ClassnotesonbasicDC DCconverters EP N4

ELEC 431 Class notes Basic DC/DC Converters P.K. Jain

1

iL

tton

iL

tTs

iL

tton

ton

Ts

Continuous conduction mode

Boundary mode

Discontinuous conduction mode

Ts

I. Converters: convert DC voltage from one level to another DC level

The three basic types of DC/DC converter:

1) Buck converter – step down input DC voltage

2) Boost converter – step up input DC voltage

3) Buck-boost converter– step up/down input DC voltage

II. Buck Converter:

L

C RD Vo

+

_

+ vL

iL

_

Vi

+

_

ioS

+_vsw

isw

iD

Figure 1

Depends on the inductance in the DC/DC converter, there are three basic modes of operation

Modes of operation:

1) Continuous conduction mode (CCM)

2) Boundary between continuous and discontinuous conduction mode

3) Discontinuous conduction mode (DCM)

Circuit Analysis

Assumptions:

1) The circuit components are ideal (i.e. lossless)

2) The output voltage is ripple free (i.e. vo = Vo and so io = Io)

3) The input voltage is ripple free

Figure 2


2

vL

vsw

iLIo

t

t

t

Vi -Vo

S on S off

Vi

-Vo

S on S off

dTs (1-d)Ts Ts

IL,max

IL,min

t

t

isw

iD

A. Analysis in continuous conduction mode:

L

C RVo

+

_

+ vL

iL

_

Vi

+

_

L

C RVo

+

_

+ vL

iL

_

(a) when S is on (b) when S is off

Figure 3

If we let: ton = turn on of S

toff = turn off of S

fs = 1/Ts = switching frequency

then

cycledutyT

td

s

on

son dTt and sssoff TddTTt 1

i. Relationship between Vi and Vo

The relationship between Vi and Vo can be obtained by applying the fact that the average

voltage across inductor (L) must be equal to zero within a switching period (Ts)

sT

L dttv0

0

offoonoi

offoonoi

tVtVV

tVtVV

0

sososi TdVdTVdTV 1

Figure 4


3

dV

V

i

o --------------------------------- (1)

ii. Relationship between Ii and Io

Let Io = average output current

Ii = input average current

For ideal circuit, input power = output power

Or Pin = Po → ViIi = IoVo

Hence: dI

I

V

V

o

i

i

o ------------------------------------ (2)

iii. Maximum and minimum inductor current

Let IL,max = maximum inductor current

IL,min = minimum inductor current

oiL

L VVdt

diLv

KtL

VV

KdtL

VVi

oi

oiL

From Figure 4, at t = 0, iL = IL,min

Hence, K = IL,min and

min,Loi

L ItL

VVi

----------------------------------- (3)

From Figure 4, iL = IL,max at t = ton = dTs

Hence, min,max, Lonoi

L ItL

VVI

onoi

LL tL

VVII

min,max, --------------------------- (4)

Since the capacitor C cannot allow any average current to flow through it, the average current of

inductor L must be equal to the output load current (Io)


4

vsw

isw

dTs

S on S off

Vi

Ts

IL,max

IL,min

t

t

o

T

L

s

IdttiT

s

0

1

From Figure 4, the area of iL over a switching cycle (Ts) is the average value of iL

o

LLI

II

2

max,min,

oLL III 2max,min, -------------------------------------------------- (5)

oon

oiL It

L

VVI

2max,

----------------------------------- (6)

on

oioL t

L

VVII

2min,

----------------------------------- (7)

iv. RMS switch current

Since the input Ii is flowing through the switch S, the RMS value of the switch current can be

obtained according to:

sT

sw

s

RMSsw dttiT

I0

2

,

1

Alternatively, the current Isw,RMS can be determined by combining the following two components

of Figure 6 as follows:

Figure 5


5

t

Isw,1 IL,min

dTs Ts

t

Isw,2 IL,max - IL,min

dTs Ts

+

Figure 6

dI

T

dTII L

s

sL

RMSsw min,

2

min,

1, --------------------------------------------- (8)

