2. DC-DC Converters
DC-DC converters are power electronic circuits that convert a DC
voltage to a different DC voltage level, often providing a regulated
output.
A BASIC SWITCHING CONVERTER
An efficient alternative to the linear regulator
Uses Power electronics switches like BJT,MOSFET IGBT…
Also known as DC Chopper
3. DC-DC Converters
Figure: (a) A basic DC-DC switching converter; (b) Switching
equivalent; ( c) Output voltage.
Assuming the switch is ideal
• The output is the same as the
input when the switch ON
• And the output is zero when the
switch OFF
Periodic opening and closing of the
switch gives the pulsed output
waveform.
The average or DC component of the
output voltage is
𝑉𝑜 =
1
𝑇
0
𝑇
𝑣 𝑜 𝑡 𝑑𝑡 =
1
𝑇
0
𝐷𝑇
𝑉𝑠 𝑑𝑡 = 𝑉𝑠 𝐷
4. DC-DC Converters
𝐷 ≡
𝑡 𝑜𝑛
𝑡 𝑜𝑛 + 𝑡 𝑜𝑓𝑓
The DC component of the output voltage will be less than or equal to
the input voltage for this circuit.
ideal switch Zero loss Zero voltage across when ON Zero
current through it when OFF
But real switch has some power loss Considerable because it
creates heat on the switch.
5. THE BUCK (STEP-DOWN)
CONVERTERApplication Example: Controlling the speed of DC Motor
Low-pass filter will be placed at the output
• The diode provides a path for the inductor
current when the switch is opened and is
reverse-biased when the switch is closed.
6. THE BUCK (STEP-DOWN)
CONVERTERVOLTAGE AND CURRENT RELATIONSHIPS:
𝑣 𝑥 in Fig. above, is 𝑉𝑠
when the switch is closed and is zero when the
switch is open, provided that the inductor current remains positive,
keeping the diode on.
If the switch is closed periodically at a duty ratio D, the average
voltage at the filter input is 𝑉𝑠
𝐷
An inductor current that remains positive throughout the switching
period is known as continuous current.
Conversely, discontinuous current is characterized by the inductor
current’s returning to zero during each period.
7. THE BUCK (STEP-DOWN)
CONVERTERBuck converters and DC-DC converters in general, have the following properties when
operating in the steady state
1. The inductor current is periodic.
𝒊 𝑳 𝒕 + 𝑻 = 𝒊 𝑳 𝒕
2. The average inductor voltage is zero
𝑽 𝑳 =
𝟏
𝑻
𝒕
𝒕+𝑻
𝒗 𝑳 𝝀 𝒅𝝀 = 𝟎
3. The average capacitor current is zero
𝑰 𝑪 =
𝟏
𝑻
𝒕
𝒕+𝑻
𝒊 𝑪 𝝀 𝒅𝝀 = 𝟎
4. The power supplied by the source is the same as the power delivered to the load. For
non-ideal components, the source also supplies the losses.
𝑷 𝒔 = 𝑷 𝒐 𝒊𝒅𝒆𝒂𝒍
𝑷 𝒔= 𝑷 𝒐 + 𝒍𝒐𝒔𝒔𝒆𝒔 𝒏𝒐𝒏𝒊𝒅𝒆𝒂𝒍
8. THE BUCK (STEP-DOWN)
CONVERTERAssumptions for Analysis of Buck converter:
• The circuit is operating in the steady state.
• The inductor current is continuous (always positive)
• 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.
• The switching period is 𝑇; the switch is closed for time 𝐷𝑇 and
open for time (1 − 𝐷)𝑇.
• The components are ideal.
The key to the analysis for determining the output 𝑉𝑜 is to
examine the inductor current and inductor voltage first for
the switch closed and then for the switch open.
