Transcript Power Electronics - Dr. Imtiaz Hussain

Power Electronics
Lecture-10
D.C to D.C Converters (Choppers)
Dr. Imtiaz Hussain
Assistant Professor
email: [email protected]
URL :http://imtiazhussainkalwar.weebly.com/
1
Power Electronic Interface
• Power Electronics is an enabling technology providing the
needed interface between the electrical source and electrical
load.
• The source and load often do differ in frequency, voltage
amplitudes and number of phases.
2
Powering the information Technology
• Most of the consumer electronics equipment supplied from
the mains, internally needed very load dc voltages.
• Figure shows the distributed
architecture typically used in
computers.
• In which incoming voltage from
the utility is converted into dc
voltage (24V).
• This semi regulated voltage is
distributed within the computer
where on-board power supplies
convert this 24V into tightly
regulated lower voltage.
3
Powering the information Technology
• Many devices such as cell phones operates from low battery
voltages.
• However, the electronic circuitry requires higher voltages.
• Thus necessitating a circuit to boost input dc to higher dc
voltages
4
Introduction to D.C to D.C Converters (Choppers)
• DC to DC converters are important in portable electronic
devices such as cellular phones and laptop computers, which
are supplied with power from batteries primarily.
• Such electronic devices often contain several sub-circuits,
each with its own voltage level requirement different from
that supplied by the battery or an external supply.
• They are also widely used in dc-motor drive applications.
• Often input to these converters is an unregulated dc
voltage, which is obtained by rectifying the line voltage.
Introduction
Battery
DC
Unregulated
AC Line
Voltage
1-Phase or
3-Phase
Uncontrolled
Diode Rectifier
Filter
DC
Unregulated
DC
Unregulated
D.C to D.C
Converter
vc
D.C to D.C Converter System
DC
Regulated
Load
Efficiency & Power Losses
• High efficiency is essential in any power processing
application.
• The efficiency of a converter is
Pout

Pin
• The power lost in converter is
1
Ploss  Pin  Pout  Pout (  1)

Efficiency & Power Losses
• Efficiency is a good measure of the success of a given
converter technology.
• With very small amount of power lost, the converter
elements can be packaged with high density, leading to a
converter of small size and weight, and of low temperature
rise.
• How can we build a circuit that changes the voltage, yet
dissipates negligible power?
Efficiency & Power Losses
• The various conventional circuit elements are illustrated in
Following figure.
• The available circuit elements fall broadly into the classes of
resistive elements, capacitive elements, magnetic devices
including inductors and transformers, semiconductor devices
operated in the linear mode and semiconductor devices operated
in the switched mode.
Types of dc-dc Converters
• Types of D.C to D.C converters
– AC Link Choppers
– Linear Converters
– Switch Mode
– Magnetic
– E.t.c
AC Link Choppers
• First dc is converted to ac with the help of an inverter.
• After that, AC is stepped-up or stepped-down by a
transformer, which is then converted back to dc by a diode
rectifier.
• Ac link chopper is costly, bulky and less efficient as the
conversion is done in two stages.
Simple dc-dc Converters
• Let us now construct a simple dc-dc converter.
The input voltage vg is 100 V. It is desired to
supply 50 V to an effective 5Ω load, such that
the dc load current is 10 A.
Resistive dc-dc Converters
• Using Voltage divided rule.
Linear dc-dc Converters
• Linear Mode dc-dc converter
Switch Mode dc-dc Converters
Conclusion
• Capacitors and magnetic devices are important elements of
switching converters, because ideally they do not consume
power.
• It is the resistive element, as well as the linear-mode
semiconductor device, that is avoided.
• Semiconductor in switch mode however dissipate
comparatively low power in either states (ON and OFF).
• So capacitive and inductive elements, as well as switchedmode semiconductor devices, are available for synthesis of
high-efficiency converters.
Switch Mode D.C to D.C Converters
• Switch-mode DC to DC converters convert one DC voltage
level to another, by storing the input energy temporarily
and then releasing that energy to the output at a different
voltage.
• This conversion method is more power efficient (often 75%
to 98%) than linear voltage regulation (which dissipates
unwanted power as heat).
• This efficiency is beneficial to increasing the running time
of battery operated devices.
Switch Mode D.C to D.C Converters
• PWM or PFM regulates the output dc voltage.
