I do not understand the PT4 Pull Up
I read the Forum and find different Reply, 5V or 12V Pull Up!
I have a 12V Hall signal (VSS1) and it works without JP8 (R12) on the Ms3X, but only up to 155km/h.
When I change to PT4 I need a 5V or 12V Pull Up Pull Up (1Kohm)?
The PT4 (Prozessor Pin) can only handel 5V, why find 12V Pull Up?
Frank
The input circuit for PT4 includes a 1k series resistor and a pair of diodes, one from ground to the signal, the other from the signal to Vcc. The series resistor limits the current, and the diodes dump the excess voltage to the Vcc rail (where it gets dumped to GND via the Zener diode). Assuming a 1k pullup to 5v, the current to the processor pin is 2.5mA at 5v.
If you look at the PT6 circuit, you'll note that the series resistor is 10k, along with a 2.2k pull-up resistor to +12v. This extra resistance limits the current even more to ~ 1mA, most of which is going to be dumped via the Vcc rail Zener.
What does all that rubbish mean? You can use a 1k pull-up to +12v to PT4, but I would suggest you add a 10k series resistor between the pull-up and the pin.
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dontz125 wrote:
You can use a 1k pull-up to +12v to PT4, but I would suggest you add a 10k series resistor between the pull-up and the pin.
OK thank you for the answer.
I wanted to build the resistor in the V3.0 housing (MS3X Board) and can not find any place?
And then
I split up the Ribbon Cables and change Pin15 and Pin1
before:
VSS1>DB37 (ms3x) Pin28>Tabel Swich Circuit>PH6 Ribbon Cable (Pin1)
new:
VSS1>DB37 (ms3x) Pin28>Tabel Swich Circuit>PT4/IGN2 Ribbon Cable (Pin15)
Frank
