The formula for the buck converter in the continuous-conduction mode, neglecting the losses, is [3]:

Chapter 4

More Accurate Formulae for a Chopper

4.1 More Accurate Formula for a Buck Converter

4.1.1 Continuous-conduction Mode

The formula for the buck converter in the continuous-conduction mode, neglecting the losses, is [3]:

V

out

= V

in

·d (4.1)

This formula is valid assuming the follow hypothesis:

- V

in

is constant

- Voltage across the inductore is constant and equal V

in

− V

^out

- No power losses in the device

- I

load

is constant

In order to obtain a more accurate formula, the circuit in figure 4.1.1 is considered . It is supposed that V

in

and V

out

are constant. The average voltage value across the inductor must be constant, thus:

1 T ·

Z

T 0

v

ind

dt = 1 T ·

Z

T^on 0

v

_ind^on

dt + 1 T ·

Z

T Ton

v

_ind^{of f}

dt = 0 (4.2) At turn on

v

_ind^on

= V

in

−I

^load

·R

^onm

−R

^ind

·I

^load

−V

^load

=

= V

in

−V

^load

·

 R

onm

+R

ind

+R

load

R

load

 =V

in

−b·V

^load

(4.3)

17

C

²

R

c2

R

_ind

R

_load

i

ind

L

ind

i

d

i

c2

i

m

R

onm

R

_ond

V

in

A

i

load

V

fd

+

-

+

- v

_load

Figure 4.1: buck circuit

and at turn off

v

_ind^{of f}

= −I

^load

·R

^ond

−R

^ind

·I

^load

−V

^load

−V

^{f d}

=

= −V

^load

·

 R

ond

+R

ind

+R

load

R

load



−V

^{f d}

= −V

^load

·a−V

^{f d}

(4.4) where I

load

= V

load

/R

load

is substituted, and:

a = R

ond

+ R

ind

+ R

load

R

load

(4.5) and

b = R

onm

+ R

ind

+ R

load

R

load

(4.6) Substituting the 4.3 and the 4.4 in the 4.2:

1 T ·

Z

T 0

v

ind

dt = 1 T ·

Z

ton

0

(V

in

−b·V

^load

)dt + 1 T ·

Z

T ton

(−V

^load

·a−V

^{f d}

)dt (4.7)

considered that all the terms are constant:

V

load

·[b·t

^on

+ (T − t

^on

)·a] = V

ⁱⁿ

·t

^on

− V

^{f d}

·(T − t

^on

) (4.8) substituting t

on

= T ·d too:

V

load

·[b·d + (1 − d)·a] = V

ⁱⁿ

·d − V

^{f d}

·(1 − d) (4.9) Finally, the follow equations are obtained:

V

load

= V

in

·d − V

^{f d}

·(1 − d)

b·d + a·(1 − d) (4.10)

d = V

load

·a + V

^{f d}

V

in

− V

^load

·(b − a) + V

^{f d}

(4.11)

0.3 0.4 0.5 0.6 0.7 0.8 0.9 1 6

8 10 12 14 16 18 20 22 24 26 28

d Vload

Vf d= 0.7 V Vin= 26 V Ronm= 4.5 ∗ 10⁻³Ω Ronm= 4.5 ∗ 10⁻³Ω Rind= 4.5 ∗ 10⁻³Ω

Rload= 1.2 Ω

Figure 4.2: Comparison between eq.4.10 and eq. 4.1

Equation 4.10 and 4.11 represent more accurate formulae for a buck converter in the continuous-conduction mode. If the diode on-resistance, the Mosfet on-resistance and the diode forward voltage are considered to equal zero, equation 4.10 then becomes equal to equation 4.1. Figure 4.2 shows the comparison between 4.10 (continuous line) and 4.1 (dashed line) in the continuous mode: It is recognized that as d becomes smaller, the error increases. This is due mainly to V

f d

·(1 − d), that is not always possible to neglect.

4.1.2 Boundary between Continuous and Discontinuous-conduction Mode

The object of this section is finding the boundary between continuous and discontinuous- conduction mode. Figure 4.3 shows the current and the voltage across the inductor at the boundary in a switching period. I

lB

is the mean value of the current at the boundary.

