Hello Spehro and Dave,

Spehro says:
> RC time constant is just R * C, your formula is correct for 1/f (Xc = R,
so
> 3dB down), or lose the 2*Pi and calculate f in radians/sec.
Dave says:
> Actually, what we refer to as the "RC time constant" is usually the
> product T = RC, without multiplying by 2*pi.

    oh, big confusion here. Now it clarified. Thanks.

Dave says:
>This product is used when calculating the time-domain response of an RC
>circuit such as above: when a voltage step is presented to the input,
>going from zero volts to some voltage Vin, the output waveform is
>described by the expression
>  Vout = Vin * (1 - (e^(-t/T))),
>where t is the time elapsed since the input voltage step and T = RC.

    Sorry for my math ignorance but what's that "e" in the formula?

>(I'm not disputing your frequency calculation, just clarifying the usage
>of the term "time constant".)

    thank you again for clarifying this.

Dave says:
>>    IN----R1----*-----*----OUT
>>                |     |
>>                C     R2
>>                |     |
>>               GND   GND
>The way to do this is to consider R1 and R2 to be in parallel with one
>another with regard to AC, since R1 is connected from C to a circuit
>node (the input) which has zero impedance with respect to ground (that
>is, the input is driven by a voltage source), just as is R2.  So in
>terms of the resistance that the capacitor "sees", R1 and R2 are in
>parallel.

    Great explanation.

Spehro says:
<SNIP>
> If you're not familiar with this method, here's a web page
> that 'splains it fairly well:
> http://hyperphysics.phy-astr.gsu.edu/hbase/electric/thevenin.html

    thank you again. Great site.

Spehro says:
> >     IN----R1----*-----*----R3----OUT
> >                 |     |
> >                 C     R2
> >                 |     |
> >                GND   GND
>
> The same as above unless there is loading on the circuit.  ;-)
> If you short the output to ground and look at the current, then the
> 3dB down frequency would be
>
> f = 1/( 2 * Pi *C * R1 || R2 || R3)
>
> where A || B || C = 1/(1/A + 1/B + 1/C)

    Good.

Dave says:
>And finally, if R3 is feeding into a load resistance Rload which is
>neither infinite nor zero, then you do the calculation as above but with
>(R3+Rload) replacing R3 in the formula.

    Actually it's connected to the non-inverting input of a comparator. I
think this makes the calculation more complex as we have the input impedance
of the comparator, the positive-feedback resistance (schmitt trigger) and
the pull-up resistor at the comparator output.

    But your lesson give me clues of what to do and what to expect.

    Thank you very much again,

    Brusque

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Edson Brusque                 C.I.Tronics Lighting Designers Ltda
Researcher and Developer               Blumenau  -  SC  -  Brazil
Say NO to HTML mail                          www.citronics.com.br
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