UNIT I ANALYSIS OF NON-ISOLATED DC-DC CONVERTERS

Basic topologies: Buck, Boost and Buck-Boost - Principles of operation – Continuous conduction mode–
Concepts of volt-sec balance and charge balance – Analysis and design based on steady state
relationships – Introduction to discontinuous conduction mode.
PART-A TWO MARKS QUESTION AND ANSWERS
1. What is difference between SMPS and inverter?

The power source generally being lead acid batteries, converting DC to AC via a circuit called the
inverter. So without going into too much detail and simplifying things to an extent, an SMPS is an AC to
DC power supply. Whereas a UPS is a DC to AC power supply.
2. Is SMPS and UPS same?

SMPS stands for Switched Mode Power Supply. UPS stands for Uninterruptible Power Supply. The
primary function of SMPS is to convert and regulate electric power as per the need. The primary function
UPS is to provide backup power supply during main power outages.
3. What is meant by isolated and Non-Isolated DC-DC converters?
In an isolated converter, the input and output stage have separate grounds whereas in a non-
isolated converter, current is able to flow directly between the two sides as they share a common
ground.

4. What is an isolated and non-isolated DC to DC converter?

Isolated: DC input isolated from output. 2. Non-isolated: DC input and output are connected to the
same potential.
5. Write the working principle of Buck converter.
Working principle A Buck converter steps down a DC voltage from the input to the output. The
circuit operation depends on the conduction state of the MOSFET: On-state: The current through the
inductor increases and the diode blocks.
6. What is the working principle of Buck boost regulator?

The special feature of the buck-boost regulator is that the output voltage can be higher or lower than the
input voltage. It operates in a non-linear regime. In other words, it receives a DC voltage at the input and,
at the output, reduces or raises it according to the needs of the application.2
7. What is the working principle of boost converter?

The boost converter is used to "step-up" an input voltage to some higher level, required by a load. This
unique capability is achieved by storing energy in an inductor and releasing it to the load at a higher
voltage. This brief note highlights some of the more common pitfalls when using boost regulators.
8. What is continuous conduction mode in boost converter?

If the MOSFET switches from t OFF to tON before the inductor are completely discharged, then the current
in the inductor is never zero. This operation is called continuous conduction mode (CCM).
9. What is continuous and discontinuous mode of buck converter?

Continuous inductor current mode is characterized by current flowing continuously in the inductor during
the entire switching cycle in steady state operation. Discontinuous inductor current mode is characterized
by the inductor current being zero for a portion of the switching cycle.
10. What are the advantages of continuous conduction mode?
There are several reasons: In discontinuous conduction mode, the output voltage depends on the duty
cycle, inductor size, PWM frequency, and the input voltage value. In continuous conduction mode, the
output voltage only depends on the PWM duty cycle.

11. Write the topology of buck converter

This report describes three DC/DC buck converter topologies: Current Mode, Current Mode-
Constant On Time and Richtek Advanced Constant On time topologies. The differencesbetween
the three topologies are explained and the advantages and disadvantages of each typeare listed
with respect to the end application.
12. What is buck converter topology?

The multiphase buck converter is a circuit topology where basic buck converter circuits are placed in
parallel between the input and load. Each of the n "phases" is turned on at equally spaced intervals
over the switching period. This circuit is typically used with the synchronous buck topology,
described above.

13.What is the topology of buck-boost?

The topology of the buck-boost converter can be set up in a variety of ways, including non-
inverting, Ćuk, and single-ended primary-inductor converter (SEPIC) configurations. Each of
these setups has unique benefits in terms of input-output isolation, voltage polarity, and
efficiency.

14. What are the topologies of converters?

These include the simplest and most fundamental types of topologies, namely the voltage buck,
boost, and buck–boost converters and their transformer-isolated versions, namely, the forward
and flyback converters.
15. Give the principle of operation of BUCK converter.
The basic concept of a buck converter is: Use the higher-than-needed voltage of the source to
quickly induce a current into an inductor ("on" in fig. 2 and 4). Disconnect the source and use the
inertia of the current in the inductor to provide more current than the source delivers.
16. What is the operation of buck-boost converter?
 A Buck-Boost converter transforms a positive DC voltage at the input to a negative DC voltage at
the output. The circuit operation depends on the conduction state of the MOSFET: On-state: The
current through the inductor increases and the diode is in blocking state.

The key principle that drives the boost converter is the tendency of an inductor to resist changes
in current by either increasing or decreasing the energy stored in the inductor magnetic field . In a
boost converter, the output voltage is always higher than the input voltage.

17. How does a boost converter work?

The boost converter is used to "step-up" an input voltage to some higher level, required by a load.
This unique capability is achieved by storing energy in an inductor and releasing it to the load at a
higher voltage. This brief note highlights some of the more common pitfalls when using boost
regulators.

