VCO design problem

Dear asrf,

asrf said:

hen I use this kind of biasing and sweep my ideal capacitor to change the frequency, freq varies but VCO output swing changes as well !!

This is not an unusual issue aarf. What you need to consider is that the steady-state amplitude of the VCO is governed by the amplitude that provides a negative real impedance that is equal to the absolute value of the resonator - in your case inductor and varactor. Hence, as you change your ideal capacitor, you are changing its impedance which will require the sustaining amplifer to adjust the amplitude of its waveform. 

asrf said:

nd also I dont want to want full swing for this VCO since if I have full swing the Vgd max will be high

I want to have swing of 1v pp so:

VD swing : (0.9 --> 1.9)

VG swing : (0.2 --> 1.2)

Without an AGC, it will be very hard to set the amplitude to a fixed amplitude (1 V in your case as you state) over all process and environments AND over all frequencies of VCO operation.

I hope this brief explanation provide some intuitive understanding of what you are observing aarf,

Shawn

Thank you for explanation but I am confused.

The voltage swing depends on your current and also the Rp of tank which is equal to Rp of inductor.

I have defined a fixed Rp for my inductor so at resonance my C and L are open and what I have is Rp.

So the swing in theory should be 4*Id*Rp/pi.

So for each transistor when it is on, it gives the Id which flows through a constant Rp. I still cannot figure out what really changes here.

Rp is defined as Lw0/Q  and this value in my circuit does not change with changing the frequency. 

It would be amazing and I would be really grateful if you explain it in details. Thank you for your time.

Dear asrf,

asrf said:

Thank you for explanation but I am confused.

Unfortunately, VCO circuits can be be very subtle and, although they appear quite simple, have kept designers entertained for just over a century now!

asrf said:

I have defined a fixed Rp for my inductor so at resonance my C and L are open and what I have is Rp.

So the swing in theory should be 4*Id*Rp/pi.

A VCO  with a "real" sustaining amplifier (such as you are using) will not operate at a frequency exactly as the resonance of discrete L and C you included in your schematic. Basically, you are neglecting the MOS capacitances and resistances of your sustaining amplifier and its trace parasitics. A way to estimate (and only estimate) the frequency your VCO will operate is to perform a set of small-signal (large signal is more accurate) negative resistance simulations where you separate the LC tank from the sustaining amplifer and drivie the nodes that the resonator is connected to with an ideal current source. Examine the impedance (real and imaginary parts) of the sustaining amplifier (i.e., the impedance it presents to the LC resonator that was connected to those nodes). You may estimate (and I stress this is only an estimate) the final operating, steady-state frequency, by changing the frequency of your analysis results to find the frequency at the the imaginary part of the sustaining amplifer is equal and opposite to the imaginary part of the LC resonator. 

You will also note that the real impedance of the sustaining amplifier is not constant with frequency. Hence, as you change your ideal C, the amount of gain the sustaining amplifier will chance with frequency. This means that the steady-state amplitude of the VCO will vary over frequency and not be simply given by the expression "4*Id*Rp/pi.

I know there is a lot I've "packed" into this response and it will take some time to digest, but I wanted to as clear as I could to help you understand the issue.

I am also attaching a response I provided a number of years back to a similarly phrased question as yours. There is a bit more background and a link to a prior post with an example test bench to study the impedance of your resonator.

Shawn

--------------Post from 8/23/2019------------

Dear Tom,

Tom Brown said:
i want to simulate the frequency response of the tank impddance.
If you are interested in the tank impedance, I might suggest that in lieu of simulated the S parameters, you simulate the impedance (i.e., real and imaginary parts). This will also provide a much more intuitive view of the required negative resistance required from the sustaining amplifier to assure steady-state oscillation. Basically, separate the sustaining amplifier from the tank impedance (the latter can include any varactors if desired). You can then directly simulate the small-signal impedance of the tank impedance over frequency. If you are including the varactors in the tank impedance, you will need to simulate the impedance over the full range of the control voltage used to vary the varactor impedance.

If you can set the operating point of the sustaining amplifier to its large signal operating point at start-up, you may then also determine its small-signal impedance (at start-up). As you correctly noted, if the sustaining amplifier's negative real impedance exceeds in magnitude the real impedance of the tank at a given frequency, then as oscillation builds, the negative impedance of the sustaining amplifier will decrease towards 0 until it is equal and opposite in sign to the tank's real impedance.

If you are having trouble devising a test bench to determine the impedance of the tank/varactor network, I might suggest you examine an example at URL:

community.cadence.com/.../capacitance-vs-bias-voltage-curve-for-ferroelectric-varactor

I hope I understood your question correctly and this is somewhat useful Tom,

Shawn

Dear asrf,

asrf said:

Thank you for explanation but I am confused.

