Lecture Note
9
DC-DC PWM Converters
Prepared by
Dr. Oday A Ahmed
Website: https://odayahmeduot.wordpress.com
Email: 30205@uotechnology.edu.iq
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Lecture Note 9: DC-DC PWM Converters
Instructure: Dr. Oday A Ahmed
DC-DC PWM Converters
Many industrial applications need a conversion of a voltage coming from a DC
source into another DC voltage. A device that performs this kind of conversion is
known as DC/DC converter or called DC Choppers. They achieve the voltage
regulation by varying the on–off or time duty ratio of the switching element using
a control technique called Pulse Width Modulation PWM.
Linear Conversion
The simplest way to obtain a DC voltage by a
DC source with a different voltage level consists
on using a voltage divider, as shown in Fig. 1.
Vout  Vs •
Fig.1
R2
R1  R2
The input and output powers are respectively
given by:

The efficiency conversion is obtained:
V
R2
 out
R1  R2 Vin
If Vin = 39V, and Vout = 13V, efficiency η is only 0.33
From what explained above, it is clear that a DC conversion by a voltage divider
presents some drawbacks:
• A DC voltage higher than the input voltage cannot be obtained;
• The output voltage depends on the load, in general;
• The efficiency is very poor.
In a linear conversion by series regulators the output voltage, lower than the input
one, is obtained subtracting a voltage by the input generator.
Once the stability is assured by suitably designing the regulator, the load current
will flow through the power BJT and the output voltage is given by:
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Lecture Note 9: DC-DC PWM Converters
Instructure: Dr. Oday A Ahmed
In general, the described circuit has the advantage of a regulation of the output
voltage; however, only a step down conversion is possible and the efficiency
remains low because all the power supplied by the source that it is not utilized by
the load have to be dissipated by the power BJT.
Linear converters are however, utilized, for example, for voltage sensing
applications where the power losses are negligible.
Switching Conversion
Switching conversion, on the contrary, is based on the use of a power electronic
switch used in switching operation, it means the presence of only two
fundamental states: the on state in which the voltage of the power switch is null
and its current is imposed by external circuitry and the off state in which the
current of the power switch is null and its voltage is imposed by external circuitry.
As mentioned earlier, the power electronic circuits that use this conversion
mode called: PWM DC-DC converters.
The output average voltage of PWM DC-DC converter is controlled by
controlling the turn-on ton and turn-off toff times of the switch S:
𝑉𝑜𝐴𝑉 = 𝑉𝑖𝑛
𝑡𝑜𝑛
𝑡𝑜𝑛 +𝑡𝑜𝑓𝑓
= 𝑉𝑖𝑛
𝑡𝑜𝑛
𝑇
= 𝑉𝑖𝑛 𝐷 where D is called a Duty Cycle
Vs
Vo
Fig.2
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Lecture Note 9: DC-DC PWM Converters
Instructure: Dr. Oday A Ahmed
Pulse Width Modulation
The output DC voltage of DC chopper can be varied by controlling the width
period (ton) of the signal applied to the switch S with constant
switching/chopping frequency fs. This method is called PWM method. Fig.3
shows the basic circuit diagram and waveforms of PWM.
Fig.3
Choppers Types
Two of the most popular categories of DC-DC converters are:
❖ Transformerless DC-DC Converters
❖ Insulated DC-DC Converters.
Three basic types of non-isolated DC–DC converters are
❖ Step-down converter
❖ Step-up converter
❖ Step-up-down converter
Step-Down Buck Chopper
The buck converter allows a DC voltage lower than the input voltage to be
obtained.
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Lecture Note 9: DC-DC PWM Converters
Instructure: Dr. Oday A Ahmed
The buck converter can operate in a continuous conduction mode CCM or in a
discontinuous conduction mode DCM, depending on inductance value and
duty cycle D. In CCM the inductor current flows during the entire cycle, whereas
in DCM the inductor current flows only during part of the cycle. In DCM it falls
to zero, remains at zero for some time interval, and then starts to increase.
Operation at the CCM/DCM boundary is called the critical mode.
The circuit can be studied in CCM as a succession of two linear circuits, one
corresponding to the on state in a time interval TON, and the other corresponding
to the off state in a time interval TOFF. It follows that TON + TOFF = Ts; the ratio
TON/Ts = D is indicated as the duty cycle of the circuit.
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Lecture Note 9: DC-DC PWM Converters
Instructure: Dr. Oday A Ahmed
By integrating during Ts the
inductance equation it follows that:
Hence the voltage conversion ratio is:
Assuming a loss-less circuit
In CCM, the current variation on the
inductance results as a variation of
the capacity charge while the DC component is given to the load. The increase of
the charge in a period corresponds to the triangle ABC and the voltage variation
on the load is given by:
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Lecture Note 9: DC-DC PWM Converters
Instructure: Dr. Oday A Ahmed
During TOFF the current variation
then the ripple voltage:
The relative voltage variation is given by:
where
The output voltage ripple could be minimized by choosing fc << fs. It depends on
the duty cycle and it is maximum when D = 0.
Examine the inductor current
Switch closed,
vL  Vin  Vout ,
diL Vin  Vout

