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nandy Junior Member Offline Posts: 29 Timbuktu	Ring VCO: gain of each stage Jul 04^th, 2007, 2:51am What happens to the frequency of oscillation of a ring oscillator if I increase the gain of each amplifier stage above the minimum required gain levels? Would the frequency increase or decrease? I would definitely prefer a time-domain explanation.
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Berti Community Fellow Offline Posts: 356	Re: Ring VCO: gain of each stage Reply #1 - Jul 4^th, 2007, 4:23am Hi, that depends on the assumption you make. Assuming a model where phase-shift is independent from the gain the frequency won't change. But since this is hardly the case for a real circuit implementation you need the analyse the relationship between phase shift and gain. Regards
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nandy Junior Member Offline Posts: 29 Timbuktu	Re: Ring VCO: gain of each stage Reply #2 - Jul 4^th, 2007, 5:20am Could you please help me to understand the relation between frequency and gain
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Berti

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Re: Ring VCO: gain of each stage
Reply #3 - Jul 4^th, 2007, 6:45am

Assuming the gain of your delay-cell depends on a gm times
an output impedance you probably increase the lengh
of the load transistor in order to increase the gain. This on
the other hand increases the capacitance seen at the output
which reduces the bandwidth of the amplifier and thus
the oscillation frequency. But this is only an example ... increasing
the gm might increase both gain and frequency ...

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nandy Junior Member Offline Posts: 29 Timbuktu	Re: Ring VCO: gain of each stage Reply #4 - Jul 4^th, 2007, 9:55pm C'mon I didn't ask you how to change gain of an amplifier. Pls desist from replying if you are unaware of the answer
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ACWWong

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Re: Ring VCO: gain of each stage
Reply #5 - Jul 5^th, 2007, 2:38am

Nandy,

I think Berti has answered quite well.

Frequency is inversely proportional to delay. The more delay per cell, the lower the frequency.

Varying the Gain per unit cell may well vary the delay, and hence the frequency, BUT this depends on how you intend to implement the "gain variation" to your ring ocsillator cell.

Regards
aw

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aamar

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Re: Ring VCO: gain of each stage
Reply #6 - Jul 5^th, 2007, 5:12am

yes I agree with Berti too, his answer was very good. Frequency means speed, speed in electronics means fast charging and discharging of capacitances, fast charging and discharging means more current to be available, more current means higher gm and normally gain, but it comes on the expense of the sizes of the transistors which has to be larger, transfering the problem to its input as capcitance that has to be charged also. So it depends, there is no direct formula to describe the relation, I have tried it myself and I got the optimum gain only through an optimization process nothing else.

aamar

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Berti Community Fellow Offline Posts: 356	Re: Ring VCO: gain of each stage Reply #7 - Jul 5^th, 2007, 6:32am Thank you guys, at least some of you can follow my thoughts Good explanations, ACWWong and aamar!
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nandy

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Re: Ring VCO: gain of each stage
Reply #8 - Jul 6^th, 2007, 10:12am

Cool. I guess it was only me who didnt get the point. Sorry Berti. :-[ But Im still groping in the dark.

Ok then let me be more specific. I have an inverting amplifier with gain and delay modeled as -A/(1+s/w), where 'w' is the 3dB BW. Say I have 3 such amplifiers in a ring oscillator. So the VCO gives a signal which is delayed by 60 degrees (or T/6) by every amplifier in the loop (to get the total delay to 180 degrees or T/2). My argument is that even if you increase the gain of the amplifiers, the signal frequency which undergoes 60 degree shift musn't change as the magnitude transfer function doesnt affect phase transfer function, or in other words the 3dB BW is still the same.

