This document discusses transformers and capacitor smoothing circuits. It explains that transformers convert alternating voltage and current from one level to another with little power loss using electromagnetic induction without moving parts. It describes the principles of an ideal transformer, including its operation and efficiency calculations. It also covers transformer losses from I2R heating and core losses. The document then discusses using capacitors to smooth the output of rectifier circuits, providing approximations to calculate the required capacitor value to limit ripple voltage.

1. 1

2. 2
 This unit explains
 Conversion of electrical power at one alternating voltage and current
level to another, with little power loss
 Conversion does not involve any moving parts
 After completing this unit you should be able to
 Describe the principles of an ideal transformer.
 Analyse the operation of such a transformer.
 Calculate the efficiency of a transformer with I2R and core losses

3. 3
 Basic Transformers
 Transformer Losses
 Transformer Efficiency

4. 4
Some other common
Applications
1. Isolation (two grounds)
2. Voltage and Current
sensing

5. 5
 A simple single phase transformer comprises a steel core onto which
two separate coils are wound.
 The core is laminated to minimise eddy current losses.
 The coil connected to an alternating current (ac) the supply, is called
the “primary winding”.
 The coil connected to the load, is termed the “secondary winding”.
 Transformers can be used as Step up or Step down transformers
Primary
steel core

6. 6
 Time-varying flux will induce a voltage, and
hence a current.
 Time-varying flux also affects the magnetic core,
and the induced voltage will cause “eddy”
currents in the core, which depend on the
resistivity of the core.
 These currents dissipate energy in the form of
heat.
 Eddy currents are reduced by selecting:
 High-resistivity core materials, or by
 Laminating the core, introducing tiny,
discontinuous air gaps between core layers.
 Therefore, lamination of the core reduces eddy
currents greatly without affecting the magnetic
properties of the core.

7. E1 E1
7
 Open circuit means that there is no load connected to the secondary
winding
 A sinusoidal voltage V1|across the primary winding|, which has N1 turns.
 A small magnetising current I1 will flow in that winding and
establish a flux  in the core - (just like the toroid coil).
 The flux induces an emf E1 in the primary coil in accordance with
Faraday’s Law.
dt
d
N
E 1
1




8. 8
 The flux (1) is common to both coils it will also induce an emf E2 in the
secondary winding, which has N2 turns
 From E1we had
 From E1 and E2
 “turns ratio” or “transformation ratio =
dt
d
N
E 2
2



2
2
1
1
N
E
N
E
dt
d



2
1
2
1
N
N
E
E










2
1
,
turns
secondary
,
turns
primary
N
N
dt
d
N
E

1
1 

9. 9
We had “turns ratio” = 







2
1
N
turns,
secondary
N
turns,
primary
2
1
2
1
N
N
E
E

&
If the primary & secondary impedances are negligible, then:
 E1=V1, applied voltage
 E2=V2, secondary terminal voltage 2
1
2
1
N
N
V
V



10. 10
In practice:
 The windings will have a small finite resistance, and
 A very small fraction of the flux will fail to link both coils. This
we refer to as Flux Leakage
Fortunately, the effects of these are generally small and may be ignored
allowing the above equations to be used with reasonable accuracy.

11. 11
N1
V1
Primary current,
I1
E1
V2
E2


Load
Secondary current,
I2
N2

 A current I2 will start to flow when a load is connected to the
secondary circuit.
 I2 will set up its own flux 2, in a direction which will oppose the
main flux , tending to reduce it and eventually reducing E1
dt
d
N
E 1
1




Since

12. 12
N1
V1
Primary current,
I1
E1
V2
E2


Load
Secondary current,
I2
N2

 The potential difference (V1 - E1) in the primary circuit increases
 Raises both the primary current I1 and the main flux. i.e. I1 & 1
 Equilibrium is achieved when the flux  (hence E1) has been restored to its
original value
 The extra flux produced by the increase in the primary current cancels out the
flux 2 created by I2.
 In the steady state the mmfs (ampere turns) in the primary and secondary
windings balance
i.e. I1N1 = I2N2 or 1
2
2
1
N
N
I
I