dt
L
Switch open,
vL  Vout ,
diL  Vout

dt
L
 Vout
A / sec
L
iL
Imax
Iavg = Iout
Imin
Vin  Vout
A / sec
L
DT
From geometry, Iavg = Iout is halfway
between Imax and Imin
ΔI
Periodic – finishes
a period where it
started
(1 − D)T
T
Because the current consists of straight line segments, it is apparent that
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Lecture Note 9: DC-DC PWM Converters
Instructure: Dr. Oday A Ahmed
Taking the derivative of above equation with respect to D and setting it to zero
shows that ΔI is maximum when D = 0.5. Thus,
the rms value inductor current:
The boundary of continuous conduction is when ΔiLmin = 0, as shown below:
As shown, when at the boundary,
vL  Vout ,
diL  Vout

dt
L
where Lboundary is the value of L that causes the circuit to operate at the boundary
of continuous
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Lecture Note 9: DC-DC PWM Converters
Instructure: Dr. Oday A Ahmed
conduction for the given values of Vout, Iout, D, and f. The maximum required value
of Lboundary
occurs when D → 0. Therefore, the value of L
will guarantee continuous conduction for all D.
Component Ratings
A. Inductor current Ratings
Max impact of ΔI on the rms current occurs at the boundary of
continuous/discontinuous conduction, where ΔI =2Iout
2Iout
iL
Iavg = Iout
0
ΔI
2
2
I Lrms
 I avg

I Lrms 
 
1 2
1
2
I pp  I out

I 2
12
12
2
2
I Lrms
 I out

1
2
2I out 2  4 I out
12
3
2
I out
3
B. Capacitor current Ratings
Max rms current occurs at the boundary of continuous/discontinuous conduction,
where ΔI =2Iout
Iout
iC = (iL – Iout)
0
ΔI
−Iout
2
2
I Crms
 I avg

I Crms 
Note – raising f or L, which lowers
ΔI, reduces the capacitor current
1
2
2I out 2  02  1 I out
12
3
I out
3
C. Transistor and diode currents and Voltage ratings
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Lecture Note 9: DC-DC PWM Converters
Instructure: Dr. Oday A Ahmed
For the transistor and diode, a conservative voltage rating is 2Vin because of the
oscillatory ringing transients that invariably occur with parasitic inductances and
capacitances
A conservative assumption for transistor and diode current is to assume small D,
so that their currents is essentially the same as the inductor current.
Impedance matching
Iout = Iin / D
Iin
+
Source
Vin
+
DC−DC Buck
Converter
Vout = DVin
−
−
V
Rload  out
Iout
Iin
+
Vin
Equivalent from
source perspective
Requiv
−
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Lecture Note 9: DC-DC PWM Converters
Instructure: Dr. Oday A Ahmed
Vout
V
Vout
Rload
Requiv  in  D 

