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Copyright © 2014 John Wiley & Sons, Inc. All rights reserved.
Chapter 2-1Transformers
Third Edition
P. C. Sen
Principles of Electric Machines
and
Power Electronics
Copyright © 2014 John Wiley & Sons, Inc. All rights reserved.
Transformer application 1: power transmission
Copyright © 2014 John Wiley & Sons, Inc. All rights reserved.
Ideal Transformer
1 1 1
2 2 2
1 1
2 2
dv e N
dt
dv e N
dt
v Na
v N
1, step-down transformer
1, step-up transformer
a
a
Assumptions:
1. Negligible winding resistance
2. No leakage flux
3. Infinite permeability of core
4. Zero core loss
𝑣1
𝑉1 𝜔 = 𝑗𝜔𝐿 ∗ 𝐼1(𝜔)
Voltage relation:
Excitation current:
⇒ 𝐼1 𝜔 =𝑉1 𝜔
𝑗𝜔𝐿
𝐿 =𝑁12𝜇0𝜇𝑟𝐴
𝑙
⇒ 𝐼1 𝜔 = 0
Copyright © 2014 John Wiley & Sons, Inc. All rights reserved.
Basic equations- general definition
1 1 2 2
1 1 2 2
1 2
2 1
net mmf=0
1
N i N i
N i N i
i N
i N a
1 1 2 2v i v i
Current relationship: Instantaneous power:
𝑁1𝑖1 𝑁2𝑖2
ℜ =𝑙
𝐴𝜇
Copyright © 2014 John Wiley & Sons, Inc. All rights reserved.
Polarity marking:
Like polarity: two entering
currents produce the same
direction of magnetic flux
Copyright © 2014 John Wiley & Sons, Inc. All rights reserved.
Voltage polarity 2
Parallel operation of two single-phase transformer
Copyright © 2014 John Wiley & Sons, Inc. All rights reserved.
Impedance Reflection:
𝑍2′ =
𝑉1𝐼1=𝑎𝑉2𝐼2𝑎
= 𝑎2𝑉2𝐼2= 𝑎2𝑍2
' 11 2
ZZ
a
Copyright © 2014 John Wiley & Sons, Inc. All rights reserved.
Example: Determine the primary and secondary currents for
the ideal transformer below if Zs = (18-j4) Ωand Z2 = (2+j1) Ω.
Copyright © 2014 John Wiley & Sons, Inc. All rights reserved.
Other single-phase transformers
• Single primary multiple secondary
windings
𝑉𝑖 =𝑉1𝑎𝑖
𝐼1 =𝑎𝑖𝐼𝑖
Copyright © 2014 John Wiley & Sons, Inc. All rights reserved.
Non- ideal transformer: model development 1
• Ideal transformer assumptions:1. Negligible winding resistance
2. No leakage flux
• Practical transformer: ideal transformer+ external
impedance
Copyright © 2014 John Wiley & Sons, Inc. All rights reserved.
• Ideal transformer assumptions:3. Infinite permeability of core
4. No core loss
• Practical transformer: Magnetizing reactance
Core resistance
Non- ideal transformer: model development 2
Copyright © 2014 John Wiley & Sons, Inc. All rights reserved.
Non- ideal transformer: model development 3
𝑉2′ = 𝑎𝑉2
𝐼2′ =
𝐼2𝑎= 𝐼1
𝑅𝑤2′ = 𝑎2𝑅𝑤2
𝑋𝑙2′ = 𝑎2𝑋𝑙2
𝑍2′ = 𝑎2𝑍2
Copyright © 2014 John Wiley & Sons, Inc. All rights reserved.
Approximate equivalent circuit
Voltage drops across
primary winding resistance
and reactance is quite small
Neglect excitation branch
Copyright © 2014 John Wiley & Sons, Inc. All rights reserved.
Determine equivalent circuit parameters- No load test
Procedure:• Apply rated voltage to either high-voltage or low-voltage side
• Primary current: exciting current
• Loss: core loss (the same for applying either high-voltage or low-voltage side)
Parameters obtained:
• Magnetizing reactance
• Core resistance
Copyright © 2014 John Wiley & Sons, Inc. All rights reserved.
Determine equivalent circuit parameters – short circuit
test
Procedure:• Short-circuiting one winding
• Apply rated current to the other winding
Parameters obtained:
• Primary and secondary resistance
• Primary and secondary leakage reactance
Current
source
Copyright © 2014 John Wiley & Sons, Inc. All rights reserved.
Transformer ratings: power, primary/secondary voltages
→ turn ratio, current ratings
Example: single-phase transformer, 10 kVA, 2200/220V, 60 Hz
(1) Determine core loss resistance and magnetizing inductance from no load
test.
(2) Derive the parameters for the approximate equivalent circuits referred to
the High voltage side
(3) Derive the parameter when refer to low voltage side
Copyright © 2014 John Wiley & Sons, Inc. All rights reserved.
Solution (practice): single-phase transformer,
10 kVA, 2200/220V, 60 Hz
Copyright © 2014 John Wiley & Sons, Inc. All rights reserved.
Voltage regulation
2 2
2
NL L
L
V VVR
V
' '
2 2
'
2
NL L
L
V VVR
V
Refer to the primary
Basic definition
Copyright © 2014 John Wiley & Sons, Inc. All rights reserved.
Load voltage is normally taken as rated voltage
'
1 2 rated
'
2 rated
100%V V
VRV
Voltage regulation
Copyright © 2014 John Wiley & Sons, Inc. All rights reserved.
Example: single-phase transformer,
10 kVA, 2200/220V, 60 Hz
Determine voltage regulation in percent for
(a) 75% full load, 0.6 power factor lagging
(b) 75% full load, 0.6 power factor leading
(c) Draw the phasor diagram for (a) and (b)
Copyright © 2014 John Wiley & Sons, Inc. All rights reserved.
Solution (practice)
Copyright © 2014 John Wiley & Sons, Inc. All rights reserved.
Efficiency of a transformer
out out out
in out loss out c cu
100%P P P
P P P P P P
2 2 2
cu 1 1 2 2 2 2w w eqP I R I R I R
•Loss of transformerCore loss (hysteresis and eddy current)
Winding (copper) resistive loss
•Core loss Almost constant
Obtain from no-load test
Copyright © 2014 John Wiley & Sons, Inc. All rights reserved.
Example (practice)
Example (practice): single-phase transformer,
10 kVA, 2200/220V, 60 Hz
Determine
Efficiency at 75% rated output and 0.6 PF