Purpose of this resistor in LC-VCO

时间：04-10 整理：3721RD 点击：

Why there is a 2.5k ohm resistor on the control line?

What is that for?

current protection
espercially when the bias voltage is switched on,
this resistor can protect the varactors

Dingjingfen,

I do not think so........
The resistor actually is in the RF path and influences the loaded Q of the coils.
It serves the purpose of applying the DC path with a suitable AC resistance..
Sometimes a R and L are in ut in series to get a bit less loading on the coils and improve phasenoise.

rgds

Mike

the bias point of the resistor is AC ground, because the circuits is symmetry.
so it will not impact the Q of the tank
you cannot use aresistor here if it is not a AC ground, because it will generate additional phase noise.

I guess this only necessary when bias voltage coming from off-chip, right?

Yes
The bias voltage is added from external
so you can measure the tuning range of this VCO

I guess this only necessary when bias voltage coming from off-chip, right?

ALthough this tap off point is THEORETICALLY at zero potential, in reality there will be some RF energy there. In a discrete oscillator you would place a whoping big inductor where the resistor is, and count on the inductance to attenuate any rf leakage (you ight also put a 220 ohm series inductor in to keep the varactors from frying if something weird happened). However, since this looks like a monolithic circuit, instead of an inductor, they put a 2.5K resistor to limit RF leakage.

One should be aware that that is a pretty big resistance, and might impact the PLL design (phase stability), as the bulk capacitance of the varactors (not the RF capacitance) is probably in the 100-1000 pF region.

Biff
max caps in this kind of circuit cannot be larger than few pF (<10!).
If it would be so easy to integrate 1nF with this small area (13u*0.35*8) the analog IC design will completely change.
I agree with the other your considerations.

Mazz

I do not necessarily agree. The RF capacitance might be a few pF. But if you want to change the tuning voltage on the varactor diode, you have to inject or remove charge. If you calculate the amount of charge needed to move in or out, you would find that the equivalent capacitance is MUCH bigger than you thought. And that "bulk" capacitance is what the op amp in your control loop has to move. They are two entirely different capacitances.

It is not clear:
you are talking about a physical capacitor to bulk (I don't think so) or to an EQUIVALENT capacitor?

And, if is an equivalent cap, would you like to give a formula to calculate it?

I usually use a charge pump and passive loop filter to drive the VCO inside the PLL and generally I do not care at all about this equivalent cap you are talking about. And PLLs works as designed. So I am interested...

Mazz

Spoken like a fellow who does not know if his PLL is stable of not. A risky existence!

More than 20M* parts in US are a good guarantee: a quite existance.

*No part with PLL unstable.