• The power flow through these converters is only in one
direction thus their voltages and currents remain unipolar.
Switch Mode D.C to D.C Converters
• A Chopper is a high speed on/off semiconductor switch.
• It connects source to load and disconnects source from load
at very high speed.
• In this manner a chopped dc voltage is obtained from a
constant dc supply Vs is obtained.
Switch Mode D.C to D.C Converters
• During the period Ton, the chopper is on and load voltage Vo
is equal to source voltage Vs.
• During the period Toff, the chopper is off and load current io
flows through the freewheeling diode.
Switch Mode D.C to D.C Converters
• The load voltage is given by
𝑇𝑜𝑛
𝑉𝑜 =
𝑉𝑠
𝑇𝑜𝑛 + 𝑇𝑜𝑓𝑓
𝑇𝑜𝑛
𝑉𝑜 =
𝑉𝑠
𝑇
𝑉𝑜 = 𝐷𝑉𝑠
𝑇𝑜𝑛
𝑤ℎ𝑒𝑟𝑒, 𝐷 =
𝑖𝑠 𝑐𝑎𝑙𝑙𝑒𝑑 𝑠𝑤𝑖𝑡𝑐ℎ𝑖𝑛𝑔 𝑑𝑢𝑡𝑦 𝑟𝑎𝑡𝑖𝑜
𝑇
• Thus the load voltage can be varied by varying the switching
duty ratio D.
𝑉𝑜 = 𝑓𝑇𝑜𝑛 𝑉𝑠
Control of D.C to D.C Converters
• Average value of output voltage Vo can be controlled by
opening and closing the semiconductor switch periodically.
• The control strategies for varying duty ratio D are
1. Constant frequency system
2. Variable frequency system
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Constant frequency system
• In this scheme Ton is varied but frequency is kept constant.
• Variation of Ton means adjustment of pulse width.
Therefore, this scheme is called PWM scheme.
𝑇𝑜𝑛
𝑇𝑜𝑛
1
= 𝑇
4
3
= 𝑇
4
1
𝑉𝑜 = 𝑉𝑠
4
3
𝑉𝑜 = 𝑉𝑠
4
Constant frequency system (PWM)
+
Vo (desired)
amplifier
Vo (actual)
Vcontrol
Comparator
-
Switch
Control Signal
Sawtooth Wave
vst
vcontrol
t
Switch
Control
Signal
Ton
Toff
Ts
t on Vcontrol
D

Ts
Vst
Variable frequency system
• In this scheme Ton is kept constant but the frequency is
varied.
Variable frequency system
• Or Toff is kept constant but the frequency is varied.
Switch Mode D.C to D.C Converters
• Types of Switch Mode D.C to D.C Converters
– Step-Down (Buck) converter
– Step-up (Boost) converter
– Step Down/Up (Buck-Boost) converter
Step-Down Converter (Buck Converter)
• As name implies a step-down converter produces a lower
average output voltage than the dc input voltage Vd.
• Its main application is in regulated dc power supplies and
dc-motor speed control.
Step-Down Converter (Buck Converter)
• Circuit operation when S is ON (closed)
– Diode is reversed biased. Switch conducts inductor current 𝑖𝐿 .
– This results in positive inductor voltage, i.e:
𝑣𝐿 = 𝑉𝑑 − 𝑉𝑜
– It causes linear increase in inductor current
1
𝑖𝐿 =
𝐿
𝑣𝐿 𝑑𝑡
Step-Down Converter (Buck Converter)
• Circuit operation when S is OFF (open)
– Because of inductive energy storage, 𝑖𝐿 continues to flow.
– Diode is reversed biased. Therefore current flows through
the diode.
𝑣𝐿 = −𝑉𝑜
Step-Down Converter (Buck Converter)
Ts
1
Vo   vo (t )dt
Ts 0
t on
t off
1
Vo   Vd (t ) dt   0 dt
Ts 0
t on
ton
Vo  Vd
Ts
Vo  DVd
Step-Down Converter (Buck Converter)
• Continuous conduction mode
– A buck converter operates in continuous mode if the current
through the inductor (IL) never falls to zero during the
commutation cycle.
Design procedure for Buck Converter
• Calculate D to obtain required output voltage.