In order to avoid the discontinuous-conduction mode, the value of the average current

through the inductor (I

ind

) must be larger than the value of the average current through

the inductor at the boundary (I

ind

> I

lB

). Assuming that all of the ripple of i

ind

flows

through the output capacitor, the average component of the current flows through the load

T ton toff

vind on

iind

t t vind

off

Ip

IlB

vind

Figure 4.3: Progress of the current and of the voltage in the inductor at the Boundary

resistor, the condition for avoiding the discontinuous conduction mode becomes:

I

lB

< I

load

= V

load

R

load

(4.12) I

lB

is half of I

p

, so it is possible to write:

I

lB

= 1

2 I

p

= 1 2·L

^ind

·

Z

ton

0

v

ind

dt = (V

in

−b·V

^load

)· d·T

2·L

^ind

(4.13)

Substituting equations 4.13 and 4.10 in equation 4.12, the follow inequality is obtained:



V

in

−  V

in

·d − V

^{f d}

·(1 − d) b·d + a·(1 − d)



· d·T

2·L

^ind

<  V

in

·d − V

^{f d}

·(1 − d) b·d + a·(1 − d)



· 1 R

load

(4.14) Developing the last inequality in order to achieve a relation depending on d:

−T·R

^load

·(V

ⁱⁿ

·a+V

^{f d}

·b)·d

²

+[T ·R

^load

·(V

^{f d}

·b+V

ⁱⁿ

·a)−2·L

^ind

(V

in

+V

f d

)]·d+ (4.15) +2·L

^ind

·V

^{f d}

< 0

or more simply stated:

y(d) = −E·d

²

+ (E − F )·d + 2·L

^ind

·V

^{f d}

< 0 (4.16) Where:

E = T·R

^load

·(V

ⁱⁿ

·a+V

^{f d}

·b) (4.17)

F = 2·L

^ind

(V

in

+V

f d

) (4.18)

Solving y(d):

d

₁

= (E − F ) + p(E − F )

²

+ 8·E·L·V

^{f d}

2·E (4.19)

d

2

= (E − F ) − p(E − F )

²

+ 8·E·L·V

^{f d}

2·E (4.20)

E and F are larger than zero, so it follows that d

2

is smaller than zero. Hence it is not a valid solution. Moreover y(d) is a concave parabola because E > 0. Hence if d is larger than d

1

, the mode of conduction is continuous, otherwise discontinuous. Figure 4.4 shows the relationship between y(d) and d, at the right side of the point d

₁

, y(d

₁

) the mode is continuous, at the left side discontinuous.

0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1

−2.5

−2

−1.5

−1

−0.5 0 0.5x 10⁻³

d

y(d)

Vin=12 V Ronm=4.5e−3 Ω Rond=4.5e−3 Ω Vfd=0.7 V R_ind=5.17e−3 Ω R_load=1.2 Ω

continuous mode (d₁,y(d₁))

Figure 4.4: Discontinuous conduction in a buck converter

4.1.3 Discontinuous-Conduction Mode

In the same way as the continuous-conduction mode, in order to find a relationship be- tween V

load

and d for the discontinuous-conduction mode, the average voltage value across the inductance is imposed to be zero. For easier understanding, figure 4.5 is considered.

Asserting condition 4.2, and considering that the length of the turn off is ∆

1

·T : 1

T · Z

T

0

v

ind

dt = 1 T ·

Z

ton

0

v

_ind^on

dt + 1 T ·

Z

∆₁·T t^on

v

_ind^{of f}

dt = 0 (4.21) Substituting 4.3 and 4.4 in equation 4.53 and taking the same route:

(V

in

− V

^load

·b)·d − (V

^{f d}

+ a·V

^load

)·∆

¹

= 0 (4.22)

Figure 4.5: discontinuous conduction in a buck converter

The unknown quantities in this case are: ∆

1

and V

load

. Another equation is needed, and is obtained by stating that the average value of the current through the inductor is equal to I

load

. Developing:

