tza3015hw NXP Semiconductors, tza3015hw Datasheet - Page 14

• Manufacturer Part Number: tza3015hw
• Description: 30 Mbit/s To 3.2 Gbit/s A-rate 4-bit Fibre Optic Transceiver
• Manufacturer: NXP Semiconductors

Philips Semiconductors
Having calculated the division factor (n.k), the values for N and K can be calculated according to the flow depicted in the
flowchart of Fig.8.
The value of the octave divider M is programmed by bits RXDIV_M[2:0] in register RXOCTDIV (C0h). The value for the
main divider N is programmed by bits RXN[8:0] in registers RXMAINDIV1 (C1h) and RXMAINDIV0 (C2h). The value for
the fractional divider K is programmed by bits RXK[21:0] in registers RXFRACN2 to RXFRACN0 (C3h to C5h). Bit
RXNILFRAC in register RXFRACN2 (C3h) must be set depending on whether there is a fractional part or not.
Example 1: An SDH or SONET link has a bit rate of 2488.32 Mbit/s (STM16/OC48) and consequently fits in octave
number 0, so M = 1. Suppose the reference frequency provided at pins CREF(Q) is 77.76 MHz. This means that the
reference division R needs to be 4. The values of n and k can be calculated from the flowchart:
Since k = 0 in this example, no fractional functionality is required, bit RXNILFRAC (register C3h), should be logic 1.
N = 2
(register C0h) and N = 256 (registers C1h and C2h).
Example 2: An SDH STM16 or SONET OC48 link with FEC has a bit rate of 2666.057143 Mbit/s
(15/14
at pins CREF(Q) is 38.88 MHz. This means that the reference division R needs to be 2. The values of n and k can be
calculated from the flowchart:
This means that n = 137, k = 0.1428571 and bit RXNILFRAC (register C3h) should be logic 0. Since k < 0.25, k is
corrected to 0.6428571, while the corrected N becomes N = 273. Consequently the appropriate values are: R = 2
(register A1h), M = 1 (register C0h), N = 273 (registers C1h and C2h) and K = 10 1001 0010 0100 1001 0011 (registers
C3h to C5h). The FEC bit rate is usually quoted to be 2666.06 Mbit/s. Due to round off errors, this leads to a slightly
different value for k than in the example.
Example 3: A Fibre Channel link has a bit rate of 1062.50 Mbit/s and consequently fits in octave number 1, so M = 2.
Suppose the reference frequency provided at pins CREF(Q) is 19.44 MHz. This means that the reference division R
needs to be 1. The values of n and k can be calculated from the flowchart:
This means that n = 109, k = 0.3107 and bit RXNILFRAC should be logic 0 (register C3h). Since k is between 0.25 and
0.75, k does not need to be corrected and N = 2
A1h), M = 2 (register C0h) and N = 218 (registers C1h and C2h). K = 01 0011 1110 0010 1000 0001 (registers C3h to
C5h).
Example 4: A non standard transmission link has a bit rate of 3012 Mbit/s and consequently fits in octave number 0, so
M = 1. Suppose the reference frequency provided at pins CREF(Q) is 20.50 MHz. This means that the reference
division R needs to be 1. The values of n and k can be calculated from the flowchart:
This means that n = 146, k = 0.9268293 and bit RXNILFRAC should be logic 0 (register C3h). Since k is larger than 0.75,
k needs to be corrected to 0.4268293 and N = 2
A1h), M = 1 (register C0h) and N = 293 (registers C1h and C2h). K = 01 1011 0101 0001 0010 1011 (registers C3h to
C5h).
If the I
applied reference frequency of 19.44 MHz (see Table 3).
2003 Dec 16
n.k
n.k
n.k
30 Mbit/s to 3.2 Gbit/s A-rate
4-bit fibre optic transceiver
=
=
=
2
bit rate M R
--------------------------------------- -
bit rate M R
--------------------------------------- -
bit rate M
--------------------------------------- -
C-bus is not used, the DCR can be set up for the eight pre-programmed bit rates by pins DR0 to DR2 with an
n and no correction is required. Consequently the appropriate values are: R = 4 (register A1h), M = 1
2488.32 Mbit/s) and consequently fits in octave number 0, so M = 1. Suppose the reference frequency provided
f
f
f
ref
ref
ref
R
=
=
=
2488.32 Mbits
-------------------------------------------------------- -
1062.50 Mbits
-------------------------------------------------------- -
3012 Mbits 1 1
----------------------------------------------- -
20.50 MHz
n.k
77.76 MHz
19.44 MHz
=
bit rate M R
--------------------------------------- -
1 4
2 1
f
ref
=
146.9268293
=
=
n + 1 = 293. Consequently the appropriate values are: R = 1 (register
n = 218. Consequently the appropriate values are: R = 1 (register
128
109.3106996
=
2666.05714283 Mbits
---------------------------------------------------------------------------- -
14
38.88 MHz
1 2
=
137.1428571
Preliminary specification
TZA3015HW