S
Sommes
Guest
I'm not very understand how does the divide by n counter work?
Thank you
Thank you
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C
Chris
Guest
"Sommes" <j@jl.com> wrote in message
news:clpihv$qta$1@news-02.connect.com.au...
frequency (i.e the input to the divder) becomes a locked stable output. (put
in simple terms)
Chris
news:clpihv$qta$1@news-02.connect.com.au...
comparator (in order to acheive the lock) and the desired multipliedI'm not very understand how does the divide by n counter work?
Thank you
the output of the the counter is fed to the 2nd input of the phase
frequency (i.e the input to the divder) becomes a locked stable output. (put
in simple terms)
Chris
S
Sommes
Guest
thx mike...but one more question...
the ref frequency would not be changed, it's always 1MHz, and i placed a
divded by n counter(*10) between vco and phase detector, therefore the 2nd
input of phase detector would be 10MHz, but the ref frequency still 1MHz.
The question, what is the result after detection? why the output the 10MHz?
why not 1MHz?
the ref frequency would not be changed, it's always 1MHz, and i placed a
divded by n counter(*10) between vco and phase detector, therefore the 2nd
input of phase detector would be 10MHz, but the ref frequency still 1MHz.
The question, what is the result after detection? why the output the 10MHz?
why not 1MHz?
M
Michael Black
Guest
"Sommes" (j@jl.com) writes:
controlled oscillator, going from 2MHz to 20MHz. You put your divider
between that oscillator and the other input of the phase detector.
The phase detector will always output voltages to try to get that second
input to match the reference oscillator, though of course it may not
always lock. In the example, the VCO can never equal the reference
frequency. But if you set the divider to divide by 2, the phase detector
will see a 1MHz signal if the VCO is set to 2MHz. The phase detector can
then output the signal to steer the VCO to exactly 2MHz, so both inputs
to the phase detector are equal (and in this case equal to 1MHz).
Move the divider to divide by three. The output from the phase detector
gets starts moving the VCO, and when the VCO hits 3MHz, again the two
inputs to the phase detector are equal.
And so on.
In effect you have a multiplier. The output of the PLL is the reference
frequency multiplied by the division factor in the variable divider chain
between the VCO ouput and the phase detector input.
Note one could get the very same effect if one started with the same
reference frequency and then had a bunch of multipliers. Indeed, that
was what was common before semiconductors made PLL synthesizers practical
for everyday use. But multipliers are much bulkier than digital dividers,
require a lot of tuned circuits, and if you're not careful you can end up
with plenty of spurious signals.
Michael
Feed that into a phase detector. You've got a wide range voltageI'm not very understand how does the divide by n counter work?
Thank you
You've got a 1MHz crystal oscillator, the "reference frequency".
controlled oscillator, going from 2MHz to 20MHz. You put your divider
between that oscillator and the other input of the phase detector.
The phase detector will always output voltages to try to get that second
input to match the reference oscillator, though of course it may not
always lock. In the example, the VCO can never equal the reference
frequency. But if you set the divider to divide by 2, the phase detector
will see a 1MHz signal if the VCO is set to 2MHz. The phase detector can
then output the signal to steer the VCO to exactly 2MHz, so both inputs
to the phase detector are equal (and in this case equal to 1MHz).
Move the divider to divide by three. The output from the phase detector
gets starts moving the VCO, and when the VCO hits 3MHz, again the two
inputs to the phase detector are equal.
And so on.
In effect you have a multiplier. The output of the PLL is the reference
frequency multiplied by the division factor in the variable divider chain
between the VCO ouput and the phase detector input.
Note one could get the very same effect if one started with the same
reference frequency and then had a bunch of multipliers. Indeed, that
was what was common before semiconductors made PLL synthesizers practical
for everyday use. But multipliers are much bulkier than digital dividers,
require a lot of tuned circuits, and if you're not careful you can end up
with plenty of spurious signals.
Michael
J
Jason Rosinski
Guest
The 2nd input to the phase detector should be the same frequency as the reference. The VCO would
run at 10MHz, which after the divide by N (/10 not *10) would match the reference frequency. Hence
PLL frequency multiplication.
Sommes wrote:
run at 10MHz, which after the divide by N (/10 not *10) would match the reference frequency. Hence
PLL frequency multiplication.
Sommes wrote:
thx mike...but one more question...
the ref frequency would not be changed, it's always 1MHz, and i placed a
divded by n counter(*10) between vco and phase detector, therefore the 2nd
input of phase detector would be 10MHz, but the ref frequency still 1MHz.
The question, what is the result after detection? why the output the 10MHz?
why not 1MHz?