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previous: Gil Ayan q{If you have a large resistor in s... -- 11/23/1999 7:19 PM

Re: This is How I Do it

11/23/1999 9:55 PM
peter	Re: This is How I Do it Respectfully, Gil, I think Dave is right here: the time constant for eg 100k resistor charging a 4.7 uF cap, is 4.7 seconds. So the charge on the cap increases until an equilibrium is reached between the charge drained off by the R to ground, and the charge delivered through the 100k. It is quite a nasty problem to calculate analytically. If you examine the Marshall circuit, considering the two extremes of the pot, you get a divider ratio of .17 to .23... so if I put in a peak voltage of say 400 volts, your method would predict a bias voltage from 68 to 92 volts. (unless I have misunderstood your calculation!!)

11/23/1999 9:55 PM

peter

Re: This is How I Do it

Respectfully, Gil, I think Dave is right here:

the time constant for eg 100k resistor
charging a 4.7 uF cap, is 4.7 seconds.
So the charge on the cap increases until
an equilibrium is reached between the
charge drained off by the R to ground,
and the charge delivered through the 100k.
It is quite a nasty problem to calculate
analytically.

If you examine the Marshall circuit,
considering the two extremes of the pot,
you get a divider ratio of .17 to .23...
so if I put in a peak voltage of say
400 volts, your method would predict
a bias voltage from 68 to 92 volts.

(unless I have misunderstood your calculation!!)

Replies:

Gil Ayan q{Respectfully, Gil, I think Dave i... -- 11/23/1999 11:43 PM