• Select a particular switching frequency:
– –preferably >20KHz for negligible acoustic noise
• Higher fs results in smaller L, but higher device losses.
– Thus lowering efficiency and larger heat sink.
Design procedure for Buck Converter
• Inductor requirement
𝐿 ≥ 𝐿𝑚𝑖𝑛
1−𝐷
=
𝑅
2𝑓
• C Calculation
1−𝐷
𝑟=
8𝐿𝐶𝑓 2
• Possible switching devices: MOSFET, IGBT and BJT. Low power
MOSFET can reach MHz range.
Example-1
• A buck converter is supplied from a 50V battery source.
Given L=400uH, C=100uF, R=20 Ohm, f=20KHz and D=0.4.
Calculate: (a) output voltage (b) output voltage ripple.
Solution
(a) output voltage
𝑉𝑜 = 𝐷𝑉𝑠
𝑉𝑜 = 0.4 × 50
𝑉𝑜 = 20𝑉
35
Example-1
(b) Output Ripple
1−𝐷
𝑟=
8𝐿𝐶𝑓 2
1 − 0.4
𝑟=
8 × 400𝜇 × 100𝜇 × (20𝐾)2
𝑟 = 4.6%
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Example-2
• A buck converter has an input voltage of 50V and output of 25V.
The switching frequency is 10KHz. The power output is 125W. (a)
Determine the duty ratio, (b) value of L to ensure continuous
current, (c) value of capacitance to limit the output voltage ripple
factor to 0.5%.
Solution
(a) output voltage
𝑉𝑜 = 𝐷𝑉𝑠
𝑉𝑜
𝐷=
𝑉𝑠
25
𝐷=
= 0.5
50
(b) Value of L
𝐿 ≥ 𝐿𝑚𝑖𝑛
1−𝐷
=
𝑅
2𝑓
𝐿 ≥ 𝐿𝑚𝑖𝑛
1 − 0.5
=
𝑅
2 × 10𝐾
𝑉2
𝑃𝑜 =
𝑅
37
Example-2
• Resistance is calculated as
𝑉 2 252
𝑅=
=
= 5Ω
𝑃𝑜 125
𝐿 ≥ 𝐿𝑚𝑖𝑛
1 − 0.5
=
×5
2 × 10𝐾
𝐿 ≥ 𝐿𝑚𝑖𝑛
2.5
=
= 125𝜇𝐻
20𝐾
• L must at least be 10 times greater than Lmin.
𝐿 = 1.25𝑚𝐻
38
(c) Value of C
Example-2
• Value of capacitance to limit the output voltage ripple factor to
0.5% can be calculated using following equation.
1−𝐷
𝑟=
8𝐿𝐶𝑓 2
1 − 0.5
0.005 =
8 × 1.25𝑚 × 𝐶 × (10𝐾)2
1 − 0.5
𝐶=
8 × 1.25𝑚 × 0.005 × (10𝐾)2
0.5
𝐶=
= 100𝜇𝐹
5000
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Example-3
• Design a buck converter such that the output voltage is 28V when
the input is 48V. The load is 8Ohm. Design the converter such
that it will be in continuous current mode. The output voltage
ripple must not be more than 0.5%. Specify the frequency and
the values of each component. Suggest the power switch also.
Solution:
• First of all determine the switching frequency.
𝑓 = 25𝐾𝐻𝑧
• Then calculate the switching duty ratio
𝑉𝑜 = 𝐷𝑉𝑠
28
𝐷=
= 0.58
48
40
Example-3
• Now value of inductor can be chosen to ensure continuous conduction
𝐿 ≥ 𝐿𝑚𝑖𝑛
1−𝐷
=
𝑅
2𝑓
𝐿 ≥ 𝐿𝑚𝑖𝑛
1 − 0.58
=
× 80
2 × 25𝐾
𝐿 ≥ 𝐿𝑚𝑖𝑛 = 0.67𝑚𝐻
𝐿 = 6.7𝑚𝐻
41
Example-3
• Value of capacitance to limit the output voltage ripple factor to
0.5% can be calculated using following equation.
