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trashbox Post subject: question about "attenuation pad"... Posted:
Mon Jul 18, 2005 11:31 pm Captain Joined: Mon May 24,
2004 11:30 pm Posts: 10 Hi all During the RF testing,such
as the path between 1.2GHz vco signal and divider, there are always
an attenuation pad(pi or Tee resistor network),generally 3~5dB.
Some articles or posts mentioned that it is for isolation and the
others think it is for impedance mismatch. What is role of such
attenuation pad? Would you recommend any URL,materials or references?
Thank you very much. Top IR Post subject:
Posted: Tue Jul 19, 2005 1:34 am Site Admin Joined:
Mon Jun 27, 2005 2:02 pm Posts: 373 Location: Germany Hello
trashbox, The functions of the attenuator are: - To adjust
the level of the signal from the VCO's output to the LO driver's input
port or mixer. In many designs, there is a direct connection between
the output of the LO driver to the mixer's LO port. Therefore, you will
need to go with a specific power level to the LO driver's input (Which
has its own gain). Thus, you will have to do adjustment by means of
an attenuator. - Impedance matching: Sometimes the VCO's output
or the input of the LO driver have poor match (Mostly it will be the
LO driver's input). The function of the attenuator is to provide matching.
Since a pad is a resistive element is provides the best matching. One
dB of attenuation takes the return loss 2dB down. - Provide additional
isolation between the output of the VCO to the LO driver stage.
_________________ Best regards, - IR Top
trashbox Post subject: Posted: Tue Jul 19, 2005 3:07 am
Captain Joined: Mon May 24, 2004 11:30 pm Posts: 10
IR, Thank you very much. Top trashbox
Post subject: Posted: Tue Jul 19, 2005 3:36 am Captain
Joined: Mon May 24, 2004 11:30 pm Posts: 10 Hi IR, Thanks
for your nice help.According to your answer,I think: -- How to
decide the input power requirement of a mixed-signal circuit? For example,
My VCO is used in a PLL,i.e.,the VCO output is connected to a current-mode
logic divider that require the input signal a specific DC voltage and
peak-to-peak value(for example,1.0v<input signal<1.4v DC voltage=1.2v).
I am puzzled how to decide the input power requirement of the divider
if I want to select an attenuator? -- As a matching pad, I think
the attenuators' leg resistors are not equal(suppose the attenuator
is Tee section).Am I right? Thanks again. Top
IR Post subject: Posted: Tue Jul 19, 2005 4:07 am Site
Admin Joined: Mon Jun 27, 2005 2:02 pm Posts: 373 Location:
Germany Hello again trashbox, I Quote: -- As a matching
pad, I think the attenuators' leg resistors are not equal(suppose the
attenuator is Tee section).Am I right? Usually the attenuator
is balanced (Has the same impedance on both sides). This is because
you would like to terminate both sides of the attenuator to the same
impedance (Normally 50 ohm) Quote: -- How to decide the input
power requirement of a mixed-signal circuit? For example, My VCO is
used in a PLL,i.e.,the VCO output is connected to a current-mode logic
divider that require the input signal a specific DC voltage and peak-to-peak
value(for example,1.0v<input signal<1.4v DC voltage=1.2v). I am
puzzled how to decide the input power requirement of the divider if
I want to select an attenuator? The input power requirement
is being calculated according to P=V^2/R (Where R is the load impedance,
again normally 50 ohm). Regarding the DC levels you mentioned, you
probably refer to the new standards as: LVPECL, PECL and similar. There
is a series of buffers, dividers etc manufactured by MICREL www.micrel.com
(And other IC manufacturers) that deal with converting sine signals
(AC-Coupled) to these standards. I think that you should check their
devices. An example to such device: SY89871U Good luck!!
_________________ Best regards, - IR Top
Guest Post subject: Posted: Tue Jul 19, 2005 6:50 am
IR,thank you sincerely. What you mentioned that the attenuator
can finish impedance match. And I also saw this conclusion on h**p://mrtmag.com/mag/radio_rf_attenuators_terminations/.What
I can not undersand is: How the pi-section pad,whose leg resistors
are equal,smooth out the impedance mismatch?(In fact,it is explained
in above URL,but I can not see it since the lack of mentioned figures)
Top IR Post subject: Posted: Tue Jul 19, 2005 7:42
am Site Admin Joined: Mon Jun 27, 2005 2:02 pm
Posts: 373 Location: Germany Hello, I don't know how familiar
you are with Smith Chart and impedance matching, so I would explain
it as follows: As I mentioned before, the pad is a resistive
element. The calculation of resistor values in the pad is done to satisfy
2 conditions: 1) The desired attenuation value (it is done by
means of a voltage divider) 2) The required impedance. By
achieving an impedance close to 50 ohm you will tighten the VSWR up
to a point which is the center of the Smith Chart: 50+j0 (namely pure
resistive 50 ohm without reactive component). The actual impedance based
on reflection coefficient can be calculated by: Zl=Z0*(1-gamma)/(1+gamma)
gamma=reflection coefficient. _________________ Best
regards, - IR Top trashbox Post subject:
Posted: Tue Jul 19, 2005 9:21 pm Captain Joined: Mon
May 24, 2004 11:30 pm Posts: 10 IR, Thanks. I think I understand
it. By the means of reducing gamma by adding an attenuator, we can
have a better impedance match.(closer to the cener of Smith Chart).
Thank you again for your nice help. Please send me a message if
you have any difficulty,I will try my best.
Posted 11/12/2012
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