Issue 138 January 2002
WHAT GOOD IS IrD,Eh?
Part 1: Cordless Protocol

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byJeff Bachiochi

Start • Standards Make It Happen • IrPHY • The Range • UART To IrDA • IR Encoder/Decoder • Gimme More • Sources & PDF

IR Encoder/Decoder

You probably know Microchip as a microprocessor manufacturer, so you can easily guess the company’s approach for the MCP2120. I considered using Microchip’s least expensive processor, the PIC12C508. Microchip was the first company to recognize the potential merit of a microprocessor with low pin count. Although this device has an internal RC oscillator, which requires no crystal for applications not requiring high accuracy timing, I chose to use an external crystal to achieve higher accuracy over temperature and voltage. This limits the I/O available to four, the minimum necessary for this project (see Figure 4).

Figure 4—The smallest available 8-pin microprocessor is used to encode and decode data between a UART and IR transceiver.

There are essentially two signals that must be looked at for this application, TX from the serial port and IRRX from the IR receiver. Because the IrDA protocol is half-duplex, I won’t have to do two things at once, however, the transmit path will have priority over the receive path. The shortest loop, which includes testing for both signals, is shown in Listing 1.

Listing 1—This minimum timing loop of 5.425 µs will most likely miss a minimum IRRX pulse of 1.41 µs.

The numbers in parenthesis show the instruction cycle count for the loop as five cycles. Using a standard crystal frequency of 3.6864 MHz (divides evenly into standard data rates), you get an execution cycle time of 1.085 µs, or:

The five-cycle loop will therefore take 5.425 µs to complete. The fastest bit time of the TX serial output is 104 µs (1/9600). So, you can see that pulse if you are in the loop watching.

The shortest IRRX pulse is 1.41 µs (the minimum acceptable pulse duration). There is a good chance you will miss an IRRX pulse while in the loop. The loop would be fast enough if you increased the crystal frequency 10 times, however, 36.864 MHz is faster than the 4-MHz maximum of the PIC12C508, so that’s out.

An interrupt could solve this problem. Unfortunately, the PIC12C508 has no interrupts available. It does, however, have a wake-on-pin-change function. Note though that by the time the oscillator starts from sleep, you’ve missed a whole bunch of IR pulses, so that won’t help either.

The saving grace here is that an IR pulse won’t repeat for at least 104 µs, which gives you plenty of time to do something after you’ve seen the IRRX pulse. About the only peripheral the 12C508 has is an 8-bit timer. Although the timer doesn’t have an overflow interrupt, it does have a clock input for the timer. This means that an external (selectable rising or falling) edge will increment the timer. You can read the cleared timer to determine if an IR pulse has been detected (timer can’t be zero). Now, if only the IRRX pulse fits within the input specifications. Let’s look at that. An external T0CKI input must be low for two clocks + 20, or:

The minimum received pulse is 1410 ns (1.41 µs), so you’re in like Flynn.

The SEND_BIT and RECEIVE_BIT routines are not fancy. I use in-line cycle counting to assure things happen at the right times. During the SEND_BIT routine, I wait 0.4375 of a bit time before enabling the IRTX pulse for 0.1875 of a bit time. At around 0.9375 of a bit time, I start watching TX for another low. If I see TX low, I jump to the top of SEND_BIT again without going back to the loop. If TX stays high, I jump back to the loop routine to look for more action.

During the RECEIVE_BIT routine, I clear the timer (to get ready for the next IRRX pulse), wait 0.4375 of a bit time, and then clear the RX output bit. I start testing the timer for zero at around 0.9375 of a bit time (no pulse yet). If it has incremented within 0.125 of a bit time, I jump back to the top of the RECEIVE_BIT routine without setting the RX output bit (for another continuous 0 bit time). If the timer doesn’t increment during the sampling period, I wait out the rest of the bit time, set the RX output bit, and jump to the loop to look for more action.