1−𝐷
𝑟=
8𝐿𝐶𝑓 2
1 − 0.58
𝐶=
8 × 6.7𝑚 × 0.005 × (25𝐾)2
0.42
𝐶=
= 2.5𝜇𝐹
167500
42
Example-3
• Selection of Power semiconductor switch
𝑓 = 25𝐾𝐻𝑧
𝑇 = 40𝜇𝑠
43
Step-Up Converter (Boost Converter)
• A boost converter (step-up converter) is a DC-to-DC power
converter with an output voltage greater than its input
voltage.
Step-Up Converter (Boost Converter)
• When the switch is closed, current flows through
the inductor in clockwise direction and the
inductor stores the energy.
• Polarity of the left side of the inductor is
positive.
𝑣𝐿 = 𝑉𝑑
𝑑𝑖𝐿 𝑉𝑑
=
𝑑𝑡
𝐿
Step-Up Converter (Boost Converter)
• When switch is opened, the output receives energy from
the input as well as from the inductor.
• Hence output is large.
𝑣𝐿 = 𝑉𝑑 − 𝑉𝑜
𝑑𝑖𝐿 𝑉𝑑 − 𝑉𝑜
=
𝑑𝑡
𝐿
Step-Up Converter (Boost Converter)
𝑣𝐿 = 𝑉𝑑
𝑑𝑖𝐿 𝑉𝑑
=
𝑑𝑡
𝐿
S Closed
𝑣𝐿 = 𝑉𝑑 − 𝑉𝑜
S Open
𝑑𝑖𝐿 𝑉𝑑 − 𝑉𝑜
=
𝑑𝑡
𝐿
Step-Up Converter (Boost Converter)
𝐷𝑇
𝑣𝐿 =
𝑇
𝑉𝑑 𝑑𝑡 +
0
0 = 𝑉𝑑 𝑡
𝑉𝑑 − 𝑉0
𝐷𝑇
𝐷𝑇
0
+ (𝑉𝑑 −𝑉𝑜 ) 𝑡
𝑇
𝐷𝑇
0 = 𝐷𝑇𝑉𝑑 + (𝑉𝑑 −𝑉𝑜 )[𝑇 − 𝐷𝑇]
0 = 𝑇𝑉𝑑 − 𝑇𝑉𝑜 + 𝐷𝑇𝑉𝑜
𝑉𝑑
𝑉𝑜 =
1−𝐷
Step-Up Converter (Boost Converter)
• Boost Converter Design
• Minimum inductor value
𝐿𝑚𝑖𝑛
𝐷(1 − 𝐷)2 𝑅
=
2𝑓
• Capacitor Value
𝐷
𝑟=
𝑅𝐶𝑓
Example-4
• The boost converter has the following
parameters: Vd=20V, D=0.6, R=12.5ohm, L=65uH,
C=200uF, fs=40KHz. Determine (a) output
voltage, (b) output voltage ripple.
Solution
(a) output voltage
𝑉𝑑
𝑉𝑜 =
1−𝐷
20
𝑉𝑜 =
= 50𝑉
1 − 0.6
(b) output voltage ripple
𝐷
𝑟=
𝑅𝐶𝑓
0.6
𝑟=
12.5 × 200𝜇 × 40𝐾
𝑟 = 0.6%
50
Example-5
• Design a boost converter to provide an output voltage of
36V from a 24V source. The load is 50W. The voltage
ripple factor must be less than 0.5%. Specify the duty cycle
ratio, switching frequency, inductor and capacitor size, and
power device.
Solution
• First of all determine the switching frequency.
𝑓 = 30𝐾𝐻𝑧
• Then calculate the switching duty ratio
𝑉𝑑
𝑉𝑜 =
1−𝐷
24
36 =
1−𝐷
⇒ 36(1 − 𝐷) = 24
⇒ 𝐷 = 0.33
51
Example-5
• Calculate the load resistance
𝑉2
𝑃𝑜 =
𝑅
362
𝑅=
= 25.92Ω
50
• Calculate inductor value to ensure continuous current
𝐿𝑚𝑖𝑛
𝐷(1 − 𝐷)2 𝑅
=
2𝑓
𝐿𝑚𝑖𝑛
0.33(1 − 0.33)2 × 25.92
=
= 63.9𝜇𝐻
2 × 30𝐾
• Inductance must be greeter than or equal to Lmin
𝐿 ≥ 𝐿𝑚𝑖𝑛 = 63.9𝜇𝐻
𝐿 = 100𝜇𝐻
52
Example-5
• Then calculate the Capacitor Value for ripple factor less than 0.5%
𝐷
𝑟=
𝑅𝐶𝑓
0.33
0.005 =
25.92 × 𝐶 × 30𝐾
0.33
𝐶=
= 84.8𝜇𝐹
25.92 × 0.005 × 30𝐾
• While selecting the power device we must take into account the switching frequency
𝑓 = 30𝐾𝐻𝑧
1
1
𝑇= =
= 33.3𝜇𝑠
𝑓 30𝐾
53
Example-5
• Selection of Power semiconductor switch
𝑓 = 30𝐾𝐻𝑧
𝑇 = 33.3𝜇𝑠
54
Buck-Boost Converter
• The buck–boost converter is a type of DC-to-DC
converter that has an output voltage magnitude that is
either greater than or less than the input voltage
magnitude.
