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zyyang

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hi everyone
now i am designing a pll,and i want to calculate that how much spur at pll'output if 1uv ripple voltage on vctrl .
and my filter is the traditional one: two order filter that has a zero and 2 pole. here someone tell me that the wave is a trianglar but why? and why not a square wave? and if my filter is 3_order which one will the wave be?
thank you very much!

Added after 2 minutes:

the circuit
 

If you look at the current fed into the Loop filter, that is a square wave of current. The loop filter capacitors are charged up by that current, and to the voltage at the output of the charge pump looks like an exponential charging voltage.

As far as the spurious components with a 1 uV noise voltage, the VCO output is given by Fout=Kvco*VTune, so the spurious is given by 1 uV * Kv.

In addition to that, Spurious output at the output of a closed loop pll is shaped by the Closed Loop Transfer function, so it is a low pass response. For the total loop response to a 1 uV spurious,

SpuriousResponse(f)=1 uV*Kvco*20Log10(|CLG(f)|)


Dave
www.keystoneradio.com
 

Here is time response for the first circuit, where Iin is pulse train of Ic, -Ic.
Circuit1:
Vout(t) = Iin*(C2*R1*exp(-(C1+C2)*t/C1/R1/C2)*C1-C1*R1*C2+t*C1+t*C2)/(C1+C2)^2
If time constant R1C1C2/(C1+C2) is much smaller than switching frequency exp=1
Vout(t) = Iin*(C2*R1*C1-C1*R1*C2+t*C1+t*C2)/(C1+C2)^2
Two pole:
v[out] = (1+s*C1*R1)*Iin/s/(C1+C2+s*C2*C1*R1)
zero=1/C1R1
p1=0, p2=(C1+C2)/C1/C2/R1

Circuit2:
"v[out1] = Iin*(1+s*C1*R1)/s/(Cout+s*C1*R1*Cout+s*C1*Cout*Rout+C1+s*C2*Cout*Rout+C2+s^2*C2*C1*R1*Cout*Rout+s*C2*C1*R1)"
 

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