18. What are the applications of boost converters?

Boost converters are used in electronics to generate a DC output voltage that is greater than the DC
input, therefore boosting up the supply voltage. Boost converters are often used in power supplies for
white LEDs, battery packs for electric automobiles, and many other applications.
19. Why is it called boost converter?

In a boost converter, the output voltage is greater than the input voltage – hence the name “boost”.
A boost converter using a power MOSFET is shown below.
20.Is a boost converter AC or DC?

A boost converter or step-up converter is a DC-to-DC converter that increases voltage, while
decreasing current, from its input (supply) to its output (load). Low-cost converter modules: two buck
and one boost. Boost converter from a TI calculator, generating 9 V from 2.4 V provided by two AA
rechargeable cells.

21. What are the different types of boost converters?

Boost converter types that will be compared are Single Switch Cascaded Converter, Multilevel Boost
Converter, Quadratic Boost Converter, Double Cascade Boost Converter and Phase Interleaved
Boost Converter.

22. What is advantage and disadvantage of boost converter?

Advantage is that output voltage can be lower and higher then input voltage. Disadvantage is that
unlike cuk convertor, source current is discontinuous and also same problem on output side (problem
related to voltage here).
23.What is continuous conduction mode?

If the MOSFET switch from tOFF to tON before the inductor is completely discharged, then the
current in the inductor is never zero. This operation is called continuous conduction mode (CCM).

24. What is CCM mode in buck converter?

 Use of unidirectional switch forces the buck converter to exhibit two different. operational modes:
the continuous conduction mode and the discontinuous conduction mode. Continuous conduction
mode (CCM) operation. • Discontinuous conduction mode (DCM) operation.

25. What is the difference between continuous and discontinuous conduction mode?

In discontinuous conduction mode, the output voltage depends on the duty cycle, inductor size,
PWM frequency, and the input voltage value. In continuous conduction mode, the output voltage
only depends on the PWM duty cycle.
26. What is CCM and DCM mode in buck converter?

Introduction. In the previous experiment, Buck converter - Continuous conduction mode (CCM), was
analyzed in continuous conduction mode (CCM), i.e., the inductor current was always greater than 0.
In this experiment, the buck converter operation under discontinuous conduction mode (DCM) is
analyzed.

27. What causes discontinuous conduction mode?

The discontinuous conduction mode arises when the switching ripple in an inductor current or
capacitor voltage is large enough to cause the polarity of the applied switch current or voltage to reverse,
such that the current- or voltage-unidirectional assumptions made in realizing the switch with
semiconductor devices

28. What is CCM for boost converter?

A PFC boost converter is in continuous conduction mode when the switching device turns on
before the inductor current drops to zero. The inductor current is continuous in the continuous
conduction mode of a PFC boost converter. In this mode, there are two states in one switching
cycle: the ON state and the OFF state.
29. What is the concept of volt sec balance?

The principle of inductor volt-second balance states that the average value, or dc component, of
voltage applied across an ideal inductor winding must be zero. This principle also applies to each
winding of a transformer or other multiple winding magnetic devices.
30. What unit is volt sec?

Magnetic flux
The fundamental unit of magnetic flux is Volt-seconds.
31. What is the principle of capacitor charge balance?

This principle also applies to each winding of a transformer or other multiple winding magnetic.
Its dual, the principle of capacitor amp-second or charge balance, states that the average current
that flows through an ideal capacitor must be zero.

32. What is buck-boost charging?

A buck-boost topology will accept input voltages above, below or equal to the battery voltage and
charge the battery with high accuracy to its final charge voltage. Product Selection Table.
33. What is the discontinuous current operation?
During discontinuous mode operation, the inductor current is discharged to zero before the end of
the switching period.
34. What is continuous and discontinuous load current?

The term continuous means that load current never ceases but continues to flow through SCR or
their combination. The term discontinuous is applied to the condition when load current reaches
zero during each half cycle before the next SCR in sequence is fired.
35. What are the advantages of discontinuous conduction mode?

Discontinuous conduction mode (left) is more efficient than continuous conduction mode (right)
at low power levels because it does not allow the current through the pFET high side switch to
flow negative.
36. Write short notes about Discontinuous conduction mode.
Occurs because switching ripple in inductor current or capacitor voltage causes polarity of applied
switch current or voltage to reverse, such that the current- or voltage-unidirectional assumptions
made in realizing the switch are violated.
● commonly occurs in dc-dc converters and rectifiers, having single quadrant switches. May also
occur in converters having two-quadrant switches.
● typical example: dc-dc converter operating at light load (small load current). Sometimes, dc-dc
converters and rectifiers are purposely designed to operate in DCM at all loads
● Properties of converters change radically when DCM is entered: M becomes load-dependent
Output impedance is increased Dynamics are altered Control of output voltage may be lost when
load is removed
37. What is the application of Buck-Boost?