9. THE BUCK (STEP-DOWN)
CONVERTERANALYSIS FOR THE SWITCH CLOSED:
• The voltage across the inductor is
• 𝑣 𝐿 = 𝑉𝑆 − 𝑉𝑜 = 𝐿
𝑑𝑖 𝐿
𝑑𝑡
• Rearranging,
•
𝑑𝑖 𝐿
𝑑𝑡
=
𝑉 𝑆−𝑉𝑜
𝐿
𝑠𝑤𝑖𝑡𝑐ℎ 𝑐𝑙𝑜𝑠𝑒𝑑
•
𝑑𝑖 𝐿
𝑑𝑡
=
∆𝑖 𝐿
∆𝑡
=
∆𝑖 𝐿
𝐷𝑇
=
𝑉 𝑆−𝑉𝑜
𝐿
• ∆𝑖 𝐿 𝑐𝑙𝑜𝑠𝑒𝑑 =
𝑉 𝑆−𝑉𝑜
𝐿
𝐷𝑇
10. THE BUCK (STEP-DOWN)
CONVERTERANALYSIS FOR THE SWITCH OPEN:
• When the switch is open, the diode becomes forward-biased to carry the inductor
current
• The voltage across the inductor when the switch is open is
𝑣 𝐿 = −𝑉𝑜 = 𝐿
𝑑𝑖 𝐿
𝑑𝑡
• Rearranging,
𝑑𝑖 𝐿
𝑑𝑡
=
−𝑉𝑜
𝐿
𝑠𝑤𝑖𝑡𝑐ℎ 𝑜𝑝𝑒𝑛
• The derivative of current in the inductor is a negative constant, and the current
decreases linearly
• The change in inductor current when the switch is open is
∆𝑖 𝐿
∆𝑡
=
∆𝑖 𝐿
1−𝐷 𝑇
=
−𝑉𝑜
𝐿
∆𝑖 𝐿 𝑜𝑝𝑒𝑛 = −
𝑉𝑜
𝐿
(1 − 𝐷)𝑇
11. THE BUCK (STEP-DOWN)
CONVERTERAt steady state
∆𝑖 𝐿 𝑐𝑙𝑜𝑠𝑒𝑑 + ∆𝑖 𝐿 𝑜𝑝𝑒𝑛 = 0
𝑉 𝑆−𝑉𝑜
𝐿
𝐷𝑇 −
𝑉𝑜
𝐿
1 − 𝐷 = 0
• Solving for 𝑉𝑜,
𝑉𝑜 = 𝑉𝑠 𝐷
• Therefore, the buck converter produces an output voltage that is less than or
equal to the input.
• An alternative derivation of the output voltage is based on the inductor voltage.
• Since the average inductor voltage is zero for periodic operation
𝑉𝐿 = 𝑉𝑆 − 𝑉𝑂 𝐷𝑇 + −𝑉𝑂 1 − 𝐷 𝑇 = 0
• Solving for 𝑉𝑂
𝑉𝑂= 𝑉𝑆 𝐷.
12. THE BUCK (STEP-DOWN)
CONVERTERThe average inductor current must be the same as the average current in
the load resistor, since the average capacitor current must be zero for
steady-state operation.
𝐼𝐿 = 𝐼 𝑅 =
𝑉𝑂
𝑅
The maximum and minimum values of the inductor current are computed
as
∆𝑖 𝐿 𝑐𝑙𝑜𝑠𝑒𝑑 =
𝑉𝑆 − 𝑉𝑜
𝐿
𝐷𝑇
∆𝑖 𝐿 𝑜𝑝𝑒𝑛 = −
𝑉𝑜
𝐿
(1 − 𝐷)𝑇
𝐼 𝑚𝑎𝑥= 𝐼𝐿 +
∆𝑖 𝐿
2
=
𝑉𝑂
+
1 𝑉𝑂
1 − 𝐷 𝑇 = 𝑉𝑂
1
+
1 − 𝐷
13. THE BUCK (STEP-DOWN)
CONVERTER
𝐼 𝑚𝑖𝑛 = 𝐼𝐿 −
∆𝑖 𝐿
2
=
𝑉𝑂
𝑅
−
1
2
𝑉𝑂
𝐿
1 − 𝐷 𝑇 = 𝑉𝑂
1
𝑅
−
1 − 𝐷
2𝐿𝑓
where 𝑓 = 1/𝑇 is the switching frequency.
• The above Eq. can be used to determine the combination of L and f
that will result in continuous current. Since 𝐼 𝑚𝑖𝑛 = 0 is the boundary
between continuous and discontinuous current,
𝐼 𝑚𝑖𝑛 = 0 = 𝑉𝑂
1
𝑅
−
1 − 𝐷
2𝐿𝑓
(𝐿𝑓) 𝑚𝑖𝑛=
(1 − 𝐷)𝑅
2
14. THE BUCK (STEP-DOWN)
CONVERTER• If the desired switching frequency is established,
𝐿 𝑚𝑖𝑛 =
(1 − 𝐷)𝑅
2𝑓
𝑓𝑜𝑟 𝑐𝑜𝑛𝑡𝑖𝑛𝑢𝑜𝑢𝑠 𝑐𝑢𝑟𝑟𝑒𝑛𝑡
where 𝐿 𝑚𝑖𝑛 is the minimum inductance required for continuous current.
