The first design of the triple resonant EOM circuit has been done.
If only EOM has a loss of 4 Ohm, the gain of the circuit is expected to be 11 at 55MHz
So far I've worked on investigation of the single resonant circuit and accumulated the knowledge about resonant circuits.
Then I started the next step, designing the triple resonant circuit.
Here I report the first design of the circuit and the expected gain.
( What I did )
At first in order to determine the parameters, such as inductors and capacitors, I have solved some equations with numerical ways (see the past entry).
In the calculation I put 6 boundary conditions as followers;
(first peak=11MHz, second peak=29.5MHz, third peak=55MHz, first valley=22MHz, second valley=33MHz, Cp=18pF)
The valley frequencies of 22MHz and 33MHz are chosen in order to eliminate the higher harmonics of 11MHz, and Cp of 18pF represents the capacitance of the EOM.
Basically the number of parameters to be determined is 6 ( L1, L2, ...,), therefore it is completely solved under 6 boundary conditions. And in this case, only one solution exists.
The point is calculation does not include losses because the loss does not change the resonant frequency.
( results )
As a result I obtained the 6 parameters for each components shown in the table below.
Then I inserted the loss into the EOM to see how the impedance looks like. The loss is 4 Ohm and inserted in series to the EOM. This number is based on the past measurement .
Let us recall that the gain of the impedance-matched circuit with a transformer is proportional to square-root of the peak impedance.
It is represented by G = sqrt(Zres/50) where Zres is the impedance at the resonance.
As you can see in the figure, Zres = 6.4 kOhm at 55MHz, therefore the gain will be G=11 at 55MHz.
Essentially this gain is the same as that of the single resonant circuit that I've been worked with, because its performance was also limited mainly by the EOM loss.
An interesting thing is that all three peaks are exactly on the EOM limited line (black dash line), which is represented by Zres = L/CR = 1/ (2pi f Cp)^2 R. Where R = 4 Ohm.
( next plan )
There are other solutions to create the same peaks and valleys because of the similar solution.
It is easy to understand if you put Cp' = Cp x N, the solutions must be scaled like L1'=L1/N, C1'=C1 x N, ..., Finally such scaling gives the same resonant frequencies.
So the next plan is to study the effect of losses in each components while changing the similar solution.
After the study of the loss I will select an optimum similar solution.