13. 13
N1
V1
Primary current,
I1
E1
V2
E2


Load
Secondary current,
I2
N2

I2 2opposing 1  1  E1  (V1 - E1)   I1 
 1counteract 2  2  E2  (V2 – E2)   I2 
 and so on  a balance is reached
 In the steady state the mmfs (ampere turns) in the primary and
secondary windings
i.e. I1N1 = I2N2 or
1
2
2
1
N
N
I
I


14. 14
 I2R Losses
 These are losses in both, the primary and the secondary
 Losses are due to the resistance of the coils, hence losses in
each of the windings is equal to I2R
 These losses translate as heat and are often referred “copper
losses”.
N1
V1
Primary current,
I1
E1
V2
E2


Load
Secondary current,
I2
N2

 Core Losses
 Two types of losses caused by:
 Magnetic hysteresis effects, and
 Eddy current losses in the steel core.
There are two types of losses in a
transformer, namely:
constant and
independent of load.

15. 15
 Efficiency,
Primary,
N1 turns
R1 
V1
Primary current,
I1
E1
V2
E2 Z
Secondary current,
I2
Secondary,
N2 turns
R2 
Core losses, Pc
1
2
,
,
P
Power
Input
P
Power
Output


Input power = Output power + (I2R losses) + (core losses)
Output power = P2 = V2 I2 cos2 where (cos2) = power factor of the load
Primary (I2 R) Losses, P11 = I1
2 R1 & Secondary (I2 R) Losses, P12 = I2
2R2
Total Core loss, Pc = hysteresis loss + eddy current loss (= constant)
Input power = P1 = P2 + (P11 + P12) + Pc

16. 16
Ideal Transformer
 The losses are zero in an ideal transformer. Therefore the
input power (VA) is equal to the output power (VA).
I1 V1 = I2 V2
 The voltage and current relations are:
a = V1/V2 = I2/I1 or V2 =V1/a & I2= I1 a
 If a transformer increases the voltage, the current decreases
and vice versa.
V1 = E1
E2 =E 1 / a
I1 I2 = aI1
E2 = V2

17. 17
A 200V:50V 60Hz transformer supplies a load comprising a 10 resistor
in series with a 100mH inductor. Calculate the primary current.
200 V
I1
N1
N2 I2
10 
100 mH

18. 18
By definition, the load impedance is equal to the ratio of
secondary phasor voltage and current: Z2 = V2/I2
To find the reflected impedance we can express the above
ratio in terms of primary voltage and current:
Find the equivalent load impedance seen by the
voltage source (i.e., reflected from secondary to
primary) for the transformer in the Figure
where the ratio V1/I1 is the impedance seen by the
source at the primary coil, that is, the reflected load
impedance seen by the primary (source) side of the
circuit.
Thus, we can write the load impedance, Z2, in terms of the primary circuit voltage and
current; we call this the reflected impedance, Z’2:

19. 19
By definition, the load impedance is equal to the ratio of
secondary phasor voltage and current: Z2 = V2/I2
To find the reflected impedance we can express the
above ratio in terms of primary voltage and current:
where the ratio V1/I1 is the impedance seen by the
source at the primary coil, that is, the reflected load
impedance seen by the primary (source) side of the
circuit.
Thus, we can write the load impedance, Z2, in terms of
the primary circuit voltage and current; we call this the
reflected impedance, Z’2:
Thus, Z’2= α2Z2, as shown in the bottom figure, which depicts the equivalent
circuit with the load impedance reflected back to the primary.

20. 20
A 60kVA 1200V:240V single phase transformer dissipates 2kW
when its secondary is open-circuited. The resistance of the primary
(high voltage) winding is 0.8 and the resistance of secondary is
0.04. Calculate the efficiency of the transformer when it supplies
full load at 0.8 power factor lagging.
[88.1%]

21. 21

22. 22
 To calculate the precise value of smoothing capacitance required for a
particular application can be difficult as
 The rise in capacitor voltage is sinusoidal, and
 The fall is exponential.
 However, a reasonably accurate approximation may be found by
assuming that both the rise and fall are linear.