Applications of Buck-Boost Converters: Power Supply Control: Buck-boost converters are

commonly used in power supply control. They ensure that even if the input voltage varies, the
output voltage remains constant. They can step up or step down voltages as per the requirement.
38. What are the applications of boost converters?
Boost converters are used in electronics to generate a DC output voltage that is greater than the
DC input, therefore boosting up the supply voltage. Boost converters are often used in power
supplies for white LEDs, battery packs for electric automobiles, and many other applications.
39. Write the applications of buck converter
Buck converters are commonly used to derive the required input voltage from a higher voltage
source. In some cases, generating a negative voltage from a positive input voltage source may be
required. These applications can include audio amplifiers, line drivers and receivers, or
instrumentation amplifiers.
40. What are the advantages of buck converter?
High Efficiency: Buck converters excel in recycling excess voltage through the inductor,
achieving efficiencies often above 90%. Minimal Heat Dissipation: The high-efficiency operation
of the Buck converter inherently curtails heat generation, mitigating the need for heat sinks or
other cooling systems.
41. Discuss the working of isolated converters
In short, an isolated power converter isolates the input from the output by electrically and
physically separating the circuit into two sections preventing direct current flow between input
and output, typically achieved by using a transformer.
42. Differentiate isolated and non-Isolated DC-DC converters
 Isolation describes the electrical separation between the input and output of a dc-dc converter. An
isolated dc-dc converter uses a transformer to eliminate the dc path between its input and output.
In contrast, a non-isolated dc-dc converter has a dc path between its input and output.
43. What are the disadvantages of Isolated DC-DC converter
As the DC/DC converter performs switching operation, switching noise occurs in principle. IC
may malfunction or it may be necessary to add noise countermeasure parts / filter may due to this
switching noise.
44. What are the disadvantages of bidirectional DC-DC converter?

However, the buck–boost bidirectional converter has some limitations, including not being
suitable for high step-up or step-down due to hard switching, high duty-cycle operation, voltage-
conversion ratio limitation and high voltage stress on the switching device
45. What's the Difference Between Isolated and Non-Isolated Power Supplies?
In short, an isolated power converter isolates the input from the output by electrically and
physically separating the circuit into two sections preventing direct current flow between input
and output, typically achieved by using a transformer. A non-isolated power converter has a
single circuit in which current can flow between the input and output.
46. Mension the benefits Isolation.
Benefits of Isolation
There are several cases where an isolated power supply may be required or provide some benefit
in an application. These include safety compliance, the breaking of ground loops, and level
shifting.

47. Discuss the safety complaints of isolated converters.

Safety Compliance

Safety requirements are a common reason to use an isolated power converter. For converters
powered from high and potentially hazardous voltages (such as ac-dc converters powered from ac
mains) isolation separates the output from dangerous voltages on the input.

48. Explain the breaking of loo[p in Isolation of DC converters

Because the input and output of isolated supplies do not share a ground, they can be used to break
up ground loops. Circuits that are sensitive to noise can benefit from this by having their ground
broken up and separated from noisy circuits that could cause problems.
49.Write about the floating output and level shifting in Isolation Dc converter
Floating Outputs and Level Shifting
Another benefit to isolated converters is a floating output. Isolated outputs, while having a fixed
voltage between output terminals, don't have a defined or fixed voltage relative to voltage nodes
in circuits from which they've been isolated, and are said to be floating. However, a floating
 output may have one of its terminals connected to another circuit node to fix it to that voltage.
This fact can be used to shift or invert the output relative to another point in a circuit.

50. What are the benefits of Non Isolation of DC-DC converter?

The choice between isolated and non-isolated converters depends on many factors. Some
applications require isolation for safety reasons, and others may benefit from a floating output by
breaking up ground loops or shifting of reference voltages. However, where isolation is not
required, a non-isolated converter may provide a decrease in cost, size, and/or an increase in
efficiency. Understanding the costs and benefits of isolation is important in choosing the right
converter for an optimized design.
51. Define Volt/second balance of a converter

It refers to flux conservation in an inductor during stable, steady state operation. Flux in the
inductor core is proportional to the current through the windings which is in turn proportional to
the time integral of voltage (“Volt-Second”) applied to the inductor.

