Thanks Harlod.

Isn't this the best place to get info from? Always great answers and
discussions! I'm 47 and I still learn every day from all you talented
and kind people out there.

Thanks!

/Ake

-----Ursprungligt meddelande-----
Fr=E5n: pic microcontroller discussion list
[mailto:PICLIST@MITVMA.MIT.EDU]F=F6r Harold Hallikainen
Skickat: den 19 april 2004 01:12
Till: PICLIST@MITVMA.MIT.EDU
=C4mne: Re: SV: SV: [EE:]Light detector


You are determining the Thevenin equivalent of the circuit driving the
A/D. The Thevenin equivalent replaces a network with many sources and
impedances with one source and impedance (the Thevenin equivalent). For
DC, the place an imaginary voltmeter between the terminals you are
trying
to find the equivalent for. What it reads is the Thevenin voltage. Next,
short all voltage sources and open all current sources. Put an imaginary
ohm meter between the two terminals. This is the Thevenin resistance.
The
Thevenin equivalent is the Thevenin voltage in series with the Thevenin
resistance.

In a simple two resistor voltage divider driving a PIC input from 5V,
say
the top resistor is R2 and the bottom one is R1.

The Thevenin voltage is 5V(R1/(R1+R2)) due to voltage dividers (multiple
applications of Ohm's law.

Now, short out the voltage source (shorting the top of R2 to ground),
and
put an imaginary ohm meter between the pic analog input and ground (this
is across R1). Once the top of R2 is connected to ground, R2 and R1 are
in
parallel, so the Thevenin resistance is now (R2*R1)/(R2+R1).

As an example, what is the maximum source resistance for a 10k pot
between
5V and 0V? The maximum source resistance will be when the wiper is in
the
middle. At that point, the Thevenin resistance will be 5k//5k or 2.5k.

Harold


> Thanks Neil, I think I get the point.
>
> /Ake
>
> -----Ursprungligt meddelande-----
> Fr=E5n: pic microcontroller discussion list
> [mailto:PICLIST@MITVMA.MIT.EDU]F=F6r PicDude
> Skickat: den 18 april 2004 21:11
> Till: PICLIST@MITVMA.MIT.EDU
> =C4mne: Re: SV: [EE:]Light detector
>
>
> On Sunday 18 April 2004 01:45 pm, Ake Hedman scribbled:
>> Doing mostly software work for the last decade... Why does the ADC
> input
>> see the two resistances in the divider in parallel as its input
>> impedance? Why isn't it just the LDR that set this? Sorry if this is
a
>> very dumb question...
>
> Same here (software for the last 12+ years) but from the remnants of
> knowledge
> from an old EE degree, an ideal voltage source has 0 resistance.  So
> from an
> impedance standpoint, the positive voltage rail is 0 ohms away from
the
> negative voltage rail (ground).  In reality it's quite low, and can be
> considered to be zero for these purposes.
>
> This is good for you, since the parallel combination brings down the
> impedance
> seen by the PIC.
>
>
>> I plan to use the 12F675 and you are right there is a 10K recommended
>> impedance of the analog voltage source. I', glad I asked the
question.
>> Nothing is ever as simple as it first look is it!
>
> No kidding!
>
> Cheers,
> -Neil.
>
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