
TC32306FTG
2015-10-01
59
6.10.3 h’0B VCO Frequency Settings 1
Table 6-58 Register (h’0B)
D7
D6
D5
D4
D3
D2
D1
D0
Name
NC3
NC2
NC1
NC0
FC11
FC10
FC9
FC8
Initial
1
0
0
1
0
0
1
1
Type
R/W
R/W
R/W
R/W
R/W
R/W
R/W
R/W
[D7:D4]NC3..0 [Local Frequency: Integer Counter]
NC = 2
3
×NC3+2
2
×NC2+2
1
×NC1+2
0
×NC0
Initial Value: NC = 9
h’0C [D7:D0]FC7..0 & h’0B [D3:D0]FC11..8 [Local Frequency: Fractional Counter]
FC = -2
11
×FC11+2
10
×FC10+2
9
×FC9+2
8
×FC8+2
7
×FC7+2
6
×FC6+2
5
×FC5+2
4
×FC4+2
3
×FC3+2
2
×FC2+2
1
×FC1+2
0
×FC0
Initial Value: FC = 1016 (Two's complement format)
Must set the FC value between -1516 to +1515.
Example1)
VCO frequency "fvco" is set as follows.
fvco = (NC+53) x fosc + FC x fstep
fvco: VCO Frequency
fosc: Reference Clock Frequency (30.32MHz)
fstep: Frequency Step (= fosc / 3032)
<Case of setting fvco = 1890MHz>
fvco = 1890MHz / fosc = 30.32MHz / fstep = 10kHz (= 3032 / 30.32MHz)
NC = fvco / fosc - 53 = 1890MHz / 30.32MHz - 53 = 9.335
à
NC = 9 (This number is rounded to an integer.)
FC = (fvco / fosc - (NC+53)) x fosc / fstep = (1890MHz / 30.32MHz - (9+53)) x 30.32MHz / 10kHz = 1016
NC = 9
à
NC = “b’1001” (Binary)
FC = 1016
à
FC = “b’001111111000” (Two's complement format)
Example2)
RX frequency "frx" and TX frequency "ftx" are set as follows.
frx = (fvco / nd) - f_if = [(NC+53) x fosc + FC x fstep] / nd - f_if
ftx = fvco/ nd = [(NC+53) x fosc + FC x fstep] / nd
frx: RX frequency
ftx: TX frequency
f_if: IF frequency (230 or 280 kHz)
nd: Division Ratio (nd = 6 at 315MHz Band, nd = 4 at 434MHz Band, nd = 2 at 868 / 915MHz Band)
<Case of setting frx = 314.94MHz>
frx = 314.94MHz / fosc = 30.32MHz / fstep = 10kHz (= 3032 / 30.32MHz) / f_if = 280kHz
In this case, nd = 6, as frequency band is 315MHz.
Actually to calculate frx, firstly calculate fvco based on frx, next determine the number of NC and FC.
fvco = (frx - f_if) x nd = (314.94MHz - 0.28MHz) x 6 = 1887.96MHz
NC = fvco / fosc - 53 = 1887.96MHz / 30.32MHz - 53 = 9.268
à
NC=9 (This number is rounded to an integer.)
FC = (fvco / fosc - (NC+53)) x fosc / fstep = (1887.96MHz / 30.32MHz - (9 + 53)) x 30.32MHz / 10kHz = 812
NC = 9
à
NC =“b’1001” (Binary)
FC = 812
à
FC =“b’00110010110” (Two's complement format)
By using those NC and FC, fvco and frx can be calculated again.
fvco = (NC+53) x fosc + FC x fstep = (9 + 53) x 30.32MHz + 812 x 10kHz = 1887.96MHz
frx = (fvco / nd) + f_if = (1887.96MHz / 6) + 280kHz = 314.94MHz