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neoflash

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Question on Ken's "Predicting the phase noise
Nov 01^st, 2005, 6:50am

I am reading Predicting the phase noise and jitter of PLL-based frequency synthesizers.

And I found in page 28, equation (57):

JeePFD/CP = T*sqrt(var(n)/2)/Kdet;

If Kdet=Ip as stated in the doc page 7, then above equation has incorrect units.

T: s
var(n): A²/s
Kdet: A

thus, all in all, Jee 's units is not s;

I think that the correct conversion equation is:
JeePFD/CP = sqrt(T*var(n)/2)/Kdet;

any body concur?

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Ken Kundert Global Moderator Offline Posts: 2386 Silicon Valley	Re: Question on Ken's "Predicting the phase n Reply #1 - Nov 4^th, 2005, 2:37pm Why do you believe that the units for var(n) are A²/s? I expect the units to be A², in which case the units work out properly for the equation given in the paper. -Ken
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neoflash

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Re: Question on Ken's "Predicting the phase n
Reply #2 - Nov 4^th, 2005, 6:30pm

[quote author=Ken Kundert link=1130856611/0#1 date=1131143846]Why do you believe that the units for �var(n) are �A²/s? I expect the units to be A², in which case the units work out properly for the equation given in the paper.

-Ken [/quote]

Basically, var(n)=rms(n), if it is root mean square, it should be averaged over time.

also from wiener-Khinchine theorem, what we get from
integrate(sphi(f)*df) is total power, should have units of A^2/s.

Only if we have different difinition on var(n)?

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Ken Kundert

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Re: Question on Ken's "Predicting the phase n
Reply #3 - Nov 5^th, 2005, 12:15am

The value var(n) represents the normalized power, or power into a 1 Ohm resistor.

Power has units of I²R. If R=1, then the units are I².

This is consistent with the definition of var(n), which is given in (57), which states that var(n) is equal to the integral of S_n over all frequencies, and S_n has the units of I²/Hz, and so var(n) has the units of I².

-Ken

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neoflash Community Fellow Offline Mixed-Signal Designer Posts: 397	Re: Question on Ken's "Predicting the phase n Reply #4 - Nov 5^th, 2005, 5:12am Oh, i see. I made the mistake somewhere. Thanks for correcting me. However, I still think that I can not understand how the Jee is get: I will calculate it in this way: JeeI^2 ------------ = var(n) T/2 Thus, Jee= Tvar(n)/(2*I^2) Why not calculate it in this way?
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Ken Kundert

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Re: Question on Ken's "Predicting the phase n
Reply #5 - Nov 5^th, 2005, 3:13pm

Your equation cannot be correct as the units are not correct.

From (3)
dt = I_max/T di
or
dt = K_det/T di

Looking at noise powers ...
var(t) = (K_det/T)² var(n)
where n is the noise in i.

Since there are two edges with J_ee,
J_ee = sqrt(2 var(t))
J_ee = K_det/T sqrt(2 var(n))

So it seem like the equation in the paper is incorrect in that it divides by two where it should multiply two. Do you agree that this equation is correct?

-Ken

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neoflash

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Re: Question on Ken's "Predicting the phase n
Reply #6 - Nov 5^th, 2005, 7:21pm

Ken:

Thanks for correcting my mistake in math. I am puzzled when doing translation between current domain and time domain.

However, I did not think the new equation you give is right.

dT*Im / (T) = in

==> sigma(dT) = sigma(in) * T / (Im)

==> sigma(dT) = sqrt(2)*J = sqrt(var(n)) * T / (Im)

The original equation on the paper is correct.
Sorry for the confusion. Thanks

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« Last Edit: Nov 6^th, 2005, 12:47am by neoflash »

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Ken Kundert

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Re: Question on Ken's "Predicting the phase n
Reply #7 - Nov 6^th, 2005, 12:51am

Whoops. I copied things down somewhat backward. I'll try to be a bit more careful this time.

From (3)
dt = T/I_max di
or
dt = T/K_det di

Inverting this to calculate the contribution to the noise on the output from jitter on one edge at the input gives
di = K_det/T dt
We cannot combine the contribution from the noise on the second edge using this formula as the jitter on the edges will be uncorrelated. So instead, we must work in noise powers.
var(n) = (K_det/T)² (var(j₁) + var(j₂))
or
var(n) = (K_det/T)² (2 var(j))
where n is the noise in i and j is the noise in t.

J_ee = sqrt(var(j))
and so
J_ee² = var(j) = (T/K_det)² var(n)/2
or
J_ee = T/K_det sqrt(var(n)/2)

Which suggests that my original equation was in fact correct.

-Ken

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« Last Edit: Nov 6^th, 2005, 5:20pm by Ken Kundert »

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neoflash Community Fellow Offline Mixed-Signal Designer Posts: 397	Re: Question on Ken's "Predicting the phase n Reply #8 - Nov 6^th, 2005, 3:18am thanks for the discussion.
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neoflash Community Fellow Offline Mixed-Signal Designer Posts: 397	Re: Question on Ken's "Predicting the phase n Reply #9 - Nov 30^th, 2005, 7:43am I found I cannot understand why there are two edges for a period? if it is a pfd, it only responds to rising edge, isn't it?
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Ken Kundert

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Re: Question on Ken's "Predicting the phase n
Reply #10 - Nov 30^th, 2005, 10:52am

The phase detector has two inputs, in and ref. A rising edge on the in input causes the output to rise, and a rising edge on the ref input causes the outptu to fall. The charge pump integrates the output, so we are interested in the time difference between the rising and falling edges on the output. Those are the two edges.

-Ken

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neoflash Community Fellow Offline Mixed-Signal Designer Posts: 397	Re: Question on Ken's "Predicting the phase n Reply #11 - Dec 1^st, 2005, 6:27am I can not agree. If we assume ref clock is clean. Then the jitter is refered to PFD input, the feedback clock. Thus J*I ----- = delta(I) T then we get the relationship between input jitter and current noise current. No other edges involved.
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