PART B & C QUESTION AND ANSWERS

1. Discuss in detail about the Isolation of converter benefits
1. Safety Compliance
Safety requirements are a common reason to use an isolated power converter. For converters
powered from high and potentially hazardous voltages (such as ac-dc converters powered from ac
mains) isolation separates the output from dangerous voltages on the input.
When safety is the concern, the insulation grade must also be considered. Safety standards should
be reviewed to determine what level of insulation is required for a given application. The
insulation grade is divided into several categories including functional, basic, supplementary, and
reinforced.
2. Functional insulation: is the most simple and, while providing isolation, it does not provide
any protection against electric shock.
3. Basic insulation: provides a single layer of protection against shock.
4. Supplementary insulation: is basic insulation plus one additional barrier for redundancy.
5. Reinforced insulation: is a single barrier equivalent to two layers of basic insulation.
Benefits of Non-Isolated
While there are many benefits to isolation, there are also reasons to use a non-isolated converter
including cost, size, and performance.
6. Cost Savings
Isolated converters tend to be more expensive than non-isolated. A major contributor to the cost
difference is the use of a transformer in place of an inductor.
 Transformers tend to be custom built as opposed to the inductor in a non-isolated converter which
can be purchased off the shelf. If a higher level of insulation is needed (such as that needed for
safety compliance) the cost will further increase. In addition to the transformer, there are
components, such as the opto-couplers, that may be added to an isolated design that wouldn't be
necessary in a non-isolated. All of this adds to increased cost compared to a non-isolated design.
7. Smaller Size
Non-isolated converters tend to be smaller than isolated. The cost-adding components mentioned
previously take up more space than those used in a non-isolated design. In addition to substituting
an inductor for a transformer, non-isolated converters tend to operate at higher switching
frequencies which will further reduce the size of the magnetic components and capacitors.
8. Efficiency
The efficiency and regulation of non-isolated converters also tend to be better than that of an
isolated converter. The transformer and opto-couplers again are major contributors to the
difference in performance. The absence of an isolation barrier allows for the output to be directly
sensed and tightly controlled for better regulation and transient performance. Their small size also
allows them to be placed closer to the load to reduce the transmission line effects.
2. Explain the principle operation of buck converters with continuous conduction mode. Also
derive the expression for the output voltage of converters.
This is a step-down circuit, where average output voltage is lesser than input DC voltage. The circuit is
shown in Fig.. In the absence of L and C, v0 is same as step down chopper which is undesirable for
powering ICs and biasing transistors. Hence L and C are added to get a ripple free DC voltage across
the load.

Fig.
Buck converter circuit
 Fig. Voltage and current waveform for Buck converter
When the switch is ON, the free-wheeling diode FD is reverse biased; current flows from Vdc to load
through inductor. The value of capacitor C is large so that v0 is assumed constant throughout. During ON
period, inductor current builds up as shown in Fig. where it is assumed that iL is continuous. Applying
KVL to the loop of Vdc, vL and v0, we get

where K is the constant of integration. At t = 0, I1 =IL and hence K= I1

When the switch is put OFF, inductor current continue to flow through FD and load. Now using KVL, it
is seen that vL =- V0.
Hence

Where t’ = t -Ton and K1 is a constant. At t’=0, iL = I2 and hence K1= I2.

It is obvious that inductor current decays linearly and is shown in Fig. 1.9. Assuming an ideal inductor
the energy stored in the inductor during ON period is transferred to the load when switch is OFF. Since
average power in an ideal inductor is zero and in the present case, average inductor current is non-zero, it
implies that time integral of inductor voltage over one cycle of output voltage should be zero.

The above equation shows that with continuous inductor current, output voltage varies linearly with duty
ratio, D and its circuit parameters. By assuming a lossless circuit.

Therefore the average input current is given by

 Is DI0
The switching period T can be calculated as
T =Ton+Toff
By substituting the values of Ton and Toff the above Eqs.

Similarly Eq. become

Substituting in above equation T

Now
CONTINUOUS CONDUCTION MODE OF BUCK CONVERTER
The waveforms for the voltages and currents are shown in Figure, for a continuous current flow in the
inductor L. It is assumed that the current rises and falls linearly. In practical circuits, the switch has a
finite, nonlinear resistance. Its effect can generally be negligible in most applications. Depending on the
switching frequency, filter inductance, and capacitance, the inductor current could be discontinuous.

Fig.1.10 Continuous conduction mode Voltage and current voltage waveform for Buck converter
Assuming a lossless circuit, VsIs = VaIa = kVsIa and the average input current
Is = kIa
A boost regulator can step up the output voltage without a transformer. Due to a single transistor, it has a
high efficiency. The input current is continuous. However, a high-peak current has to flow through the
power transistor.
The output voltage is very sensitive to changes in duty cycle k and it might be difficult to stabilize the
regulator. The average output current is less than the average inductor current by a factor of (1 – k), and a
much higher rms current would flow through the filter capacitor, resulting in the use of a larger filter
capacitor and a larger inductor than those of a buck regulator.
3. A boost regulator in Figure 5.18a has an input voltage of Vs = 5 V. The average output voltage Va
= 15 V and the average load current Ia = 0.5 A. The switching frequency is 25 kHz. If L = 150 μH
and C = 220 μF, determine (a) the duty cycle k, (b) the ripple current of inductor ∆I, (c) the peak
current of inductor I2, (d) the ripple voltage of filter capacitor ∆Vc, and (e) the critical values of L
and C.
Solution
Vs = 5 V, Va = 15 V, f = 25 kHz, L = 150 μH, and C = 220 μF.
a. , 15 = 5/(1 – k) or k = 2/3 = 0.6667 = 66.67,.

