On constant impedance passive attenuators

it is impossible to construct an attenuator to the following specifications:

1) constructed from an arbitrary number of resistors and potentiometers where the potentiometers are ganged together on a simple shaft

2) attenuator driven by a voltage source

3) output voltage varies with shaft position

4) the output impedance of the network is constant with shaft position

Proof:

I believe it's possible to construct an attenuator of this type with a constant input impedance? But not output. It's possible to do the above if negative resistors are allowed but I have none handy.

The proof is from the IEEE Transactions on Circuit Theory from June of

1964, a lot of interesting papers in that one. Wish I could get hardcopies of those at reasonable price
Reply to
bitrex
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should be _linear_ potentiometers, there, it is perhaps not true for other tapers?

Reply to
bitrex

Think of a single stage Kelvin Varley divider without the pot on the end; just a resistor to satisfy the divider. I guess you could do it with a couple of ganged pots, the shaft of the second pot offset by the proper angle to satisfy the KV ratio? A kluged arramgement for sure, but would seem to satisy your requirements. Coupling the shafts together with some gears would also be a possibility. Or a specially constructed pot with two wipers inside.

Dave M

Reply to
Dave M

That's a truth, but for practical purposes, a fixed attenuator of 1000 ohms and 10.101 ohms has 10 ohms output impedance, +1%/-0% impedance tolerance, and can be appended to any non-ideal attenuator such as the author describes.

Yes, it also attenuates by a factor of a hundred, but... gain is cheap.

Reply to
whit3rd

Definately not true for other tapers, especially if arbitrary tapers are allowed.

eg: use three potentiometers to form a pi or tee network, one will, I think, need a reciprocal taper.

--
  When I tried casting out nines I made a hash of it.
Reply to
Jasen Betts

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