Re: Steal RAM -- Question

["sicklittlemonkey" <Nick.Westgate@gmail.com>]

Michael J. Mahon wrote:
> Prior to the //c, the toggle was AC-coupled to the Darlington driver
> through an RC circuit with a time constant of about 5 milliseconds.
>
> The effect is that no matter which way the toggle goes, after 5-10
> milliseconds, it's back to its "rest" state: no current through speaker.
>
> As long as the toggle is changing at least every 5-10 milliseconds
> (100-200Hz), the AC-coupling makes little difference, and the speaker
> sees each toggle as a full-scale current transition.
>
> But if more than 5-10 ms. elapses between toggles, then what happens
> on the *next* toggle depends on whether it is high-to-low or low-to-
> high.  If it is high-to-low, it will simply try to turn off the already
> off Darlington driver, producing essentially no sound.  If it is a
> low-to-high transition, then it will produce a maximum "tick".
>
> As a result, sounds with toggle intervals longer than 5-10 ms. or--
> very importantly--short sounds with asymmetric pulses *separated* by
> periods of silence *will* be affected, since the "attack" caused by
> the first speaker pulse will either be heard or not, depending on the
> initial state of the speaker toggle.
>
> In the //c, the actual circuit of the audio op amp is not given, but
> the diagram suggests that it is DC-coupled to the toggle, so it shows
> no time-constant effects.

The above explanation, while excellent, is suffiently dense to those of
us without formal electronics training (I had to look up AC-coupled,
for instance) that I went back to Sather just to remind myself how
little he had written about the speaker. In his //e book the nearly
obsolete cassette port gets considerably more converage!

A version of the above explanation - expanded for the attentive layman
in the style of Sather - would form an invaluable supplement to his
works.

Still, I basically got that the AC-coupling acts as a high-pass filter,
and that its response time means that after 5-10ms the voltage on both
sides is equalized, resulting in the speaker "rest" state. I don't
pretend to understand the analog circuit, but the implications you
mention follow nicely.

I can't think of any topic that he left so undone. Can you?

Cheers,
Nick.