How will the FOM of an LC oscillator change?

[liusupeng]

suppose i have a LC oscillator and there is a series R with the inductor. If i simply half the capacitor, how will the FOM of the LC oscillator change?
assume that the oscillation amplitude is kept the same and ignore all other sources of noise. only noise from the series R is considered.
FOM=L(f)*fm*fm*Power/(fout*fout)

 

[KerimF]

Re: FOM of LC oscillator

Since long time I didn't think of our dear FOM for RLC circuit.
Let us think together to solve this ambiguity.
First FOM is a ratio of energies and not powers.

By assuming that L and C are parallel and Rs is in series with L, we can say:
FOM = (maximum energy stored in C or L ) / (energy dissipated in Rs in one cycle)
Using abbreviation:
FOM = MES / ED

Case 1:
MES_1 = 1/2 * C1 * Vp^2
Where Vp is the tank peak voltage

Case 2:
MES_2 = 1/2 * C2 * Vp^2 = 1/2 * C1 / 2 * Vp^2
Where Vp is the same as in case 1

Now let us find out the energy dissipated in Rs.
Since we fixed here the tank voltage, it is better to calculate first the equivalent parallel resistance Rp of Rs.
As we know:
(wL) / Rs = Rp / (wL) => Rp = (wL)^2 / Rs

Case 1:
Rp1 = (w1*L)^2 / Rs
P_Rp1 = 1/2 * V^2 * 1/Rp1 = 1/2 * V^2 * Rs / [2*PI*L / T1]^2
ED_1 = P_Rp1 * T1 = 1/2 * V^2 * Rs * T1 / [2*PI*L / T1]^2

Case 2:
Rp2 = (w2*L)^2 / Rs
P_Rp2 = 1/2 * V^2 * 1/Rp2 = 1/2 * V^2 * Rs / [2*PI*L / T2]^2
ED_2 = P_Rp2 * T2 = 1/2 * V^2 * Rs * T2 / [2*PI*L / T2]^2


FOM_2 / FOM_1 = (MES_2/MES_1) / (ED_2/ED_1)

MES_2/MES_1 = [1/2 * C1 / 2 * Vp^2] / [1/2 * C1 * Vp^2]
MES_2/MES_1 = 1/2

ED_2/ED_1 = {1/2 * V^2 * Rs * T2 / [2*PI*L / T2]^2} / {1/2 * V^2 * Rs * T1 / [2*PI*L / T1]^2}
ED_2/ED_1 = ( T2 / T1 ) ^ 3

I expect some help on your side... so please let me know if anything doesn't sound logical to you so far.

It is time to talk about the resonance frequencies f1 and f2 (w1 and w2):
As you know:
w = 2*PI*f = 1 / SQRT(L*C)
so
w1 = 1 / SQRT(L*C1)
w2 = 1 / SQRT(L*C2) = 1 / SQRT(L*C1/2)
This gives:
w2/w1 = SQRT(2)
but
w2/w1 = 2*PI*f2 / 2*PI*f1 = (2*PI/T2) / (2*PI/T1) = T1/T2
or
T2/T1 = 1 / SQRT(2)
ED_2/ED_1 = ( T2 / T1 ) ^ 3 = 1/2*1/SQRT(2)

Finally:
FOM_2 / FOM_1 = (MES_2/MES_1) / (ED_2/ED_1)
FOM_2 / FOM_1 = 1/2 / [1/2*1/SQRT(2)]
FOM_2 / FOM_1 = SQRT(2)

Please remember to check this work with yours and other's work... This is just what came to my mind so... :wink:

Kerim

 

[liusupeng]

Re: FOM of LC oscillator

Since a long time I didn't think of our dear FOM for RLC circuit.
Let us think together to solve this ambiguity.
First FOM is a ratio of energies and not powers.

