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Oct 14th, 2019, 4:54pm
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Ring Oscillator (Tuned Load stage) (Read 805 times)
cktdesigner
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Ring Oscillator (Tuned Load stage)
Mar 12th, 2019, 8:55pm
 
Hi,

I am trying to implement a 13-stage ring oscillator with each inverter stage as shown.

What I observe is that when the fixed resistor is changed from 42-Ohms to 43-Ohms the frequency of oscillation changes from 550-MHz to 13.9-GHz abruptly.

i.e. a mere change of 1-Ohm results in the frequency of oscillation changing by more than 10-GHz.

Can someone (preferably people closely familiar with Ring VCOs) please explain the reason for such an abrupt/huge change in the oscillation frequency?

Thank you,

Best regards.
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« Last Edit: Mar 13th, 2019, 10:05am by cktdesigner »  

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Jacki
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Re: Ring Oscillator (Tuned Load stage)
Reply #1 - Mar 13th, 2019, 7:44am
 
Do you check if the output signal is stable ringing or the frequency is updating. Maybe your settling time is long, and in fact your oscillator has reached the steady state.
also I don't understand why you add a zero at your inverter.
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cktdesigner
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Re: Ring Oscillator (Tuned Load stage)
Reply #2 - Mar 13th, 2019, 8:13am
 
Jacki wrote on Mar 13th, 2019, 7:44am:
Do you check if the output signal is stable ringing or the frequency is updating. Maybe your settling time is long, and in fact your oscillator has reached the steady state.
also I don't understand why you add a zero at your inverter.


Yes, I checked the output signal for >1.5usec (for the 550-MHz case)
and for ~130nsecs (for the 13-GHz case which itself takes a very long time for the 13GHz case with maxstep=1psec) in a simple transient analysis with maxstep for time set to 1psec. The output signal is stable and so is the frequency of oscillation (vs. time) when plot using the freq() function in the ADE calculator.

I used the following:
freq(VT("/out") "rising" ?xName "time" ?mode "user" ?threshold (0.5 * VAR("pvdd")))

where pvdd (=800mV) is the supply voltage of the inverter/entire VCO.

In fact the plot of the Fosc vs. time curve stabilizes pretty fast soon after oscillations start, within 10nsecs of start of simulation and does not change or droop or rise w.r.t time after 10nsecs or so for both cases.

I am trying to implement the ring oscillator topology shown in the attached picture. I observed the cliff effect (w.r.t varying Vc) when I was originally using xtors in place of the ideal 42ohm resistor.

I decided to replace the xtor to simplify things by replacing with a resistor.

This is a vco where the frequency of oscillation is controlled by the voltage applied to the gate of the nmos in the attached picture which in turn varies the effective load cap seen by each inverter and hence Fosc.

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« Last Edit: Mar 13th, 2019, 10:01am by cktdesigner »  

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cktdesigner
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Re: Ring Oscillator (Tuned Load stage)
Reply #3 - Mar 13th, 2019, 8:21am
 
See the attached picture which explains how this works as a VCO, supposedly.

My problem is that as per the picture on how the topology is supposed to work as a VCO there should be a smooth Fosc vs. Vtune frequency tuning curve.

There should not be a cliff effect which I am observing.

The Fosc vs. Vtune curve should be a smooth monotonically decreasing curve, as per theory.

By the way this is supposed to be a large signal swing ring oscillator and not a small swing RO.

The Vc varies the Ron of the xtor and hence is supposed to vary the Ceff seen by the inverter in every stage and hence Vc is supposed to control the inverter delay of each stage and hence in turn control the Fosc of the overall ring VCO.

I simplified things by replacing the xtor with an ideal resistor and varying the value of the resistor manually and measuring the Fosc.

I am doing this experiment of using an ideal analogLib cap and analogLib res to eliminate any other issues related to non-idealities or parasitics w.r.t the xtor.

I observe this cliff effect, obviously, even when using a real xtor in place of the resistor and varying the VGS (i.e. Vc from the picture) of the xtor.