2
 2 5
2
 4 9
2

3
(t) rads
Vm
VR
discharge time
charge time
exponential fall
sinusoidal rise
linear approximations
ripple voltage, Vr

23. 23
VS VR
C
D
VS
VR
R
C
 
t
Vm
VS
 
t
Vm
VR
 
t
Vm
VS
 
t
Vm
VR

24. 24

2
 2 5
2
 4 9
2

3
(t) rads
Vm
VR
discharge time
charge time
exponential fall
sinusoidal rise
linear approximations
ripple voltage, Vr

25. 25
 If maximum peak to peak ripple voltage = Vr
 Average DC output voltage
 During the discharge time from , the only current that flows
is that through the capacitor and load resistor.
2
V
-
V
=
V r
m
DC
 
1
to
2






 

2
 2 5
2
 3
Vm
VR
discharge time
charge time
exponential fall
VS VR
C
D
VR

26. 26
 We know that ,
 however, by assuming that the drop in voltage is linear,
 and the current I is constant, this simplifies to
dt
dv
C
dt
dq
i 

t
v
C
t
q
I 

where v = total change (fall) in voltage
t = discharge time.
In this case: Fall in voltage = ripple voltage, Vr
Capacitor current, IC = load resistor current =
R
VDC
VS VR
C
D

2
 2 5
2
 3
Vm
VR
discharge time
charge time
exponential fall
1

27. 27

2
 2 5
2

Vm
VR
discharge time
charge time
exponential fall
t
v
C
dt
dv
C
dt
dq
IC 







 
2





 
2
5
The discharge time stretches from to

 2
t
 discharge time,
R
t
V
V
C
r
DC

Substituting this back in gives:
R
2
V
V
C
r
DC 


t
CV
R
V r
DC

R
t
V
V
C
r
DC

If the ratio 0.05
<
V
V
DC
r





 


2
5
1
IC =
R
VDC
Substituting in
1
Ripple-less
output voltage
Large Capacitor

28.  
t
VR
Vm
28
In full wave bridge rectifier there are twice as many pulses, therefore,
 Halving the capacitor discharge time.
 As a result the required smoothing capacitance Farads
R
V
V
C
r
DC 


VS VR
C
D
R
2
V
V
C
r
DC 


R
V
V
C
r
DC 


VS
VR
R
C
Smaller Capacitor

29. 29
Today we studied:
 Transformers:
 Conversion of ac|voltage -current from one level to another, with little
power loss
 Conversion does not involve any moving parts
 Described the principles of an ideal transformer:
 Analysed its operation.
 Looked at I2R and core losses
 We will look at Core losses in more details in the future
 Calculated the efficiency
 Revised RC Smoothing from Last week

30. 30
1. A 240V:12V 50Hz transformer supplies a secondary (low voltage) load comprising 3
resistor and 12.7mH inductor. Calculate the primary current. [0.12 –53.1 A]
2. A 240V:120V 60Hz transformer has its low voltage winding connected to 120V supply.
Across the high voltage winding is a resistor in series with a capacitor. The supply current is
(0.8 + j0.6)A. Calculate the resistance and capacitance. [384, 9.2F]
3. When a 100kVA 3.3kV : 240V 50Hz single phase transformer is operated at full load, 0.8
power factor, the total (I2R) loss and core loss are both 1kW. Calculate the efficiency when
the transformer operates at half load 0.8 power factor. [97.0%]
4. When a 15kVA 1500V : 150V 50Hz single phase transformer is operating at full load unity
power factor the total (I2R) loss and core loss are equal and the efficiency is 98%.
a) Calculate the core loss: [153W]
b) During one day the transformer loading cycle is
12 hours at 4kW output, 0.5 power factor
8 hours at 12kW output, 0.98 power factor
4 hours at 9kW output, 0.9 power factor.
Calculate
i.total output energy in the day [180kWh (=648MJ)]
ii.input energy during each of the three periods and total input energy in the
day.[50.33kWh (181MJ), 98.04kWh (353MJ), 36.88kWh (132.7MJ),
185.27kWh (666.9MJ)]
iii.Given that all day efficiency = , find the all day efficiency.
[97%] energy
input
daily
energy
output
daily