I

load

= I

p

· (∆

1

+ d)·T

2 (4.23)

considering that:

I

p

= V

in

− b·V

^load

L

ind

·d·T (4.24)

the following system is obtained:



 

 

 

 

−a·V

^load

·∆

¹

− V

^load

·b·d − V

^{f d}

·∆

¹

+ V

in

·d = 0

−b·V

^load

·∆

1

− V

^load

·



b·d +

Rload^2·Lind·d·T²



+ V

in

·∆

1

+ V

in

·d = 0

Solving the system in the unknown quantities V

load

and ∆

₁

, a relationship between V

load

,

∆

₁

and d is obtained. The final formulae as they are too lengthy, and are outside of the

scope of this Thesis.

4.2 Initial Condition in a Buck Converter

Assuming the same hypothesis of paragraph 4.1 is valid, the formulae of the initial values of the current flowing in the main inductor and of the voltage across the capacitor will be stated.

4.2.1 Initial condition in the main inductance

Observing fig. 4.6, the follow equation is valid:

i

ind

= 1 L

ind

·

Z

t

v

ind

dt (4.25)

Iload=Im

T ton toff

vind on

vind off

I2

I1

iind

t t

Figure 4.6: Progress of the current and of the voltage in the inductance at the steady-state

Solving equation 4.25:

I

2

= v

_ind^on

L

ind

·t

^on

+ I

1

(4.26)

Substituting equation 4.3 in the 4.26 equation:

I

1

− I

²

= − (V

in

− b·V

^load

)·d·T L

ind

(4.27) A further equation obtained is that the average inductor current is equal to the load current, and hence:

I

1

+ I

2

2 = I

m

= V

load

R

load

(4.28)

Equations 4.27 and 4.28 make a system of equations, where I

1

and I

2

are the unknown quantities. Solving the system of equations the following results are obtained:

I

1

= V

load

R

load

− V

in

− b·V

^load

2·L

^ind

·d·T (4.29)

I

2

= V

load

R

load

+ V

in

− b·V

^load

2·L

^ind

·d·T (4.30)

I

1

represents the initial value of the current in the main inductor, I

2

represents the peak value of the current.

4.2.2 Initial Condition in the Output Capacitor

Considering that the average current through the main inductor is equal to the load current, it follows that in the capacitor only the fluctuating component of i

ind

is flowing. Stating the second Kirchhoff low at the output part of the circuit:

v

c2

= V

load

− i

^c2

·R

^c2

(4.31)

where i

c2

= i

ind

− I

^load

and at the beginning (t = 0) i

c2

= I

1

− I

^load

. Hence:

v

_c2

(t = 0) = V

load

− (I

1

− I

^load

)·R

c2

(4.32)

4.2.3 Output voltage ripple

In order to calculate the output voltage ripple in the continuous-conduction mode of oper- ation, the current in through the load resistance is considered constant. It follows that the ripple component of the i

ind

flows through the output capacitor. The dashed area in figure (4.7) is the charge that causes the ripple of the output voltage. Therefore the peak-to-peak output voltage can be written as:

∆V

load

= ∆Q C

2

= ∆I

C₂

·T

8·C

²

(4.33)

Considering that ∆I

C₂

= I

₂

− I

¹

, from the equation 4.27:

∆I

C₂

= I

2

− I

¹

= (V

in

− b·V

^load

)·d·T L

ind

and considering from equation 4.10 that:

∆V

in

= V

load

·[d·b + a·(1 − d) + V

^{f d}

·(1 − d)]

d

t

^on

v

ind

on

v

ind off

T/2

v

load

v

load

Q v

ind

i

C2

i

C2

/2

t

off

T

t t

t

Figure 4.7: Output voltage ripple in a buck converter

Substituting the previous equations in the 4.33 the follow equation is obtained:

∆V

load

= [V

load

·a + V

^{f d}

]·(1 − d)·T

²

8·C

²

·L

^ind

(4.34)

Dividing by V

load

, the ripple is obtained:

∆V

load

V

load

= [V

load

·a + V

^{f d}

]·(1 − d)·T

²

8·V

^load

·C

²

·L

^ind

= π

²

·[V

^load

·a + V

^{f d}

]·(1 − d)

2·V

^load

·

 f

c

f

s



2

(4.35) Where:

f

c

= 1

2·π· √

C

₂

·L

^ind

(4.36)

and f

s

=

_T¹

is the switching frequency.

