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Fundamentals of Power Electronics Chapter 8: Converter Transfer Functions45

8.1.7. The low-Q approximation

G(s) = 11 + a1s + a2s

2 G(s) = 1

1 + s

Q0+ s

0

2

Given a second-order denominator polynomial, of the form

or

When the roots are real, i.e., when Q < 0.5, then we can factor the

denominator, and construct the Bode diagram using the asymptotes

for real poles. We would then use the following normalized form:

G(s) = 1

1 + s

1

1 + s

2

This is a particularly desirable approach when Q


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An example

A problem with this procedure is the complexity of the quadratic

formula used to find the corner frequencies.

R-L-C network example:

+ –

L

C Rv1(s)

+

v2(s)

–

G(s) = v2(s)

v1(s)= 1

1 + s L R

+ s2 LC

Use quadratic formula to factor denominator. Corner frequencies are:

1, 2 = L / R !

L / R

2

– 4 LC 2 LC


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Factoring the denominator

1, 2 = L / R ! L / R

2 – 4 LC

2 LC

This complicated expression yields little insight into how the cornerfrequencies ω1 and ω2 depend on R, L, and C .

When the corner frequencies are well separated in value, it can be

shown that they are given by the much simpler (approximate)

expressions

1 R L

, 2 1 RC

ω1 is then independent of C , and ω2 is independent of L.

These simpler expressions can be derived via the Low-Q Approximation.


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Derivation of the Low-Q Approximation

G(s) = 1

1 + sQ0

+ s0

2

Given

Use quadratic formula to express corner frequencies ω1 and ω2 interms of Q and ω0 as:

1 = 0

Q

1 – 1 – 4Q2

2 2 =

0

Q

1 + 1 – 4Q2

2


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Corner frequency ω2

2 = 0

Q

1 + 1 – 4Q2

2

2 = 0

Q F (Q)

F (Q) = 12

1 + 1 – 4Q2

2 0

Q for Q


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Corner frequency ω1

F (Q) = 12

1 + 1 – 4Q2

F(Q)

0 0.1 0.2 0.3 0.4 0.5

Q

0

0.25

0.5

0.75

1

can be written in the form

where

For small Q, F(Q) tends to 1.

We then obtain

For Q < 0.3, the approximation F(Q)

=

1 iswithin 10% of the exact value.

1 = 0

Q

1 – 1 – 4Q2

2

1 = Q 0

F (Q)

1 Q 0 for Q


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The Low-Q Approximation

f 2 = f 0F (Q)Q

f 0Q

–40dB/decade

f 0

0dB

|| G ||dB

–20dB/decade

f 1 = Q f 0F (Q)

Q f 0


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R-L-C Example

1 Q 0 = R C L

1

LC = R L

2 0

Q = 1

LC 1

R

C

L

= 1 RC

G(s) = v2(s)

v1(s)= 1

1 + s L R

+ s 2 LC

f 0 = 0

2= 1

2 LC

Q = R

C

L

For the previous example:

Use of the Low-Q Approximation leads to

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