neoflash,

I do not have a paper to reference but from experience the Q will be well below 10.  Many factors play into this like  metal thickness of metal , type of metal and physical size of the inductor.   Simply the Q is set by the (imaginary series impedance)/(real series impedance) so for 450MHz the imaginary part of the impedance will be around 30 Ohms.  

The following is a very coarse approximation of resistance:
assume for example metal resistance of the conductor is around .05 Ohms per square.  a straight line of on chip metal has around 1nH per mm.  so around 10mm for 10nH if the conductor is straight, but if it is coiled then the overall length will reduce by say a factor of 4 (to ~2500uM) due to the magnetic flux concentration.  So calculate the number of squares for example if the conductor width is 10uM 0.05OhmsPerSq*2500uM/10uM equates to 12.5Ohms at DC. now include skin effect and this will increase the effective reisitance.   But even at 12.5Ohms the Q is only a little greater than 2.


http://rfcooltools.com

vp1953,

I agree a Q of 2 does seem a little low.

Neoflash,

why are you wondering about the Q?

http://rfcooltools.com

jayce3390 wrote on Dec 2^nd, 2010, 12:23pm:
I can't see how possibly how this could be true.... unless "usually" the inductor design is very poor...

Anyway the Q value at 450MHz for a 10nH will tend to be low (maybe upto 5 in most technologies/designs), but its heavily dependant on your design. The only real way to find out do is to do some analysis. Either hand calculations like in Tom Lee's RFIC design book or via a quick simulation. I would recommend using ASITIC (free from the web if you google it) for a quick simulation. After you enter the technology information, then draw your spirals and simulate the L & Q. So if you use a large area/fewer turns/more hollow inductor for your 10nH design you will tend to get a better Q, buts a trade off of area for modest increases in Q may not be worth it.