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

56/4 = 14 so a 14 ohm resistor can be made with 4 ea 56 ohm resistors.

Now lets check the work:
If the exciter is 20 watts and will see 50.04 ohms.
The 50 of the amp input + 14 ohms is 64 ohms.
If we parallel that with the 82.5 ohm...
For 2 resistors in parallel R=( r1 x r2)/ (r1+ r2) = 36.04 ohm plus the 14
ohm in series =50.04 ohm to the input of the network.

So the above values are correct for a T pad.

Go to Ohms and Kirchoff's laws to see the current, voltage and power
dissipation of the components:
P=I squared R there is 632.5 ma through the 1st 14 ohm resistor for 5.6
watts,

E=IR= 8.855 volts drop for the first 14 ohm resistor.

The voltage for 20 watts at 50 ohms is SqRt of PR so
31.62 volts. The rest of the network after the 14 ohm resistor is 22.77v.
The 82.5 watt resistor will see 6.283 watts of dissipation.

Now current through the 14 ohm output resistor and 50 ohm input of the amp
will be .3558amp

By now you should have a copy of ohms law around.

1.77 watts is dissipated through the second 14 ohm resistor and 6.33 to the
amp. You will notice all the dissipation of the amp and resistors adds up
to 20 watts.

The identical resistors in parallel will divide the power dissipation
equally.

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

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.