ABSTRACT
A
In this circuit the
transistor turning ON will put voltage Vin
on one end of the inductor. This voltage will tend to cause the inductor
current to rise. When the transistor is OFF, the current will continue flowing
through the inductor but now flowing through the diode. We initially assume
that the current through the inductor does not reach zero, thus the voltage at Vx will now be only the voltage across the conducting diode
during the full OFF time. The average voltage at Vx
will depend on the average ON time of the transistor provided the inductor
current is continuous.

Fig. 1: Buck Converter

Fig. 2: Voltage and current changes
To analyse
the voltages of this circuit let us consider the changes in the inductor
current over one cycle. From the relation
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the change of current satisfies

For steady state
operation the current at the start and end of a period T will not change. To
get a simple relation between voltages we assume no voltage drop across
transistor or diode while ON and a perfect switch change. Thus during the ON
time Vx=Vin and in the OFF Vx=0.
Thus
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which simplifies to
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or
![]()
and defining "duty ratio"
as
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the voltage relationship becomes Vo=D
Vin Since the circuit is lossless and the input and
output powers must match on the average Vo* Io = Vin* Iin.
Thus the average input and output current must satisfy Iin
=D Io These relations are based on the assumption that the inductor
current does not reach zero.
When the current in the
inductor L remains always positive then either the transistor T1 or the diode
D1 must be conducting. For continuous conduction the voltage Vx is either Vin or 0. If the inductor current ever
goes to zero then the output voltage will not be forced to either of these
conditions. At this transition point the current just reaches zero as
seen in Figure 3. During the ON time Vin-Vout
is across the inductor thus
|
|
(1) |
The average current which
must match the output current satisfies
|
|
(2) |

Fig. 3: Buck Converter at Boundary
If the input voltage is
constant the output current at the transition point satisfies
|
|
(3) |
As for the continuous
conduction analysis we use the fact that the integral of voltage across the
inductor is zero over a cycle of switching T. The transistor OFF time is now
divided into segments of diode conduction ddT
and zero conduction doT. The
inductor average voltage thus gives
|
(Vin
- Vo ) DT + (-Vo) ddT = 0 |
(4) |

Fig. 4: Buck Converter - Discontinuous Conduction
|
|
(5) |
for the case
.
To resolve the value of
consider
the output current which is half the peak when averaged over the conduction
times ![]()
|
|
(6) |
Considering the change of
current during the diode conduction time
|
|
(7) |
Thus from (6) and (7) we
can get
|
|
(8) |
using the relationship in (5)
|
|
(9) |
and solving for the diode conduction
|
|
(10) |
The output voltage is
thus given as
|
|
(11) |
defining k* = 2L/(Vin
T), we can see the effect of discontinuous current on the voltage ratio of the
converter.

Fig. 5: Output Voltage vs
Current
As seen in the figure,
once the output current is high enough, the voltage ratio depends only on the
duty ratio "d". At low currents the discontinuous operation tends to
increase the output voltage of the converter towards Vin.
The schematic in Fig. 6
shows the basic boost converter. This circuit is used when a higher output
voltage than input is required.

Fig. 6: Boost Converter Circuit
While the transistor is
ON Vx =Vin, and the OFF state the inductor
current flows through the diode giving Vx
=Vo. For this analysis it is assumed that the inductor current
always remains flowing (continuous conduction). The voltage across the inductor
is shown in Fig. 7 and the average must be zero for the average current to
remain in steady state
![]()
This can be rearranged as

and for a lossless circuit the power
balance ensures
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Fig. 7: Voltage and current waveforms (Boost Converter)
Since the duty ratio
"D" is between 0 and 1 the output voltage must always be higher than
the input voltage in magnitude. The negative sign indicates a reversal of sense
of the output voltage.
3. BUCK-BOOST
CONVERTER

Fig. 8: schematic for buck-boost converter
With continuous
conduction for the Buck-Boost converter Vx
=Vin when the
transistor is ON and Vx =Vo
when the transistor is OFF. For zero net current change over a period the
average voltage across the inductor is zero

