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Autotransformer

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**Autotransformer connected for step-down operation**

NHS = # of turns on the High Side NLS = # of turns “embraced by the Low Side

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**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

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**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”

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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.

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**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

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**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

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**Two-Winding Transformer connected as an Autotransformer**

Reconnected as Autotransformer

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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.

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**As a two-winding transformer**

Example 3.2 continued As a two-winding transformer

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Example 3.2 continued As an autotransformer

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Example 3.2 Simulation

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**Buck-Boost Transformer**

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

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**Buck-Boost Transformer voltages**

120 X 240 V primary 12 X 24 V or 16 X 32 V secondary

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

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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

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**Available Buck-Boost Voltage Ratios**

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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.

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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

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

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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

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**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

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