buck converter

Guru/DC2DC/Buck/ February 16, 2006 Buck Converter 1

Buck Converter

Buck converter is a dc-to-dc converter that steps down the dc voltage from its fixed high

level to a desired low level. It is also called the forward converter and its circuit topology

is given in Figure 1.

Figure 1: Buck Converter

The switch S is usually an electronic device that operates either in the conduction

mode (on) or the cut-off mode (off). The on and off time-periods are controlled by the

suitably designed gating circuits, which are usually not shown. The on time of the switch

is a fraction of its time period T such that TDTON = , where D is the duty cycle. During

the off time, ,T)D1(TOFF −= the freewheeling diode FWD provides a path to maintain

the continuity of the current through the inductor. The inductor controls the percent of the

ripple and determines whether or not the circuit is operating in the continuous mode. The

capacitor C provides the filtering action by providing a path for the harmonic currents

away from the load. In addition, its value is large enough so that voltage ripple is very

small.

It must be clear from Figure-1 that there are two energy storing elements L and C

in a sequentially-switching circuit. These elements results in a second order differential

equation either in terms of the inductor current or the capacitor voltage. The differential

equation in terms of the capacitor voltage, when the switch S is closed, may be written as

SCC

2

C

2

V)t(vdt

)t(dv

R

L

dt

)t(vdLC =++

The solution of the above second-order differential yields the voltage during the

time switch is closed. A similar equation may be written when the switch is in its open

position.


The above differential equation can be simplified if we make an assumption that

the voltage across the load, and thereby across the capacitor, is fairly constant. The

differential equation in terms of the current through the inductor, when the switch is

closed, may now be written as

oSL VVdt

)t(diL −=

Let us assume that the circuit has been operating for quite some time and has

attained its steady state. In other words, there may already be some current in the

inductor, min,LI , just prior to the closing of switch S. With this understanding, the solution

of the above equation, for DTTt0 ON =≤≤ , yields

min,LoS

L ItL

VV)t(i +

−= (1)

The inductor current increases linearly with time and attains its maximum value max.LI as

DTTt ON =→ such that

min,LoS

max,L IDTL

VVI +

−= (2)

Defining the change in the current from its minimum to maximum value as the

peak-to-peak current ripple LI∆ , the above equation yields an expression for LI∆ as

DTL

VVIII oS

min,Lmax,LL

−=−=∆ (3)

Note that the current ripple is directly proportional to D, the duty cycle, upon

which we may not have any control because of the output voltage requirement. However,

it is inversely proportional to the inductance L upon which we can exert some control.

Thus, the current ripple can be maintained within its bounds by a proper selection of the

inductor.

Let us now analyze the circuit when the switch is in its open position. The

inductor current completes its path through the freewheeling diode and the corresponding

differential equation, for OFFTt0 ≤≤ , is

oL Vdt

)t(diL −=


From the solution of the above first-order differential equation, we obtain

max,Lo

L ItL

V)t(i +

−= (4)

Where max,LI is the maximum value of the current in the inductor at the opening of the

switch or the beginning of the off period. As ,T)D1(Tt OFF −=→ the inductor current

decreases to its minimum value min,LI such that

max,Lo

min,L IT)D1(L

VI +−−=

This equation yields another expression for the peak-to-peak current ripple as

T)D1(L

VIII o

min,Lmax,LL −=−=∆ (5)

The current ripple as given by (3) must be the same as given by (5). Therefore, equating

the two equations, we get

T)D1(L

VDT

L

VV ooS −=−

This equation upon simplification yields

So VDV = (6)

Equation (6) states that the output voltage of the buck converter is directly

proportional to the duty cycle and the source voltage. Since the duty cycle is usually less

than unity, the output voltage is smaller than the source (applied or input) voltage. This is

the reason why a buck converter is commonly called the step-down converter. Since the

power flow is from the source to the load and there is no power flowing back to the

source, it is also called the forward converter.

Figure 2: Inductor current


The current through the inductor as given by (1) during the on time and by (4)

during the off time is sketched in Figure 2.

The average current in the inductor must be equal to the dc current through the

load. That is,

R

VII o

oavg,L == (7)

The expressions for the maximum and minimum currents through the inductor

may now be written as

T)D1(L2

V

R

V

2

III ooL

avg,Lmax,L −+=∆

+= (8)

T)D1(L2

V

R

V

2

III ooL

avg,Lmin,L −−=∆

−= (9)

The current supplied by the source varies from min,LI to max,LI during the time the

switch is closed and is zero otherwise as shown in Figure 3.

Figure 3: The source current

When the switch, the inductor, and the capacitor are treated as ideal elements, the

average power dissipated by them is zero. Consequently, the average power supplied by

the source must be equal to the average power delivered to the load. That is,

oSooSS IDVIVIV ==

This equation helps us express the average source current in terms of the average

load current as

oS IDI = (10)

The current through the freewheeling diode is shown in Figure 4. Its average

value is

oD I)D1(I −= (11)


Figure 4: The freewheeling diode current

When the load current is subtracted from the inductor current, we obtain the time-

varying current through the capacitor. Its minimum and maximum values from (8) and

(9) are

T)D1(L2

V

2

II oL

max,C −=∆

= (12)

T)D1(L2

V

2

II oL

min,C −−=∆

−= (13)

The corresponding waveform for the capacitor current is given in Figure 5. The

average current through the capacitor is zero.

