How does the integer-N PLL work?

Hello!

Could you tell me how is it working?(Integer-N PLL)
For example if you have a 4/5 divider and then you divide 4 times with 4 why will be N=16 and not 256?

So, N=PA + (P+1)(M-A) = 4*4+5*0=16 but why?
Why not: N=P^A*(P+1)^(M-A) ???

Thanks,

Re: Integer-N PLL

Hi Drabos

Well the thing is that it is NOT the output from the first division you divides again (and again up til 4) cause that what of course be 256. No the way an Integer PLL works is (nearly always) with the help of a Dual-Modulus divider/counter.

The VCO signal is the input to the divider, and this VCO signal is the one that is either divided by 4 or by 5. So when you say you are dividing the signal 4 times by 4 this just means that you are counting the number of times you are dividing this signal by 4. As illustrated (well lousy illustrated..);
Number of VCO cycles:.............................1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19
Counting how many times you divide by 4 :......1...........2..............3................ 4 etc
The M and A counter is just counting the output from this divider. So in this way you will get N=16.

The generel functionality using a 4/5 dual modulus structure is like this:
The vco divider starts with division by 5. Both the A counter and the M counter counts the divider output until A has reached its programmed value.
Then the divider starts dividing by 4 until it reaches the total count number M (Subtracted the times it has already counted when it divided by 5)

In this way you will se that the formula for a dual modulus counter is (where the modulus is P (eg: P=4 / 5).

Total number of VCO cycles N= 5*A + 4*(M-A) which is equal to N=5*A+4*M-4*A = 4*M + A (or N=P*M + A)

Is it possible to multiply using such PLL?