• In practice inductance is chosen greater than 𝐿 𝑚𝑖𝑛
• The peak-to-peak variation in the inductor current is often used as a design
criterion in buck converter.
• to determine the value of inductance for a specified peak-to-peak inductor
current for continuous-current operation:
∆𝑖 𝐿 =
𝑉𝑠 − 𝑉𝑜
𝐿
𝐷𝑇 =
𝑉𝑠 − 𝑉𝑜
𝐿𝑓
𝐷 =
𝑉𝑂 1 − 𝐷
𝐿𝑓
𝑜𝑟 𝐿 =
𝑉𝑠 − 𝑉𝑜
∆𝑖 𝐿 𝑓
𝐷 =
𝑉𝑂 1 − 𝐷
∆𝑖 𝐿 𝑓
15. THE BUCK (STEP-DOWN)
CONVERTER• Since the converter components are assumed to be ideal,
the power supplied by the source must be the same as the
power absorbed by the load resistor.
𝑃𝑆 = 𝑃𝑂
𝑉𝑠 𝐼𝑠 = 𝑉𝑜 𝐼 𝑜 𝑜𝑟
𝑉𝑜
𝑉𝑠
=
𝐼𝑠
𝐼 𝑜
• Basically from this expression we can say that DC-DC
converter is DC transformer.
16. THE BUCK (STEP-DOWN)
CONVERTER• OUTPUT VOLTAGE RIPPLE
• In the preceding analysis, the capacitor was assumed to be very
large to keep the output voltage constant.
• The variation in output voltage is computed from the voltage-
current relationship of the capacitor.
• The current in the capacitor is
𝑖 𝐶 = 𝑖 𝐿 − 𝑖 𝑅
• While the capacitor current is positive, the capacitor is charging.
From the definition of capacitance,
𝑄 = 𝐶𝑉𝑜
∆𝑄 = 𝐶∆𝑉𝑜
∆𝑉𝑜 =
∆𝑄
𝐶
17. THE BUCK (STEP-DOWN)
CONVERTER• The change in charge ∆𝑄 is
the area of the triangle
above the time axis
∆𝑄 =
1
2
𝑇
2
∆𝑖 𝐿
2
=
𝑇∆𝑖 𝐿
8
• resulting in
∆𝑉𝑜 =
𝑇∆𝑖 𝐿
8𝐶
18. THE BUCK (STEP-DOWN)
CONVERTER
• Using ∆𝑖 𝐿 𝑜𝑝𝑒𝑛 = −
𝑉𝑜
𝐿
(1 − 𝐷)𝑇 = ∆𝑖𝐿
∆𝑉𝑜 =
𝑇𝑉𝑜
8𝐶𝐿
1 − 𝐷 𝑇 =
𝑉𝑜(1 − 𝐷)
8𝐿𝐶𝑓2
• It is also useful to express the ripple as a fraction of the
output voltage
∆𝑉𝑜
𝑉𝑜
=
1 − 𝐷
8𝐿𝐶𝑓2
• 𝐶 =
1−𝐷
8𝐿(∆𝑉𝑜/𝑉𝑜)𝑓2
19. THE BUCK (STEP-DOWN)
CONVERTER
• Using ∆𝑖 𝐿 𝑜𝑝𝑒𝑛 = −
𝑉𝑜
𝐿
(1 − 𝐷)𝑇 for ∆𝑖𝐿
∆𝑉𝑜 =
𝑇𝑉𝑜
8𝐶𝐿
1 − 𝐷 𝑇 =
𝑉𝑜(1 − 𝐷)
8𝐿𝐶𝑓2
• It is also useful to express the ripple as a fraction of the
output voltage
∆𝑉𝑜
𝑉𝑜
=
1 − 𝐷
8𝐿𝐶𝑓2
• 𝐶 =
1−𝐷
8𝐿(∆𝑉𝑜/𝑉𝑜)𝑓2
20. CAPACITOR RESISTANCE—THE
EFFECT ON RIPPLE VOLTAGE
The ESR may have a significant effect on the output voltage rippleA
real capacitor can be modelled as a capacitance with an equivalent
series resistance (ESR) and an equivalent series inductance (ESL).
ESR, often producing a ripple voltage greater than that of the ideal
capacitance.
The inductance in the capacitor is usually not a significant factor at
typical switching frequencies.