Figure 5: Current through the capacitor

The current waveform of Figure 5 helps us determine the change in the voltage

across the capacitor. During one-half the time period, the current is charging the capacitor

and the increase in the charge can be computed from Figure 5 as

TI8

1

2

T

2

I

2

1Q L

L ∆=∆

=∆

Substituting for LI∆ from (5), we obtain an expression for the increase in charge on the

capacitor as

2o T)D1(L8

VQ −==∆ (14)


Therefore, increase in the capacitor voltage is

o2

2

oo VLCf8

D1TV

LC8

D1

C

QV

−=

−=

∆=∆ (15)

When we define the capacitor ripple as the ratio of the increase in the capacitor

voltage to its average value, it can then be expressed as

2

o

o

LCf8

D1

V

V −=

∆ (16)

Note that the capacitor ripple define by (16) is not the same as the peak-to-peak

voltage ripple for the rectifiers. The peak-to-peak voltage ripple for the buck converter

will be twice of that given by (16). Equation (16) may be viewed as one-sided voltage

ripple.

The buck converter can operate either in its continuous conduction mode or

discontinuous conduction mode. When it operates in the continuous conduction mode,

there is always a current in the inductor. The minimum current in the continuous

conduction mode can be zero at the time of switching. Consequently, there is a minimum

value of the inductor that ensures its continuous conduction mode. It can be obtained

from (9) by setting min,LI to zero as

0T)D1(L2

V

R

V

min

oo =−−

Hence,

Rf2

D1RT

2

D1Lmin

−=

−= (17)

From the peak-to-peak current ripple, we can obtain an expression for the percent current

ripple as

RLf

)D1(100100

I

ICR%

avg,L

L −=×

∆= (18)

Example 1: ___________________________________________________________

A buck converter operating at a frequency of 20 kHz is used to step-down a 480-V dc

supply to 120-V so that it can provide the rated voltage to a 120-V, 1000-W heater. To

ensure a reliable operation, inductor must at least be 20% greater than its minimum value

and the voltage ripple should be within 1%. Design the buck converter.


Solution:

From the given data, the duty cycle, from (6), is

25.0480

120D ==

The time period, the on time and off times of the switch are

s50000,20

1

f

1T µ===

s5.12105025.0DTT 6

ON µ=××== −

s5.37)1050)(25.01(T)D1(T 6

FF µ=×−=−= −

The equivalent resistor of the heater is

Ω=== 4.141000

120

P

VR

22

o

The minimum value of the inductor for the continuous conduction mode, from (17), is

mH875.1)4.14(000,202

25.01R

f2

D1Lmin =

×

−=

−=

The minimum value of the inductor to satisfy the design requirement must be

mH25.210875.12.1L2.1L 3

min =××== −

To allow for inductor-to-inductor variations, let us select

mH5.2L =

The average current through the heater (or the inductor), from (7), is

333.84.14

120

R

VII o

oavg,L ==== A

The peak-to-peak current ripple from (5) is

8.1)1050)(25.01(105.2

120T)D1(

L

VI 6

3

oL =×−

×=−=∆ −

−A

Hence, the maximum and minimum currents through the inductor, from (8) and (9), are

233.92

8.1333.8

2

III L

avg,Lmax,L =+=∆

+= A

433.72

8.1333.8

2

III L

avg,Lmin,L =−=∆

−= A


Let us now select the capacitor based upon the desired voltage ripple of 1%. Using (16)<

we determine C as

F38.9000,2001.0105.28

25.01

fV

VL8

D1C

232

o

o

µ=××××

−=

∆

−=

−

For this application, let us select a standard 10-µF capacitor.

The percent current ripple of the inductor current can be computed as

%6.21100333.8

8.1100

I

ICR%

avg,L

L =×=×∆

=

In some applications, the percent current ripple as high as 21.6% may not be acceptable.

See the following example which limits the percent current ripple to 5%.

Example 2: ___________________________________________________________

Redesign the buck converter of Example-1 for a 5% current ripple. All other

requirements are the same.

Solution:

From (18), we determine the new value of L, when %CR = 5%, as

8.10)4.14(000,205

)25.01(100R

fCR%

)D1(100L =

×

−=

−= mH

The average value of the inductor current is 8.333 A. The peak-to-peak ripple is

416.005.0333.8IL =×=∆ A

The maximum and minimum inductor currents are

541.8208.0333.8I max,L =+= A

125.8208.0333.8I min,L =−= A

The new capacitor value is

F17.2000,2001.0108.108

25.01

fV

VL8

D1C

232

o

o

µ=××××

−=

∆

−=

−

In this case, we can use a 2.2-µF capacitor. Since the choice of standard inductors is very

limited, it is a good practice to select the inductor first. Then select the appropriate

standard capacitor based upon voltage ripple requirements.

buck converter

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