1
\$\begingroup\$

Supposing I have an repeatedly object passing through IR sensors and I am looking at the output waveform. What is the maximum frequency the output waveform can have?
I think this maybe related to the frequency of IR rays used for detection but I am unable to come up with a concrete logic as to why it may depend on that.
PS: ignore other limitations such as the speed of the object to give rise to such high frequency and measuring instrument limitations.

\$\endgroup\$
2
  • \$\begingroup\$ Photodiodes have a fast response time on the order of nanoseconds. \$\endgroup\$ Commented Aug 9, 2018 at 19:31
  • 1
    \$\begingroup\$ What does the datasheet for your sensor say? \$\endgroup\$
    – The Photon
    Commented Aug 9, 2018 at 19:40

3 Answers 3

2
\$\begingroup\$

From a fundamental point of view there is practically no speed limit for IR detection (i.e., convert light intensity to an electrical signal). The frequency of the light is so high (in the 100 THz range) that there is practically no limit from that, given typical light modulation frequencies that can be electrically handled.

The fastest photodetectors for the IR that I know of, have cutoff frequencies in the 100 GHz range. This corresponds to rise times in the order of picoseconds.

Well this is an extreme example. Practically you have to take into account other constraints, e.g., the size of the detector (smaller is faster) which has an impact on the field of view with optics applied, the cutoff wavelength (different detector materials have different "speeds"), the detector noise level (there are detector types that are very fast but not low noise), the desired amplification gain (the gain bandwidth product is usually somehow limited) or, the detector price.

Often the amplifier imposes the bandwidth limitation, and for the high speeds, an integrated module (i.e. detector and amplifier) is required.

\$\endgroup\$
1
\$\begingroup\$

In addition to the other answer, the practical limitations include current levels, diode junction capacitance and load resistance as the breakpoint. Thus some have a rise time >>10x decay time.

A typical 5mm PD like LTR-323DB has a junction capacitance of 50pF and useable up to 3MHz.

By converting the PD current to voltage the trans impedance amplifier, (TIA) which is an OpAmp Vin-/Rf and making Rf low (xxx Ohms) you can cascade more gain to get a more useable voltage to detect.

If you have a narrow optical beam restricted by a PD aperture of 5mm and a recessed PD and narrow emitter Beamwidth of <10 deg. , it is possible to define a small path in between , you can now compute speed. But since emitter power is the biggest variable you need AGC. This poses a problem with detection but is easily solved by having 2 equal emitters , pulsed alternately , one blocked and the other not so that AGC always has a stable input to compare the dropout . Then you need a simple sync pattern to discern 1 source from the other such as time duration.

Now you “can” detect the object relative transmission loss not only in short pulses ~1us but at high velocity in a very small aperture such as 1mm in the path between using a recessed 5mm 10 deg PD and detecting a 50% drop in transmission loss. 1mm/us is 1km/s which is pretty fast assuming high SNR with daylight blocking lens on PD.

Anecdotal

Now that does seem too fast to me but I have only done this with 1m path and detected a 1mm wire passing fast across the middle using 100us pulses to verify the optical design. It worked. With multiple detectors I detected direction.

\$\endgroup\$
0
\$\begingroup\$

Here are some facts about silicon.

The thermal timeconstant of a cubic meter of silicon is 11,400 seconds.

The thermal timeconstant of a cubic centimeter of silicon is 100*100 faster, or 11,400 / 10,000 = 1.14 seconds

The thermal timeconstant of a cubic millimeter of silicon is yet 10*10 faster, or 1.14 seconds / 100 = 11.4 milliSeconds.

The thermal timeconstant of a cubic micron of silicon (perhaps the junction depth of some detector) is 11.4 nanoSeconds.

\$\endgroup\$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.