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I'm trying to figure out the input capacitance of a VCO I am looking to use. The datasheet of the device does not specify an input capacitance. However, it does include the following. illustrationenter image description here Looking at the interface schematic of the Vtune pin, I see two possible figures I could use for input capacitance. One is 5.2pF, the given capacitance of the capacitor in parallel with the Vtune pin. The other is 16-18 pF, given as C; to the right of the schematic. From this drawing, which one of these figures seems more likely to be the input capacitance of the VCO. Here are some more datasheet images for context. enter image description here enter image description here enter image description here

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  • \$\begingroup\$ plan on 25pf; if you need the value to be more precise, then explain what loop bandwidth you need. \$\endgroup\$ – analogsystemsrf Aug 31 '19 at 3:51
  • \$\begingroup\$ Could you justify your anwser based on the datasheet? Other anwsers said 5.2 pF. \$\endgroup\$ – Saunders Aug 31 '19 at 3:59
  • \$\begingroup\$ depends on your control bandwidth; the time constant of 150 ohm and 20pf is 3 nanosecond or 50 MHz. \$\endgroup\$ – analogsystemsrf Aug 31 '19 at 4:32
  • \$\begingroup\$ I am using a TI simulator to help me with loop design. The simulator is reccomending a 500 khz bandwidth for generating a ramp signal with 150mhz bandwidth. I am a pll beginner so this bandwidth might be inappropriate for this application. \$\endgroup\$ – Saunders Aug 31 '19 at 4:34
  • \$\begingroup\$ depends on how fast the FM slide occurs \$\endgroup\$ – analogsystemsrf Aug 31 '19 at 6:23
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Although a 5pF capacitor dominates the shunt reactance at 5GHz, that's irrelevant. You want to know the reactance at your modulation frequency.

The 150ohm + 18pF varactor has a time constant of 2.7nS, or a -3dB point of 58MHz. The 7.5nH resonates with the C at over 400MHz, so you can ignore the L. Anything much below 58MHz you will see substantially both capacitors in parallel, so plan for driving a maximum of 23pF from your loop amplifier.

It's important to be able to drive the higher capacitance, because it will affect two things. One, the speed with which you can modulate the VCO, and two, the phase shift which will affect PLL stability. The fact that you get an unavoidable 45 degrees shift at 58MHz from the RC, however you drive it, sets an upper limit to your stable loop bandwidth. As you are kicking around sub-MHz figures for bandwidth around in the comments, this should not be a problem.

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The data sheet shows a model of the input circuitry as seen by the VCO control voltage. It consists of a 5.2 pF capacitor in parallel with a series RLC. A quick calculation shows that the reactance of the 5.2 pF capacitor is only about 6 ohms at 5 GHz. This is considerably less than the impedance of the RLC network at the same frequency. Thus, for most practical purposes, you can consider the input capacitance of the VCO to be solely due to the 5.2pF capacitor.

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    \$\begingroup\$ Is 5GHz a relevant frequency for computing the reactance? \$\endgroup\$ – Neil_UK Aug 31 '19 at 14:44
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As Barry says, 5.2pF is almost a short circuit @ 5GHz, you can forget about the L, the R and the C beyond them. It is an internal decoupling capacitor, not a part of the LC oscillating circuit. Pin 22 is for applying DC, and eventually a modulation with a frequency much less than VCO's.

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    \$\begingroup\$ Is 5GHz a relevant frequency for computing the reactance? \$\endgroup\$ – Neil_UK Aug 31 '19 at 14:44
  • \$\begingroup\$ Two frequencies that I feel relevant to compute Xc here, are: - VCO frequency: Xc should be close to a short circuit to avoid VCO signal from exiting through the modulation port. - Maximum modulation frequency: should have a high Xc. \$\endgroup\$ – LW1ECP Sep 5 '19 at 14:26

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