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[SOLVED] Negative VCO transfer function

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CataM

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Hello everyone,

The VCO XR2206 shows at page 11 that the kVCO is negative. That introduces 180º phase in the transfer function of a PLL making it unstable. How can PLLs built around XR2206 work ?

What am I missing ? Obivously if kVCO was positive, we had no problem, but how is that justified ?

Any comment is appreciated.
 

Interchange the reference and feedback clock inputs of your PFD.
 

XOR phase detectors do not capture the frequency information but only the phase information. You can also align the phase by lowering the frequency for a certain amount of time. (Phase is periodic) Thus in this case, it does not matter. [See Razavi's PLL chapter in "Design of Analog CMOS IC" ]
 

vivekroy said:
Thus in this case, it does not matter.
Click to expand...
Well yes it does. As I mentioned, the loop gain is phase shifted with additional 180º making it unstable.

PLL is implemented as: XOR phase detector kpd>0 --> LPF --> kVCO<0 (for XR2206) can you see it ? (unity gain negative feedback)
 

Hi,

Well yes it does. As I mentioned, the loop gain is phase shifted with additional 180º making it unstable.

PLL is implemented as: XOR phase detector kpd>0 --> LPF --> kVCO<0 (for XR2206) can you see it ? (unity gain negative feedback)
Click to expand...
I don't think so.
Please check low pass filtered XOR output vs phase_shift.
--> You will see it has equal positive as well as negative slope.
Thus the input polarity does not care. Even loop gain will be the same.

The only thing that happens is that the resulting phase alignment changes.

Klaus
 
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    CataM

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Right. XOR detector transfer function is a triangle for square wave signals. Depending on the VCO slope sign, the PLL either locking on the rising or falling triangle part.
 
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Alright. That was it. I was missing the triangle wave of the XOR phase detector. Since kVCO<0, the PLL will lock in the [Π, 2*Π] interval where kPD<0.
 

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