Which is why I decided to replace the xtor with an ideal resistor and vary its resistance manually in the first place to verify if the "cliff" effect is still present even when using ideal cap and res.
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cktdesigner
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Re: Ring Oscillator (Tuned Load stage)
Reply #4 - Mar 13th, 2019, 9:46am
 
I am attaching the plots of the Fosc vs. Time and that of the output signal itself for the 42-Ohms case here (Zoomed-in view).
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cktdesigner
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Re: Ring Oscillator (Tuned Load stage)
Reply #5 - Mar 13th, 2019, 9:48am
 
I am attaching the plots of the Fosc vs. Time and that of the output signal itself for the 42-Ohms case here (Zoomed-out view).
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cktdesigner
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Re: Ring Oscillator (Tuned Load stage)
Reply #6 - Mar 13th, 2019, 9:59am
 
I am attaching the plots of the Fosc vs. Time and that of the output signal itself for the 43-Ohms case here (Zoomed-in view).
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cktdesigner
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Re: Ring Oscillator (Tuned Load stage)
Reply #7 - Mar 13th, 2019, 10:00am
 
I am attaching the plots of the Fosc vs. Time and that of the output signal itself for the 43-Ohms case here (Zoomed-out view).
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Re: Ring Oscillator (Tuned Load stage)
Reply #8 - Mar 13th, 2019, 12:08pm
 
Hi,

After a suggestion from my former colleague to plot the delay of each inverter stage vs. the load resistance, I figured out after going through a slew of thought process an idea for a solution to my problem.

I tried to set the "Initial Condition" on my output node to "Zero" volts.

Now, I observe that at 63-Ohms the frequency switches from what it is supposed to be to a high frequency of 16-GHz.
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Jacki
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Re: Ring Oscillator (Tuned Load stage)
Reply #9 - Mar 14th, 2019, 1:27am
 
when we look at the oscillator, we just look at the gain and phase shift. since you add a zero at the output, it will compensate your phase shift. since you have 13 inverters, if no zero, each stage will shift the phase by 180/13, to reach this phase shift, your frequency doesn't need to be high. but when you add the phase compensation, the lower frequency of the zero, the more phase compensation, hence I guess you will need to find the new phase shift/delay, so your oscillating frequency is becoming higher. But I don't know why your oscillating frequency is so sensitive when you change the resistor from 42Ohm to 43Ohm. Have you tried to remove the resistor, and look at the oscillating frequency with capacitor only? Also how big your capacitor is?
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Re: Ring Oscillator (Tuned Load stage)
Reply #10 - Mar 14th, 2019, 10:20pm
 
Here is a plot of the delay through each inverter stage with varying value for the fixed resistor.

The curve is smooth. There exists no cliff.

However when this same inverter is used in the RO the frequency does not vary smoothly with resistance.
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cktdesigner
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Re: Ring Oscillator (Tuned Load stage)
Reply #11 - Mar 14th, 2019, 10:27pm
 
Here is the plot of the frequency of oscillation as it is supposed to vary with the resistance.

Here F_osc = 1/(2*T_Delay)
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Jacki
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Re: Ring Oscillator (Tuned Load stage)
Reply #12 - Mar 17th, 2019, 1:35am
 
I have still the same question, when you remove your resistor, what is the oscillating frequency? Based on your delay plot, when you increase the resistor value, like 250Ohm, what is your oscillating frequency?
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Re: Ring Oscillator (Tuned Load stage)
Reply #13 - Mar 17th, 2019, 1:40am
 
Jacki wrote on Mar 17th, 2019, 1:35am:
I have still the same question, when you remove your resistor, what is the oscillating frequency? Based on your delay plot, when you increase the resistor value, like 250Ohm, what is your oscillating frequency?


When I set the resistor to zero (i.e. remove the resistor and have only the capacitor) the oscillation frequency varies smoothly with varying capacitor value.

When I increase the resistor value above 43ohms the frequency of oscillation stays at 13GHz. For 42ohms and below it stays at a low value of 550MHz.
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Re: Ring Oscillator (Tuned Load stage)
Reply #14 - Jun 20th, 2019, 5:07am
 
I think you add the zero to compensate your original pole, hence to get enough phase for the feedback, the frequency has to be higher.
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