[ARC5] [Milsurplus] Smart People: Attenuators

[J. Forster]

Two more comments:

First, you can make very good, low inductance resistors by using many
resistors cordwood style between brass or copper plates...  for example 20
ea 1K 1/2 W.

-----o----o----o----
     |    |    |    |
     R    R    R    R
     |    |    |    |
-----o----o----o-----

Second, a sheild between the input and output sections of a mini-box will
help on feedthrough. Drill a small hole for the series R.

              |shield
              |
IN >----o---RRRRRR------o-------> OUT
        |     |         |
        R     |         R
        |     |         |
--------o-----o---------o-------- GND


FWIW,

-John

===============



> 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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