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Design Article

Charge-Pump Phase-Locked Loop—A Tutorial—Part II

Jeffrey S. Pattavina

7/21/2011 1:14 PM EDT

4.    Frequency Response

 

In this section we examine the frequency response of the CP-PLL. The effects of zero-pole placement in the loop-filter are examined in relationship to stability, peaking and jitter transfer.

 

4.1. General PLL Gain

The following model for the PLL in Fig. 4-1 explicitly shows the loop constants.

 

 

A simplified model for the PLL (Fig. 4-2) consolidates the individual loop constants into a single loop constant k.

 

 

Note that the transfer function g(s) in Fig. 4-2 is the open-loop gain of the system.

 

 

The open-loop gain is

 

The closed-loop gain is defined as

 

 

Substituting (4.2) into (4.3) yields

 

 

The phase-error gain is defined as

 

 

From which

 

 

4.2 Desirable Filter Characteristics

In section 3.3, we discussed the loop-filter transfer function (3.12). Here we further determine the suitable characteristics of the loop-filter regarding its ability to track changes on the input clock. First, we consider the response to a step change in phase of magnitude ΔØ given by

 

Taking the Laplace Transform of Øi(t) gives

 

 

The phase-error transfer function for this particular input is given by

 

 

Canceling terms yields

 

 

Applying the final value theorem yields

 

 

Thus, regardless of the type of loop-filter, there is no steady state phase-error when a step change in phase is applied to the input.

 

Now consider the response to a step change in frequency. This is of a practical concern for the following reason: when the input is removed from the loop, the VCO will operate at the VCO’s free running frequency wo. The application of an input constitutes a step change in frequency of magnitude Δw.

 

 

The time-domain response to a step change in frequency of Δw U(t)

 

 

Taking the Laplace Transform of Øi(t) gives

 

 

The associated phase-error gain is

 

 

Factoring yields

 

 

Applying the final value theorem yields

 

 

The resulting steady-state phase-error is

 

 

Equation (4.19) indicates that the steady-state phase-error for a step change in frequency does depend on the characteristics of ƒ(s). In particular, the phase error can be made arbitrarily small if the D.C. gain of ƒ(s) is very large. Thus the condition for zero steady-state error is given by

 

 

Example:

 

Consider the following two filters driven from a voltage source (Fig. 4-3). The D.C. gain of the passive filter is unity, indicating a nonzero steady- state error for a step in frequency. On the other hand, the active filter has infinite gain at D.C., ensuring zero steady state error.

 

 

It is for this reason that active filters are frequently employed in PLL designs where the loop-filter is driven from a voltage source. In the following section we consider a passive filter that is driven by a current source.

Example:

Now consider the passive filter driven by a current source (Fig. 4-4). The D.C. gain in this case is infinite. This is one advantage of the charge-pump PLL; that the D.C. gain can be made infinite using only a passive filter.

 

 

4.3. CP-PLL Gain

 

From (4.2) the open-loop gain for CP-PLL is

 

 

Substituting (3.13) for ƒ(s) into (4.2) yields

 

 

The corresponding closed-loop gain is

 

 

Substituting (4.21) into (4.22) yields

 

 

Equation (4.23) can also be expressed in standard form (4.24), however, we will find it more convenient to work with (4.23) instead.

 

 

Where the damping constant is

 

 





ARNAV

7/23/2011 5:27 AM EDT

Very helpfull

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ptvn

8/15/2011 4:31 PM EDT

The article has some transcribing errors that the editor still needs to fix. I can send you a copy of the original article if you like.

Jeff Pattavina



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SYakov

3/14/2012 8:46 PM EDT

Hi Jeff,
Could you please send me a copy of the original article.
Thanks in advance.
Sergei.

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WKetel

7/30/2011 8:07 PM EDT

Unfortunately it is difficult to save a copy of this to my HDD.

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goranisaev

8/4/2011 6:26 AM EDT

Here is a simple way to save a copy of this article:
1. Mark the text of page 1 of the article and copy it.
2.Paste it in an empty .doc file.
3.Mark, copy and paste page 2 and 3.
4. Save the full document to HDD.

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ptvn

8/15/2011 4:32 PM EDT

The article has some transcribing errors that the editor still needs to fix. I can send you a copy of the original article if you like.

Jeff Pattavina

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dyos_60

4/4/2012 4:24 AM EDT

Hi Jeff

May I have a copy too?

Tx
Pramod

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