4.3 A More Accurate Formula for a Boost-Converter

4.3.1 continuous-conduction mode

Like the buck converter, the boost converter is analyzed. The boost-converter formula in the continuous-conduction mode without losses is [3]:

V

out

= V

in

· 1

1 − d (4.37)

For a more accurate formula, the circuit in figure 4.8 is considered . It is supposed that V

in

C

2

R

ind

i

ind

L

ind

R

onm

i

load

+

-

v

_load

i

d

i

m

R

_ond

V

_in

V

fd

+

- R

c2

R

_load

i

c2

Figure 4.8: Boost circuit

and V

out

are constant. The average voltage value across the inductor must be zero, thus:

1 T ·

Z

T 0

v

ind

dt = 1 T ·

Z

Ton

0

v

_ind^on

dt + 1 T ·

Z

T Ton

v

_ind^{of f}

dt = 0 (4.38) At this point a further simplification is made: the current in the inductor (I

ind

) is fixed as I

ind

= I

load

/(1 − d). The previous simplification is given from the dissertation of the boost- converter without losses . It gives a large simplification for the calculations, but the errors are relatively small. Observing at the formulae 4.39 and 4.40, the error is the fluctuating component of i

ind

multiplied by the parasitic resistances that are small. Hence at turn on:

v

_ind^on

= V

in

−R

^onm

·I

^ind

−R

^ind

·I

^ind

=

= V

in

−V

^load

·

"

R

onm

+R

ind

R

load

·(1 − d)

#

=V

in

− c·V

^load

(1 − d) (4.39)

and at turn off:

v

_ind^{of f}

= V

in

−R

^ond

·I

^ind

−R

^ind

·I

^ind

−V

^load

−V

^{f d}

=

= V

in

−V

^load

·

"

R

ond

+R

ind

R

load

·(1 − d) + 1

#

−V

^{f d}

= V

in

−V

^load

·

h e

(1 − d) + 1 i

−V

^{f d}

(4.40) where:

c = R

onm

+ R

ind

R

load

(4.41)

and

e = R

ond

+ R

ind

R

load

(4.42) Substituting the 4.39 and the 4.40 in the 4.38:

1 T ·

Z

T 0

v

ind

dt = (4.43)

= 1 T ·

Z

t^on 0

"

V

in

− c·V

^load

(1 − d)

#

dt + 1 T ·

Z

T ton

(

V

in

−V

^load

·

h e

(1 − d) + 1 i

−V

^{f d}

)

dt considered that all the terms are constant and substituting t

on

= T ·d:

V

load

·

"

(1 − d)

²

+ (c − e)·d + e (1 − d)

#

= V

in

− V

^{f d}

·(1 − d) (4.44)

V

load

·[−c·d − (1 − d)·a] = −V

ⁱⁿ

+ V

f d

·(1 − d) (4.45) and finally

V

load

= V

in

− V

^{f d}

·(1 − d)

(1 − d)

²

+ (c − e)·d + e ·(1 − d) (4.46) Equation 4.46 represents a more accurate formula for a boost converter in the continuous- conduction mode. If the diode on -resistance the diode forward voltage and Mosfet on- resistance are considered equal to zero, equation 4.46 become equal to equation 4.37. Figure 4.9 shows the comparison between 4.46 (continuous line) and 4.37 (dashed line) in the continuous mode: The larger is d, the larger is the error.

