On Sep 8, 12:41*pm, "David Looser"
wrote:
"Ian Bell" wrote in message
...
John Byrns wrote:
There is only one logical way to connect the first two resistors, and
then seemingly only one place to connect the odd resistor, but what is
the point of including the third 10k resistor?
To provide some attenuation to get the 2V rms output closer to the 0.5V
rms input requirement.
If the idea is to prevent overload of the amplifier then the shunt resistor
should be a lot lower, 1.67k to be exact.
David.
I agree with Ian, with John, *and* with you! Savour the moment!
See. Ian is right, the third resistor shunted to ground across the
junction of the other two is the traditional way to do it on 600 ohm
networks, where everything was matched, but the pot was at the
beginning of the chain, under the engineer's control.
See, next, John is right too, because the form of the question is:
There's an unknown pot following (for whatever reason). Maybe in this
case better to leave off the shunt resistor and content yourself with
the two series resistors.
See, finally, you're right, because that shunt resistor is in parallel
with the unknown pot. But rather than make it very low, I'd make it
very high.
On the whole, I think I'd leave off the shunt resistor and avoid the
uncertainty.
BTW, cutting the signal level isn't primarily to protect the amp from
overload but to limit the output beyond the poweramp to protect the
Lowther driver which unloads right smartly below about 32Hz. That's
another reason why John's suggestion about putting the monoing network
in the amp end of the cable is smart: it's not the high frequency we
want to roll off... Limiting the signal level before the amp also
keeps the output centred in the lowest distortion section of the
transfer curve.
Andre Jute
Visit Jute on Amps at
http://www.audio-talk.co.uk/fiultra/
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