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.

Ian Jackson wrote:
In message , dave
writes
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.

**** you.

Dave, that's a very strange response. Do you have a problem with me
suggesting that you probably had a reason for asking if WW resistors
could be used? I've already said that you will probably 'get away with
it' up to 14MHz, especially if you mount the resistors as close as
possible to a ground plane.

You may not realise that lot of 'engineering' is the art of 'getting
away with it' (for whatever reason). For example, the reason might be
economics, it might be practicality, it might be urgency, or it might be
availability. The list is endless.

I'm just putting myself in the situation where someone needs to lose 5dB
of RF drive, and (say) all he has immediately at hand is a selection of
WW resistors. Does he try them (knowing that they are not recommended
for RF work)? Or does he take advice from the experts who, without a
second thought, will probably say "No way!".

The point is, do you want to try the WW resistors, knowing full well
that they are not really the right thing to use, but will probably work
well enough for what you want? Or do you want to do things 'correctly',
knowing that what you have done will be almost perfect? You have
received various bits of advice (some conflicting). It's now for you to
decide what suits you best in the circumstances.

I know how to think. I have no resistors, period, except for the 1/4
Watt ones I use at work. Mr. Jim has pointed me in the right direction.

All "amateur" means in the context of Amateur Radio is that we do not
get paid for doing our thing. It doesn't imply that we are stupid or
incompetent.

I am fully cognizant of the possibility of meaningful reactance being
introduced by a coil of Nichrome, and was hoping that someone who deals
with passive components more regularly than I could point me in the
right direction, and that has happened.

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.

In message , dave
writes



I am fully cognizant of the possibility of meaningful reactance being
introduced by a coil of Nichrome, and was hoping that someone who deals
with passive components more regularly than I could point me in the
right direction, and that has happened.

From your other post, I see that you are going to use metal oxide
resistors. In the circumstances, that's obviously the correct decision.

But your question did not ask about what were the best resistors to use
for the attenuator. It was "Can I use wirewound resistors to build an HF
attenuator?" I assumed that you knew that WW would be inductive, but
might have some reason for wanting to use them.

Ralph Mowery replied that the "Simple answer is no" etc. You responded
with "Can the inductive reactance be cancelled with capacitors? I'm
mainly concerned with 40, 30 and 20 meters" which again made me think
that you wanted to use the WW, if possible. I suggested that the effects
of the inductance might be reduced if you mounted the resistors hard
down against a ground plane. [This is a frequently-used technique for
preserving the characteristic impedance, especially at higher
frequencies.]

JB then suggested "a bunch of chip resistors". I commented that you
would need quite a lot of them. It was only then that I actually
suggested that there might be a reason why you were asking about WW, and
that the application was 'only' for amateur radio. Your response was
"**** you". "How very strange", I thought!

Anyway, I'm glad you have got yourself sorted out. I hope - no, I know -
that the attenuator will definitely work as expected, with the advantage
that it will probably work way up into the VHF region.
--
Ian (licensed for 'amateur' radio 49 years ago this week)

Ian Jackson wrote:
In message , dave
writes



I am fully cognizant of the possibility of meaningful reactance being
introduced by a coil of Nichrome, and was hoping that someone who
deals with passive components more regularly than I could point me in
the right direction, and that has happened.

From your other post, I see that you are going to use metal oxide
resistors. In the circumstances, that's obviously the correct decision.

But your question did not ask about what were the best resistors to use
for the attenuator. It was "Can I use wirewound resistors to build an HF
attenuator?" I assumed that you knew that WW would be inductive, but
might have some reason for wanting to use them.

Ralph Mowery replied that the "Simple answer is no" etc. You responded
with "Can the inductive reactance be cancelled with capacitors? I'm
mainly concerned with 40, 30 and 20 meters" which again made me think
that you wanted to use the WW, if possible. I suggested that the effects
of the inductance might be reduced if you mounted the resistors hard
down against a ground plane. [This is a frequently-used technique for
preserving the characteristic impedance, especially at higher frequencies.]

JB then suggested "a bunch of chip resistors". I commented that you
would need quite a lot of them. It was only then that I actually
suggested that there might be a reason why you were asking about WW, and
that the application was 'only' for amateur radio. Your response was
"**** you". "How very strange", I thought!

Anyway, I'm glad you have got yourself sorted out. I hope - no, I know -
that the attenuator will definitely work as expected, with the advantage
that it will probably work way up into the VHF region.

That's way revisionist, but I am used to it. I apologize for the
vulgarity.

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.

On Aug 18, 8:32*am, dave wrote:
I need to lose about 5 dB from a 15 Watt exciter. *Thanks.

You could build the attenuator so it takes the power down in steps
perhaps 2 1db attenuators followed by a 3db attenuator. I didnt run
then numbers but I think this is doable with 2 watt resistors.

Jimmie

JIMMIE wrote:
On Aug 18, 8:32 am, dave wrote:
I need to lose about 5 dB from a 15 Watt exciter. Thanks.

You could build the attenuator so it takes the power down in steps
perhaps 2 1db attenuators followed by a 3db attenuator. I didnt run
then numbers but I think this is doable with 2 watt resistors.

Jimmie

Thanks. I have some honkin' metal oxide resistors that I'm going to use.

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.


Thanks, Jim. The thing works like a champ. I added a couple of $2.00
relays to bypass the pad when receiving.