# 3.6864 MHz

Discussion in 'Electronic Basics' started by pawihte, Sep 15, 2010.

1. ### pawihteGuest

The answer to this question will probably make me feel stupid,
but here goes.......

I know of common applications for crystals and oscillators
operating at 32.768 kHz, 2.048 MHz, 3.579 MHz, 4.194 MHz, 4.433
MHz, 27 MHz, etc etc.(and in general why those figures are
chosen), but I don't know the significance of 3.6864 MHz which,
judging from its ubiquitousness on shop inventories, must have a
common application. There must be many more standard frequencies
that I don't know about, but I keep coming across this one in
recent times. Is it in audio? Data transmission? Please enlighten
me.

2. ### Rich WebbGuest

One reason (there may be others) is that this freq and its close
relatives 7.3728 and 14.7456 MHz divide-down nicely to standard serial
baud rates with integer divisors.

3. ### pawihteGuest

Thanks to all those who took the time to reply.

So that number is based on an integral multiple of standard baud
rates like 19.2k. Still, I can't help wondering why they simply
didn't standardize on binary multiples like 2.4576 MHz, etc.

4. ### Rich GriseGuest

There is no "stupid" on sci.electronics.basics, the operative word
being "basics," which start at zero.

There are some stupid people who troll the group, but just ignore
them - it won't take long to figure out which they are.
The only one that jumps off the tip of my frontal lobe is 3.579545,
the color burst for NTSC color. Could it be PAL?

It could also be in the 80 meter ham band, which last time I
looked was something like 3.5 - 4.0 MHz.

Good Luck!
Rich

5. ### pawihteGuest

Nope. I'm familiar with the chroma sub-carrier freqs of 3.58 and
4.433 MHz (rounded off here) for NTSC and PAL respectively. I
accept the other replies saying that it's an integral multiple of
standard baud rates.

6. ### Rich GriseGuest

and that seems to be the best answer.

Cheers!
Rich

7. ### NobodyGuest

Binary multiples of *what*, though? The ratios of the standard baud rates
aren't all powers of two; some involve a factor of 3 (14400 is 4800*3).

Using 300 multiplied by a power of two wouldn't give you any easy way to
obtain 14400, 28800, 57600 or 115200 (you would need to multiply
by 3, which is much harder than dividing by 3). Including the factor of 3
means that you can get all of the standard rates by dividing by
(2^N * 3^M) where M is 0 or 1:

3686400 / 16 = 230400
3686400 / 32 = 115200
3686400 / 64 = 57600
3686400 / 32 / 3 = 38400
3686400 / 128 = 28800
3686400 / 64 / 3 = 19200
3686400 / 256 = 14400
3686400 / 128 / 3 = 9600
3686400 / 256 / 3 = 4800
3686400 / 512 / 3 = 2400
3686400 / 1024 / 3 = 1200
3686400 / 2048 / 3 = 600
3686400 / 4096 / 3 = 300
3686400 / 8192 / 3 = 150
3686400 / 16384 / 3 = 75

8. ### pawihteGuest

I'm not sure I get it. Is there something special about operating
close to 4 MHz but not exceeding it?

9. ### pawihteGuest

Now that you've listed out those numbers, they're not unfamiliar
to me, but they're not something I'm regularly involved with
either.

Out of the 15 you listed, two-thirds involve dividing by 3 and
one-third can be obtained by dividing with a binary integral
which IMHO is technically easier. The reverse is true with a
starting frequency of 2.4576, 4.9152 MHz, etc.

10. ### pawihteGuest

I understand the principle. To put my earlier question another
way: Is 4 MHz a common limit of processors for this type of work,
somewhat like the 4 GHz barrier faced by desktop processors?

11. ### Jasen BettsGuest

no it's not
3.6864 MHz get you all the standard baud rates with an integer divisor
4.9152 MHz only gets you most of them but with a power 2 divisor,

12. ### Bob MastaGuest

No, there is no particular bleeding edge technology barrier
like that, but even at lower speeds the chip cost may rise
with higher speed ratings.

The original IBM PC/XT used a 4.77 MHz CPU clock to stay
under the 5 MHz limit of the chips that were readily (and
cheaply) available at the time. It derived that from a
14.31818 MHz crystal that it divided by 3. It then further
divided by 4 to get 1.1931817 MHz to use for the system
timer. The system timer counted up 65536 cycles of that
clock to get the famous 18.2 Hz timer interrupts. These
values all seem to be rather arbitrary, until you note that
counting up 65536 of those timer interrupts gives 3600
seconds... one hour... almost exactly. (The PC/XT only had
16-bit internal registers, so the count went from 0-65535
and then overflowed into the hours counter.)

Best regards,

Bob Masta

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www.daqarta.com
Scope, Spectrum, Spectrogram, Sound Level Meter
Frequency Counter, FREE Signal Generator
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13. ### Michael BlackGuest

CPUs rarely need an exact clock frequency, either because it's not doing
something that requires it, or because one can redo the software to
accomodate a different clock frequency.

You asked for reasons why there'd be more than one crystal frequency
to do the same thing. The answer is because it's better, design wise,
to use the same crystal for clocking the CPU and for generating the baud
rate.

So if you had a CPU that had a maximum clocking rate of 1MHz (and
that was certainly common in the old days, when this particularly
crystal frequency was chosen), you'd have to have a divider before
the CPU on either crystal frequency. But you'd get closer to the maximum
1MHz with the 3.6864/4 (.9216MHz) than with 2.4576/2 (1.2288), well,
the latter would be too high, though of course one could often get
away with a higher clocking frequency. It became more straightforward
with some CPUs, such as the Z80, that used a crystal frequency four
times the actual clock frequency. The latter crystal would have
the CPU running at .6144MHz, significantly less than the maximum
1MHz clocking frequency. Whether or not one really needed to
run the CPUs at the maximum clock frequency, it wsa as much
a speed thing as anything else. The exact frequency meant little
in most cases, running it as fast as possible was.

At that time, small increments were important. I remember when I
changed my Ohio Superboard from the usual 1MHz CPU clock to 2MHz,
it suddenly showed a snappy improvement, yet nobody would bother
at this point to make a 1MHz change in clock frequency.

It matters less these days, since most CPUs run at a much faster
frequency. But it mattered back then, enough to warrant adding
yet another commodity crystal frequency to the parts bins.

Michael

14. ### pawihteGuest

Uh, not exactly. What I meant was, why involve division by a
non-binary integral? Dividing by multiples of 2 would have made
things simpler, if not by very much. This was clarified to some
extent by someone else who pointed out that standard baud rates
do not go up by factors of 2 throughout the whole range.
The operative word for me is (or rather was) the 'if' at the
beginning of the para. Why was it felt necessary to stay at or
near 1, 2 or 4 MHz clock frequency? But I'm beginning to see the
light. Early CPUs were usually rated for maximum clock freqs in
multiples of 1 MHz, so their utilisation would be optimum at or
just below those frequencies, right? This was not immediately
obvious from the first mention of the 4 MHz figure.
My first computer was a 7 MHz Amiga A500, but I was only an end
user. I've dabbled in electronics for a long time, but was very
late in getting directly into the digital world. Hence my
ignorance of how such standards came to be adopted.

15. ### NobodyGuest

No it isn't. As I said before:
E.g. 2457600 / 14400 = 170.6666...

If you start with 2.4576MHz, two thirds can be obtained by dividing by a
power of two and the other third can't be obtained without a frequency
multiplier (VCO + divide-by-3 + PLL).

Divide-by-3 isn't exactly hard (if you want 50% duty cycle, divide-by-6 is
easier, and just as useful in this context).