In rec.radio.amateur.equipment Ian Jackson wrote:
In message , JB
writes

"Ian Jackson" wrote in message
...
In message , dave
writes
Ralph Mowery wrote:
"dave" wrote in message
news:0vqdnaEyq4zYAhfXnZ2dnUVZ_hJi4p2d@earthlink. com...
I need to lose about 5 dB from a 15 Watt exciter. Thanks.
Simple answer is no.
While you can build one with wirewound resistors , the normal
formulars will not usually work at RF and you will have a lot of
inductance to deal with.
Even the socalled non-inductive resistors are only so at the audio
frequencies.


Can the inductive reactance be cancelled with capacitors? I'm mainly
concerned with 40, 30 and 20 meters.

If your highest frequency is only 14MHz, and you're not trying to obtain
a sooper-dooper perfect match for your exciter, then I'd say that you
would certainly get away with using wirewound resistors.

To minimise the effects of the inductance, you could try mounting the
resistors 'hard down' against a ground plane. You might need a sheet of
some insulating material to avoid any danger of shorting to it.

And presumably, to get rid of 5dB (nearly 3/4 of your original 15W) you
intend to use paralleled-up resistors. This in itself will help minimise
the inductance.
--
Ian

A bunch of chip resistors in parallel to ultimately make a T-atten.

Yebbut....
12W is a fair bit to dissipate with chip resistors (even a lot of them).
However, the question was if you could use WW resistors, not 'how to do
it properly'. I presume there was a reason why this was asked! After
all, we ARE talking 'amateur' radio.

A 5db T attenuator with 15 W in has the maximum dissipation in the shunt
resistor at about 5 W.

The input series resistor dissipates about 4 W and the output series
resistor dissipates a little over 1 W.

I see no reason to go to wirewound resistors at these power levels.


--
Jim Pennino

Remove .spam.sux to reply.

In message ,
writes
In rec.radio.amateur.equipment Ian Jackson
wrote:
In message , JB
writes

"Ian Jackson" wrote in message
...
In message , dave
writes
Ralph Mowery wrote:
"dave" wrote in message
news:0vqdnaEyq4zYAhfXnZ2dnUVZ_hJi4p2d@earthlink. com...
I need to lose about 5 dB from a 15 Watt exciter. Thanks.
Simple answer is no.
While you can build one with wirewound resistors , the normal
formulars will not usually work at RF and you will have a lot of
inductance to deal with.
Even the socalled non-inductive resistors are only so at the audio
frequencies.


Can the inductive reactance be cancelled with capacitors? I'm mainly
concerned with 40, 30 and 20 meters.

If your highest frequency is only 14MHz, and you're not trying to obtain
a sooper-dooper perfect match for your exciter, then I'd say that you
would certainly get away with using wirewound resistors.

To minimise the effects of the inductance, you could try mounting the
resistors 'hard down' against a ground plane. You might need a sheet of
some insulating material to avoid any danger of shorting to it.

And presumably, to get rid of 5dB (nearly 3/4 of your original 15W) you
intend to use paralleled-up resistors. This in itself will help minimise
the inductance.
--
Ian

A bunch of chip resistors in parallel to ultimately make a T-atten.

Yebbut....
12W is a fair bit to dissipate with chip resistors (even a lot of them).
However, the question was if you could use WW resistors, not 'how to do
it properly'. I presume there was a reason why this was asked! After
all, we ARE talking 'amateur' radio.

A 5db T attenuator with 15 W in has the maximum dissipation in the shunt
resistor at about 5 W.

The input series resistor dissipates about 4 W and the output series
resistor dissipates a little over 1 W.

I see no reason to go to wirewound resistors at these power levels.

Oh, I quite agree. I wouldn't really choose to use WW myself, but I
think they would work in this application. It's just that I usually try
first to answer the question as asked, and, if necessary, start
embellishing things from then on!
http://www.youtube.com/watch?v=k7vMvlRio5Y&feature=related
--
Ian

In rec.radio.amateur.equipment Ian Jackson wrote:
In message ,
writes
In rec.radio.amateur.equipment Ian Jackson
wrote:
In message , JB
writes

"Ian Jackson" wrote in message
...
In message , dave
writes
Ralph Mowery wrote:
"dave" wrote in message
news:0vqdnaEyq4zYAhfXnZ2dnUVZ_hJi4p2d@earthlink. com...
I need to lose about 5 dB from a 15 Watt exciter. Thanks.
Simple answer is no.
While you can build one with wirewound resistors , the normal
formulars will not usually work at RF and you will have a lot of
inductance to deal with.
Even the socalled non-inductive resistors are only so at the audio
frequencies.


Can the inductive reactance be cancelled with capacitors? I'm mainly
concerned with 40, 30 and 20 meters.

If your highest frequency is only 14MHz, and you're not trying to obtain
a sooper-dooper perfect match for your exciter, then I'd say that you
would certainly get away with using wirewound resistors.

To minimise the effects of the inductance, you could try mounting the
resistors 'hard down' against a ground plane. You might need a sheet of
some insulating material to avoid any danger of shorting to it.

And presumably, to get rid of 5dB (nearly 3/4 of your original 15W) you
intend to use paralleled-up resistors. This in itself will help minimise
the inductance.
--
Ian

A bunch of chip resistors in parallel to ultimately make a T-atten.

