Re: Question about laser detector circuit - rejecting 'daylight'
- From: Nobody <nobody@xxxxxxxxxxx>
- Date: Fri, 23 Nov 2007 22:05:48 +0000
On Thu, 22 Nov 2007 18:44:27 -0800, Kasterborus wrote:
I get to talk to the guy once in a while, and during our last
conversation he said that the detector circuit is tuned to the laser
modulation, and so can reject daylight quite well. Does anyone know
how this is achieved?
I would guess that it's the caps in the opamp feedback - but how are
the values calculated?
Any form of high-pass filter should suffice.
The simplest form is just a resistor and a capacitor, as shown at:
http://en.wikipedia.org/wiki/High-pass_filter
You can get a sharper cut-off with an active filter (e.g. Sallen-Key), but
you really don't need it for this application.
The exact values depend upon topology, but the general rule is that the
cut-off frequency is inversely proportional to both resistance and
capacitance. For the passive RC filter, you get -3dB at 1/(2.pi.R.C).
In general, you would prefer larger R and smaller C (lower-value
capacitors tend to be cheaper, smaller and have lower parasitic R and L,
whereas resistors tend to differ only in their resistance value). In
practice, the impedance of the following stage puts an upper bound on R
(or may even *be* the R component), and thus a lower bound on C.
Note that an active filter using an op-amp and negative feedback would
have the capacitor in the input and the resistor in the feedback loop.
Putting the capacitor in the feedback path would give you a low-pass
filter.
.
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- From: Kasterborus
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