This would mean the frequency of oscillation cannot vary. But I know I'm wrong because I simulated a VCO with ideal amplifiers and I found the frequency to increase proportionally with the gain of the amplifiers. I'm unable to explain this phenomenon

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nandy Junior Member Offline Posts: 29 Timbuktu	Re: Ring VCO: gain of each stage Reply #9 - Jul 6^th, 2007, 10:16am Just to add to the above point, while simulating I only changed the gain without varying the delay. For example -A/(1+s/w) becomes -2A/(1+s/w).
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boe

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Re: Ring VCO: gain of each stage
Reply #10 - Jul 6^th, 2007, 11:03am

nandy wrote on Jul 6^th, 2007, 10:12am:

I have an inverting amplifier with gain and delay modeled as -A/(1+s/w), where 'w' is the 3dB BW. Say I have 3 such amplifiers in a ring oscillator. So the VCO gives a signal which is delayed by 60 degrees (or T/6) by every amplifier in the loop (to get the total delay to 180 degrees or T/2). My argument is that even if you increase the gain of the amplifiers, the signal frequency which undergoes 60 degree shift musn't change as the magnitude transfer function doesnt affect phase transfer function, or in other words the 3dB BW is still the same.
...
Just to add to the above point, while simulating I only changed the gain without varying the delay. For example
-A/(1+s/w) becomes -2A/(1+s/w).

If you increase the gain you also increase the unity gain bandwidth of your amplifier, so, yes, it gets faster...
And the amplifiers in a VCO always clamp, so you cannot derive the oscillation frequency from the phase shift of the small-signal transfer function.

BOE

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Prabhu

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Re: Ring VCO: gain of each stage
Reply #11 - Jul 6^th, 2007, 11:28am

Hi all,

This is an hypothetical experiment related to Nandy's doubt?

Consider a 3 stage ( 3 inverting amps modelled as -A/(1+s/wp) ) ring oscillator.
For the time being assume no non-linearity in the amps.
Now if we close the loop there wont be a steady state oscillation.
But the oscillation will continue to build up for eternity.

Mine and Nandy's question is how can we justify the fact, in time domain (preferably),
that the oscillation frequency ( of the ever increasing amplitude signal ) increases if we increase the gain 'A' beyond the minimum value.
Any suggestions to gain better perspective of the phenomena will be appreciated.

Prabhu

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boe

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Re: Ring VCO: gain of each stage
Reply #12 - Jul 6^th, 2007, 12:01pm

Basically, the delay of your amplifier is defined by the unity gain bandwith (or gain bandwith product), not the 3dB bandwidth...
Look at it in the frequency domain:
Vout = Vin * [-A0/ (1+s/w)]^3.
You force Vin = Vout, so [-A0/(1+s/w)]^3 must be 1, or 1+s_osc/w = A0 * (-1)^(1/3) => s_osc = w* [A0 * (-1)^(1/3) - 1]. For A0 >> 1 this is approx. s_osc = w* A0 * (-1)^(1/3).
And the signal you observe is exp(s_osc * t), with s_osc ~ A0 * w. So w_osc ~ A0 * w!

BOE

PS: The solution of the ODE system you get will always be at gain 1 (due to your feedback), so you are looking for a solution with gain 1 & the phase shift of 60 deg (over all complex frequencies).

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nandy

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Re: Ring VCO: gain of each stage
Reply #13 - Jul 8^th, 2007, 5:38am

Hi boe
the delay depends on only the phase transfer function of the system. all frequencies before the 3db BW undergo 0 degree phase shift while those beyond undergo 90 degree shift. But if i change the gain of the system w/o touching its 3dB BW then the phase transfer function remains the same despite the increase in UGB. i dont see how the delay varies by merely changing the gain (and hence the UGB) of the amplifier.

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boe

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Re: Ring VCO: gain of each stage
Reply #14 - Jul 9^th, 2007, 1:55am

Hi nandy,
Suppose a linear system, each of your amplifiers has a transfer function of H(s), and your VCO oscillates at s_osc = a + jw (it will have a real component).
Then (for a 3-stage ring oscillator) H(s_osc)^3 must be = 1 (forced by your feedback). If you now take H2(s) = 2*H(s), Then H2(s_osc)^3 = 8, which is incompatible with your feedback: you must increase the complex frequency (both real and imag. component) to get the gain back to one, while keeping the phase shift the same.

If you look at it in the time domain: If you increase the gain, you increase the output signal of your amplifier chain compared to the input signal. To get the gain back to one, you must increase the complex frequency.
BOE

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