4. Explain the principle operation of boost converters with continuous conduction mode. Also
derive the expression for the output voltage of converters.
In a boost regulator the output voltage is greater than the input voltage—hence the name “boost.” A boost
regulator using a power MOSFET is shown in Figure a. Transistor M1 acts as a controlled switch and
diode Dm is an uncontrolled switch. The circuit in Figure. a is often represented by two switches as
shown in Figure b.

The circuit operation can be divided into two modes. Mode 1 begins when transistor. M1 is switched on
at t = 0. The input current, which rises, flows through inductor L and transistor Q1. Mode 2 begins when
transistor M1 is switched off at t = t1. The current that was flowing through the transistor would now
flow through L, C, load, and diode Dm. The inductor current falls until transistor M1 is turned on again in
the next cycle. The energy stored in inductor L is transferred to the load. The equivalent circuits for the
modes of operation are shown in Figure c. The waveforms for voltages and currents are shown in Figure
d for continuous load current, assuming that the current rises or falls linearly.
Assuming that the inductor current rises linearly from I1 to I2 in time t1,
CONTINUOUS CONDUCTION MODE OF BOOST CONVERTER
Peak-to-peak inductors ripple current. The switching period T can be found from
Peak-to-peak capacitors ripple voltage. When the transistor is on, the capacitor supplies the load
current for t = t1. The average capacitor current during time t1 is Ic = Ia and the peak-to-peak ripple
voltage of the capacitor is

Condition for continuous inductor current and capacitor voltage.

If IL is the Average inductor current, at the critical condition for continuous conduction the inductor
ripple current ∆I = 2IL.
We get

If Vc is the average capacitor voltage, at the critical condition for continuous conduction the capacitor
ripple voltage ∆Vc = 2Va. Using Eq., we get

A boost regulator can step up the output voltage without a transformer. Due to a single transistor, it has a
high efficiency. The input current is continuous. However, a high-peak current has to flow through the
power transistor. The output voltage is very sensitive to changes in duty cycle k and it might be difficult
to stabilize the regulator.
The average output current is less than the average inductor current by a factor of (1 – k), and a much
higher rms current would flow through the filter capacitor, resulting in the use of a larger filter capacitor
and a larger inductor than those of a buck regulator.
5. A boost regulator in Figure a has an input voltage of Vs = 5 V. The average output voltage Va =
15 V and the average load current Ia = 0.5 A. The switching frequency is 25 kHz. If L = 150 μH and
C = 220 μF, determine (a) the duty cycle k, (b) the ripple current of inductor ∆I, (c) the peak
current of inductor I2, (d) the ripple voltage of filter capacitor ∆Vc, and (e) the critical values of L
and C.

6. Explain the principle operation of buck converters with continuous conduction mode. Also
derive the expression for the output voltage of converters.
A buck–boost regulator provides an output voltage that may be less than or greater than the input voltage
—hence the name “buck–boost”; the output voltage polarity is opposite to that of the input voltage. This
regulator is also known as an inverting regulator. The circuit arrangement of a buck–boost regulator is
shown in Figure a.
Fig. Transistor Q1 acts as a controlled switch and diode Dm is an uncontrolled switch. They operate as two
SPST current-bidirectional switches. The circuit in Figure a is often represented by two switches as
shown in Figure b.
The circuit operation can be divided into two modes. During mode 1, transistor Q1 is turned on and diode
Dm is reversed biased. The input current, which rises, flows through inductor L and transistor Q1. During
mode 2, transistor Q1 is switched off and the current, which was flowing through inductor L, would flow
through L, C, Dm, and the load. The energy stored in inductor L would be transferred to the load and the
inductor current would fall until transistor Q1 is switched on again in the next cycle.

The equivalent circuits for the modes are shown in Figure c. The waveforms for steady state voltages and
currents of the buck–boost regulator are shown in Figure d for a continuous load current.
Assuming that the inductor current rises linearly from I1 to I2 in time t1,
Where ∆I = I2 - I1 is the peak-to-peak ripple current of inductor L.

CONTINUOUS CONDUCTION MODE OF BUCK-BOOST CONVERTER

Condition for continuous inductor current and capacitor voltage. If IL is the average inductor current,
at the critical condition for continuous conduction the inductor ripple current ∆I = 2IL.

If Vc is the average capacitor voltage, at the critical condition for continuous conduction the capacitor
ripple voltage ∆Vc = -2Va.
A buck–boost regulator provides output voltage polarity reversal without a transformer. It has high
efficiency. Under a fault condition of the transistor, the di/dt of the fault current is limited by the inductor
L and will be Vs/L. Output short-circuit protection would be easy to implement. However, the input
current is discontinuous and a high peak current flows through transistor Q1.

