SP8854E Ver la hoja de datos (PDF) - Mitel Networks

Número de pieza

componentes Descripción

Fabricante

SP8854E

0.3-6.0V; 2-7GHz synthesiser. For complete 2-7GHz single chip system

Mitel Networks 

FROM

CHARGE PUMP

FROM

CHARGE PUMP

REFERENCE

C2

C1

R2

−

+

TO VCO

Fig. 8 Third order loop filter circuit diagram

Loop Filter Design

Generally, the third order filter configuration shown in Fig. 8

gives better results than the more commonly used second order

because the reference sidebands are reduced. Three equations

are required to determine values for the three constants, where

The equations are:

t1 = C1R1

tt23

=

=

R2 (C11C2)

C2R2

t1 =

KfK0

vn2N

11vn2

t22



1

2

11vn2 t32 

…(1)

t2 =

1

vn2t32

…(2)

t3

=

2tan

F0 1cos1F0

vn

…(3)

where

Kf is the phase detector gain factor in mA/radian

K0 is the VCO gain factor in radians/seconds/V

N is the division ratio from VCO to reference frequency

vn is the natural loop frequency

F0 is the phase margin, normally set to 45°

Since the phase detector used is linear over a range of 2p

radians, the phase detector gain is given by:

Phase comparator current setting

Kf =

2p

mA/radian

These values can now be substituted in equation (1) to obtain

a value for C1 and in equations (2) and (3) to determine values

for C2 and R2.

Example

Calculate values for a loop with the following parameters:

Frequency to be synthesised

Reference frequency

Division ratio

K0 VCO gain factor

F0 phase margin

Phase comparator current

1000MHz

10MHz

1000MHz/100MHz = 100

2p310MHz/V

45°

6·3mA

The phase detector gain factor Kf = 6·3/2p = 1mA/radian

SP8854E

From equation (3):

t3

=

2tan 45°1cos145°

100kHz32p

= 0·4142

628319

∴t3 = 65931029

From equation (2):

t2 =

1

(100kHz32p)2365931029

∴t2 = 3·84431026

Using these values in equation (1):

t1 =

13102332p310MHz/V

1

3[A]2

1003(100kHz32p)2

where

A

=

11vn2

11vn2

t22

t32





=11(100kHz32p)23(3·84431026)2

11(100kHz32p)23(65931029)2

1

t1 =

62832

39·48310212

 6·833  2

1·1714 

= 1·593102932·415

∴t1 = 3·8431029

Now, t1 = C1 ∴ C1 = 3·84nF

t2 = R2 (C11C2)

t2 = C2R2

Substituting for C2:

t2

=

R2

C11Rt32





or, R2=

t22t3

C1

=

3·84431026265931029

0·015331026

∴R2 = 829·4Ω

t3

=

C2R2

= t3

R2

= 65931029

829·4

∴C2 = 0·794nF

11

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