Hi Sapphire,

Yes; it's true that having a higher Q means faster oscillation ramp-up (if everything else is equal).  This can be analyzed in many different ways, but here's how I do it:

Oscillator tank actually consists of a parallel network of L, C, Rtank and Rneg.  Rtank represents the tank Q - the higher the tank Q, the higher Rtank is (approximately, Rtank=Q*wL=Q/wC for high-Q tanks).  Rneg is the "negative resistance" - the mechanism through which the active device provides energy into the tank, keeping the oscillation going.  The parallel combination of the two resistances is Rtot=Rtank||Rneg - if the parallel Rneg is smaller than Rtank, Rtot is negative (meaning the oscillation amplitude grows over time).

If you write down the differential equation for the tank and solve the transient response, you get something like:

v(t)=e^-(1/(2RtotC))*(c1cos(w0t)+c2sin(w0t))

where w0 is the oscillation frequency and c1 and c2 are constants (dependent on initial conditions).  Note that in our case, Rtot is negative (for growing oscillation amplitude), so e^-(1/(2RtotC))=e^(1/(2|Rtot|C)), indicating exponential growth in amplitude.

To make the amplitude ramp up faster, we need to increase the value of 1/(2|Rtot|C)=0.5*(1/C)*(|1/Rneg|-1/Rtank).  If the only knob we have to turn is the tank Q, increasing it is the best way to improve the ramp-up speed - in the extreme case of infinite Q, 1/Rtank=1/(QwL) approaches zero, so 0.5*(1/C)*(|1/Rneg|-1/Rtank) approaches 0.5*(1/C)*|1/Rneg|.  If Rtank is anything less than infinite, it would only reduce the value of 0.5*(1/C)*(|1/Rneg|-1/Rtank), reducing the oscillation ramp-up speed.

(Another way to improve the ramp-up is to make the value of Rneg really small while keeping it negative... that would really increase the value of 0.5*(1/C)*(|1/Rneg|-1/Rtank).  This is another unintuitive result, but it's all coming from the way negative resistances behave.)

One way to look at this is to consider the ideal case of an infinite tank Q of the RLC tank.  Here, your Rtank would be infinite (since it's connected in parallel it would be like an open circuit), and all you would have left is the L and the C.  Then, you connect a "negative R" component in parallel with it to provide some energy into the tank that gets the oscillation going.  If your tank wasn't ideal (i.e., finite Q), you would lose some of that energy in the non-idealities (parasitic resistance...).  If it is ideal (infinite Q), the negative resistance can do its job to increase the oscillation amplitude without any losses.

To me, these things are always a bit confusing - parallel resistances, series resistances... tank Q, inductor Q, capacitor Q, negative/positive resistance...  resistance/admittance...  All of it can be analyzed through math and network theory, but for quick roundabout answers, I rather go through energy: negative resistance supplies energy into the tank and positive resistance sucks it away (and converts it into heat).  The more effective the negative resistance is (the smaller the parallel negative resistance is), the faster energy is supplied into the tank, and the more "effective" the positive resistance is (the smaller the parallel positive resistance is), the faster energy is sucked away.

The explanation above is an incoherent mess.  :o  But if you have any questions, let me know, and I'll try to elaborate.

Hyvonen wrote on Jun 6^th, 2008, 10:09pm:


Hyvonen: There seems to be a typing error as your position is just the opposite: Q prop. to speed rather than time.

But it is to be mentioned that - as suspected by me formerly (my replies 1, 4 and 6) - it can be proven that the start-up time of oscillators is proportional to the network Q. However, it seems that this holds only for four-pole oscillators with an active feedback element.

If Hyvonen is correct in his argumentation (and I cannot argue against it) there is obviously a difference between two-pole negative-resistance and four-pole oscillator circuits - as far as the start-up time ic concerned. That seems to be a fact worth mentioning. I think it was a very interesting question not answered up to now in any textbook.

Hyvonen wrote on Jun 7^th, 2008, 11:22am:


Four-pole oscillators consist of a frequency selective network (bandpass, lowpass, highpass, notch, crystal) and an amplifier, all connected in a closed loop to produce a loop gain Aloop=1 (Barkhausen condition).
The well known WIEN oscillator belongs to this group.

You can set up a differential equation for the parallel LRC circuit (where R=Rneg||Rtank) by using Kirchoff's current law:

ic(t)+ir(t)+il(t)=0
<=> C∂v(t)/∂t+1/R*v(t)+1/L∫v(t)∂t=0
<=> C*D^2+1/R*D+1/L=0 (where D=∂v(t)/∂t)

You can solve this second-order equation (see
http://www.analyzemath.com/calculus/Differential_Equations/solve_second_order_3.... for help), with the assumption that the voltage response is oscillatory (i.e., second order solution is a complex pair).  This yields the result

v(t)=e^-(t/(2RtotC))*(c1cos(w0t)+c2sin(w0t))    (note that I forgot the "t" from the original equation - that was a typo)

Now, we should be able to determine one of the two constants by defining the initial voltage at t=0 to be something (some amount of noise); that would give us c1 (sin0=0).  I'm not sure how to obtain the second constant, though... if anyone has a suggestion, that would be great!

Sapphire,

1.  By 'initial damping voltage', do you mean the tank voltage at t=0 (initial condition)?  The word 'damping' confused me.  Initial voltage is undefined, but the oscillator will start up due to noise in the circuit - the exact ramp time is hard to predict, though.  A couple of posts ago I suggested providing a well-defined initial voltage (e.g., charging up the tank capacitor to some voltage before turning on the diff. pair); this would give you an initial voltage to start from and the ramp would be possible to calculate.

2.  This is true; the negative-R value will get larger until the oscillation amplitude doesn't increase anymore; this is much harder to analyze.  You'd need to analyze this using nonlinear differential equations (as opposed to the simple linear differential equation I used), and I cannot help you there.  What is it that you're aiming to solve?  The time it takes for the oscillator to reach a certain oscillation amplitude?

[quote author=sapphire link=1212177977/15#23 date=1213159181]T
Hi buddypoor,  what's the differenence between 2-port oscillator and 4-port oscillator?  Is the common differential LC VCO 2-port or 4-port oscillator?
[quote author=Hyvonen link=1212177977/15#22 date=1213137185]

Watch your typo: Not 2-port and 4-port, but instead 2-pole resp. 4-pole topology.

Four-pole oscillators consist of a frequency selective network (bandpass, lowpass, highpass, notch, crystal) and an amplifier, all connected in a closed loop to produce a loop gain Aloop=1 (Barkhausen condition).
The well known WIEN oscillator belongs to this group.