The ripple due to ESR can be approximated by first determining the
ripple current assuming the capacitor ideal
∆𝑉𝑜,𝐸𝑆𝑅 = ∆𝑖 𝐶 𝑟𝐶 = ∆𝑖 𝐿 𝑟𝐶
∆𝑉𝑜 < ∆𝑉𝑜,𝐶 + ∆𝑉𝑜,𝐸𝑆𝑅
21. CAPACITOR RESISTANCE—THE
EFFECT ON RIPPLE VOLTAGE
• The ripple voltage due to the ESR can be much larger than the ripple
due to the pure capacitance.
• In that case, the output capacitor is chosen on the basis of the
equivalent series resistance rather than capacitance only.
∆𝑉𝑜 ≈ ∆𝑉𝑜,𝐸𝑆𝑅 = ∆𝑖 𝐶 𝑟𝐶
22. SYNCHRONOUS RECTIFICATION FOR
THE BUCK CONVERTER
Figure: A synchronous buck converter. The MOSFET S2 carries the inductor current
when S1 is off to provide a lower voltage drop than a diode.
23. THE BOOST CONVERTER
Assumptions
1. Steady-state conditions exist.
2. The switching period is 𝑇, and the
switch is closed for time 𝐷𝑇 and open
for (1 − 𝐷)𝑇.
3. The inductor current is continuous
(always positive).
4. The capacitor is very large, and the
output voltage is held constant at
voltage 𝑉𝑜.
5. The components are ideal.
24. THE BOOST CONVERTER
• ANALYSIS FOR THE SWITCH CLOSED:
𝑣 𝐿 = 𝑉𝑆 = 𝐿
𝑑𝑖 𝐿
𝑑𝑡
𝑜𝑟
𝑑𝑖 𝐿
𝑑𝑡
=
𝑉𝑆
𝐿
∆𝑖 𝐿
∆𝑡
=
∆𝑖 𝐿
𝐷𝑇
=
𝑉𝑆
𝐿
• Solving for ∆𝑖 𝐿 for the switch closed,
∆𝑖 𝐿 𝑐𝑙𝑜𝑠𝑒𝑑 =
𝑉𝑆 𝐷𝑇
𝐿
28. THE BOOST CONVERTER
A condition necessary for continuous inductor current is for 𝐼 𝑚𝑖𝑛 to be
positive.
Therefore, the boundary between continuous and discontinuous inductor
current is determined from
𝐼 𝑚𝑖𝑛 = 0 =
𝑉𝑠
1 − 𝐷 2 𝑅
−
𝑉𝑠 𝐷𝑇
2𝐿
𝑜𝑟
𝑉𝑠
1 − 𝐷 2 𝑅
=
𝑉𝑠 𝐷𝑇
2𝐿
=
𝑉𝑠 𝐷
2𝐿𝑓
𝐿𝑓 𝑚𝑖𝑛 =
𝐷 1−𝐷 2 𝑅
2
𝑜𝑟 𝐿 𝑚𝑖𝑛 =
𝐷 1−𝐷 2 𝑅
2𝑓
From a design perspective, it is useful to express 𝐿 in terms of a desired ∆𝑖 𝐿,
𝐿 =
𝑉𝑠 𝐷𝑇
∆𝑖 𝐿
=
𝑉𝑠 𝐷
∆𝑖 𝐿 𝑓
29. THE BOOST CONVERTER
The peak-to-peak output voltage ripple can be calculated from the
capacitor current waveform.
The change in capacitor charge can be calculated from
∆𝑄 =
𝑉𝑜
𝑅
𝐷𝑇 = 𝐶∆𝑉𝑜
An expression for ripple voltage is then
∆𝑉𝑜=
𝑉𝑜 𝐷𝑇
𝑅𝐶
=
𝑉𝑜 𝐷
𝑅𝐶𝑓
𝑜𝑟
∆𝑉𝑜
𝑉𝑜
=
𝐷
𝑅𝐶𝑓
where 𝑓 is the switching frequency.
𝐶 =
𝐷
𝑅(∆𝑉𝑜/𝑉𝑜)𝑓
30. THE BOOST CONVERTER
As Buck Converter the voltage ripple due to the ESR is
∆𝑉𝑜,𝐸𝑆𝑅= ∆𝑖 𝐶 𝑟𝐶 = 𝐼𝐿,𝑚𝑎𝑥 𝑟𝐶
Effects of Inductor Resistance
• Inductors should be designed to have small resistance to minimize power loss and
maximize efficiency.
• Inductor resistance affects performance of the boost converter, especially at high duty
ratios.
