[ARC5] [Milsurplus] Smart People: Attenuators

[Dennis Monticelli]

FYI.  If you parallel R's to get the right value, the parasitic L goes down
and the parasitic C goes up.  The reverse happens for putting R's in
series.  Low value R elements in your attenuator are more impacted by stray
L and high value R elements are more impacted by stray C.  So just plan
accordingly.

I also second the recommendation of avoiding taking too much attentuation in
one bite.  However, the lower the frequency the greater the chance that you
will be successful with a simple one stage solution.

Dennis AE6C

On Wed, Mar 2, 2011 at 8:19 PM, mac <w7qho at aol.com> wrote:

> Agree with John, 10W for the input shunt, 1/2 watt units OK for the
> other two resistors but I (personally) would use 1 or 2W units.  All
> resistors must be non-inductive, of course.
>
> Dennis D. W7QHO
> Glendale, CA
>
> ****************
> On Mar 2, 2011, at 7:51 PM, David Stinson wrote:
>
> > Need the help of you smart people again.
> > I built a "utility" linear HF amplifier to help put QRP transmitters
> > on the air.  Works great with the correct drive,
> > which is 10 milliwatts.
> > I need to attenuate drive levels from a couple of hundred milliwatts
> > up to about 6-7 watts, so need a variety of pi-type attenuators.
> > The resistance values are no problem,
> > because someone wrote a Java app:
> >
> > http://chemandy.com/calculators/matching-pi-attenuator-calculator.htm
> >
> > My question:
> > What's the formula for the Wattage rating of each resistor?
> > For instance:  If I have a 4-watt drive signal and need to get it down
> > to 10 milliwatts (assuming 50 Ohms in and out),
> > I need to attenuate 36 dB.
> > That's shunt in 52 Ohms, series 1580 Ohms, shunt out 52 Ohms.
> >
> > How to calculate the appropriate wattage of the three resistors?
> > Common sense says the shunt input should be big enough to
> > sink the entire carrier power, while the series and shunt out
> > could be smaller.  What's "right?"
> >
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