Hi,

I'm designing an injection-locked LC-VCO with f0=2GHz. The circuit is shown bellow, this is a PMOS-NMOS crossed coupled LC-VCO, the injected signal Vij is put through M1/M2 differential pair. The Vij comes form a LNA in the previous part and a capacitor is in between the LNA and the VCO. I have some questions:

(1) The coupled capacitor between LNA and the VCO blocks the DC part, that means the Vij is a sine-wave with DC=0. Is that mean the M1/M2 need bias circuit?

(2) Why M3/M4 is put there?

(3) My design procedure is: first finish the LC-VCO, then design the injection differential pair. Am i right?

(4) I know that increasing the injected current will increase the locking range wL (let's say the input amplitude is constant), but no matter how big the current source (for the injection pair) is, or how large the W/L of the injection pair is (big gm), the locking range is always below 10MHz and looks like constant --- the locking range didn't increase. I even changed to another low-Q inductor, but the range still didn't increase. I don't know why this happen? and How to design this injection part properly?

(5) Is it true that if the VCO(freq is f0) is locked, when the Vij (freq is f1) comes, the output frequency is f1, but if Vij disappear, the output frequency will immediately come back to f0?

Does anyone know this and can help me out? Thank you very much!

My thoughts:

1) Yes you need to establish a bias on the VCO side of the AC coupling caps
2) Looks like a cascode buffer to provide isolation
3) You need to account for the capacitance added to the tank by the injection diff pair, I would not ignore it during the design of the VCO
4) You say you decreased the Q by reducing the Q of the inductor, but you do not say what the loaded Q of the tank actually is.  I have never designed an injection locked oscillator but I have designed several VCO's using the cross coupled N/PMOS architecture you have here.  The objective was always to maximize the loaded Q for best phase noise.  Maybe you need to reduce the Q even further to increase the locking range, but at the expense of phase noise.
5) As Aaron said there will be some settling time but yes if the injection signal is removed the VCO should return to its free running frequency.

tm123 wrote on Nov 11^th, 2013, 9:51am:


Thank you! What you said is right, I want to increase the locking range a little bit, I have tried to increase the injected amplitude but the range doesn't change. So, for confirmation purpose, I decrease the Q as low as possible to check whether the range will increase, but still it doesn't. Or maybe the bias condition is not correct?

For the inductor's Q, I get it from simulation. I modeled the inductor simply as Z=Rs+jwLs, than do the AC simulation to find the Z at f=2GHz, the I got the Q=wLs/Rs. Initially, I chose the biggest Q=15, no matter how big the I_inj is, the locking range is around 10MHz. Then I decrease the Q to about 8, and the locking range didn't change a lot (12MHz).

The locking range is: wL=(w0/2Q)*(I_inj/I_osc)/(1-(I_inj/I_osc)^2).  where the I_inj and I_osc is the current amplitude of injected signal and the VCO, respectively.

You mentioned that the total tank Q. I know that put tuning cap bank to the VCO will decrease the Q, but how can I test and know the total tank Q during the design?

Any suggestions will help a lot! Thank you!

ccarrot,

Loaded Q is basically the summation of the Q of the individual components.  Just like you know the Q of the inductor, all the other components have their own Q.  This includes the varactor, any fixed capacitance (maybe a MIM capacitor you have to help set the oscillation frequency), even the MOS devices that are part of the amplifier.  One way to find the loaded Q is to find the Q of the individual components and add them together as Q_loaded=(1/Q1+1/Q2+...)^-1.  In your equation for locking range wL I assume Q is the loaded Q not the Q of the inductor by itself.  So if you say reducing the Q of the inductor didn't have much impact perhaps some other component is dominating the loaded Q.

Hope this helps.

Tim