Autotransformer
Autotransformer connected for stepdown 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
S  (V  V )  I
at
1
S V I

2w
2
2
2
S  (a  1)  S
at
2w
2
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
As a two-winding transformer
10kV A
I 
 41.67A
240V
41.67A
I 
 4.167A
10
LS
HS
Example 3.2 continued
As an autotransformer
2400
S  (a  1)  S  (
 1)  10  110kV A
240
at
2w
Example 3.2 Simulation
XWM1
T1
10
V
I
Iinput
+
45.859
A
AC 1e-009Ohm
LOAD
63.35 Ohm
V1
2400 V
60 Hz
0Deg
+
4.357
+
U1
2.640k V AC 1MOhm
-
Itransform ed
A AC 1e-009Ohm
+
41.672
-
Iconducted
A AC 1e-009Ohm
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