Question about negative Q in a resonator
regards
Where did you get this effective Q number? There is loaded and unloaded Q. I would suspect that some negative Q number was from there being a negative resistance in the tuned circuit.
this is a calculated value as i am still calculating the values of the components of the resonator. sorry -125.89 was when i took Qu=47 not 70 (another coil not 0603cs)
Qe=1/(1/QL-1/Qu)=1/(1/75-1/47) = -125.89
where Qu=47, QL=75
i have calculated Cseries =3.164pF so, there will be more approximations, i will use 3.3pF
Your numbers are wrong. Loaded Q is always lower valued than unloaded Q in circuits without feedback. In oscillators the loaded Q is higher due to the negative resistance of the active part of the circuit. This is where the negative Q comes from.
In a passive circuit the Q does have an effect on the resonant frequeny. In an active circuit the oscillator adjusts its negative resistance to cancel the loss in the tuned circuit so the tuned circuit original resistance has no effect.
the unloaded Q of the inductor is 47, the loaded Q is the Q that i want for the whole oscillator which is 75,i am still calculating the values of the resonator itself without a reflection amplifier ,should i find a coil that has a Q more than 75? the maximum Q i found for an inductor at 950MHz is about 70 which is also less than 75.
The inductor Q will affect the short term frequency stability. You should use the highest value available. Air core inductors with large diameter wire are best.
ok, what about the component values? for example Cshunt=59.51pF so i used 56p+3.3p and Cs=3.097pF i used 3.3pF. when simulating the resonator in ADS , the circuit didn't resonate at 950M or even near it. i then used the calculated values, the resonant frequency was right. so, how to make it with practical values?
Your shunt capacitors seem very large. They are in series with each other and in parallel with the series LC network. You will end up with something like a quartz crystal response. A pole and a zero.
How is this network used in a circuit?
it is the same as the one used in this tutorial:
http://www.odyseus.nildram.co.uk/RFM...Oscillator.pdf
Notice that your shunt capacitors are 3x higher than the tutorial and your frequency is within a few percent of the tutorial.
There is an equation on the top right of the first page. Try it with the tutorial values of parts. You probably made a math error.
Also notice that this LC network is used as a two port transmission line in the oscillator circuit. Not as the usual tuned circuit bandpass filter.
the values are right, the problem was the capacitor 3.3pF Cs was originally 3.097pF so the difference was large so, i used 2 series capacitors 6.8, 5.6pF and it worked well,but when i used the inductor's model (i was using ideal inductor) the output was a linear line, i think the value Qu that i used was wrong (as i approximated it from the curve in the datasheet), is there a way to use the inductor's model to know the exact value of Qu at this frequency? what did you mean by:
regards
Some oscillators use a reflection coefficient of a network. Others use the transmission coefficient. The pi nature of the circuit is more like a transmission line equivalent than a tuned circuit such as used in a Hartley oscillator.
when using models of existing components (not ideal) of coil and capacitors the response attached or a linear curve is the output. although i used Q unloaded from the curve in the datasheet to calculate the values of the components. i tried more than one coil, ofcourse with changing the values of other component values which depend on Q.
Something is very wrong. The -30 dB value of S21 is one clue.
Try simulating the parameters shown in the paper and see if you get the same results that it shows. This will determine if the problem is in your simulations.
i tried the values in the tutorial and it works,but the example in the tutorial uses ideal components not manufactured. i have made two simulations one with ideal components and it worked fine the other with manufactured components (from AVX and Microelectronics) and it didn't work the output was what i sent before although they have the same values and all my calculations depends on Qu which i have approximated from the datasheet, also i have tried more than one Qu
The options you have is to use a larger physical size inductor which will have a better Q or to use another circuit entirely.