and leakage effects, determine
(a) the load current,
(b) the incoming line current,
(c) the transformed current,
(d) the apparent power conducted and the apparent power transformed.

6. Example 3.1 part a a = NHS / NLS = 400/(0.25)(400) = 4
VLS = VHS / a = 2400 / 4 = 600 V
ILS = 4800 VA / 600 V = 8 A = ILOAD

7. Example 3.1 parts b, c, d (b) ILINE = IHS = ILS / a = 8 A / 4 = 2 A
(c) ITR = ILS – IHS = (8 – 2) A = 6 A
(d) Scond = IHSVLS = (2 A)(600 V) = 1200 VA Strans = ITRVLS = (6 A)(600 V) = 3600 VA

8. Two-Winding Transformer connected as an Autotransformer
Reconnected as Autotransformer

10. Example 3.2
A 10-kVA, 60-Hz, 2400—240-V distribution transformer is reconnected for use as a step-up autotransformer with a 2640-V output and a 2400-V input.
Determine
(a) the rated primary and secondary currents when connected as an autotransformer;
(b) the apparent-power rating when connected as an autotransformer.

11. As a two-winding transformer
Example 3.2 continued
As a two-winding transformer

12. Example 3.2 continued
As an autotransformer

13. Example 3.2 Simulation

14. Buck-Boost Transformer
“Buck”>Subtract the low-voltage output from the line voltage
“Boost” >>> Add the low-voltage output to the line voltage

15. Buck-Boost Transformer voltages
120 X 240 V primary
12 X 24 V or 16 X 32 V secondary

16. 120/240 V operation For 120 V operation, connect H1 to H3 and H2 to H4

17. 12/24 V or 16/32V operation
For 12 V or 16 V operation, connect X1 to X3 and X2 to X4
For 24 or 32 V operation, connect X2 to X3

18. Available Buck-Boost Voltage Ratios

19. Example 3.3
The rated voltage of an induction motor driving an air conditioner is 230-V. The utilization voltage is 212-V.
(a) Select a buck-boost transformer and indicate the appropriate connections that will closely approximate the required voltage.
(b) Repeat (a), assuming the utilization voltage is 246-V.

20. Example 3.3 continued
The required step-up voltage ratio is a’=VHS / VLS = 230/212 = 1.085
Choose the best available voltage ratio from Table 3.1 as a’=1.100.
Need a 240-V primary and 12-V secondaries
Connect the 120-V primaries in series
Connect the 12-v secondaries in series

21. Example 3.3 (a)
Output Voltage = a’VLS = (1.100)(212) = V

22. Example 3.3 part (b)
The required step-down voltage ratio is a’ = VHS / VLS = 246/230 = 1.070
Choose a’ = from Table 3.1
Need a 240-V primary and 16-V secondaries
Connect the 120-V primaries in series
Connect the 16-V secondaries in parallel

23. Example 3.3 (b) Output voltage = VHS / a’ = 246/1.0667 = 230.6 V
Check this connection
Page 102 of the text by
Hubert
Output voltage = VHS / a’ = 246/ = V