7. The buck–boost regulator in Figure 5.19a has an input voltage of Vs = 12 V. The duty cycle k =
0.25 and the switching frequency is 25 kHz. The inductance L = 150 μH and filter capacitance C =
220μF. The average load current Ia = 1.25 A. Determine (a) the average output voltage, Va; (b) the
peak-to-peak output voltage ripple, ∆Vc; (c) the peak-to-peak ripple current of inductor, ∆I; (d) the
peak current of the transistor, Ip; and (e) the critical values of L and C.
Solution
Vs = 12 V, k = 0.25, Ia = 1.25 A, f = 25 kHz, L = 150 μH, and C = 220μF.
a. Va = -12 * 0.25/11 - 0.252 = -4 V.
b. the peak-to-peak output ripple voltage is
8. Explain the basic concepts of volt-sec balance and charge balance in converter.
Let us more closely examine the inductor and capacitor waveforms in the buck converter of Fig.. It is
impossible to build a perfect low-pass filter that allows the dc component to pass but completely removes
the components at the switching frequency and its harmonics. So the low-pass filter must allow at least
some small amount of the high-frequency harmonics generated by the switch to reach the output. Hence,
in practice the output voltage waveform v(t) appears as illustrated in Fig. 1.14, and can be expressed as

So the actual output voltage v(t) consists of the desired dc component V, plus a small undesired ac
component Vripple (t) arising from the incomplete attenuation of the switching harmonics by the low-pass
filter.

The magnitude of Vripple (t) has been exaggerated in Fig. 1.14. The output voltage switching ripple should
be small in any well-designed converter, since the object is to produce a dc output. For example, in a
computer power supply having a 3.3 V output, the switching ripple is normally required to be less than a
few tens of milli volts, or less than 1% of the dc component V.

So it is nearly always a good approximation to assume that the magnitude of the switching ripple is much
smaller than the dc component:
Therefore, the output voltage v(t) is well approximated by its dc component V, with the small ripple term
Vripple (t) neglected:

This approximation, known as the small-ripple approximation, or the linear-ripple approximation, greatly
simplifies the analysis of the converter waveforms and is used throughout this book.

Next let us analyze the inductor current waveform. We can find the inductor current by integrating the
inductor voltage waveform. With the switch in position 1, the left side of the inductor is connected to the
input voltage Vg and the circuit reduces to Fig. 1.15(a). The inductor voltage Vt (t) is then given by

As described above, the output voltage v(t) consists of the dc component V, plus a small ac ripple term
Vripple (t). We can make the small ripple approximation here, Eq. , to replace v(t) with its dc component V:

So with the switch in position 1, the inductor voltage is essentially constant and equal to V g -V as shown
in Fig. 1.15. By knowledge of the inductor voltage waveform, the inductor current can be found by use of
the definition

Thus, during the first interval, when VL(t) is approximately (Vg –V) the slope of the inductor current
waveform is
which follows by dividing Eq. by L, and substituting Eq. . Since the inductor voltage is essentially
constant while the switch is in position 1, the inductor current slope is also essentially constant and the
inductor current increases linearly.
Similar arguments apply during the second subinterval, when the switch is in position 2. The left side of
the inductor is then connected to ground, leading to the circuit of Fig. 1.15(b). It is important to
consistently define the polarities of the inductor current and voltage; in particular, the polarity of is
defined consistently in Figs. 1.14, 1.15(a), and 1.15 (b). So the inductor voltage during the second
subinterval is given by

Use of the small ripple approximation, Eq. leads to

So the inductor voltage is also essentially constant while the switch is in position 2, as illustrated in Fig.
1.16. Substitution of Eq. into Eq. and solution for the slope of the inductor current yields

Hence, during the second subinterval the inductor current changes with a negative and essentially
constant slope.
We can now sketch the inductor current waveform (Fig. 1.17). The inductor current begins at
some initial value iL(0) . During the first subinterval, with the switch in position 1, the inductor current
increases with the slope given in Eq. At time t=DT s the switch changes to position 2. The current then
decreases with the constant slope given by Eq. At time t=T s the switch changes back to position 1, and
the process repeats.

It is of interest to calculate the inductor current ripple ∆iL. As illustrated in Fig. 1.17, the peak
Inductor current is equal to the dc component I plus the peak-to-average ripple ∆iL. This peak current
flows through not only the inductor, but also through the semiconductor devices that comprise the switch.
Knowledge of the peak current is necessary when specifying the ratings of these devices. Since we know
the slope of the inductor current during the first subinterval, and we also know the length of the first
subinterval, we can calculate the ripple magnitude. The i L(t) waveform is symmetrical about I, and hence
during the first subinterval the current increases by (since is the peak ripple, the peak-to-peak ripple is
2∆iL ). So the change in current ∆i L, is equal to the slope (the applied inductor voltage divided by L) times
the length of the first subinterval (DTs):

Solution for ∆iL yields

Typical values of ∆iL lie in the range of 10% to 20% of the full-load value of the dc component I. It is
undesirable to allow ∆iL to become too large; doing so would increase the peak currents of the inductor
and of the semiconductor switching devices, and would increase their size and cost. So by design the
inductor current ripple is also usually small compared to the dc component I. The small-ripple
approximation iL(t) =I is usually justified for the inductor current.

