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 4.8-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

Two-Winding Transformer Reconnected as Autotransformer

1 2 2

2 2 2

2

( )

( 1)

at

w

at w

S V V I

S V I

S a S

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.

Example 3.2 continued

1041.67

24041.67

4.16710

LS

HS

kVAI A

VA

I A

As a two-winding transformer

Example 3.2 continuedAs an autotransformer

2

2400( 1) ( 1) 10 110

240at wS a S kVA

Example 3.2 Simulation

V1

2400 V 60 Hz 0Deg

LOAD63.35 Ohm

U1AC 1MOhm 2.640k V

+

-

T110

ItransformedAC 1e-009Ohm 4.357 A

+

-

IconductedAC 1e-009Ohm 41.672 A

+

-

XWM1

V IIinput

AC 1e-009Ohm

45.859 A+ -

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

For 240 V operation, connect H2 to H3

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) = 233.2 V

Example 3.3 part (b)

• The required step-down voltage ratio is a’ = VHS / VLS = 246/230 = 1.070

• Choose a’ = 1.0667 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)

Check this connection

Page 102 of the text by

Hubert

Output voltage = VHS / a’ = 246/1.0667 = 230.6 V