# Autotransformer.

## Presentation on theme: "Autotransformer."— Presentation transcript:

Autotransformer

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

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

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”

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.

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

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

Two-Winding Transformer connected as an Autotransformer
Reconnected as Autotransformer

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.

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

Example 3.2 continued As an autotransformer

Example 3.2 Simulation

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

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

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

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

Available Buck-Boost Voltage Ratios

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.

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

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

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

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