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Question About "attenuation Pad"... - RF Cafe Forums

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

Amateur Radio | Antennas | Circuits & Components | Systems | Test & Measurement


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.


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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


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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.


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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.


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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


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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)


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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


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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.



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