The inductor value can be chosen such that a desired current ripple ∆iL is attained. Solution of
Eq. For the inductance L yields

This equation is commonly used to select the value of inductance in the buck converter.
It is entirely possible to solve converters exactly, without use of the small-ripple approximation. For
example, one could use the Laplace transform to write expressions for the waveforms of the circuits of
Figs. 1.15(a) and 1.15(b). One could then invert the transforms, match boundary conditions, and find the
periodic steady-state solution of the circuit. Having done so, one could then find the dc components of
the waveforms and the peak values. But this is a great deal of work, and the results are nearly always
intractable. Besides, the extra work involved in writing equations that exactly describe the ripple is a
waste of time, since the ripple is small and is undesired. The small-ripple approximation is easy to apply,
and quickly yields simple expressions for the dc components of the converter waveforms.

The inductor current waveform of Fig. 1.17 is drawn under steady-state conditions, with the converter
operating in equilibrium. Let’s consider next what happens to the inductor current when the converter is
first turned on. Suppose that the inductor current and output voltage are initially zero, and an input
voltage is then applied. As shown in Fig. 1.18, is zero. During the first subinterval, with the switch in
position 1, we know that the inductor current will increase, with a slope of and with v initially zero. Next,
with the switch in position 2, the inductor current will change with a slope of – v/L; since v is initially
zero, this slope is essentially zero. It can be seen that there is a net increase in inductor current over the
first switching period, because is greater than. Since the inductor current flows to the output, the output
capacitor will charge slightly, and v will increase slightly.
The process repeats during the second and succeeding switching periods, with the inductor current
increasing during each subinterval 1 and decreasing during each subinterval 2.

As the output capacitor continues to charge and v increases, the slope during subinterval 1 decreases
while the slope during subinterval 2 becomes more negative. Eventually, the point is reached where the
increase in inductor current during subinterval 1 is equal to the decrease in inductor current during
subinterval 2. There is then no net change in inductor current over a complete switching period, and the
converter operates in steady state.
The requirement that, in equilibrium, the net change in inductor current over one switching period be zero
leads us to a way to find steady-state conditions in any switching converter: the principle of inductor volt-
second balance. Given the defining relation of an inductor:

Integration over one complete switching period, say from t = 0 to Ts yields

This equation states that the net change in inductor current over one switching period, given by the left
hand side of Eq. is proportional to the integral of the applied inductor voltage over the interval. In steady
state, the initial and final values of the inductor current are equal, and hence the left-hand side of Eq. is
zero. Therefore, in steady state the integral of the applied inductor voltage must be zero:
The right-hand side of Eq. has the units of volt-seconds or flux-linkages. Equation states that the total
area, or net volt-seconds, under the waveform must be zero. An equivalent form is obtained by dividing
both sides of Eq. by the switching period Ts.

The right-hand is recognized as the average value, or dc component, of Equation

states that, in equilibrium, the applied inductor voltage must have zero dc component. The inductor
voltage waveform of Fig. is reproduced in Fig. 1.20, with the area under the curve specifically identified.
The total area is given by the areas ƛ of the two rectangles, or

The average value is therefore

By equating vL to zero, and noting that D+D’=1one obtains

Solution for V yields

which coincides with the result obtained previously, Eq.. So the principle of inductor volt-second balance
allows us to derive an expression for the dc component of the converter output voltage.

An advantage of this approach is its generality—it can be applied to any converter. One simply sketches
the applied inductor voltage waveform, and equates the average value to zero. This method is used later
in this chapter, to solve several more complicated converters.
Similar arguments can be applied to capacitors. The defining equation of a capacitor is
Integration of this equation over one switching period yields

In steady state, the net change over one switching period of the capacitor voltage must be zero, so that the
left-hand side of Eq. is equal to zero. Therefore, in equilibrium the integral of the capacitor current over
one switching period (having the dimensions of amp-seconds, or charge) should be zero.
There is no net change in capacitor charge in steady state. An equivalent statement is

The average value, or dc component, of the capacitor current must be zero in equilibrium. This should be
an intuitive result. If a dc current is applied to a capacitor, then the capacitor will charge continually and
its voltage will increase without bound. Likewise, if a dc voltage is applied to an inductor, then the flux
will increase continually and the inductor current will increase without bound. Equation, called the
principle of capacitor amp-second balance or capacitor charge balance, can be used to find the steady-
state currents in a switching converter.

