IR reciever's transmission distance

Question

I am trying to understand what it means by an IR receiver that has transmission range up to, say, for TSOP1738, 35m. For example, if the receiver's working at 38 kHz, how much sensitive would it be to IR LED transmitters that are way further away but working at the same modulated frequency with a much lower emitting range of a few centimeters? Does it ever detect them? I guess since it has a broad range, it should be able to detect any (even weak) 38 kHz wave circling around? In case it's gonna detect them at a long distance, how actually it does? Please clarify these for me.

Answer 1

This chip uses AGC with a BPF with a Q of ~10 to improve the Signal/Noise Ratio and thus has a wide dynamic range for distance and rejection of noise. A simple output comparator detects the pulsed Off duration of light relative to the peak which is regulated by AGC. i.e. steady carrier =0. Pulsed off =1

The necessary conditions for a "0" active output are the IR wavelength ought to be 950nm +/-100nm (50% loss), modulated at 38kHz ( 5% error reduces signal 50%) and sustained initially for 0.8ms then may be further modulated on/off with 600us bits or 1.67kbaud roughly, although 1200baud is ideal for range or >10 carrier cycles per bit.

The wide range of reception for distance is owing to the wide automatic gain control range (AGC) of burst pulses in a narrow bandwidth (Q=10) of pulses, wavelength of light and built-in electromagnetic shielding of the Photo Detector (PD) and amplifier. A clean power supply is essential.

The lens gives some gain in the horizontal axis on center but still a total reception angle of ~130 deg at half power.

Path loss (like Friis loss for RF) follows an inverse squared dynamic loss measured with signal power sensitivity of 0.35 (typ) mW/m².

The threshold for detection is fixed such that ambient noise is unlikely to trigger the duration of a start pattern needed to decode a command if there was any noisy output. This requires a sustained modulation of 38kHz pulsed IR for at least 0.8ms to allow AGC to capture the signal before the data bit follow by varying the duty cycle.

It has high rejection of 100/120Hz flicker in addition to the black tinted "Daylight Blocking filter" which is transparent at IR wavelengths.

Answer 2

This device

http://www.batronix.com/pdf/tsop17xx.pdf

has an IR photodiode behind an IR filter. If some IR radiation is present, the diode starts to conduct and generates a voltage to "input circuit." That voltage is analyzed - has it remarkable enough percentage of 38 kHz fluctuation. If it has, then a positive detection is outputted as the LOW logic state.

35 meters is not the range of the receiver, it is the distance, how far away TSOP1738 and IR diode TSIP5201 are promised to work together. 38 kHz version of TSOP1738 detects a proper 38 kHz pulsed IR radiation if its intensity at the receiver's aperture ( = the observed sector) is 0,35 mW /square meter or more and no other IR source rolls over the proper signal. TSOP1738 does not care (or more exactly: it doesn't know) how far or near the transmitter is. Detection can happen as well at 1m or 100 m, if there only is a transmitter powerful enough. TSOP1738 is promised to be so sensitive that IR diode TSIP5201 with 400 mA drive is promised to be powerful enough to be detectable at all distances from 0 to 35 m. If distance is longer than 35 m, some lucky individual (receiver or transmitter) can be in such way out of the tolerances that the detection still is possible without any additional optics, but not guaranteed. Adding some reflectors or lenses or changing a more powerful transmitter can easily increase the operating range.

What is strong enough to roll the right signal down. It can be read from the datasheet at least to some degree. For ex, figure 3 shows how the minimum detectable intensity of the proper signal rises when the ambient same wavelength IR radiation intensity grows.

This device gives no other information than "Is there a detectable amount of at the right frequency pulsed IR and at the same time low enough total intensity of disturbing IR signals" - Yes or no?

This device cannot make any difference between two sources of rightly pulsed signal. If you have two different 38 kHz pulsed IR sources and they send simultaneously, their interference can be unpredictable. It is possible that you see a rapid series of positive and negative detections which changes to continuous positive when one of the sources is stopped.

How a pulsed IR that has wrong frequency affects? And is it also detectable?. This demands complex calculations or testing. The latter is surely more reliable method due to the sparse available data about the filtering and detection process. For exact calculations this datasheet is not detailed enough, one need to make much assumptions. But a genuine communication theory engineer can well guess the details close enough.

You should note that the required intensity 0,35 mW/square meter can be produced from far away or from near. Transmitter's output power and how the radiation is directed by using reflectors or lenses affect. IR does not go thru walls, but an ordinary glass is still transparent enough for this device and mirrors work, too.

The radiation diverges to a larger area when it propagates. To double the detection distance without inserting any lenses or concave mirrors for bigger receiver's catching area or more narrow transmitting beam, you should quadruple the transmitting power. It can be surprising, how weak signal is detectable from near and how high transmitting power is needed from far away.

Answer 3

Assume 10 milliWatts of IR, emitted over +-19 degree cone; at 10meters, this +-19degree (1/3 of a radian) give +- 1/3*10m coverage in X and in Y, or 6.6meters in X and 6.6meters in Y, or area of 43 square meters.

Assume the receiving photodiode (sensitivity 1amp/watt, 1ua/uW) has lense of 5mm by 5mm. The photodiode energy is 10mW*(0.005m * 0.005m)/43meter^2 or 6 nanoWatts. Thus current is 6 nanoAmps.

Into a 100Kohm resistor, Vout is 600 microVoltsPP or 200uVrms. Assume 10pF capacitance at that node, with Vc=sqrt(K*T/C) telling us the noise floor is 20 microVolts due to 100Kohm. Photodiode shot noise is extra.

How good a signal is this? Bandwidth 1/2*piRC = 159KHz, certainly adequate for 38KHz "square wave" and some casual OOK on-off-keying data. The Signal Noise Ratio (SNR) is 200uVrms / 20uVrms = 10:1 or 20dB. For on-off modulation, your BitErrorRate BER gonna be very low, maybe 1 in a Trillion bits at 20dB SNR.

But you wanted 35 meters, not 10 meters. That reduces the power by 1/(3.5 * 3.5) or another factor of 10; the photodiode current reduces a factor of 10, and the voltage into the receiver also drops 10:1, and the SNR becomes 0 dB. What is to be done? Narrow the bandwidth.

Consider using the Signetics/Philips/??? NE567 tone-decoder with internal Phase Locked Loop.

And have as your photodetector a photo-transistor, biased so the sunlight and 60Hz get nulled out, while the 38KHz comes thru strongly.