dr. oday a. ahmed - lecture note 9...1 2 2 if v in = 39v, and v out = 13v, efficiency η is only...

Lecture Note

9

DC-DC PWM Converters

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Lecture Note 9: DC-DC PWM Converters

Instructure: Dr. Oday A Ahmed

1

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.

21

2

RR

RVV sout

•

The input and output powers are respectively

given by:

The efficiency conversion is obtained: in

out

V

V

RR

R

21

2

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:

Fig.1



2

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 = 𝑉𝑖𝑛𝑡𝑜𝑛

𝑡𝑜𝑛+𝑡𝑜𝑓𝑓= 𝑉𝑖𝑛

𝑡𝑜𝑛

𝑇= 𝑉𝑖𝑛𝐷 where D is called a Duty Cycle

Vs

Vo

Fig.2



3

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.

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.

Fig.3



4

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.



5

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:



6

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,

L

VV

dt

diVVv outinL

outinL

,

Switch open,

L

V

dt

diVv outL

outL

,

Because the current consists of straight line segments, it is apparent that

T

From geometry, Iavg = Iout is halfway

between Imax and Imin

sec/ AL

VV outin

DT (1 − D)T

Imax

Imin

Iavg = Iout

sec/ AL

Vout

ΔI

iL

Periodic – finishes

a period where it started



7

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,

L

V

dt

diVv outL

outL

,

where Lboundary is the value of L that causes the circuit to operate at the boundary

of continuous



8

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

22222

12

1

12

1IIIII outppavgLrms

2222

3

42

12

1outoutoutLrms IIII

outLrms II3

2

B. Capacitor current Ratings

Max rms current occurs at the boundary of continuous/discontinuous conduction,

where ΔI =2Iout

22222

3

102

12

1outoutavgCrms IIII

3

outCrms

II

C. Transistor and diode currents and Voltage ratings

2Iout

0

Iavg = IoutΔI

iL

Iout

−Iout

0ΔI

iC = (iL – Iout) Note – raising f or L, which lowers

ΔI, reduces the capacitor current



9

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

out

outload

I

VR

DC−DC Buck Converter

+

Vin

−

+

Vout = DVin

−

Iout = Iin / DIin

Source

equivR

+

Vin

−

Iin

Equivalent from

source perspective



10

22 D

R

DI

V

DI

D

V

I

VR load

out

out

out

out

in

inequiv

•

•

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.



11

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:



12

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 will guarantee

continuous conduction for all D.

Current Waveforms in Continuous Conduction

The current waveforms of the boost converter in CCM shows below :



13

Current Ratings in Continuous Conduction

for the inductor in continuous conduction

as explained in an analogous fashion in the Buck Converter experiment, yields



14

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

outCrms II

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



15

Impedance matching

load

out

out

out

out

in

inequiv RD

I

VD

D

I

VD

I

VR

2211

1

1

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:

Average output voltage =

out

outload

I

VR

equivR

DC−DC Boost Converter

+

Vin

−

+

−

Iin

+

Vin

−

Iin

Equivalent from

source perspective

Source D

VV in

out

1

inout IDI 1

Vin-2 Vin-2



16

𝑉𝑜𝐴𝑉 = (𝑉𝑖𝑛 − 2)𝑡𝑜𝑛

𝑇= (𝑉𝑖𝑛 − 2)𝐷 = (230 − 2)×0.4 = 91.2𝑉

RMS value of output voltage

𝑉𝑜𝑟𝑚𝑠 = [(𝑉𝑖𝑛 − 2)2𝑡𝑜𝑛

𝑇]

1/2

= (𝑉𝑖𝑛 − 2)√𝐷 = (230 − 2)√0.4 = 144.19𝑉

b) Power delivered to the load

𝑃𝑜𝑑 =𝑉𝑜𝑟𝑚𝑠

2

𝑅=

1442

10= 2079.36𝑊

c) Chopper efficiency

𝜂 =𝑃𝑜𝑑

𝑃𝑖𝑛=

2079.36

𝑉𝑖𝑛×𝐼𝑜=

2079.36

230×𝑉𝑜𝐴𝑉

𝑅

=2079.36

230×91.210

= 99.12%

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

𝑓=

1

1×103= 1𝑚𝑠𝑒𝑐

𝐷 =𝑡𝑜𝑛

𝑇⟹ 𝑡𝑜𝑛 = 𝐷𝑇 = 0.667×1×10−3 = 0.667 𝑚𝑠𝑒𝑐

⟹𝑡𝑜𝑓𝑓 = 𝑇 − 𝑡𝑜𝑛 = 1×10−3 − 0.667×10−3 = 0.333 𝑚𝑠𝑒𝑐

c) 𝐼𝑜𝐴𝑉 =𝑉𝑜𝐴𝑉

𝑅=

60

10= 6𝐴



17

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

From the figure above Imax and Imin can be found as:

Imax = 𝐼𝑖𝐴𝑉 +1

2∆𝐼𝐿

Imin = 𝐼𝑖𝐴𝑉 −1

2∆𝐼𝐿

∆𝐼𝐿 =𝑉𝑖𝐴𝑉

𝐿𝑡𝑜𝑛 =

20

10×10−3×0.667×10−3 = 1.334 𝐴

Thus,

Imax = 18 +1

2×1.334 = 18.667𝐴

Imin = 18 −1

2×1.334 = 17.33 𝐴

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,

Imax

Imin



18

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.



19

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)

dr. oday a. ahmed - lecture note 9...1 2 2 if v in = 39v, and v out = 13v, efficiency η is only...

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