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what is the relationship between Kvco and jitter?

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jasonxilion

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i was asked the question when i was interviewed. my answer is for the noise of the input signal, Kvco the smaller ,the better; for noise of vco, the answer is converse

it that right ?
thx
 

What do you mean by input signal? PLL input signal?

For high KVco, it will be fast settling time, but higher jitter.
 

What do you mean by input signal? PLL input signal?
Click to expand...

i mean the reference signal, like generated by crystal

For high KVco, it will be fast settling time, but higher jitter
Click to expand...
can you explain it in detail?

thx
 

ok, for Vco ouput, ωout = ωo + Kvco(Vcont). Here, we need to have low Kvco to minimizing the noise effects of Vcont.
 

faizalism said:
ok, for Vco ouput, ωout = ωo + Kvco(Vcont). Here, we need to have low Kvco to minimizing the noise effects of Vcont.
Click to expand...

so the latter of my answer is right?
 

faizalism said:
my answer is for the noise of the input signal, Kvco the smaller ,the better
Click to expand...
The answer is correct

for noise of vco, the answer is converse
Click to expand...
I can't understand
Click to expand...


can i explain it in this way？
the transer function of the vco noise to the output is high-pass ,do u agree with me? so the bandwidth is the bigger ,the better
because the Kvco is proportional to bandwidth, in my opinion bigger kvco does restrict the noise caused by the vco

please point out any mistakes in my description, thx
 

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