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Crystal Equivalent Circuit - RF Cafe Forums
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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.
Thanks for your help
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 ?
Thanks for your help
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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