Specifically in reference to the MCX-90, it is very similar to the MCX-100, with 16 channels instead of 32. The information below for the MCX-100 should also work for the MCX-90.
The VHF radio uses this programming directly. For UHF use, divide all
frequencies FIRST by three. Also, Syntor radios use this same programming
algorythm but have a different IF freq on rcv.
I . FREQ. PROMS
Frequency PROMs will be programmed as follows:
1. All locations in the PROM will be filled. Undefined channels will
be strapped to previous channels.
2. "Blank" channels will exist on transmitter only and the PROM word will
contain all "F" hex on that transmitter channel.
(the read order of the prom is byte 7,6,4,0,1,2 : 2&5 not read)
3. Types of frequency PROMs will be:
A. Case 1 256 X 4 for < 16 channels (N82S129)
B. Case 2 512 X 4 for > 16 channels (N82S131)
Original Field proms were MMI 5301-1J for 16 ch, and MMI 5306-1J
for 32Ch. Note: the R1800 field programmer software/firmware
was upgraded to use the Signetics N82S style proms. Other proms,
once programmed with the correct data (i.e. 74s288) don't use the
same programming algorythms, and will be destroyed by the
programmer.
4. Frequency information for the PROM fills eight consecutive locations
in the PROM for each channel.
5. Channel locations in the proms are as follows:
16 ch RF prom:
Addr Ch Addr Ch Addr Ch
50-57 R10 A8-AF T5
00-07 R16 58-5F R11 B0-B7 T6
08-0F R1 60-67 R12 B8-BF T7
10-17 R2 68-6F R13 C0-C7 T8
18-1F R3 70-77 R14 C8-CF T9
20-27 R4 78-7F R15 D0-D7 T10
28-2F R5 80-87 T16 D8-DF T11
30-37 R6 88-8F T1 E0-E7 T12
38-3F R7 90-97 T2 E8-EF T13
40-47 R8 98-9F T3 F0-F7 T14
48-4F R9 A0-A7 T4 F8-FF T15
32 channel RF prom:
Addr Ch Addr Ch Addr Ch
000-007 R32 0B0-B7 T6 160-167 R28
008-00F R1 0B8-BF T7 168-16F R29
010-017 R2 0C0-C7 T8 170-177 R30
018-01F R3 0C8-CF T9 178-17F R31
020-027 R4 0D0-D7 T10 180-187 T16
028-02F R5 0D8-DF T11 188-18F T17
030-037 R6 0E0-E7 T12 190-197 T18
038-03F R7 0E8-EF T13 198-19F T19
040-047 R8 0F0-F7 T14 1A0-1A7 T20
048-04F R9 0F8-FF T15 1A8-1AF T21
050-057 R10 100-07 R16 1B0-1B7 T22
058-057 R11 108-0F R17 1B8-1BF T23
060-067 R12 110-17 R18 1C0-1C7 T24
068-06F R13 118-1F R19 1C8-1CF T25
070-077 R14 120-27 R20 1D0-1D7 T26
078-07F R15 128-2F R21 1D8-1DF T27
080-087 T32 130-37 R22 1E0-1E7 T28
088-08F T1 138-3F R23 1E8-1EF T29
090-097 T2 140-47 R24 1F0-1F7 T30
098-09F T3 148-4F R25 1F8-1FF T31
0A0-0A7 T4 150-57 R26
0A8-0AF T5 158-5F R27
6. The frequency band to be determined by the customer frequencies.
Range Table
Range I 136 < Fc < 162 Mhz
Range II 146 < Fc < 174 Mhz
7. The following steps will be performed to generate the eight PROM
words for each channel:
A. Determine range from Table in 6.
B. Subtract 21.4 Mhz from all Receiver freqs.
C. Determine the reference freq.
a. Divide frequency to synthesize by 5Khz
If the result is an integer, then D0,D1 = 1 of word 7
b. If the result is a real number, try dividing by 6.25 Khz.
If the result isn't an integer, then the frequency cant
be synthesized. If it is an interger then D0=1,D1=0 of word 7
D. Determine N
N = Freq/REF freq (typical ref freq is .005 Mhz [5 Khz]
or .00625 Mhz [6.25 kcs]
E. Calculate Integer NA,NB
NB=INT(N/63)
NA=N-(63*NB)
NB=NB-NA
IF NA=0; then NA=63; NB=NB-64
F. Convert NA, NB to binary bits.
32 16 8 4 2 1
NA= A5 A4 A3 A2 A1 A0
512 256 128 64 32 16 8 4 2 1
NB = B9 B8 B7 B6 B5 B4 B3 B2 B1 B0
G. Construct PROM words 0,1,3,4 from the bits in NA & NB
(note a * means this bit is inverted from the value you got above.
If it was a 1, enter 0 - if 0 then enter 1)
A3 A2 A1 *A0 --> W0
B1 *B0 A5 A4 --> W1
B5 B4 B3 B2 --> W3
B9 B8 B7 B6 --> W4
H. Set PROM word 2 and 5 to 1111 ( F hex)
I. Determine PIN shift bit S0 VCO1 VCO2
Range 1 Synthesizer D2 word 6 D2 word 7 D3 word 7
Below 120.6 Mhz 1 0 1
120.6-126.6 Mhz 1 1 1
126.6-132.5 Mhz 1 0 0
132.5-138.4 Mhz 1 1 0
138.4-144.3 Mhz 0 0 1
144.3-150.2 Mhz 0 1 1
150.2-156.1 Mhz 0 0 0
Above 156.1 Mhz 0 1 0
Range 2 Synthesizer
Below 130.6 Mhz 1 0 1
130.6-136.8 Mhz 1 1 1
136.8-143.0 Mhz 1 0 0
143.0-149.2 Mhz 1 1 0
149.2-155.4 Mhz 0 0 1
155.4-161.6 Mhz 0 1 1
161.6-167.8 Mhz 0 0 0
Above 167.8 Mhz 0 1 0
J. Dual front end bit
Bit D3 of word 6 is 0 when FC < (fmax+fmin)/2
A good way to remember this, is, the lower front end is selected with
the bit being 0, and the high front end is selected with a 1. Try to
split the total freqs between the two front ends, if your spread is
greater than 6 mhz.
Assemble the front end bits:
Word 6 = D3(front end bit),D2(S0),D1(1),D0(1)
( D1 and D0 spare - set high)
Word 7= D3(VCO2),D2(VCO1),D1,D0
(1,1 if 5 kc channel, or 0,1 if a 6.25kc channel)
hence VCO2,VCO1,1,1 - if a 5 kc channel
VCO2,VCO1,0,1 - if a 6.25kc channel