Hyvonen
Junior Member
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Posts: 15
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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.
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