RF Design 1

[actlitedeuter]

Imagine this circuit, a source in series w/ a 3db pad (50 ohm) connected to a resistor (50ohm) via ground. Will the attenuation be 3 dB? How will I compute for the attenuation taking account the 50 ohm resistor?

 

[VEBill]

Attuators are normally calibrated for a certain input and output impedance (often 50 ohms). Your description sounds like the output is properly loaded with 50-ohms, so the attenuation of the pad should be as rated (3dB).

This all assumes that your description is accurate and complete.

If you're misusing the word 'pad' to describe a simple 50-ohm series resistor, then no. In that case you've built a 2:1 voltage divider.

 

[logbook]

Whilst I agree with VE1BLL, your question raises concerns.

You are going from a 3dB pad to a resistor. How? If this is at 1kHz fine. If it is at >100MHz then not fine.

If you are taking an output from this construction how are you doing it? Again this pickoff is critical at higher frequencies. And then of couse, what is the actual load? If the load has significant capacitance, then the attenuation will not be as expected.

People normally use 50 ohm attenuators when working at high frequencies. Using discrete resistors as terminations at high frequencies is ill-advised.

 

[actlitedeuter]

I'm talking of a 3 dB attenuator. Isn't the 3 dB attenuation increase to 6 dB because of the presence of the resistor to ground? Is there a way to compute this?

 

[VEBill]

If the 50-ohm load resistor is the ONLY load, then the 3dB (50-ohm) pad will act like a 3dB pad. If there is another load that you haven't mentioned, then your concern is justified.

Is there another load beyond the one 50-ohm load resistor that you've mentioned?

 

[jimkirk]

Presumably your source has a 50 ohm output impedance. When you connect it to a matched load (50 ohms), that's when you get the specified output level. Inserting a 3 dB pad of the proper impedance (50 ohms, as seen looking into each port) will reduce the signal at the load impedance by 3 dB from what it was without the pad.

RFcafe has a handy collection of online calculators you might find useful.

http://www.rfcafe.com/references/calculator_links.htm

 

[pstuckey]

There are multiple attenuator topologies. For a 3 dB Tee type pad you would have

in 8.55 ohm----------8.55 ohm out
|
|
141.93 ohm
|
ground

Peter

 

[Higgler]

If this is really the RF world, you have to mention if the 50 ohm resistor is the end of the transmission line, or is there another 50 ohm line that continues past this 50 ohm resitor. Big difference.

If you just end with the 50 ohm resitor, then the power supplied to the resistor is 3 dB down from the full power available from the source (assumes low source impedance).

If you have a circuit trace/transmission line that extends beyond the 50 ohm resistor, that's different. The whole circuit is a source, with 3 dB pad (3 resistors), a 50 ohm transmission line with a load at the end of it of; 50 ohm resistor in parallel with a transmission line & end load (probably another 50 ohms). Hence you sort of have two 50 ohm loads, or 25 ohms. In reality, your end section of a transmission line plus load has to be transformed (with a simple formula) to an equivalent impedance and placed in parallel with your 50 ohm resistor, then that is the load for your 50 ohm transmission line.

Another way to look at it, if you have a 50 ohm transmission line and place a 50 ohm resistor at the end, that's very different from a 50 ohm resistor followed by another transmission line. Just putting a 50 ohm resistor somewhere in a transmission line doesn't absorb all the power. Seems like it might, but it doesn't. Placing a short circuit instead of the 50 ohm resistor will block any power from going down the transmission line, hence a 50 ohm replacing this short circuit can't also block all the power from going down the transmission line too.

kch