I in I out • D I out• D 2
D2
So, the buck converter makes the load resistance look larger to the source
Step-Up Boost Chopper
The DC/DC boost converter allows an output voltage higher than the input one to
be achieved.
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Lecture Note 9: DC-DC PWM Converters
Instructure: Dr. Oday A Ahmed
When the switch is closed, the diode is
reverse biased and open, and iL increases
at the rate of
and the inductor is charging. When the
switch is open, the diode is forward
biased, and iL decreases at the rate of
and the inductor is discharging. The
inductor voltage is shown below
Because of the steady-state inductor
principle, the average voltage vL across L
is zero. Since at any time vL takes on
one of two constant values, its average
value is
the final input-output voltage expression
Inductor Current in Continuous Conduction
The graph of iL is shown
During turn ON period one can obtained:
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Lecture Note 9: DC-DC PWM Converters
Instructure: Dr. Oday A Ahmed
The boundary of continuous conduction is when iLmin = 0
it is evident that at the boundary of continuous conduction:
The minimum value of inductance, Lboundary, needed ensure the inductor current
operates in the CCM as
Because the maximum value of D is 1, then
continuous conduction for all D.
will
guarantee
Current Waveforms in Continuous Conduction
The current waveforms of the boost converter in CCM shows below :
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Lecture Note 9: DC-DC PWM Converters
Instructure: Dr. Oday A Ahmed
Current Ratings in Continuous Conduction
for the inductor in continuous conduction
as explained in an analogous fashion in the Buck Converter experiment, yields
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Lecture Note 9: DC-DC PWM Converters
Instructure: Dr. Oday A Ahmed
current ratings for the MOSFET and diode are when D is large
the rms current through
C,
consider
the
capacitor current in
Figure below
Max
rms
current
occurs at the boundary
of
continuous/discontinuous conduction, where ΔI =2Iout
I Crms  I out
Voltage Ratings for Continuous Conduction
Because of the usual double-voltage switching transients, the MOSFET should
therefore be rated 2Vout. when the MOSFET is closed, the diode is subjected to
Vout . The diode should be conservatively rated 2Vout
Capacitor Voltage Ripple
From figure below it can be seen that the amount of charge taken from C when
the switch is closed is represented by the dotted area.
As 1 → D , the width of the dotted area increases to fill almost the entire cycle,
and the maximum peak-to-peak ripple becomes
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Lecture Note 9: DC-DC PWM Converters
Instructure: Dr. Oday A Ahmed
Impedance matching
Requiv 
Vin 1  D Vout
2 V
2