Fig. 9: Waveforms for buck-boost converter
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which gives the voltage ratio
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and the corresponding current
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Since the duty ratio
"D" is between 0 and 1 the output voltage can vary between lower or
higher than the input voltage in magnitude. The negative sign indicates a
reversal of sense of the output voltage.
The voltage ratios
achievable by the DC-DC converters is summarised in
Fig. 10. Notice that only the buck converter shows a linear relationship
between the control (duty ratio) and output voltage. The buck-boost can reduce
or increase the voltage ratio with unit gain for a duty ratio of 50%.

Fig. 10: Comparison of Voltage ratio
The buck, boost and
buck-boost converters all transferred energy between input
and output using the inductor, analysis is based of voltage balance across the
inductor. The CUK converter uses capacitive energy transfer and analysis is
based on current balance of the capacitor. The circuit in Fig. 11 is derived
from DUALITY principle on the buck-boost converter.

Fig. 11: CUK Converter
If we assume that the
current through the inductors is essentially ripple
free we can examine the charge balance for the capacitor C1. For the transistor
ON the circuit becomes

Fig. 12: CUK "ON-STATE"
and the current in C1 is IL1.
When the transistor is OFF, the diode conducts and the current in C1 becomes IL2.

Fig. 13: CUK "OFF-STATE"
Since the steady state
assumes no net capacitor voltage rise ,the net current
is zero
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which implies
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The inductor currents
match the input and output currents, thus using the power conservation rule
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Thus the voltage ratio is
the same as the buck-boost converter. The advantage of the CUK converter is
that the input and output inductors create a smooth current at both sides of
the converter while the buck, boost and buck-boost have at least one side with
pulsed current.
4. Isolated DC-DC
Converters
In many DC-DC applications,
multiple outputs are required and output isolation may need to be implemented
depending on the application. In addition, input to output isolation may be
required to meet saftey standards and / or provide
impedance matching.
The above discussed DC-DC
topologies can be adapted to provide isolation between input and output.
The flyback
converter can be developed as an extension of the Buck-Boost converter. Fig 14a
shows tha basic converter; Fig 14b replaces the
inductor by a transformer. The buck-boost converter works by storing energy in
the inductor during the ON phase and releasing it to the output during the OFF
phase. With the transformer the energy storage is in the magnetisation
of the transformer core. To increase the stored energy a gapped core is often
used.
In Fig 14c the isolated
output is clarified by removal of the common reference of the input and output
circuits.

Fig. 14(a): Buck-Boost Converter

Fig. 14(b): Replacing inductor by transformer

Fig. 14(c): Flyback converter
re-configured
The concept behind the foward converter is that of the ideal transformer
converting the input AC voltage to an isolated secondary output voltage. For
the circuit in Fig. 15, when the transistor is ON, Vin appears across the primary and then generates
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The diode D1 on the
secondary ensures that only positive voltages are applied to the output circuit
while D2 provides a circulating path for inductor current if the transformer
voltage is zero or negative.

Fig. 15: Forward Converter
The problem with the
operation of the circuit in Fig 15 is that only positive voltage is applied
across the core, thus flux can only increase with the application of the
supply. The flux will increase until the core saturates when the magnetising current increases significantly and circuit
failure occurs. The transformer can only sustain operation when there is no
significant DC component to the input voltage. While the switch is ON there is
positive voltage across the core and the flux increases. When the switch turns
OFF we need to supply negative voltage to rset the
core flux. The circuit in Fig. 16 shows a tertiary winding with a diode
connection to permit reverse current. Note that the "dot" convention
for the tertiary winding is opposite those of the other windings. When the
switch turns OFF current was flowing in a "dot" terminal. The core
inductance act to continue current in a dotted terminal, thus

Fig. 16: Forward converter with tertiary winding
For further
reading:
[1]
"Power Electronics: Converters, Applications and Design",
Mohan, Undeland and Robbins, Wiley, 1989.