Lecture slides_Ch_6 power electronics daniel hart.pdf

6.2 THE BASIC SWITCHING CONVERTER
In Class → LTSpice simulation of buck converter and assignment

● The average capacitor current is zero:
● When circuit is energize and switch starts
switching, at the rate of f, inductor and
capacitor starts storing current and voltage
respectively.
6.3 THE BUCK (STEP-DOWN) CONVERTER:
After some time, circuit will come to steady-state condition.Buck converters and dc-dc converters in
general, have the following properties when operating in the steady state:
● The inductor current is periodic:
● The average inductor voltage is zero:

Analysis of the buck converter begins by making following assumptions
1. The circuit is operating in the steady state.
2. The inductor current is continuous (always positive).
3. The capacitor is very large, and the output voltage is held constant at
voltage ≈ Vo. This restriction will be relaxed later to show the effects of
finite capacitance.
4. The switching period is T; the switch is closed for time DT and open for
time (1-D)T.
5. The components are ideal, no power loss in the circuit, means:

MODE#1: Analysis of the buck converter when switch is closed:
The the diode is reverse-biased and the voltage
across the inductor is:
Since switching frequency is very
high (100 kHz or above) therefore
we can assume linear increase and
decrease in current.

MODE#1: Analysis of the buck converter when switch is closed:

MODE#2: Analysis of the buck converter when switch is open:

Analysis of the buck converter:
The net change in inductor current over one period must be zero for steady
state operation.
● The buck converter produces an output voltage that is less than or equal to the input.
● Note that the output voltage depends on only the input and the duty ratio D. If the
input voltage fluctuates, the output voltage can be regulated by adjusting the duty ratio
appropriately

The average inductor current is:
Since for steady state condition, capacitor
voltage is periodic and therefore, average
capacitor current is zero

Since the change in inductor current is known, the maximum and minimum values of
the inductor current are computed as
Example 6.1

Design of the buck converter:
In the preceding analysis, the capacitor was assumed to be very large to keep the
output voltage constant. In practice, the output voltage cannot be kept perfectly
constant with a finite capacitance. The ratio of variation in output voltage, or ripple, to
output voltage is computed as: (Derivation is in textbook)
If we know the switching frequency:
Do not use these
equations when values of
L and C are given
This equation is used to find critical value of inductance C, when ∆Vo is very small.
● If customer output voltage (Vo) requirement is variable, which requires D to varies, then use
minimum value of D.

This equation is used to find critical value of inductance Lmin for CCM.
● If customer load (R) is variable, then use maximum value of R.
● If customer output voltage (Vo) requirement is variable, which requires D to varies, then use
minimum value of D.
Design of the buck converter:
Since in our analysis we assume that inductor current is always positive (for CCM)
therefore minimum inductor current equation can be used to find Lmin. The boundary
condition for positive inductor current is when Imin = 0.
If we know the switching frequency:

6.3 DESIGN CONSIDERATION:
● Most buck converters are designed for CCM.
● Note that as the switching frequency increases, the minimum size of the inductor to produce
continuous current and the minimum size of the capacitor to limit output ripple both decrease.
● Therefore, high switching frequencies are desirable to reduce the size of both the inductor and
the capacitor.
● The tradeoff for high switching frequencies is increased power loss in the switches. Increased
power loss in the switches means that heat is produced. This decreases the converter’s
efficiency and may require a large heat sink, offsetting the reduction in size of the inductor and
capacitor.
● Typical switching frequencies are above 20 kHz to avoid audio noise, and they extend well into
the 100s of kilohertz and into the megahertz range. Some designers consider about 500 kHz to
be the best compromise between small component size and efficiency.

6.3 DESIGN CONSIDERATION:
● Other designers prefer to use lower switching frequencies of about 50 kHz to keep switching
losses small, while still others prefer frequencies larger than 1 MHz.
● As switching devices improve, switching frequencies will increases.
● The inductor value should be larger than Lmin to ensure continuous current operation (CCM).
● Some designers select a value 25% larger than Lmin.
● Other designers use different criteria, such as setting the inductor current variation, ∆iL to a
desired value, such as 40% of the average inductor current.
● A smaller ∆iL results in lower peak and rms inductor currents and a lower rms capacitor current
but requires a larger inductor.
Example 6.2 Example 6.3

Analysis of the boost converter begins by
assuming that the circuit is operating in
the steady state condition such that:
6.5 THE BOOST (STEP-UP) CONVERTER:
1. The inductor current is continuous (always positive).
2. The capacitor is very large, and the output voltage is held
constant at voltage ≈ Vo. This restriction will be relaxed
later to show the effects of finite capacitance.
3. The switching period is T; the switch is closed for time DT
and open for time (1-D)T.
4. The components are ideal, no power loss in the circuit.

MODE#1: Analysis of the boost converter when switch is closed:
The diode is reverse-biased and the voltage

MODE#1: Analysis of the boost converter when switch is closed:
Since switching frequency is very high (100
kHz or above) therefore we can assume linear
increase and decrease in current. Therefore:
How Inductor
voltage looks
like!

MODE#2: Analysis of the boost converter when switch is open:
When the switch is opened, the inductor current cannot
change instantaneously, so the diode becomes
forward-biased to provide a path for inductor current.
Assuming that the output voltage Vo
is a constant, the
voltage across the inductor is:
How Inductor
voltage looks
like!

