[sdiy] Impedance / equivalent resistance calculation

[Roman]

your DC impedance at IRIN1 is parallel connection of IR5 and IR9,
that's 750ohms. If you expect to detect the light at frequency highier than,
say, 40Hz (twice corner frequency of filter made of IR5,9, IC9),
cou could consider the cap as short.
Your node with IR1 and IQ1 has equivalent resistance of 690 to 2.1k
(again, parallel of IQ1 and IR1 in light and dark), so overal AC
impedance is more or less (690-2100)||750 = 360 - 550 ohms

hope that helps
Roman

----- Original Message -----
From: "Bert Schiettecatte" <bert at percussa.com>
To: <synth-diy at dropmix.xs4all.nl>
Sent: Friday, September 02, 2005 10:34 PM
Subject: [sdiy] Impedance / equivalent resistance calculation


> Hi All!
>
> I have another (probably silly) question about this circuit:
>
> http://www-ccrma.stanford.edu/~bschiett/impedance.png
>
> I need to find out what the impedance at the IRIN1 net is, since
> It will be connected to an A/D pin. I suppose I need to calculate
> The equivalent series resistance of this circuit.
>
> I measured the phototransistor to be 75K under dark conditions and
> 1K under light conditions.
>
> What is the correct way to calculate the equivalent resistance?
>
> I calculated the eq. resistance for the transistor and IR5:
>
> Rd = RDark * IR5 / (RDark + IR5)
>
> And
>
> Rl = Rlight * IR5 / (Rlight + IR5)
>
> And then for IR1 and IR9:
>
> Re = IR1 * IR9 / (IR1 + IR9)
>
> But then... There is a capacitor in there and that's where my
> Knowledge of calculating equivalent resistances ends :-)
>
> Thanks in advance,
>
> Bert
>
>