Question about laser detector circuit - rejecting 'daylight'


I've been working on creating a laser harp, following the posts made by 'Genesis' here:

(In French)

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His beam detector circuit works with a photodiode sensing the reflected light from a laser beam - basically when the performers hand breaks a beam, it is reflected back at the diode. His detector is here:

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...and the timing on his laser blanking signal is shown here:

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(The blanking turns the laser on and off, and is the lower trace on the scope - based on what I see I would say the frequency of the laser modulation is about 500Hz).

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?


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It's not all that different from using a capacitor to block DC.

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Any form of high-pass filter should suffice.

The simplest form is just a resistor and a capacitor, as shown at:

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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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