The RMS value of Isw,2 is given by

3/

3

1

3

1

1

min,max,

2

min,max,

0

3

2

2

min,max,

2

2

min,max,

0

2,

dII

tII

T

t

t

II

T

dtt

tII

TI

LL

on

LL

s

t

on

LL

s

on

LL

t

s

RMSsw

on

on

----------------------------------- (9)

Hence, the RMS switch current is then given by:

3

2

min,max,

2

min,

2

2,

2

1,,

dIIdIIII LLLRMSswRMSswRMSsw -------- (10)

If the output inductor (L) is assumed to be very large, then the ripple current is zero (i.e.

0min,max, LL II ).

Hence: dIdIII oLRMSswRMSsw min,1,, ------------------------------ (11)

v. Average diode current

The average value of the diode current


6

vD

iD

ton

S on S off

Vi

Ts

IL,max

IL,min

toff t

t

dI

T

dTTII

T

tIII

o

s

ssLL

s

offLL

D

1

2

2

min,max,

min,max,

------------ (12)

vi. Converter power losses and efficiency

a) Conduction loss in the switch: 2

,, RMSswswScon IRP

b) Conduction loss in the diode: DFDDcon IVP ,,

c) Turn-on switching loss: srLiton ftIVP min,

3

1

d) Turn-off switching loss: sfLitoff ftIVP max,

3

1

Hence, total power losses: tofftonDconSconloss PPPPP ,,

Output power: ooo IVP

Input power: lossoin PPP

Efficiency: in

o

P

P

vii. Output voltage ripple

It is known that capacitance is defined as the rate of change of the electrical charge with respect

to the electrical potential (voltage)

→ oVCQ

→ C

QVo

idtQ = (Area of the triangle as marked in Figure. 8)*time

Figure 7


7

ic

t

t

ΔVo

Ts /2

IL,max

IL,min

ΔI

vo

min,max,

min,max,

8

2

1

4

22

1

LLs

LLs

sL

IIT

IIT

TI

Hence,

sis

onois

LLs

o

dTVdLC

T

tL

VV

C

T

IIC

TV

18

8

8min,max,

o

s

VdLCf

18

12

--------------------------------------- (13)

If we define the cut-off frequency of the LC output filter as: LC

fc2

1

Then 2

2

12

s

coo

f

fdVV

------------------------------------ (14)

Therefore, in order to have ΔV to be very small,

i. 1

s

c

f

f

ii. d is very high

Figure 8


8

vL

vsw

iLIo

t

t

t

Vi -Vo

S on S off

Vi

-Vo

S on S off

dTs (1-d)Ts Ts

IL,max

IL,min

B. Analysis at the boundary mode:

Since 0 Lv

→ onsoonoi tTVtVV

Or dT

t

V

V

s

on

i

o

Also, since 0min LI

**All the analysis performed for the continuous conduction mode (CCM) is applicable for the

boundary mode if we substitute IL,min = 0 in all the equations**

Minimum inductance for the boundary mode:

For the converter to operate at the boundary, eq. (7) should be set to zero.

→

onoi

o tL

VVI

20

Or

on

o

oi tI

VVL

2

------------------------------------------- (15)

If the converter has an inductance value equal to the value given by eq. (15), the converter will

operate in the boundary mode.

If the converter has an inductance which is greater than the value given by eq. (15), the converter

will operate in CCM, otherwise, it will operate in the discontinuous conduction mode (DCM)

Figure 9


9

vL

vsw

iL

Io

t

t

t

Vi -Vo

S on S off

Vi

-Vo

S on S off

dTs Δ1Ts Δ2Ts

IL,max

Vi -Vo

C. Analysis at the discontinuous conduction mode (DCM):

If we let: dT

t

s

on ,

1

1

s

off

T

t and 2

2

s

off

T

t

For 0 Lv

Hence, 1offoonoi tVtVV

sosoi TVdTVV 1

Hence 1

d

d

V

V

i

o ----------------------------------------------------- (16)

