Transcript Impedance Transformation - Amirkabir University of Technology

Real Transformer
1- There are flux leakages
2- Some power loss
(copper, core losses)
3- Limited permeability
Let’s see what effects these realities would have on voltage and
current relation of the primary and the secondary
Q:Can we modify the ideal transformer model to take these
realities into account?
A: Yes
1-Flux leakage
With no flux leakage
d
vp  N p
dt
d
vs  N s
dt
vp
vs

Np
Ns

Now with flux leakage
d p
dt
 p   M   Lp
d s
vs  N s
dt
 s   M   Ls
d Lp
d M
vp  N p
 Np
dt
dt
v p  e p  eLp
d M
d Ls
vs  N s
 Ns
dt
dt
vs  es  eLs
vp  N p
ep
es

Np
Ns

Q: Is there a way to model the flux leakage:
A: Yes, we can count for flux leakage using this
equivalent circuit
 p   M   Lp
d Lp
d M
vp  N p
 Np
dt
dt
v p  e p  eLp
eLp  N p
d Lp
dt
Manipulations of formulas
2
dip
d
d Fp
d N p .i p N p dip
eLp  N p  Lp  N p
 Np

 Lp
dt
dt Lp
dt Lp
Lp dt
dt
Lp 
Np
2
Lp
 eLp  Lp
dip
dt
v p  ep  eLp
The same for secondary
2
Ns
dis
Ls 
 eLs  Ls
Ls
dt
2- Some power loss (copper, core losses)
2-1- Copper loss
Pcu, p  R p . I
2
p
Pcu,s  Rs . I s2
2-2- Core (eddy and hysteresis) loss
3- Limited permeability
In ideal transformer:
 
0
F  
N p .i p  N s .is    0
ip
Ns 1


is N p 
That means in ideal transformer, if the current in the secondary
is zero (no load), the current in the primary should be zero as well
In real transformer:
F  
N p .i p  N s .is    0
That means in real transformer, if the current in the secondary
is zero (no load), the current in the primary is not zero
Magnetization current
Ideal Transformer Model
Real Transformer Model
Using phasor
Other Equivalent Circuits
The original equivalent circuit
Referred to the primary
The original equivalent circuit
Referred to the secondary
Some Simplifications
If we assume that
the current in Rc
and XM would not
change much if we
change its place,
If we ignore the core losses and magnetization current
These models are too simplified, the original model is too
complicated, we usually work with these models
Referred to the primary
Referred to the secondary
Determining the Parameters
Open-circuit test
Short-circuit test
Q: Given six values, how can we determine the
parameters of the equivalent circuit
Poc ,Voc , I oc
Psc ,Vsc , I sc
From open-circuit test,
Poc ,Voc , I oc
1
1
YE 
j
 Gc  jBM  YE   
Rc
XM
I oc
YE 
,
Voc
Poc
Poc  Voc I oc . cos    cos
Voc I oc
1
Short circuit test
Psc ,Vsc , I sc
Z SE  Req  jX eq  Z SE 
Z SE
Vsc

,
I sc
Psc
Psc  Vsc I sc . cos    cos
Vsc I sc
1
Note: We cannot split the series impedance into primary
and secondary components
In Summary
Given:
Poc ,Voc , I oc
Psc ,Vsc , I sc
If the test is done on the primary
For the equivalent circuit referred to the primary
I oc
1
1
1 Poc
j

  cos
Rc
X M Voc
Voc I oc
Vsc
1 Psc
Req  jX eq 
 cos
I sc
Vsc I sc
Two Obtainable Equivalent Circuit
To obtain the equivalent circuit referred to secondary,
Divide the parameters by the turn ratio square
Example
Equivalent circuit
Referred to the primary
Given:
8000/240
Equivalent circuit
Referred to the secondary
0.034 
143 
34 
0.172 