The mu follower has a relatively high DC impedance, but it is an active
component with an AC negative resistance. It contributes to raising the
apparent output impedance of the amplifier.
Silk is a protein and a poor conductor of electricity; albeit polar [compared to polypropylene], and thus polarizable [via voltage]. It has amino acids in its primary polymeric structure which do not have extra [basic] amine and [acid] carboxylic groups. These groups react chemically with each other to generate zwitter ions [intra acid-base neutralization] to make the ionic ammonium carboxylate. These ions are electrical dipoles which will cause the dreaded dielectric absorption [of acoustic energy] which reverts them to the un-neutralized parent. Silk is a great eco-dielectric for high quality electrolytic capacitors.Picky Picky....
It's a silk electrolytic. Measures and sounds better than a
lot of film caps.
And you can't get a tube or a transistor to give you voltage gain and
not have inversion somewhere unless you introduce the dreaded
transformer. Non-inverting amplifiers with voltage gain simply find
some way to flip the phase an even number of times. In this amp,
the second time is at the output terminals.
Of additional interest is an answer/explanation for the amplifier without loop feedback, driven by one complete sinusoidal signal into a load of 8 Ohms. What is its output impedance it experiences along the signal trajectory? Is is a constant 10 Ohms like in the quiescent mode?I am confused by these statements
If it is "an active component with an AC negative resistance" how can it's contibution be higher impeadance? It's negative, simply speaking, thats lower.
We have 2 sources of gain totalling 29db loaded with 8 ohms. We have at this point neglected the Drain impeadance and the Mu follower contibution to the impeadance the drain see's. And the feedback....But it's more complicated than that. The Mu follower current source on
top of the circuit also contributes to gain, adding maybe 5 dB. So now we estimate that with infinite Drain resistance the gain is 29 dB, which is
OK so far?
So the impeadance value of the Mu follower can be calculated to exhibit 10 ohms of paralell load to the 8 ohms?But we actually experience 23 dB gain due to the finite Drain resistance(s) and the contribution of the DC feedback loop. This 5 dB drop is a factor of 0.56X, and is equivalent to our high Drain resistance example driving 8 X 0.56, which is 4.5 ohms
So now we have an apparent Drain resistance which is equivalent to that value which paralleled with 8 ohms gives us 4.5 ohms. That resistor is close to 10 ohms.
Wait a minute, The -5db that reduced our theoritical gain from 29-23 included feedback and Drain resistance(s) and was used to determine the actual Drain load was 10 ohms?Without the R12/R11 loop, the output impedance of the amplifier is 10
Thanks for making the connection to my previous employer Rohm and Haas. This pair of entreupreuners started selling chemicals to the leather industry in 1907. I am still chasing new tanning agents. Collagen [skin fiber] is a poor eco-dielectric. It has the dreaded zwitter ions which ironically are needed for tanning. Amine groups for aldehyde and carboxylic for chrome tannages. I may be able to get the dielectric properties of silk [or better] with leather as we know it. Food for my thought.You should have been into capacitor designing, instead of tanning leather strings.
So the impeadance value of the Mu follower can be calculated to exhibit 10 ohms of paralell load to the 8 ohms?