MAX2620EUA-T Maxim Integrated Products, MAX2620EUA-T Datasheet - Page 9

• Manufacturer Part Number: MAX2620EUA-T
• Description: RF Wireless Misc 10MHz to 1050MHz VCO with Differential-O
• Manufacturer: Maxim Integrated Products

According to the electrical model shown in Figure 5, the
resonance frequency can be calculated as:
[Equation 1]
Figure 4. 10MHz Crystal Oscillator
f
where C
O
X = STATEK AT-3004 10MHz
FUNDAMENTAL MODE CRYSTAL SURFACE MOUNT
C
LOAD
120pF
120pF
30pF
2
= 20pF
SHDN
n
L1 C
V
CC
C
C
10
3
1
2
3
4
0.01 F
STRAY
3
+ C
_______________________________________________________________________________________
V
TANK
FDBK
SHDN
+ C
CC
1
03
MAX2620
03
+
+ C
C
C
C
17
17
4
V
4
GND
RF Oscillator with Buffered Outputs
OUT
OUT
CC
Sample Calculation
+ C
x C
2
+ C
+ C
10 H
1
8
7
6
5
04
D1
D1
51
04
V
CC
V
V
CC
CC
0.01 F
C
0.01 F
0.01 F
0.01 F
10MHz to 1050MHz Integrated
0.01 F
27pF
6
C
C x C
OUT
OUT
5
5
+ C
n
n
R
and is approximately:
[Equation 2]
where g
Using the circuit model of Figure 5, the following exam-
ple describes the design of an oscillator centered at
900MHz.
Using Equation 1, solve for varactor capacitance (C
C
age applied to the varactor is approximately at half-
supply (the center of the varactor’s capacitance range).
Assume the following values:
C
C
The value of C
mance of the MAX2620 EV kit. Values of C3 and C4 are
chosen to minimize R
the resonant circuit with excessive capacitance. C
and C
The varactor’s capacitance range should allow for the
desired tuning range. Across the tuning frequency
range, ensure that R
The MAX2620’s oscillator is optimized for low-phase-
noise operation. Achieving lowest phase-noise charac-
teristics requires the use of high-Q (quality factor)
components such as ceramic transmission-line type
n
STRAY
03
D1
, the negative real impedance, is set by C3 and C4
= 2.4pF, C
is the capacitance of the varactor when the volt-
Calculate: R
04
Choose: L1 = 5nH ±10%
m
= 2.7pF, C17 = 1.5pF, C6 = 1.5pF, C5 = 1.5pF,
are parasitic capacitors.
R
= 18mS.
n
g
04
m
STRAY
Q = 140
p
= 2.4pF, C3 = 2.7pF, and C4 = 1pF
2
= Q
f C
s
< 1/2 R
is based on approximate perfor-
n
3
1
(Equation 2) while not loading
C
2
03
n
f
2
L1
f C
4
1
C
04
D1
03
).
9