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a simple question on PLL open loop transfer function

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eejli

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If I write PLL open loop transfer function as T(jw)=Kpd*Zlpf(jw)*Kvco(jw)/S/N,
does it means Kvco must be in unit rad/s/volt rather than Hz/volt?
And then if I make |T(jw)|=1, what I get is Wu,i.e., loop bandwidth with unit rad/s rather than Hz.


Thanks.
 

Yes, you do it all in rad/s, so multiply your Hz by 2Π.

Not sure about your terminology, but what do you mean by "S/N". If you mean "1/S" as the integrator in your VCO, then don't you mean "1/S*N"?
 

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    eejli

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Yes you are right S=jw, and N is divider ratio.

One popular formula to calculate R1(the resistor to create the zero in the open loop transfer function) is:
R1=loop BW(in unit Hz)/Ktotal
while Ktotal=(Iqp/2pi)*Kvco(in hz/volt)/N

According our discussion the above equation is right. Do you agree? Otherwise, I should conver the Kvco to unit rad/second/volt instead of hz/volt.

Thanks.
 

What I am saying is that it looks like in your open loop gain equation, you are multiplying by N. You instead want to be dividing by N. This is possibly the source of your problem.

N is the divider ratio. If you are dividing by 10, for instance, then any phase deviation will be divided by 10 also, and you will only get 1/10 of the voltage out of the phase detector. Even for just a N=10, that will give you a 40 dB error (20 +20 dB going from multiplying to dividing by N=10).
 

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