Yebbut....
12W is a fair bit to dissipate with chip resistors (even a lot of them).
However, the question was if you could use WW resistors, not 'how to do
it properly'. I presume there was a reason why this was asked! After
all, we ARE talking 'amateur' radio.

A 5db T attenuator with 15 W in has the maximum dissipation in the shunt
resistor at about 5 W.

The input series resistor dissipates about 4 W and the output series
resistor dissipates a little over 1 W.

I see no reason to go to wirewound resistors at these power levels.

Oh, I quite agree. I wouldn't really choose to use WW myself, but I
think they would work in this application. It's just that I usually try
first to answer the question as asked, and, if necessary, start
embellishing things from then on!
http://www.youtube.com/watch?v=k7vMvlRio5Y&feature=related

In general I agree with that, but if I ask a question based on a shaky
or incorrect premise, i.e. "special" resistors would be required for
a 15 W attenuator, I'd prefer to be told my premise was foobar rather
than waste my time chasing after something less than desirable.


--
Jim Pennino

Remove .spam.sux to reply.

Can the inductive reactance be cancelled with capacitors? I'm
mainly
concerned with 40, 30 and 20 meters.

If your highest frequency is only 14MHz, and you're not trying to
obtain
a sooper-dooper perfect match for your exciter, then I'd say that you
would certainly get away with using wirewound resistors.

To minimise the effects of the inductance, you could try mounting the
resistors 'hard down' against a ground plane. You might need a sheet
of
some insulating material to avoid any danger of shorting to it.

And presumably, to get rid of 5dB (nearly 3/4 of your original 15W)
you
intend to use paralleled-up resistors. This in itself will help
minimise
the inductance.
--
Ian

A bunch of chip resistors in parallel to ultimately make a T-atten.

Yebbut....
12W is a fair bit to dissipate with chip resistors (even a lot of
them).
However, the question was if you could use WW resistors, not 'how to do
it properly'. I presume there was a reason why this was asked! After
all, we ARE talking 'amateur' radio.

A 5db T attenuator with 15 W in has the maximum dissipation in the shunt
resistor at about 5 W.

The input series resistor dissipates about 4 W and the output series
resistor dissipates a little over 1 W.

I see no reason to go to wirewound resistors at these power levels.

Oh, I quite agree. I wouldn't really choose to use WW myself, but I
think they would work in this application. It's just that I usually try
first to answer the question as asked, and, if necessary, start
embellishing things from then on!
http://www.youtube.com/watch?v=k7vMvlRio5Y&feature=related
--
Ian

He obviously has a source of WW on hand. I hate to place an order
when I find stuff in the junk box.

So now the question is:
Will it be a significant problem? If he really knew what the inductance
values were, then it would be a simple matter to make sure it is
insignificant or at least minimal compared to the resistive component.
If the resultant SWR is low enough, the 5 dB return loss will help.
After that, the question would be "How will the amp act with
whatever input load is there?

Some resistors have a few laser cut turns that reverse themselves to
cancel the inductance. Some are a Carbon pack. A true WW
is either made by several turns of resistance wire over carbon or
ceramic for larger power ratings. Those have way too many turns.
It was mentioned that these even have a significant inductance at
audio frequencies. Now I ask you, is 5k inductive reactance
significant on a 50 ohm resistor?

You can always build one and see, if the exciter SWR is under
2:1, the amp input would see that too and you might be OK!

The other part of this is: How much power gain is the amp if
you need to kill 5 dB. Another thought is to incorporate
the inductance of the resistors to design a bandpass filter.
With many solid state exciters, some of the spurious outputs
don't reduce when you turn down the power.

http://www.repeater-builder.com/tech...ndbook-4th.pdf
All might find it helpful.
A more legible html version:
http://mikeyancey.com/files/Other%20...ion/title.html

All food for thought.

wrote:


A 5db T attenuator with 15 W in has the maximum dissipation in the shunt
resistor at about 5 W.

The input series resistor dissipates about 4 W and the output series
resistor dissipates a little over 1 W.

I see no reason to go to wirewound resistors at these power levels.



What kind of 5 Watt resistors should I use?

In rec.radio.amateur.equipment dave wrote:
wrote:


A 5db T attenuator with 15 W in has the maximum dissipation in the shunt
resistor at about 5 W.

The input series resistor dissipates about 4 W and the output series
resistor dissipates a little over 1 W.

I see no reason to go to wirewound resistors at these power levels.



What kind of 5 Watt resistors should I use?

Just about anything you can get other than wire wound.

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.



--
Jim Pennino

Remove .spam.sux to reply.

wrote:
In rec.radio.amateur.equipment dave wrote:
wrote:

A 5db T attenuator with 15 W in has the maximum dissipation in the shunt
resistor at about 5 W.

The input series resistor dissipates about 4 W and the output series
resistor dissipates a little over 1 W.

I see no reason to go to wirewound resistors at these power levels.


What kind of 5 Watt resistors should I use?

Just about anything you can get other than wire wound.

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.

dave wrote:
wrote:
In rec.radio.amateur.equipment dave wrote:
wrote:

A 5db T attenuator with 15 W in has the maximum dissipation in the
shunt
resistor at about 5 W.

The input series resistor dissipates about 4 W and the output series
resistor dissipates a little over 1 W.

I see no reason to go to wirewound resistors at these power levels.


What kind of 5 Watt resistors should I use?

Just about anything you can get other than wire wound.

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.

I got 10 of each, so I'll have some in the "miscellaneous" tackle box
next time. Thanks.

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.