The reverse formula of equation 4.46 is formed by two equation: d

1

which is valid for d < ¯ d, d

₂

for d > ¯ d (figure 4.9).

d1 = −Vload c + 2 Vload − Vin + Vload e + 2 Vfd

2 (Vload + Vfd ) + (4.47)

+ (2 Vload cVin − 2 Vin Vload e − 4 Vload cVfd+

2 (Vload + Vfd )

−2 Vload

²

ce − 4 Vload

²

cVin

²

+ Vload

²

e

²

+ Vload

²

c

²

)

¹²

2 (Vload + Vfd )

d2 = −Vload c + 2 Vload − Vin + Vload e + 2 Vfd

2 (Vload + Vfd ) + (4.48)

− (2 Vload cVin − 2 Vin Vload e − 4 Vload cVfd+

2 (Vload + Vfd )

−2 Vload

²

ce − 4 Vload

²

cVin

²

+ Vload

²

e

²

+ Vload

²

c

²

)

¹²

2 (Vload + Vfd )

0 0.2 0.4 0.6 0.8 1 0

5 10 15 20 25 30 35 40 45 50 55

d Vload

d¯ Vf d= 0.7 V

Vin= 12 V Ronm= 4.5 ∗ 10⁻³Ω Ronm= 4.5 ∗ 10⁻³Ω Rind= 4.5 ∗ 10⁻³Ω

Rload= 1.2 Ω

Figure 4.9: Comparison between eq.4.46 and eq. 4.37

4.3.2 Boundary between Continuous and Discontinuous-Conduction Mode

The Objective of this section is finding the Boundary between continuous and discontinuous- conduction mode. Figure 4.10 shows the current and the voltage across the inductor at the boundary in a switching period. I

lB

is the mean value of the current at the boundary.

T ton toff

vind on

iind

t t vind

off

Ip

IlB

vind

Figure 4.10: Progress of the current and of the voltage in the inductor at the Boundary

In order to avoid the discontinuous-conduction mode, the value of the average current through the inductor (I

ind

) must be larger than the value of the average current through the inductance at the boundary (I

ind

> I

lB

). Assuming that I

ind

= I

load

/(1 − d), the condition for avoiding the discontinuous conduction mode becomes:

I

lB

< I

load

(1 − d) = V

load

R

load

·(1 − d) (4.49)

I

lB

is half of I

p

, hence:

I

lB

= 1

2 I

p

= 1 2·L

^ind

·

Z

ton

0

v

ind

dt =

"

V

in

− c·V

^load

1 − d

#

· d·T

2·L

^ind

(4.50)

Substituting equations 4.50 and 4.46 in equation 4.49, the follow inequality is obtained :



V

in

−c·

 V

in

− V

^{f d}

·(1 − d) (1 − d)

²

+ (c − e)·d + e



· d·T 2·L

^ind

<

 V

in

− V

^{f d}

·(1 − d) (1 − d)

²

+ (c − e)·d + e



· 1 R

load

(4.51) Developing the last inequality in order to achieve a relationship depending on d:

V

in

·T ·R

^load

·d

³

+ T·R

^load

·[V

ⁱⁿ

·(c − 2 − e) −V

^{f d}

·c]·d

²

+ (4.52) +[ T ·R

^load

·V

ⁱⁿ

·(1 + e − c) −(c·T ·R

^load

+2·L

^ind

)·V

^{f d}

]·d−2·L

^ind

·(V

ⁱⁿ

− V

^{f d}

) < 0 Inequality 4.52, has three solutions, two complex and one real. The solutions are not reported because they are too lengthy and are outside the scope of this thesis. If d is larger than the real solution the mode of conduction is continuous. Care must be taken, because for small values of d, the output voltage is smaller than the input. In this case the diode is conducting and the mode of conduction is continuous.

4.3.3 Discontinuous-Conduction Mode

In the same way as the continuous-conduction mode, in order to find a relationship between V

load

and d for the discontinuous-conduction mode, the voltage average value across the inductance is imposed to be zero. For simpler understanding, figure 4.11 is considered.

Asserting condition 4.38 and considering the length of the turn off equal to ∆

1

·T : 1

T · Z

T

0

v

ind

dt = 1 T ·

Z

ton

0

v

_ind^on

dt + 1 T ·

Z

∆₁·T ton

v

_ind^{of f}

dt = 0 (4.53) Substituting equations 4.39 and 4.40 in equation 4.53 and taking the same route:

(V

in

− V

^load

·c)·d + (V

ⁱⁿ

−V

^load

·a−V

^{f d}

)·∆

¹

= 0 (4.54) The unknown quantities in this case are: ∆

1

and V

load

, hence another equation is needed.