• For the boost converter, recall that the output voltage for the ideal case is
𝑉𝑜 =
𝑉𝑠
1 − 𝐷
• The power supplied by the source must be the same as the power absorbed by the
load and the inductor resistance, neglecting other losses.
𝑃𝑠= 𝑃𝑜 + 𝑃𝑟𝐿
𝑉𝑠 𝐼𝐿 = 𝑉𝑜 𝐼 𝐷 + 𝐼𝐿
2
𝑟𝐿
where 𝑟𝐿 is the series resistance of the inductor.
33. BUCK-BOOST CONVERTER
• The output voltage of the buck-boost converter can be either
higher or lower than the input voltage.
• Assumptions :
1. The circuit is operating in the steady state.
2. The inductor current is continuous.
3. The capacitor is large enough to assume a constant
output voltage.
4. The switch is closed for time 𝐷𝑇 and open for (1 − 𝐷)𝑇.
5. The components are ideal.
37. BUCK-BOOST CONVERTER
• For steady-state operation, the net change in inductor current
must be zero over one period.
(∆𝑖 𝐿) 𝑐𝑙𝑜𝑠𝑒𝑑 + (∆𝑖 𝐿) 𝑜𝑝𝑒𝑛= 0
𝑉𝑠 𝐷𝑇
𝐿
+
𝑉𝑜 1 − 𝐷 𝑇
𝐿
= 0
• Solving for 𝑉𝑜,
𝑉𝑜 = −𝑉𝑠
𝐷
1 − 𝐷
• The required duty ratio for specified input and output voltages can be
expressed as
𝐷 =
𝑉𝑜
𝑉𝑠 + 𝑉𝑜
38. BUCK-BOOST CONVERTER
• The average inductor voltage is zero for periodic operation,
resulting in
𝑉𝐿 = 𝑉𝑠 𝐷 + 𝑉𝑜 1 − 𝐷 = 0
• Solving for 𝑉𝑜 yields
𝑉𝑜 = −𝑉𝑠
𝐷
1 − 𝐷
• The output voltage has opposite polarity from the source
voltage.
• If 𝐷 > 0.5, the output voltage is larger than the input; and
if 𝐷 < 0.5, the output is smaller than the input.
39. BUCK-BOOST CONVERTER
• Power absorbed by the load must be the same as that supplied by the source, where
𝑃0 =
𝑉𝑜
2
𝑅
𝑃𝑠 = 𝑉𝑠 𝐼𝑠
𝑉𝑜
2
𝑅
= 𝑉𝑠 𝐼𝑠
• Average source current is related to average inductor current by
𝐼𝑠 = 𝐼𝐿 𝐷
• resulting in
𝑉𝑜
2
𝑅
= 𝑉𝑠 𝐼𝐿 𝐷
• Substituting for 𝑉𝑜 and solving for 𝐼𝐿, we find
𝐼𝐿 =
𝑉𝑜
2
𝑉𝑠 𝑅𝐷
=
𝑃0
𝑉𝑠 𝐷
=
𝑉𝑠 𝐷
𝑅(1 − 𝐷)2
41. BUCK-BOOST CONVERTER
• OUTPUT VOLTAGE RIPPLE
• The output voltage ripple for the buck-boost converter is computed
from the capacitor current .
∆𝑄 =
𝑉𝑜
𝑅
𝐷𝑇 = 𝐶∆𝑉𝑜
• Solving for ∆𝑉𝑜,
∆𝑉𝑜 =
𝑉𝑜 𝐷𝑇
𝑅𝐶
=
𝑉𝑜 𝐷
𝑅𝐶𝑓
𝑜𝑟
∆𝑉𝑜
𝑉𝑜
=
𝐷
𝑅𝐶𝑓
42. BUCK-BOOST CONVERTER
• As is the case with other converters, the equivalent series resistance of
the capacitor can contribute significantly to the output ripple voltage.
The peak-to-peak variation in capacitor current is the same as the
maximum inductor current.
∆𝑉𝑜,𝐸𝑆𝑅= ∆𝑖 𝐶 𝑟𝐶 = 𝐼𝐿,𝑚𝑎𝑥 𝑟𝐶
43. BUCK-BOOST CONVERTER
•As is the case with other converters, the
equivalent series resistance of the capacitor can
contribute significantly to the output ripple
voltage. The peak-to-peak variation in capacitor
current is the same as the maximum inductor
current.
∆𝑉𝑜,𝐸𝑆𝑅= ∆𝑖 𝐶 𝑟𝐶 = 𝐼𝐿,𝑚𝑎𝑥 𝑟𝐶
