RF Design
RF Design
(OP)
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?
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RE: RF Design
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.
RE: RF Design
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.
RE: RF Design
RE: RF Design
Is there another load beyond the one 50-ohm load resistor that you've mentioned?
RE: RF Design
RFcafe has a handy collection of online calculators you might find useful.
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RE: RF Design
in 8.55 ohm----------8.55 ohm out
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141.93 ohm
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ground
Peter
RE: RF Design
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