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 40m Log Not logged in  Tue Jan 19 02:39:57 2010, kiwamu, Update, Electronics, design complete --- triple resonant circuit for EOM ---   Tue Jan 19 03:20:28 2010, Koji, Update, Electronics, design complete --- triple resonant circuit for EOM ---
Message ID: 2525     Entry time: Tue Jan 19 02:39:57 2010     Reply to this: 2528
 Author: kiwamu Type: Update Category: Electronics Subject: design complete --- triple resonant circuit for EOM ---

The design of the triple resonant circuit has been fixed.

I found the optimum configuration, whose gain is still 11 at 55MHz even if there are realistic losses. As I mentioned in the last entry, there are infinite number of the similar solutions to create the same resonant frequencies.

However owing to the effect of the losses, the resultant gain varies if the similar solution changes

The aim of this study is to select the optimum solution which has a maximum gain ( = the highest impedance at the resonance ).

In order to handle the losses in the calculation, I modeled the loss for both inductors and the capacitors.

Then I put them into the circuit, and calculated the impedance while changing the solutions.

(method)

1). put the scaling parameter as k in order to create the similar solution.

2). scale the all electrical parameters (L1, L2,...) by using k, so that C1'=C1 x k, L1'=L1/k ,...

3). Insert the losses into all the electrical components

4). Draw the impedance curve in frequency domain.

5). See how the height of the impedance at the resonance change

6). Repeat many time this procedure with another k.

7). Find and select the optimum k There is a trick in the calculation.

I put a capacitor named Cpp in parallel to the EOM in order to scale the capacitance of the EOM (see the schematic).

For example if we choose k=2, this means all the capacitor has to be 2-times larger.

For the EOM, we have to put Cpp with the same capacitance as Cp (EOM). As a result these two capacitors can be dealt together as 2 x Cp.

So that Cpp should be Cpp = (k-1) Cp, and Cpp vanishes when we choose k=1.

The important point is that the scaling parameter k must be greater than unity, that is k > 1.

This restriction directly comes from Cp, the capacitance of the EOM, because we can not go to less than Cp.

If you want to put k < 1, it means you have to reduce the capacitance of the EOM somehow (like cutting the EO crystal ??)

(loss model)

I've modeled the loss for both the inductors and the capacitors in order to calculate the realistic impedance.

The model is based on the past measurements I've performed and the data sheet.

Loss for Capacitor :  R(C) = 0.5 (C / 10pF)^{-0.3} Ohm

Loss for Inductor :    R(L) = 0.1 ( L / 1uH) Ohm

Of course this seems to be dirty and rough treatment.

But I think it's enough to express the tendency that the loss  increase / decrease monotonically as  L / C get increased.

These losses are inserted in series to every electrical components.

( Note that: this model depends on both the company and the product model. Here I assume use of Coilcraft inductors and mica capacitors scattered around 40m )

( results )

The optimum configuration is found when k=1, there is no scaling. This is the same configuration listed in last entry

Therefore we don't need to insert the parallel capacitor Cpp in order to achieve the optimum gain.

The figure below shows the some examples of the calculated impedance. You can see the peak height decrease by increasing the scale factor k. The black dash line represents the EOM-loss limit, which only contains the loss of the EOM.

The impedance at the resonance of 55MHz is 6.2 kOhm, which decreased by 3% from the EOM-loss limit. This corresponds to gain of G = 11.

The other two peaks, 11MHz and 29.5MHz dramatically get decreased from EOM-loss limit.

I guess this is because the structure below 50MHz is mainly composed by L1, L2, C1, C2.

In fact these components have big inductance and small capacitance, so that it makes lossy.

( next step )

The next step is to choose the appropriate transformer and to solder the circuit.

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