Transformer Practical
Considerations
UCF
Magnetizing Current
N pi p − N s=
is H ( B )l=
F
C
N pi p − N sis = N p (i p −
Define: iM= i p −
Ns
is )
Np
Ns
is
Np
Φ
N piM= H ( )lC= F( Φ )
Ac
UCF
Effects of Peak Flux on Magnetizing Current
Saturated!
I M ,rms
1 T 2
=
iM (t )dt
∫
0
T
Current Inrush at Start-up (1)
UCF
Transient
=
vP (t )
2Vrms ,max=
cos(ωt + α ) N P
dΦ
dt
2Vrms ,max
1 t
=
Φ (t )
=
vP (t )dt
[sin(ωt + α ) − sin(α )]
∫
0
ωNP
NP
2Vrms ,max
2Vrms ,max
=
Φ max . No problem!
If α =
0, Φ (t) =
sin(ωt ), peak is
ω NP
ω NP
2Vrms ,max
π
π
− or vP (t ) =2Vrms ,max sin(ωt ), Φ (t) =
If α =
[cos(ωt ) + sin( )].
ω NP
2
2
at ω t = 0, Φ = 2 ×
2Vrms ,max
ω NP
= 2 × Φ max
Will induce very
high inrush current!
UCF
Current Inrush at Start-up (2)
vP (t ) = 2Vrms ,max sin(ωt )
Maximum Voltage
UCF
dΦ
Steady State
2V cos(ωt ) N p
v p (t ) =
=
dt
2V p ,rms
1
sin(ωt )
v p (t )dt
=
Φ
=
∫
Np
ωN p
2V
Φ max = p ,rms
2π fN p
V p ,rms=
,max
2π fN p Φ max ≈ 4.44 fN p Bmax Ac
Vs ,rms=
,max
2π fN s Φ max ≈ 4.44 fN s Bmax Ac
UCF
Voltage and Frequency Ratings
Volt-Second Balance
=
Vrms ,max
2π fN Φ max ≈ 4.44 fNBmax Ac
Vrms ,max
=
or:
f
2π N Φ max ≈ 4.44 NBmax Ac
UCF
Number of Turns and Cross Section of Core
From
Vrms ,max
=
2π fN Φ max ≈ 4.44 fNBmax Ac
⇒=
NAc
Vrms ,max
2π fBmax
Vrms ,max
≈
4.44 fBmax
UCF
Apparent Power Rating
Magnetic Loss (1)
UCF
Pm = pm (γ m ⋅ Vol m )
where pm = core loss power density at rated flux density (W/kg)
γ m = core material density (kg/m 3 )
Vol m = volume of the core
Core-form
Vol m = Ac ⋅ lc
Ac = ld ⋅ SF, SF is stacking factor
lc = 2(l + w + l + h)
⇒ Vol m = 2ld (2l + w + h) ⋅ SF
Pm = 2γ m pmld (2l + w + h) ⋅ SF
Magnetic Loss (2)
UCF
Shell-form
Based on what we got from core form (like 2 core together)
l
l
Vol m = 2 × 2 d (2 + w + h) ⋅ SF
2
2
⇒ Vol m = 2ld (l + w + h) ⋅ SF
Pm = 2γ m pmld (l + w + h) ⋅ SF
Copper Loss (1)
UCF
PCu = I P2 RP + I S2 RS
where I P = J Cu AwireP
I S = J Cu AwireS
J Cu is rated current density of conductor
RP = N P ρ Cu
l P ,average
AwireP
,
RS = N S ρ Cu
lS ,average
AwireS
ρ Cu is equivalent AC resistivity of copper
ρ Cu ≈ 1.05ρ Cu , DC at 60Hz
2
ρ Cu ( N P AwireP lP ,average + N S AwireS lS ,average )
PCu = J Cu
Since the primary and secondary are interleaved l P ,average ≈ lS ,average
2
2
⇒ PCu = J Cu
ρ Cu laverage ( N P AwireP + N S AwireS ) = J Cu
ρ Cu laverage Awindow kCu
where Awindow is the area of the window;
kCu is coil window conductor filling factor (needs insulation, cooling)
Copper Loss (2)
UCF
Core-form
w+l
2
ρ Cu laverage Awindow kCu
PCu = J Cu
Awindow = wh
laverage =
w+l
linner louter 2(l + d ) 2( w + l + d )
w
+
=
+
= 2(l + d + )
2
2
2
2
2
2
⇒ PCu = 2 J Cu
ρ Cu kCu wh(l + d +
w
)
2
Copper Loss (3)
UCF
Shell-form
2w+l
2
ρ Cu laverage Awindow kCu
PCu = J Cu
Awindow = wh (one window!)
laverage =
2w+l
linner louter 2(l + d ) 2(2 w + l + d )
+
=
+
= 2(l + d + w)
2
2
2
2
2
⇒ PCu = 2 J Cu
ρ Cu kCu wh(l + d + w)
UCF
Instrument Transformers (1)
Potential transformer
UCF
Instrument Transformers (2)
Current transformer (1)
UCF
Instrument Transformers (3)
Current transformer (2)
shorting interlock
UCF
Pole Transformer Fabrication
http://www.youtube.com/watch?v=tUO3o5JTGhQ