9. Describe the Analysis and Design Based on Steady State Relationships of basic converter
topology.
The buck converter was introduced as a means of reducing the dc voltage, using only non dissipative
switches, inductors, and capacitors. The switch produces a rectangular waveform V s(t) as illustrated in
Fig.1.20. The voltage Vs(t) is equal to the dc input voltage V g when the switch is in position 1, and is
equal to zero when the switch is in position 2. In practice, the switch is realized using power
semiconductor devices, such as transistors and diodes, which are controlled to turn on and off as required
to perform the function of the ideal switch.

The switching frequency fs equal to the inverse of the switching period T s generally lies in the range of 1
kHz to 1 MHz, depending on the switching speed of the semiconductor devices. The duty ratio D is the
fraction of time that the switch spends in position 1, and is a number between zero and one.
The complement of the duty ratio D’, is defined as (1 – D). The switch reduces the dc component of the
voltage: the switch output voltage has a dc component that is less than the converter dc input voltage
From Fourier analysis, we know that the dc component of is given by its average value or

As illustrated in Fig. 1.21, the integral is given by the area under the curve, DT sVg or The average value
is therefore

So the average value, or dc component, of Vs(t) is equal to the duty cycle times the dc input voltage V g.
The switch reduces the dc voltage by a factor of D.

What remains is to insert a low-pass filter as shown in Fig. 1.21. The filter is designed to pass the dc
component of Vs(t) but to reject the components V s(t) of at the switching frequency and its harmonics.
The output voltage v(t) is then essentially equal to the dc component of the converter of Fig. 1.22 has
been realized using lossless elements. To the extent that they are ideal, the inductor, capacitor, and switch
do not dissipate power. For example, when the switch is closed, its voltage drop is zero, and the current is
zero when the switch is open. In either case, the power dissipated by the switch is zero. Hence,
efficiencies approaching 100% can be obtained. So to the extent that the components are ideal, we can
realize our objective of changing dc voltage levels using a lossless network.

The network of Fig. 1.23 also allows control of the output. Figure 1.23 is the control characteristic of the
converter. The output voltage, given by Eq, is plotted vs. duty cycle. The buck converter has a linear
control characteristic.

10. Explain the concepts of discontinuous conduction mode of converter.

When the ideal switches of a dc-dc converter are implemented using current-unidirectional and/or
voltage- unidirectional semiconductor switches, one or more new modes of operation known as
discontinuous conduction modes (DCM) can occur.

The discontinuous conduction mode arises when the switching ripple in an inductor current or capacitor
voltage is large enough to cause the polarity of the applied switch current or voltage to reverse, such that
the current- or voltage-unidirectional assumptions made in realizing the switch with semiconductor
devices are violated. The DCM is commonly observed in dc–dc converters and rectifiers, and can also
sometimes occur in inverters or in other converters containing two quadrant switches. The discontinuous
conduction mode typically occurs with large inductor current ripple in a converter operating at light load
and containing current-unidirectional switches. Since it is usually required that converters operate with
their loads removed, DCM is frequently encountered. Indeed, some converters are purposely designed to
operate in DCM for all loads.

The properties of converters change radically in the discontinuous conduction mode. The conversion ratio
M becomes load-dependent and the output impedance is increased. Control of the output may be lost
when the load is removed. We will see in a later chapter that the converter dynamics are also significantly
altered.

Let us consider how the inductor and switch current waveforms change as the load power is reduced.
Let’s use the buck converter (Fig.1.24) as a simple example. The inductor current i L(t) and iD(t) diode
current waveforms are sketched in Fig. 1.25 for the continuous conduction mode.

The inductor current waveform contains a dc component I, plus switching ripple of peak amplitude ∆iL
during the second subinterval, the diode current is identical to the inductor current. The minimum diode
current during the second subinterval is equal to (I-∆i L). Since the diode is a single-quadrant switch,
operation in the continuous conduction mode requires that this current remain positive. The inductor
current dc component I is equal to the load current:

Since no dc current flows through capacitor C. It can be seen that I depends on the load resistance R.
The switching ripple peak amplitude is:

The ripple magnitude depends on the applied voltage (V g-V) on the inductance L, and on the transistor
conduction time DTs but it does not depend on the load resistance R. The inductor current ripple
magnitude varies with the applied voltages rather than the applied currents.
Suppose now that the load resistance R is increased, so that the dc load current is decreased. The dc
component of inductor current I will then decrease, but the ripple magnitude ∆i L will remain unchanged.
If we continue to increase R, eventually the point is reached where I=∆i L illustrated in Fig. 1.26. It can be
seen that the inductor current iL (t) and the diode current iD (t) are both zero at the end of the switching
period. Yet the load current is positive and nonzero.

The diode current cannot be negative; therefore, the diode must become reverse-biased before the end of
the switching period. As illustrated in Fig. 1.27, there are now three subintervals during each switching
period Ts during the first subinterval of length D 1Ts the transistor conducts, and the diode conducts during
the second subinterval of length D2Ts At the end of the second subinterval the diode current reaches zero,
and for the remainder of the switching period neither the transistor nor the diode conduct. The converter
operates in the discontinuous conduction mode.