The average output current:

1max,

1

max,2

1

2

1

dI

T

ttII L

s

offon

Lo -------------------------------------------- (17)

L

TV

L

tVI sooffo

L11

max,

Hence, so

o TdL

VI 11

2

Or 11

1

22

o

so

so

o

V

fLI

TV

LId -------------------------------------------- (18)

From eq. (16) and eq. (18)

dV

fLI

V

V

o

so

i

o 21

---------------------------------------- (19)

From eq. (16) and eq. (19)

dV

fLI

V

Vd

V

Vd

o

so

i

o

i

o 22

--------------------------- (20)

Figure 10


10

Example 1:

A buck converter has Vi = 15V, Vo = 5V, Io = 10A, fs = 100kHz.

a) Calculate the minimum inductance value for the converter to operate in the CCM;

b) Calculate the value of capacitor C to keep the output voltage ripple to be 50mV;

c) Calculate ΔIL

Solution:

33.015

5

i

o

V

Vd

1010100

113

xfT

s

sµs

3.31033.0 sxdTt son µs

66.1

102

103.333.0115

2

1 6

min

x

xxx

I

tdVL

o

oniµH

2

1

8

1

s

oo

LCf

dVV

→

500

101001066.11050

33.015

8

11

8

123632

xxxxx

xx

LfV

dVC

so

oµF

min,max, LLL III

201066.1

103.35156

6

x

xxt

L

VVon

oi A


11

Example 2:

If the switching frequency in example 1 is increased to 250kHz,

a) Calculate the voltage ripple (ΔVo);

b) Calculate the current ripple (ΔIL);

c) Comment on the effect of increasing frequency on the converter performance

Solutions:

8

102501066.110500

33.015

8

11

8

123662

xxxxx

x

LCf

dVV

s

oo mV

1.8

1066.1

10433.05156

6

x

xxx

L

tVVI onoi

L A

For the same converter components, increasing the frequency reduces the output voltage ripple.

Example 3:

If the output current of the converter in example 1 has dropped to 5A, in which mode the

converter will now operate? Calculate the new duty cycle to keep the output voltage constant.

Solutions:

Since the converter was operating at the boundary at 10A load current, it will operate in the

discontinuous conduction mode at 5A.

From eq. (20):

dxxdx

xxxd

6

62

10105

51066.12

15

5

15

5

→ dd

d 27

133.0

→ d

d27

167.0

→ 234.067.027

1

xd


12

Example 4:

In example 1, calculate:

i. RMS switch current

ii. average diode current

iii. peak voltage across the switch

Solutions:

1010101066.12

103.3515

2

6

6

max,

x

x

ItL

VVI oon

oiL

= 20A

1010

2min,

on

oioL t

L

VVII

= 0A

From eq. (10), IL,min = 0

3

33.020

3max,,

dII LRMSsw = 6.63A

From eq. (12)

33.01101 dII oD = 6.67A

Peak switch voltage is 15V


13

Example 5:

In example 4, if the switch is a MOSFET and has Rsw = 10mΩ, tr = tf = 100ns, the diode has a forward

voltage drop of 0.8V. Calculate

i. total losses

ii. efficiency

44.063.6101023

, xP Scon W

36.567.68.0, DconP W

010100101000153

1

3

1 39

min, xxxxxxftIVP srLiton

1101001010020153

1 39 xxxxxxPtoffW

Hence 8.61036.544.0 lossP W

50510 xPo W

888.56

50

losso

o

PP

P %


14

L

+ vL

iL

_

Vi

+

_

L

C R Vo

+

_

+ vL

iL

_

Vi

+

_

io

III. Boost Converter:

L

C R

D

Vo

+

_

+ vL

iL

_

Vi

+

_

io

S

+

_

vsw

Figure 11

Similar to the buck converter, the boost converter has the following three modes of operation:

1) Continuous conduction mode (CCM)

2) Boundary between continuous and discontinuous conduction mode

3) Discontinuous conduction mode (DCM)

