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For an Arduino project that I'm building, I'd like to read codes from an IR remote, operating at 455 kHz. There is a combined IR eye and demodulator TSOP7000 that I think would fit the bill perfectly, but unfortunately it is discontinued and appears to be difficult acquiring today.

Hence, I figured it might be possible to buy the modulator separately from the IR eye, so basically my circuit would be [IR eye] - [IR modulator] - Arduino digital In. Alternatively build one, but the only I have seen is this that does not really seem to work that well, judging from the authors closing remarks.

Any ideas where I could buy one, or build my own in order to get the same result as with the discontinued TSOP7000?

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    \$\begingroup\$ Why does it have to be 455 kHz? Can't you do anything about the transmitter? There are lots of IR receiver modules at 36 kHz or 38 kHz available. \$\endgroup\$
    – stevenvh
    Commented Jul 23, 2012 at 10:39
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    \$\begingroup\$ Are you really sure this remote uses a carrier of 455 kHz? There is a reason that demodulator is dicontinued. The "normal" range is 35-50 kHz. You will see that the same Vishay TSOP line has a number of units accross that frequency range. \$\endgroup\$
    – Olin Lathrop
    Commented Jul 23, 2012 at 11:32
  • \$\begingroup\$ IF you can tolerate other carrier frequencies then you will find a very large number of other receivers here \$\endgroup\$
    – Russell McMahon ♦
    Commented Jul 23, 2012 at 11:33
  • \$\begingroup\$ If you want to use a TSOP7000, Farnement 14 claim to have them in stock - these would be available internationally. \$\endgroup\$
    – Russell McMahon ♦
    Commented Jul 23, 2012 at 11:36
  • \$\begingroup\$ Making IR receivers that work well using basic components is usually MUCH harder than may be apparent. Signal levels and noise effects and data separation can all prove quite challenging in practice. If you can use an off the shelf unit you can concentrate on the project and not the data link design. \$\endgroup\$
    – Russell McMahon ♦
    Commented Jul 23, 2012 at 11:38

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Like Russell says a receiver is not easy, that's also the reason why I asked if you couldn't work on the transmitter. If you don't mind tampering with that I'd suggest open-transmitter surgery. The idea is to intercept the 455 kHz output from the controller to the LED's transistor, and place a small microcontroller like the PIC10LF320 in between the controller's output and the transistor's input. It's just a SOT23-6 package, so it will probably not be a problem to include it in the remote control.

Let the PIC wake up from sleep upon a rising edge on its input. That means the transmitter starts sending a code. At that moment the PIC starts a PWM output at 36 kHz, which drives the LED. Also start a 3 µs timer. That can be a soft timer, you don't have anything else to do anyway. Restart the timer on each new interrupt. At 455 kHz that will be every 2.2 µs, so that's shorter than the timeout. As long as the timer doesn't timeout you're receiving the 455 kHz carrier. If the timer times out it means the 455 kHz carrier has stopped, either because of a 0 bit ("space") is being sent, or because of the end of the code. In that case stop the PWM output and go to sleep.

That's it. This way you detect the 455 kHz carrier and replace it with a 36 kHz (or 38 kHz, whatever), so you can use a regular IR RC receiver. Vishay has lots of them.

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  • \$\begingroup\$ Cool idea - although I don't get all the details. However, I need to stick with the transmitter as it is (it's a B&O remote) and just work on the receiving end of things. I'm reaching out to a few distributors to see if they can find alternatives unless they somehow manage to scrap up a TSOP7000 or two. \$\endgroup\$
    – Musslan
    Commented Jul 23, 2012 at 15:57
  • \$\begingroup\$ Thanks a lot Steven for taking the time to reply, greatly appreciated! \$\endgroup\$
    – Musslan
    Commented Jul 23, 2012 at 15:59
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    \$\begingroup\$ @Markus - You're welcome. I kind of like small challenges like this. I'm always curious myself how I'm going to solve it. :-) \$\endgroup\$
    – stevenvh
    Commented Jul 23, 2012 at 16:02

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