The second equation is obtained observing at figure 4.12, that shows the behavior of the

Figure 4.11: Discontinuous conduction in a buck converter

Figure 4.12: Progress of the current in the inductance, the diode, the load resistance and the output capacitor

current through the inductance, the diode, the load resistance and the output capacitor.

Load current is supposed to be constant. The average value of the current through the

output capacitor I

C2

is equal to zero at the steady-state, hence:

I

load

·d·T + I

^load

·∆

²

·T + y·I

^load

2 = (I

p

− I

^load

)x

2 (4.55)

Where I

p

:

I

p

= V

in

− c·V

^load

L

ind

·d·T (4.56)

Making such a geometric considerations the value of x and y is obtained:

y = V

load

R

load

·I

^p

·∆

¹

·T (4.57)

x = R

load

·I

^p

− V

^load

R

load

·I

^p

·∆

¹

·T (4.58)

considering that I

load

= V

load

/R

load

, ∆

2

= 1 − ∆

¹

− d and substituting equations 4.56, 4.57 and 4.58 in equation 4.55 an ulterior equation in the unknown quantities V

load

and ∆

1

is obtained. Thus equations 4.54 and 4.55 make an algebra system. Solving the system in the unknown quantities V

load

and ∆

1

a relation between V

load

, ∆

1

and d is obtained. The solutions are not reported because they are too lengthy and are outside the scope of this thesis.

4.4 Initial Condition in a Boost-converter

Assuming that the same hypothesis of paragraph 4.1 are valid, the formulae of the initial values of the current flowing in the main inductor and of the voltage across the capacitor will be stated.

4.4.1 Initial condition in the main inductance

Looking at fig. 4.13, the follow equation is valid:

i

ind

= 1 L

ind

·

Z

t

v

ind

dt (4.59)

developing equation 4.59:

I

2

= v

_ind^on

L

ind

·t

^on

+ I

1

(4.60)

Substituting equation 4.39 in the previous equation:

I

1

− I

²

= −

"

V

in

− c· V

load

1 − d

# d·T L

ind

(4.61)

Iload=Im

T ton toff

vind on

vind off

I2

I1

iind

t t

Figure 4.13: Progress of the current and of the voltage in the inductance at steady-state

A further equation that the average inductance current, is equal to the load current divided by (1 − d), hence:

I

1

+ I

2

2 = I

m

= V

load

R

load

·(1 − d) (4.62)

Equations 4.61 and 4.62 make an equations system, where I

1

and I

2

are the unknown quantities. Solving the system the following results are obtained:

I

1

= V

load

R

load

·(1 − d) −

"

V

in

− c· V

load

1 − d

# d·T

2·L

^ind

(4.63)

I

2

= V

load

R

load

·(1 − d) +

"

V

in

− c· V

load

1 − d

# d·T

2·L

^ind

(4.64)

I

1

represents the initial value of the current in the main inductance, I

2

represents the peak value of the current.

4.4.2 Initial Condition in the Output Capacitor

At the turn on the output capacitor and the load resistance are isolated: it follows that the load current (I

load

) and the capacitor current (I

c2

) have the same magnitude and the opposite sign. Writing the second Kirchhoff low at the output part of the circuit

v

c2

= V

load

− i

^c2

·R

^c2

= V

load

+ I

load

·R

^load

(4.65)

4.4.3 Output Voltage Ripple

As in section 4.2.3, in order to calculate the output voltage ripple in the continuous- conduction mode of operation, the current through the load resistance is considered con- stant. It follow that the ripple component of i

d

flow all through the output capacitor, and the mean value in the output. Therefore it is possible to write:

∆V

load

= ∆Q C

2

= I

load

·T ·d C

2

= V

load

·T ·d

R

load

·C

²

(4.66)

And dividing by V

load

, the ripple is obtained [3]:

∆V

load

V

load

= T ·d

τ (4.67)

Where:

τ = R

load

·C

²

(4.68)