Circuit Analysis

Assumptions:

1) The circuit components are ideal (i.e. lossless)

2) The output voltage is ripple free (i.e. vo = Vo)

3) The input voltage is ripple free

A. Analysis in continuous conduction mode (CCM):

For steady-state operation:

00

sT

Ldtv and s

on

T

td

0 offoioni tVVtV

onsoioni tTVVtV

dVVdV oii 1

dVdVdV oii 11

dVV oi 1

(a) when S is on

(b) when S is off

Figure 12


15

Hence, dV

V

i

o

1

1 ------------------------------------------ (21)

For d = 0; → Vo = Vi

For d = 1; → Vo = ∞

vL

vsw

iLIo

t

t

t

Vi

S on S off

Vo

Vi -Vo

S on S off

dTs (1-d)Ts Ts

IL,max

IL,min

Figure 13

Theoretically, the output voltage of the boost converter can be increased from Vi to infinity by changing

the duty cycle from d = 0 to 1.


16

vL

vsw

iL

Io

t

t

t

Vi

S on S off

Vo

S on S off

dTs Δ1Ts Δ2Ts

IL,max

Vi

Vi -Vo

B. Analysis in discontinuous conduction mode (DCM):

Again, if we define the following:

s

on

T

td

s

off

T

t 1

1 s

off

T

t 2

2

and d 121 ---------------------- (22)

For steady-state operation: 00

sT

Ldtv

Hence, 01 offoioni tVVtV

01 soisi TVVdTV

1

1

d

V

V

i

o --------------------------------------------------------------- (23)

When the switch S is on, the equivalent circuit of the boost converter is shown in Figure 12 (a).

dt

diLV L

i

→ KdtL

Vi

t

iL

0

→ KtL

Vi i

L --------------------------------------------------------------- (24)

At t = 0; iL = 0 → K = 0

Eq. (24) now becomes: → tL

Vi i

L ------------------------------------------- (25)

At t = ton, iL = IL,max → oni

L tL

VI max, ------------------------------------------- (26)

The average diode current is the same as the area of the triangle:

1

1

max,2

1

2

1 on

i

s

off

LD tL

V

T

tII -------------- (27)

Figure 14


17

Since the output capacitor cannot carry any average current, the diode current ID then must be

equal to the average load current Io.

→ 12

1 on

iDo t

L

VII

Or 111222

s

isionio

Lf

dV

L

dTV

L

tV

R

V

Or

i

os

V

V

Rd

Lf21 ----------------------------------------------------- (28)

Substituting Δ1 in eq. (23), we have

12

2

2

o

i

so

i

V

V

Lf

Rd

V

V ------------------------ (29)

C. Analysis in boundary mode:

For the boundary condition, toff2 in Fig. 13 must be zero. From eq. (22)

→

i

os

V

V

Rd

Lfd

211 [from eq. (28)]

Or ddV

V

R

Lf

i

os

1

2 ------------------------------------------------------ (30)

For boundary condition, eq. (30) should be satisfied.


18

Example 6:

In a boost converter Vi = 100V, Vo = 200V, fs = 100kHz, output power Po = 1 kW. What is the minimum

value of the inductance L for the converter to operate at the boundary condition?

Solutions:

At the boundary,

100

200

1

1

dV

V

i

o

Or 122 d → d = 0.5

From eq. (30), the value of inductance L is

o

i

s V

V

f

RddL

2

)1(

25)200)(10100(2

)100)(40)(5.01(5.03

µH where

40

101

)200(3

22

o

o

P

VR

Example 7:

In example 6, if the output power is reduced to 100W, calculate the new value of duty-cycle, d, to

maintain the same output voltage of 200V.

Solutions:

400100

)200( 22

o

o

P

VR

Since the converter of example 6 was already at the boundary condition, this converter will now operate

in the discontinuous mode as the load current is now lower than before.

