# Crystal Equivalent Circuit - RF Cafe Forums

The original RF Cafe Forums were shut down in late 2012 due to maintenance issues. Original posts:

gui076
Post subject: T Divider after VCO Posted: Tue Jul 08, 2008 6:13 am

Captain

Joined: Mon May 05, 2008 9:56 am
Posts: 15
Location: France
Hello all,

I have seen on the evaluation board schematics of analog device synthesizer that there was a T network with three 18R.

After a little search i have seen that it seems to be a T power divider.

Datasheet page number 4:

http://www.analog.com/Analog_Root/stati ... 8EB1_a.pdf

The inputs to the synthesizer have a maximum level so my question is.
How is distributed the power on this T junction ? after some research i have supposed it was a 6dB power divider but i'm not sure.

Guillaume

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nubbage
Post subject: Posted: Tue Jul 08, 2008 11:15 am

General

Joined: Fri Feb 17, 2006 12:07 pm
Posts: 218
Location: London UK
Hi Guillaume
I think this is sometimes called a pseudo-Wilkinson combiner.
If you imagine the R11 and the 50 ohms Z at J4 are in effect series connected, although grounded at the centre, then it can be re-examined as a combiner where the voltage across R11 and the input at J4 are combined to give the output as the sum at J5.
Just why the resistors are 18 ohms beats me for the moment.

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yendori
Post subject: Posted: Tue Jul 08, 2008 4:39 pm

General

Joined: Thu Sep 25, 2003 1:19 am
Posts: 50
Location: texarcana
Zo=50 R=18

18+ 50>
> 18
18+50 >

(50+18//50+18)+18=50ish

6dB Loss = 3 dBdiss + 3dBsplit

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gui076
Post subject: Posted: Wed Jul 09, 2008 2:49 am

Captain

Joined: Mon May 05, 2008 9:56 am
Posts: 15
Location: France
I have not well understood the calculation

If i have this network after my VCO does it mean that i will have 6 db of attenuation on each output ?

Guillaume

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nubbage
Post subject: Posted: Wed Jul 09, 2008 4:07 am

General

Joined: Fri Feb 17, 2006 12:07 pm
Posts: 218
Location: London UK
Hi Gui
The network operates in the reverse: 10mW from the VCO is added to say 10mW coming into J4, and the result is 20mW at J5 less 3dB of dissipation loss (assuming Yendori's analysis is OK, which it always is) = 10mW out of J5.

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gui076
Post subject: Posted: Wed Jul 09, 2008 9:17 am

Captain

Joined: Mon May 05, 2008 9:56 am
Posts: 15
Location: France
Hi,

Ok so if i have my VCO with 4 dBm at the output, i will have 4 dBm at J5 ?

Guillaume

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yendori
Post subject: Posted: Wed Jul 09, 2008 11:31 am

General

Joined: Thu Sep 25, 2003 1:19 am
Posts: 50
Location: texarcana
What Nubbage said is correct, but I think maybe there is confusion about the input and outputs.

The VCO is not shown in the schematic. The VCO output applied to J4. which is then split equally by 6dB. One continues to the output, the other applied to the PLL/Divider.

Its true if you combined two 0dBm signals in-phase with a resistive power, the power would add by 3dB. You still have your resistive dissipatation of 3dB, totalling 0dBm. If it was a Wilkinson you would get +3dBm.

The equation for a resistive power splitter is very simple, but not easy to show via blog window.

Draw a resistive power splitter, three 18ohms, including the 50ohm output load resistors.

Start from one output, add the load resistor (50) + the resistor in series with it(18)=68.

Now start from the other output, 50+18=68. Now add them in parallel, 68//68=34.

Now you are left with the last 18ohm (input resistor), which adds in series. 34+18=52. So the input looking into the power splitter will see a good load.

Rod

Posted  11/12/2012

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