MODE#2: Analysis of the boost converter when switch is open:
The rate of change of inductor current is a constant, so the
current must change linearly while the switch is open. The
change in inductor current while the switch is open is
How Inductor
voltage looks
like!

For steady-state operation, the net change in inductor current
must be zero, therefore:
Solving for Vo
:
The boost converter produces an output voltage that is greater than or
equal to the input voltage. However, the output voltage cannot be less
than the input, as was the case with the buck converter

Average input current is same as average inductor current,
assuming the ideal components, input power = output power:

A condition necessary for continuous inductor current is for ILmin to be positive. Therefore, the boundary
between continuous and discontinuous inductor current is determined by:
The minimum combination of inductance and
switching frequency for continuous current in the
boost converter is therefor
From a design perspective, it is useful to express
L in terms of a desired iL,
Expressing capacitance in terms of output
voltage ripple yields
Example 6.4

Effect of Inductor Resistance:
The existence of a small inductor resistance does not substantially change the analysis of the buck converter as
presented previously. However, inductor resistance affects performance of the boost converter, especially at high
duty ratios (D). The output voltage of Boost converter when inductor is consider non-ideal (or having resistance rL) is
given as:

Analysis of the buck-boost converter begins
by assuming that the circuit is operating in
the steady state condition such that:
6.6 THE BUCK-BOOST CONVERTER:
1. The inductor current is continuous or CCM (always
positive).
2. The capacitor is very large, and the output voltage is held
constant at voltage ≈ Vo. This restriction will be relaxed
later to show the effects of finite capacitance.
3. The switching period is T; the switch is closed for time DT
and open for time (1-D)T.
4. The components are ideal, no power loss in the circuit.

MODE#1: Analysis of the buck-boost converter when switch is closed:
The diode is reverse-biased and the voltage

MODE#1: Analysis of the buck-boost converter when switch is closed:
Since switching frequency is very high (100
kHz or above) therefore we can assume linear
increase and decrease in current. Therefore:

MODE#2: Analysis of the buck-boost converter when switch is open:
When the switch is open, the current in the inductor
cannot change instantaneously, resulting in a
forward-biased diode and current into the resistor and
capacitor. In this condition, the voltage across the
inductor is:

MODE#2: Analysis of the buck-boost converter when switch is open:
The rate of change of inductor current is a
constant, so the current must change linearly while
the switch is open. The change in inductor current
while the switch is open is

For steady-state operation, the net change in inductor current must be zero, therefore:
Solving for Vo
:
Output voltage magnitude of the buck-boost converter can be less than that of the source or greater than the source,
depending on the duty ratio of the switch. If D>0.5, the output voltage is larger than the input; and if D<0.5, the output
is smaller than the input
The required duty ratio for specified input and output
voltages can be expressed as:

Note that the source is never connected directly to
the load in the buck-boost converter. Energy is
stored in the inductor when the switch is closed and
transferred to the load when the switch is open.
Hence, the buck-boost converter is also referred to
as an indirect converter.
Assuming the ideal components, input power = output power:

What is the relationship between average inductor current and average source current:
Therefore average inductor current can be given as:

If we know the average inductor current then we can find maximum and minimum
inductor current as:

Also value of capacitor can be found using the equation of ripples in output
voltage to output voltage as:
Likewise we can find the inductor value by putting as IL(min) = 0.

● Interleaving, also called multiphasing,
is a technique that is useful for
reducing the size of filter components.
● The current entering the capacitor and
load resistance is the sum of the
inductor currents, which has a smaller
peak-to-peak variation and a frequency
twice as large as individual inductor
currents.
● Each inductor supplies one-half of the
load current and output power, so the
average inductor current is one-half of
what it would be for a single buck
converter.
6.9 THE INTERLEAVED CONVERTERS:

6.9 THE INTERLEAVED CONVERTERS:

6.11 DISCONTINUOUS CURRENT OPERATION:
Buck converter with discontinuous current:
1. The inductor current is NOT continuous .
1. The capacitor is very large, and the output
voltage is held constant at voltage ≈ Vo. This
restriction will be relaxed later to show the
effects of finite capacitance.
1. The switching period is T; the switch is closed for
time DT and open for time (1-D)T.
1. The components are ideal, no power loss in the
circuit, means:

Computing the average inductor current from the
figure:
Therefore, by comparing:

Solving for Imax from the figure:
Substituting Imax in the equation to find D1 :

Putting the equation of D1:
Remember for DCM the following condition must be
satisfied:
Example 6.9

Boost converter with discontinuous current:
1. The inductor current is NOT continuous .
1. The capacitor is very large, and the output
voltage is held constant at voltage ≈ Vo. This
restriction will be relaxed later to show the
effects of finite capacitance.
1. The switching period is T; the switch is closed for
time DT and open for time (1-D)T.
1. The components are ideal, no power loss in the
circuit, means:

Since average voltage is zero, therefore:

To find Imax we can use inductor voltage
equation when switch is charging.

Since therefore, average diode current is equal to average resistor
current.
Solving for D1
The boundary between continuous and discontinuous current
occurs when D1 = 1 - D.
Example 6.10

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