JB wrote:
Mouser has metal oxide 5 W resistors for $0.49 quantity 1.

You don't really want to run them at full rating but you will likely
have to parallel two to get the values you need anyway unless you have
a cheap source of better than 5 W precision resistors.

You need two 14.01 Ohm resistors and one 82.24 Ohm resistor to build
the attenuator.

If you parallel two 27 Ohm for 13.5 Ohm, and two 150 ohm for 75 Ohm, you
wind up with an attenuation of 4.95 db and a VSWR of 1.04 for a 50 Ohm
system.



Thank-you. That's probably how I'll do it.

Pretty good choice. I hate to buy from mail order suppliers because they
often have a minimum order or minimum handling charge. I prefer to use what
I can, especially for evaluation of a circuit.

The formula for resistors in parallel is R = reciprocal of :
sum of reciprocals of the selected resistors.

So: R= reciprocal of 1/330+1/330+1/330+1/33=
reciprocal of 4/330= 330/4 = 82.5 ohm

for 4 ea 330 ohm 1 watt resistors for an 82.5 ohm 4 watt resistor

Or for equal value resistors: The individual resistor values divided by
the number of resistors

Thanks for the exercise! I leave it to the rest to critique and cross check
my math.



If you go to RFcafe.com they have lots of on line calculators. I keep a
scientific calculator on my desk, but the web is much faster. I have
shortcuts to cable attenuation calculators, Ohm's Law calculators, LED
resistor calculators, etc. I have a wall chart of dBm vs Volts at 50 Ohms.

The above formula also works for capacitors in series. I knew that in
my head.