For the real enthousiasts, here is how it works 'inside'.

At the electric level, the Aladin port behaves like this:

• when the computer is idle, it sends a 100Hz +/-3V square signal on the black port. This is used to trigger the different functions when shorting the contacts with your fingers.
• just before the computer enters divelog mode, the logbook information is sent as a 0V/+3V square signal.
• once the logbook has been sent, the Aladin listen and optionally receives the data to be written: ppO2, metric/imperial,...
• the Aladin then enters the divelog mode.

The driver inside the Aladin is detailed below. This is just a (good) guess of how it looks like, it's not hard data from Uwatec!

The impedance of the Aladin port is around 1M. Observations of the port signals leads to the conclusion that data signals are sent via a 1M resistor (hence the 1M impedance). It has been suggested that the circuit was like the one shown on the figure below, but this is not correct because rise-time and fall-time of the signals are equal while the circuit shown would result in near-zero fall-time. The voltage applied to the output when listening to incoming data is not known (it is probably zero). It is also not known whether it is necessary to send symetric-voltage data or if a positive (or negative) signal is sufficient.

This driver characteristics is important for good operation of the interface. It means that if the line connected is too long (hence it has a high capacitance), its charge time via a high-impedance circuit will be too high compared to the signal frequency and the output will stay around 0V (or at least will not trigger the our external interface circuit). Also, pumping too much current from the port will very quickly result in the same effect of flat output. Let us quantize those effects to see what is the acceptable limits for the characteristics of the aladin interface we want to plug.

• Acceptable current drain:

  Since, by Ohm's Law,

  V = I * R,
  

  the voltage drop at the port will be

  drop = I * 1e6,
  

  assuming that the impedance is 1MOhm. The minimal voltage to drive the interface is around 0.7+0.8=1.5V (for the bi-trs version), so let's say we want a minimum of 2V, hence a maximal drop of 1V. This results in a maximal current drain of 1µA:

  Idrain = Vdrop * 10e-6 = 1µA.
  

• Acceptable wire capacitance:

  Charge/discharge of voltage of a capacitor vary following the equation

      i * t
  V = -------.
        C
  

  Imagine we use the values of V and I found above. Morover, the charge time t must be lower than 1/f=1/19200=52µs. We can thus set it safely to 25µs. This results in the following maximal capacitance:

          i * t     1e-6 * 25e-6
  Cmax = ------- = -------------- = 12.5pF.
            V             2
  

  This is extremely small, and explains why the length of the cables from the interface to the Aladin must be minimal. For example, a typical twisted-pair telecom cable has a capacitance of 60pF/m. The length is thus limited to 20cm. If you use a simple shielded audio cable, the capacitance can easily reach 200pF/m, limiting the length to... 6cm!! It is thus necessary to take extra care when selecting the cable type.