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The Way to make 61.44MHz clock from 10MHz and 1PPS!

J

Jay

Jan 1, 1970
0
Hello,

I would like to know the way to make 61.44MHz clock which has to
synchronized with 10MHz
periodically.
Let me say... 10MHz Clock Input(1PPS Sync. Aligned) -> PLL or
Something -> 61.44MHz Output!(1PPS Sync. Aligned)
If you may have a good solution, Please let me know that!

Thank you.
Jay
 
P

Phil Allison

Jan 1, 1970
0
** Groper Alert !!!

Hello,

I would like to know the way to make 61.44MHz clock which has to
synchronized with 10MHz periodically.



** Just like a wall clock that does not work still shows the right time
twice a day - your two will be in perfect synch quite regularly.

And every bit as usefully.




........ Phil
 
E

Eeyore

Jan 1, 1970
0
Jay said:
Hello,

I would like to know the way to make 61.44MHz clock which has to
synchronized with 10MHz periodically.

By periodically do you mean there's only a possibility of synching them
occasionally and you expect them to track inbetween times ?

Graham
 
T

Tam/WB2TT

Jan 1, 1970
0
Jay said:
Hello,

I would like to know the way to make 61.44MHz clock which has to
synchronized with 10MHz
periodically.
Let me say... 10MHz Clock Input(1PPS Sync. Aligned) -> PLL or
Something -> 61.44MHz Output!(1PPS Sync. Aligned)
If you may have a good solution, Please let me know that!

Thank you.
Jay
I think what you mean is that you want the two frequencies to be coherent.
61.44MHz is 61,440 kHz. If you divide the 10 MHz by 1000, you will have 10
KHz. Now, in a phase locked loop put a div by 6144 in the feedback loop, and
it will run at 61440 KHz. You could also use ratios of 125 and 768 for
better performance and fewer parts; in this case the reference frequency
will be 80 KHz, instead of 10 KHz.

Tam
 
P

Phil Hobbs

Jan 1, 1970
0
Jay said:
Hello,

I would like to know the way to make 61.44MHz clock which has to
synchronized with 10MHz
periodically.
Let me say... 10MHz Clock Input(1PPS Sync. Aligned) -> PLL or
Something -> 61.44MHz Output!(1PPS Sync. Aligned)
If you may have a good solution, Please let me know that!

Thank you.
Jay
Your requirements aren't very clear. Is the 1 Hz derived from the 10
MHz? How does the 1 Hz sync help you?

Cheers,

Phil Hobbs
 
T

Tom Bruhns

Jan 1, 1970
0
Hello,

I would like to know the way to make 61.44MHz clock which has to
synchronized with 10MHz
periodically.
Let me say... 10MHz Clock Input(1PPS Sync. Aligned) -> PLL or
Something -> 61.44MHz Output!(1PPS Sync. Aligned)
If you may have a good solution, Please let me know that!

Thank you.
Jay

"THE" way??? "A" way is a PLL. Find the largest frequency which
multiplies by integers to 10MHz and 61.44MHz, since you already have
the 10MHz to work with. Set those integers into the reference and VCO
divisors of a programmable PLL. You may even be able to find one with
an integrated silicon VCO that will do the job for you, but if you
want low phase noise/low jitter, plan on using a VCXO in the loop.

Quiz: what's the frequency, and what are the integers? Why use the
highest such frequency?

If you don't always have the 10MHz available, use a VCXO that's stable
enough to hold your desired tolerance for the time the 10MHz is
unavailable.

Or would you rather lock to the 1PPS? (Why, if 10MHz is available??)

Cheers,
Tom


Quiz: what is the frequency and what are the integers?
 
D

Dr. Honeydew

Jan 1, 1970
0
** Groper Alert !!!



** Just like a wall clock that does not work still shows the right time
twice a day - your two will be in perfect synch quite regularly.

And every bit as usefully.

....... Phil

And more useful by far than Philthy Phil.
 
Hello,



I think what you mean is that you want the two frequencies to be coherent.
61.44MHz is 61,440 kHz. If you divide the 10 MHz by 1000, you will have 10
KHz. Now, in a phase locked loop put a div by 6144 in the feedback loop, and
it will run at 61440 KHz. You could also use ratios of 125 and 768 for
better performance and fewer parts; in this case the reference frequency
will be 80 KHz, instead of 10 KHz.

This would work. As Tom Bruhns says, you'd want to use a VCXO as your
61.44MHz oscillator if you needed to minimise the jitter on your
61.44MHz output.

A DDS chip doesn't have any problem with non-integral frequency
ratios. Check out the Analog Devices AD9859 and comparable parts

http://www.analog.com/en/prod/0,,770_843_AD9859,00.html


This has a built in phase-locked loop controlled reference oscillator,
which you could run at 200MHz locked to your 10MHz reference
oscillator, and use to generate a 61.44MHz output. It would only be
producing three samples per cycle of the 61.44MHz output
(3.255308 ...) so you'd need a good anti-aliasing filter on the
output.

If you look further down their product list you may be able to find
something that can lock a faster clock to your 10MHz reference, which
would give you more samples per cycle at 61.44 MHz and let you get
away with a less elaborate anti-aliasing filter.
 
M

MooseFET

Jan 1, 1970
0
Hello,



I think what you mean is that you want the two frequencies to be coherent.
61.44MHz is 61,440 kHz. If you divide the 10 MHz by 1000, you will have 10
KHz. Now, in a phase locked loop put a div by 6144 in the feedback loop, and
it will run at 61440 KHz. You could also use ratios of 125 and 768 for
better performance and fewer parts; in this case the reference frequency
will be 80 KHz, instead of 10 KHz.