Unfortunately, VCO circuits can be be very subtle and, although they appear quite simple, have kept designers entertained for just over a century now!

asrf said:

I have defined a fixed Rp for my inductor so at resonance my C and L are open and what I have is Rp.

So the swing in theory should be 4*Id*Rp/pi.

A VCO  with a "real" sustaining amplifier (such as you are using) will not operate at a frequency exactly as the resonance of discrete L and C you included in your schematic. Basically, you are neglecting the MOS capacitances and resistances of your sustaining amplifier and its trace parasitics. A way to estimate (and only estimate) the frequency your VCO will operate is to perform a set of small-signal (large signal is more accurate) negative resistance simulations where you separate the LC tank from the sustaining amplifer and drivie the nodes that the resonator is connected to with an ideal current source. Examine the impedance (real and imaginary parts) of the sustaining amplifier (i.e., the impedance it presents to the LC resonator that was connected to those nodes). You may estimate (and I stress this is only an estimate) the final operating, steady-state frequency, by changing the frequency of your analysis results to find the frequency at the the imaginary part of the sustaining amplifer is equal and opposite to the imaginary part of the LC resonator. 

You will also note that the real impedance of the sustaining amplifier is not constant with frequency. Hence, as you change your ideal C, the amount of gain the sustaining amplifier will chance with frequency. This means that the steady-state amplitude of the VCO will vary over frequency and not be simply given by the expression "4*Id*Rp/pi.

I know there is a lot I've "packed" into this response and it will take some time to digest, but I wanted to as clear as I could to help you understand the issue.

I am also attaching a response I provided a number of years back to a similarly phrased question as yours. There is a bit more background and a link to a prior post with an example test bench to study the impedance of your resonator.

Shawn

--------------Post from 8/23/2019------------

Dear Tom,

Tom Brown said:
i want to simulate the frequency response of the tank impddance.
If you are interested in the tank impedance, I might suggest that in lieu of simulated the S parameters, you simulate the impedance (i.e., real and imaginary parts). This will also provide a much more intuitive view of the required negative resistance required from the sustaining amplifier to assure steady-state oscillation. Basically, separate the sustaining amplifier from the tank impedance (the latter can include any varactors if desired). You can then directly simulate the small-signal impedance of the tank impedance over frequency. If you are including the varactors in the tank impedance, you will need to simulate the impedance over the full range of the control voltage used to vary the varactor impedance.

If you can set the operating point of the sustaining amplifier to its large signal operating point at start-up, you may then also determine its small-signal impedance (at start-up). As you correctly noted, if the sustaining amplifier's negative real impedance exceeds in magnitude the real impedance of the tank at a given frequency, then as oscillation builds, the negative impedance of the sustaining amplifier will decrease towards 0 until it is equal and opposite in sign to the tank's real impedance.

If you are having trouble devising a test bench to determine the impedance of the tank/varactor network, I might suggest you examine an example at URL:

community.cadence.com/.../capacitance-vs-bias-voltage-curve-for-ferroelectric-varactor

I hope I understood your question correctly and this is somewhat useful Tom,

Shawn

"you are neglecting the MOS capacitances and resistances of your sustaining amplifier and its trace parasitics"

The capacitor I have at the node is not a total capacitor. So I am aware of that Cload + Cparasitic is at resonance with L and I am changing Cload.

I think I understood some part of it. So what you claim is the real part of sustaining amplifier varies with frequency and this changes the overal gain. 

Basically everything changes with frequency and there are too many non ideal effects in this circuit. Thank you very much.

Dear asrf,

asrf said:

I think I understood some part of it. So what you claim is the real part of sustaining amplifier varies with frequency and this changes the overal gain. 

Exactly - very good! Therefore, the steady-state amplitude (which is the point at the the overall gain is exactly unity) will end up being different as the frequency varies.

Hopefully, our discussion has provided a bit more intuitive feel as to what you observing. Good luck,

Shawn

I separated tank and amplifier and checked the impedance.

at 2.86 GHz the imaginary parts are equal with opposite signs and real part of tank is much higher than amplifier negative impedance.

I deliberately multiplied the real part of tank by -1 to have better comparison.  Rp >> |Ramp|

when I simulate this i get 2.4 GHz ! and my swing is max ! I think because the difference between Rp and |Ramp| is huge 

Dear asrf,

asrf said:

when I simulate this i get 2.4 GHz ! and my swing is max ! I think because the difference between Rp and |Ramp| is huge 

The small -signal methodology provides only an approximate estimate of the operating frequency since it does not know about the saturating effects of your sustaining amplifier as the signal amplitude grows. Hence, it is not surprising that your estimate of the steady-state frequency is not very accurate. In addition, the amplitude will be large as the difference in real impedances is large. The large signal amplitude must increase significantly in order for the average overall gain to be equal to unity.

Shawn