 1  D  out  1  D  Rload
I out
I in
I out
1 D
I out  1  DIin
Iin
+
+
Source
DC−DC Boost
Converter
Vin
Vin
1 D
−
Vout 
−
V
Rload  out
I out
Iin
+
Equivalent from
source perspective
Vin
Requiv
−
Example 1: Step-Down DC-DC Converter supplied by 230V DC voltage. The
load resistance equal to10Ω. Voltage drop across the chopper when it is ON equal
to 2V. For a duty cycle of 0.4, calculate:
a) Average and RMS values of output voltage
b) Power delivered to the load and
c) Chopper efficiency.
Solution:
a) When chopper is ON, Vo = (Vin-2) and during OFF time Vo =0 as
shown below:
Vin-2
Vin-2
Average output voltage =
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Lecture Note 9: DC-DC PWM Converters
Instructure: Dr. Oday A Ahmed
𝑉𝑜𝐴𝑉 = (𝑉𝑖𝑛 − 2)
𝑡𝑜𝑛
= (𝑉𝑖𝑛 − 2)𝐷 = (230 − 2)×0.4 = 91.2𝑉
𝑇
RMS value of output voltage
𝑉𝑜𝑟𝑚𝑠 = [(𝑉𝑖𝑛 − 2)
𝑡
2 𝑜𝑛
𝑇
1/2
]
= (𝑉𝑖𝑛 − 2)√𝐷 = (230 − 2)√0.4 = 144.19𝑉
b) Power delivered to the load
𝑃𝑜𝑑
2
𝑉𝑜𝑟𝑚𝑠
1442
=
=
= 2079.36𝑊
𝑅
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c) Chopper efficiency
𝑃𝑜𝑑 2079.36
2079.36
2079.36
𝜂=
=
=
=
= 99.12%
𝑉𝑜𝐴𝑉
91.2
𝑃𝑖𝑛
𝑉𝑖𝑛 ×𝐼𝑜
230×
230×
𝑅
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Example 2: A boost chopper has input voltage of 20 V with switching frequency
equal to 1 kHz. Calculate:
a) The required duty cycle that can be applied to the switch to boost the input
voltage to 60V.
b) The ON and OFF period for the constant switching frequency operation.
c) Output current if the resistance load equal to 10 Ω.
d) Average input inductor current.
e) The maximum and minimum currents via the input inductor if the
inductance is 10mH.
Solution:
a) 𝑉𝑜𝐴𝑉 =
𝑉𝑖𝑛
1−𝐷
⟹𝐷 = 1 −
𝑉𝑖𝑛
𝑉𝑜𝐴𝑉
=1−
20
60
= 0.667
b) 𝑇 = 𝑡𝑜𝑓𝑓 + 𝑡𝑜𝑛
𝑇=
1
1
=
= 1𝑚𝑠𝑒𝑐
𝑓 1×103
𝑡𝑜𝑛
⟹ 𝑡𝑜𝑛 = 𝐷𝑇 = 0.667×1×10−3 = 0.667 𝑚𝑠𝑒𝑐
𝑇
= 𝑇 − 𝑡𝑜𝑛 = 1×10−3 − 0.667×10−3 = 0.333 𝑚𝑠𝑒𝑐
𝐷=
⟹𝑡𝑜𝑓𝑓
c) 𝐼𝑜𝐴𝑉 =
𝑉𝑜𝐴𝑉
𝑅
=
60
10
= 6𝐴
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Lecture Note 9: DC-DC PWM Converters
Instructure: Dr. Oday A Ahmed
d) 𝐼𝐿𝐴𝑉 = 𝐼𝑖𝐴𝑉 =
𝐼𝑜𝐴𝑉
(1−𝐷)
=
6
(1−0.667)
= 18𝐴
e) The maximum Imax and minimum Imin input currents via the inductor are
shown in the figure below
Imax
Imin
From the figure above Imax and Imin can be found as:
1
Imax = 𝐼𝑖𝐴𝑉 + ∆𝐼𝐿
2
1
Imin = 𝐼𝑖𝐴𝑉 − ∆𝐼𝐿
2
𝑉𝑖𝐴𝑉
20
∆𝐼𝐿 =
𝑡𝑜𝑛 =
×0.667×10−3 = 1.334 𝐴
−3
𝐿
10×10
Thus,
1
Imax = 18 + ×1.334 = 18.667𝐴
2
1
Imin = 18 − ×1.334 = 17.33 𝐴
2
From above the input ripple current percentage is equal to 7.4%
Example 3: Design a buck converter to produce an output voltage of 18 V across
a 10Ω load resistor. The output voltage ripple must not exceed 0.5 percent. The
dc supply is 48 V. Design for continuous inductor current. Specify the duty ratio,
the switching frequency, the values of the inductor and capacitor, the peak voltage
rating of each device, and the RMS current in the inductor and capacitor. Assume
ideal components.
Solution
The circuit diagram of the buck converter is shown below,
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Lecture Note 9: DC-DC PWM Converters
Instructure: Dr. Oday A Ahmed
The switching frequency and inductor size must be selected for continuous-current operation. Let the
switching frequency arbitrarily be 40 kHz, which is well above the audio range and is low enough to
keep switching losses small.
The minimum inductor size:
Let the inductor be 25 percent larger than the minimum to ensure that inductor current is continuous:
Average inductor current and the change in current are
The maximum and minimum inductor currents are
The capacitor is
► Peak capacitor current is ΔiL/2 = 1.44 A, and
► RMS capacitor current for the triangular waveform is 1.44/ √3 =0.83 A.
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Lecture Note 9: DC-DC PWM Converters
Instructure: Dr. Oday A Ahmed
► The maximum voltage across the switch and diode is Vs, or48 V. The inductor voltage when
the switch is closed is Vs - Vo = 48 - 18 = 30 V.
► The inductor voltage when the switch is open is Vo = 18 V. Therefore, the inductor must
withstand 30 V.
► The capacitor must be rated for the 18V output.
Exercises
1) A DC-DC converter used to step up the solar cell DC voltage from 12V to
24V.
a) Name the converter and draw its schematic circuit.
If the non-conducting time equal to 100µsec,
b) Determine the required on-time and switching frequency.
Also compute the average output current if the converter connected to resistance
load with R = 10Ω.
2) For a boost DC-DC Converter supplied by 100V, switching frequency
500Hz, on-period = 600μsec, and load resistance 1Ω. Compute
a) Average output voltage
b) Average output current
c) Input DC current
d) Average inductor current
e) The required input inductance L and output filter capacitance C so that
reduce the input ripple current to 20% of average input current and output
ripple voltage to 10% of output average voltage.
3)
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