By assuming that L and C are parallel and Rs is in series with L, we can say:
FOM = (maximum energy stored in C or L ) / (energy dissipated in Rs in one cycle)
Using abbreviation:
FOM = MES / ED

Case 1:
MES_1 = 1/2 * C1 * Vp^2
Where Vp is the tank peak voltage

Case 2:
MES_2 = 1/2 * C2 * Vp^2 = 1/2 * C1 / 2 * Vp^2
Where Vp is the same as in case 1

Now let us find out the energy dissipated in Rs.
Since we fixed here the tank voltage, it is better to calculate first the equivalent parallel resistance Rp of Rs.
As we know:
(wL) / Rs = Rp / (wL) => Rp = (wL)^2 / Rs

Case 1:
Rp1 = (w1*L)^2 / Rs
P_Rp1 = 1/2 * V^2 * 1/Rp1 = 1/2 * V^2 * Rs / [2*PI*L / T1]^2
ED_1 = P_Rp1 * T1 = 1/2 * V^2 * Rs * T1 / [2*PI*L / T1]^2

Case 2:
Rp2 = (w2*L)^2 / Rs
P_Rp2 = 1/2 * V^2 * 1/Rp2 = 1/2 * V^2 * Rs / [2*PI*L / T2]^2
ED_2 = P_Rp2 * T2 = 1/2 * V^2 * Rs * T2 / [2*PI*L / T2]^2


FOM_2 / FOM_1 = (MES_2/MES_1) / (ED_2/ED_1)

MES_2/MES_1 = [1/2 * C1 / 2 * Vp^2] / [1/2 * C1 * Vp^2]
MES_2/MES_1 = 1/2

ED_2/ED_1 = {1/2 * V^2 * Rs * T2 / [2*PI*L / T2]^2} / {1/2 * V^2 * Rs * T1 / [2*PI*L / T1]^2}
ED_2/ED_1 = ( T2 / T1 ) ^ 3

I expect some help on your side... so please let me know if anything doesn't sound logical to you so far.

It is time to talk about the resonance frequencies f1 and f2 (w1 and w2):
As you know:
w = 2*PI*f = 1 / SQRT(L*C)
so
w1 = 1 / SQRT(L*C1)
w2 = 1 / SQRT(L*C2) = 1 / SQRT(L*C1/2)
This gives:
w2/w1 = SQRT(2)
but
w2/w1 = 2*PI*f2 / 2*PI*f1 = (2*PI/T2) / (2*PI/T1) = T1/T2
or
T2/T1 = 1 / SQRT(2)
ED_2/ED_1 = ( T2 / T1 ) ^ 3 = 1/2*1/SQRT(2)

Finally:
FOM_2 / FOM_1 = (MES_2/MES_1) / (ED_2/ED_1)
FOM_2 / FOM_1 = 1/2 / [1/2*1/SQRT(2)]
FOM_2 / FOM_1 = SQRT(2)

Please remember to check this work with yours and other's work... This is just what came to my mind so... :wink:

Kerim

Dear Kerim,
Thanks a lot for your detailed analysis. It seems that you are deriving the Q of the LC tank. But for FOM of an oscillator, is it the same as the Q of the LC tank?

 

[KerimF]

Re: FOM of LC oscillator

Q = wL/Rs = Rp/wL

It seems you are right. In this case:
Q2/Q1 = SQRT(2)

Is it perhaps just a coincidence for this particular case?
I am glad it is your homework not mine :grin:

 

[liusupeng]

Re: FOM of LC oscillator

Q = wL/Rs = Rp/wL

Dear Kerimf,
I understand Q = wL/Rs = Rp/wL. but according to my knowledge, FOM=L(f)*fm*fm*Power/(fout*fout). where L(f) is the phase noise power spectral density, fm is the frequency offset where the phase noise is measure and fout is the output frequency.
what i am interested is actually FOM of the oscillator.

 

[KerimF]

Re: FOM of LC oscillator

To be honest... when I was a graduate student 35 years ago, I didn't have the chance to be asked about FOM... you have more luck than I, these days

Hope someone familiar with this noisy formula be ready to help you. On my side, by curiosity I will try to read about it :wink:

Did you read the thread:
https://www.edaboard.com/threads/54065/

 

[liusupeng]

Re: FOM of LC oscillator

To be honest... when I was a graduate student 35 years ago, I didn't have the chance to be asked about FOM... you have more luck than I, these days

Hope someone familiar with this noisy formula be ready to help you. On my side, by curiosity I will try to read about it :wink:

Thanks kerimf

 

[KerimF]

Re: FOM of LC oscillator

So far... I got 4 definitions for FOM (related to electronics)... but I keep trying

Just in case you didn't notice it on the previous post:
https://www.edaboard.com/threads/54065/