Tam

If you have too very stable frequencies, you can remove one divider
chain. A simple flip-flop clocked from the output of a divider on one
signal and sampling the other will work as a phase detector. You end
up with some "phase ripple". Think of it like this: ASCII art.

"Late case"

--- --- --- --- ---
--- --- --- --- --- --- Undivided

--------------- Divided
---------------

???????????HHHHHHHHH Flip-flop output



"Early case"

--- --- --- --- ---
--- --- --- --- --- --- Undivided

----------------- Divided
-------------

?????????LLLLLLLLLLLLL Flip-flop output

In this case I'd leave out the 768 divide to save the most flip-flops,
if the timing will allow.
 
T

Tam/WB2TT

Jan 1, 1970
0
MooseFET said:
If you have too very stable frequencies, you can remove one divider
chain. A simple flip-flop clocked from the output of a divider on one
signal and sampling the other will work as a phase detector. You end
up with some "phase ripple". Think of it like this: ASCII art.

"Late case"

--- --- --- --- ---
--- --- --- --- --- --- Undivided

--------------- Divided
---------------

???????????HHHHHHHHH Flip-flop output



"Early case"

--- --- --- --- ---
--- --- --- --- --- --- Undivided

----------------- Divided
-------------

?????????LLLLLLLLLLLLL Flip-flop output

In this case I'd leave out the 768 divide to save the most flip-flops,
if the timing will allow.
Yeah, that should usually work, provided the VCO pulling range is less than
80 KHz (or is it 40?). Div by 768 is no big deal though, just 16X16X3.

Tam
 
T

Tam/WB2TT

Jan 1, 1970
0
Tom Bruhns said:
"THE" way??? "A" way is a PLL. Find the largest frequency which
multiplies by integers to 10MHz and 61.44MHz, since you already have
the 10MHz to work with. Set those integers into the reference and VCO
divisors of a programmable PLL. You may even be able to find one with
an integrated silicon VCO that will do the job for you, but if you
want low phase noise/low jitter, plan on using a VCXO in the loop.

Quiz: what's the frequency, and what are the integers? Why use the
highest such frequency?

If you don't always have the 10MHz available, use a VCXO that's stable
enough to hold your desired tolerance for the time the 10MHz is
unavailable.

Or would you rather lock to the 1PPS? (Why, if 10MHz is available??)

Cheers,
Tom


Quiz: what is the frequency and what are the integers?
See 4/25/2007 9:56 AM
 
M

MooseFET

Jan 1, 1970
0
Yeah, that should usually work, provided the VCO pulling range is less than
80 KHz (or is it 40?). Div by 768 is no big deal though, just 16X16X3.

Some reasons to consider doing it with a 22CV10 :

Div 125 may fit into 8 or 9 sections of a 22V10 and the phase detector
into what is left of it.

If you use the "zero power" CMOS version of the 22V10, the current
draw will be low.

The swing on the flip-flop is nearly 0-5V

Using a part that is not shared with anything else prevents a route
for noise getting to the VCO.

Using one easy to get part to doa job will often score you attaboys.
 
T

Tom Bruhns

Jan 1, 1970
0
This would work. As Tom Bruhns says, you'd want to use a VCXO as your
61.44MHz oscillator if you needed to minimise the jitter on your
61.44MHz output.

A DDS chip doesn't have any problem with non-integral frequency
ratios. Check out the Analog Devices AD9859 and comparable parts

http://www.analog.com/en/prod/0,,770_843_AD9859,00.html

This has a built in phase-locked loop controlled reference oscillator,
which you could run at 200MHz locked to your 10MHz reference
oscillator, and use to generate a 61.44MHz output. It would only be
producing three samples per cycle of the 61.44MHz output
(3.255308 ...) so you'd need a good anti-aliasing filter on the
output.

If you look further down their product list you may be able to find
something that can lock a faster clock to your 10MHz reference, which
would give you more samples per cycle at 61.44 MHz and let you get
away with a less elaborate anti-aliasing filter.


Yes, a DDS will give you very good phase noise performance--but at the
expense of spurs. An advantage of the DDS is that you don't need
another good oscillator (the 61.44MHz one). A disadvantage is that if
the 10MHz is not continuous (as might be implied by the base note), it
won't do you much good.

As yet another alternative, from the analog camp (and a bit tongue-in-
cheek ;-), if the 10MHz is continuous: multiply 10MHz by 768 and
divide by 125; then filter. You'd do it in steps: multiply by 4 a
couple times to get to 160; divide by 5 to get to 32; multiply by 4 to
get to 128; divide by 5 to get to 25.6; multiply by 4 to get to 102.4;
divide by 5 to get to 20.48; multiply by 3 to get to 61.44.

But the OP also indicated that the 61.44 should be somehow "aligned"
with the 1pps. If he really means an edge of the 61.44 must be
aligned with an edge of the 1pps, and if the edges of the 10MHz are
not already aligned with the 1pps, he'll need to factor that 1pps in
there somewhere. Does he really have a 1pps whose edge is jitter-free
to within a fraction of a cycle of 61.44MHz?? Or is it a GPS output
that could have tens of nanoseconds error in each pulse? As is often
the case, the basenote was somewhat ambiguous about the real
requirements and what is available to work from.

Cheers,
Tom
 
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