how simulate Q factor in ADS simulator?

Dear pancho_hideboo,

Quote pancho_hideboo: Return "No" multiplied by hundred to you. You can't understand what is his confusion's ponit.

Thank you for your nice and polite response. I''ll try to improve my understanding with your help.

In your post #16 you have mentioned your possible "point of confusion":
You are still confusing Q of resonator and Q of inductor.


I have to excuse myself not to share your opinion.
Perhaps I may explain my position to you (if you are interested):
The originator of the thread has presented an active complex two-pole to us (with L, C, R) and he has asked for the quality factor of the whole circuit, which - as you have stated correctly - can be seen as a black box.
Now my question: Do you think there is any definition for a Q-factor other than Q=Im(Zin)/Re(Zin) - as mentioned in my most recent posting? I think, there can be no "confusion" since it is not possible to measure anything else - regarding a possible quality factor - than this ratio.
By the way: What do you mean with "Q of inductor"?
I think that there is no inductor accessible in the active circuitry that could be measured or simulated.
And I repeat: I cannot see any reason for a possible confusion regarding the Q-factor since there is only one single definition that makes sense.
In case I am wrong I am grateful to you if you could improve my knowledge in this area.

With regards
LvW

Remark: I just have noticed that pancho_hideboo has removed some of his former postings. Therefore the number of my above reference (#16) is not valid anymore. Sorry.

LvW said:

By the way: What do you mean with "Q of inductor"?

Click to expand...

Surely read my append.

pancho_hideboo said:

You have to evaluate Q of inductor.
Strictry speaking, you have to evaluate Q of effective inductor or Q of equivalent series inductive component of DUT.

Click to expand...

LvW said:

Now my question: Do you think there is any definition for a Q-factor other than Q=Im(Zin)/Re(Zin) - as mentioned in my most recent posting?

Click to expand...

(1) Q as resonator which is defined as ratio of reactive energy and loss energy at resonance.
This Q is calulated by f0/BW3dB.


(2) Q of equivalent series inductive component which is calculated by Q=imag(Zin)/real(Zin).
The Designer's Guide Community Forum - Inductance extraction: L reduce with frequency?

For rezaee's purpose, solution has to be (2).

pancho_hideboo, that's your answer?
To be honest: For my opinion, it was you who was the cause of some confusion.
You have asked "Q of the resonator or Q of the inductor?"
And this was the background for rezaee's confusio and his helpless question "what shall I do?"
I still hope that you tell us how to find the Q factor for the L-part within the active circuit.

To rezaee: I still recommend to follow my former posting (after re-numering now #19)

LvW said:

pancho_hideboo, that's your answer?
To be honest: For my opinion, it was you who was the cause of some confusion.
You have asked "Q of the resonator or Q of the inductor?"
And this was the background for rezaee's confusio and his helpless question "what shall I do?"
I still hope that you tell us how to find the Q factor for the L-part within the active circuit.

Click to expand...

No, still you can't understand his confusion.

See his following append.

rezaee said:

@ E-design : it is just an equation my question is how I simulate in ADS!?

@ pancho_hideboo: firstly I want to use this equation (Q=imag(Zin)/real(Zin)) but I saw it is usable just for RL circuit beacuse Zin=R+LWj and then Q=image(Zin)/real(Zin)=WL/R but we know in a pure parallel RLC circuit Q is equal to W0/2alfa or better I say Q=RCW0 , but if we calculate Q with image(Zin)/real(Zin) causes to different result!!

Click to expand...

He compares Q of equivalent series inductive component with Q of resonator.

LvW said:

I still hope that you tell us how to find the Q factor for the L-part within the active circuit.

Click to expand...

I'm optimistic, nevertheless.

rezaee said:

my final goal is designing an active inductor with high Q, but now I am getting started to do that , firstly I read several thesis,conf and journal papers and patents about this subject and see at first all of them introduced this basic circuit as a conventional active inductor but each of them present different definition of Q , for this reason I got confused beacuse at all of them the circuit is one thing and the goal is one thing (designing active inductor with high Q ) but different definition of Q are used!!

Click to expand...

Hello rezaee,

is your problem solved in the mean time or do you need further advice?
I am sorry for the discussion that took place because I am afraid it was not really helpful for you.

LvW

Thanks in advance LvW and pancho_hideboo, I read your conversation and I learn many things, at first I had confused beacuse I mixed up definition of Q but now I learn that the Q-factor of active inductor circuit(black box) can be calculated as Image(Zin)/Real(Zin), that this Q at resonance frequency is equal to zero
LvW yes you tell true this circuit cant provide very high Q but I am newbie and its basic circuit and after study this I want to apply a technique to this circuit to increase the Q.

rezaee said:

Thanks in advance LvW and pancho_hideboo, I read your conversation and I learn many things, at first I had confused beacuse I mixed up definition of Q but now I learn that the Q-factor of active inductor circuit(black box) can be calculated as Image(Zin)/Real(Zin), that this Q at resonance frequency is equal to zero
LvW yes you tell true this circuit cant provide very high Q but I am newbie and its basic circuit and after study this I want to apply a technique to this circuit to increase the Q.

Click to expand...

Rezaa, your fault was to present a passive LRC circuit to us without mention that this was an equivalent to an active inductor.
In such a passive circuit - of course - one can discuss if the Q shall be applied to the inductive path only or something else.
However, only in posting #14 you told us about the active inductor.
Because this active circuit does not allow any access to the corresponding components of the equivalent passive circuit it is clear that you can use only the two terminals that are available for some measurements resp. simulations.
More than that, we do not need the equivalent passive circuit at all.
The situation is as follows: You have an active circuit that shall simulate the behaviour of an inductor (as good as such a simple circuit can fulfill this task). From this, it is clear that you have to use the only Q definition that
can be applied and that makes sense:Q=Im(Zin)/R(Zin).
This can be simulated easily by grounding the circuit at one node and feeding the circuit with an ac current source (Iac=1A).
Then, the voltage Vo that is created across the circuit is identical to Zin. And you can display the ratio Im(Vo)/R(Vo).
Just one remark: Why do you expect that Q=0 at the resonance frequency? This is not the case.
As a result, you will see that the circuit may simulate an inductor (that means Zin with inductive behaviour) for lower frequencies and will deviate from this inductive behaviour for rising frequencies.

LvW

LvW said:

Just one remark: Why do you expect that Q=0 at the resonance frequency?
This is not the case.

Click to expand...

No.

Even for active inductor, self resonance will occur.
Q will be zero at self resonance frequency.

pancho_hideboo said:

No.

Even for active inductor, self resonance will occur.
Q will be zero at self resonance frequency.

Click to expand...

Yes, you are right. The inductor Q - other tan the Q of the resonant circuit - goes to zero.