– If D>0.5, output is higher
– If D<0.5, output is lower
• Output voltage is always negative
Buck-Boost Converter
• while in the On-state, the input voltage source is
directly connected to the inductor (L). This results in
accumulating energy in L. In this stage, the capacitor
supplies energy to the output load.
Buck-Boost Converter
• In Off-state, the inductor is connected to the output
load and capacitor, so energy is transferred from L to C
and R.
Buck-Boost Converter
• In ON-state (Switch Closed)
𝑑𝑖𝐿
𝑣𝐿 = 𝑉𝑑 = 𝐿
𝑑𝑡
𝑑𝑖𝐿 𝑉𝑑
⇒
=
𝑑𝑡
𝐿
Buck-Boost Converter
• In OFF-state (Switch Opened)
𝑑𝑖𝐿
𝑣𝐿 = 𝑉𝑜 = 𝐿
𝑑𝑡
𝑑𝑖𝐿 𝑉𝑜
⇒
=
𝑑𝑡
𝐿
Buck-Boost Converter
• Steady state operation
𝑉𝑑 𝐷𝑇 𝑉𝑜 1 − 𝐷 𝑇
+
=0
𝐿
𝐿
𝐷
𝑉𝑜 = −𝑉𝑑
1−𝐷
Buck-Boost Converter
𝐷
𝑟=
𝑅𝐶𝑓
𝐿𝑚𝑖𝑛
(1 − 𝐷)2 𝑅
=
2𝑓
Example-6
• Determine the switching duty ratio of a buck-boost converter
such that the output voltage is -28V when the input is 100V. The
load is 1Ohm. Design the converter such that it will be in
continuous current mode. The output voltage ripple must not be
more than 0.5%. Specify the frequency and the values of each
component. Suggest the power switch also.
Solution:
• First of all determine the switching frequency.
𝑓 = 50 𝐾𝐻𝑧
• Then calculate the switching duty ratio
𝐷
𝑉𝑜 = −𝑉𝑑
1−𝐷
𝑉𝑜
𝐷
⇒
=
−𝑉𝑑 1 − 𝐷
0.28
⇒𝐷=
= 0.2
1.28
𝐷
⇒ 0.28 =
1−𝐷
62
Example-6
• Then calculate the Capacitor Value for ripple factor less than 0.5%
𝐷
𝑟=
𝑅𝐶𝑓
0.2
0.005 =
10 × 𝐶 × 50𝐾
0.2
𝐶=
= 80𝜇𝐹
10 × 0.005 × 50𝐾
• Value of inductor can be calculated as
(1 − 𝐷)2 𝑅
𝐿𝑚𝑖𝑛 =
𝐿 > 𝐿𝑚𝑖𝑛
2𝑓
(1 − 0.2)2 × 10
=
2 × 50𝐾
6.4
𝐿>
100𝐾
𝐿 > 64𝜇𝐻
63
Example-6
• While selecting the power device we must take into account the switching frequency
𝑓 = 50𝐾𝐻𝑧
𝑇=
1
1
=
= 20𝜇𝑠
𝑓 50𝐾
64
Switch mode vs Linear Power Supplies
• One of the major applications of switch mode dc-dc
converters is in switch mode power supplies (SMPs).
• SMPs offer several advantages over linear mode
power supplies.
– Efficient (70-95%)
– Weight and size reduction
• They also have some disadvantages
– Complex design
– EMI problems
65
Linear Mode Power Supply
66
Switch Mode Power Supply
67
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END OF LECTURE-10
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