From equation (29),

1200

100

)10100)(1025(2

400

200

1002

36

2

d

Or 15.0)20(2 d → 16.020

5.0d


19

vL

vsw

iLIo

t

t

t

Vi

S on S off

Vi +Vo

-Vo

S on S off

dTs (1-d)Ts Ts

IL,max

IL,min

IV. Buck-Boost Converter:

Modes of operation

Similar to the buck converter, the buck-boost converter also operates in the three modes. These

modes are defined in Fig. 2.

L C R

D

Vo

+

_

vL

iL

Vi

+

_

ioS

+

_

vsw+_

Figure 15

Circuit Analysis

The analysis of the converter is carried out by making the following assumptions

1) The circuit components are ideal

2) The output voltage is negative with respect to the input voltage and is ripple free

3) The input voltage is ripple free.

A. Analysis in the continuous mode:

For steady-state operation,

0

00

offooni

T

L

tVtV

dtVs

or sosi TdVdTV )1(

d

d

V

V

i

o

1 -------------------- (31)

For d = 0, Vo = 0

For d = 1, Vo = ∞

For d = 0.5, Vo = Vi Figure 16


20

L+

vL

iL

_

Vi

+

_

vL

vsw

iL

Io

t

t

t

S on S off

-Vo

S on S off

ton toff1

IL,max

Vi

Vi +Vo

toff2

Vi

voltageouputthedownsteporupsteptoconvertersameoperationBoostd

operationBuckd

15.0

5.00

B. Analysis at the discontinuous conduction mode:

Again, if we define:

d= ton/Ts

Δ1= toff1/Ts

Δ2= toff2/Ts

d 121 -------- (32)

For steady-state operation:

0

00

offooni

T

L

tVtV

dtVs

or sosi TVdTV 1

1

d

V

V

i

o ------------------------------- (33)

When switch S is on, the equivalent circuit of Fig.15 is given in Fig. 18 (i.e. when S is on, D is off). From

Fig. 18,

i

i

L Vdt

dLv L

or KdtL

Vi

sT

iL

0

KtL

Vi

At t = 0; iL = 0 K = 0

tL

Vi i

L

Figure 17

Figure 18


21

At t = ton, iL = ILmax

si

oni

L dTL

Vt

L

VI max -------------------------- (34)

The average diode current is:

s

off

LDT

tII

1

max2

1

12

1 on

i tL

V ---------------------------------------------- (35)

Since no DC current can flow through the output capacitor, the average diode current is equal to the load

current.

12

1 on

iDo t

L

VII

Or 12

1 s

io dTL

V

R

V

si

o

dTR

L

V

V 21

i

os

V

V

dR

Lf2 ------------------------------------ (36)

From eq. (33) and (36), we have:

i

os

o

i

V

V

dR

Lf

V

Vd

21

Or R

Lf

V

Vd s

i

o 2

------------------------------------ (37)


22

C. Analysis at the boundary mode:

In Fig. 17, toff2 = 0 at the boundary condition. Therefore, from eq. (32), we have:

d 11 ----------------------------------------------- (38)

From equations (36) and (38):

i

os

V

V

Rd

Lfd

21 ----------------------------------------------- (39)

For the boundary mode, eq. (39) must be satisfied. The minimum value of inductance which will keep

the operation of the converter in the continuous mode is given by:

o

i

s V

V

f

dRdL

2

1 ---------------------------------- (40)

A converter having lower inductance value than given by eq. (40) will operate in discontinuous

conduction mode.


23

Example 8:

A buck-boost converter has Vi = 50 V, Vo = 25 V, Po = 50W, fs = 100 kHz, L = 10 μH. The converter is

operating in the discontinuous mode, calculate the duty-cycle d.

Solutions:

5.1250

2522

o

o

P

VR

2.0

5.12

)10100)(1010(2

50

25

2

36

R

Lf

V

Vd s

i

o

Example 9:

What is the value of fs in example 8 to achieve d = 0.5 and still maintaining the same output voltage and

power?

Solutions:

5.12

)1010(2

50

255.0

6

sf

6251020

5.126

sf kHz

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ClassnotesonbasicDC DCconverters EP N4

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