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Home/ Questions/Q 681593
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Editorial Team
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Editorial Team
Asked: May 14, 20262026-05-14T01:30:02+00:00 2026-05-14T01:30:02+00:00

As far as I understand the serial port so far, transferring data is done

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As far as I understand the serial port so far, transferring data is done over pin 3. As shown here:
Serial Port Pinout

There are two things that make me uncomfortable about this. The first is that it seems to imply that the two connected devices agree on a signal speed and the second is that even if they are configured to run at the same speed you run into possible synchronization issues… right? Such things can be handled I suppose but it seems like there must be a simpler method.

What seems like a better approach to me would be to have one of the serial port pins send a pulse that indicates that the next bit is ready to be stored. So if we’re hooking these pins up to a shift register we basically have: (some pulse pin)->clk, tx->d

Is this a common practice? Is there some reason not to do this?

EDIT

Mike shouldn’t have deleted his answer. This I2C (2 pin serial) approach seems fairly close to what I did. The serial port doesn’t have a clock you’re right nobugz but that’s basically what I’ve done. See here:

private void SendBytes(byte[] data)
{
    int baudRate = 0;
    int byteToSend = 0;
    int bitToSend = 0;
    byte bitmask = 0;

    byte[] trigger = new byte[1];
    trigger[0] = 0;

    SerialPort p;
    try
    {
        p = new SerialPort(cmbPorts.Text);
    }
    catch
    {
        return;
    }

    if (!int.TryParse(txtBaudRate.Text, out baudRate)) return;
    if (baudRate < 100) return;
    p.BaudRate = baudRate;

    for (int index = 0; index < data.Length * 8; index++)
    {
        byteToSend = (int)(index / 8);
        bitToSend = index - (byteToSend * 8);
        bitmask = (byte)System.Math.Pow(2, bitToSend);

        p.Open();
        p.Parity = Parity.Space;
        p.RtsEnable = (byte)(data[byteToSend] & bitmask) > 0;

        s = p.BaseStream;
        s.WriteByte(trigger[0]);

        p.Close();
    }
}

Before anyone tells me how ugly this is or how I’m destroying my transfer speeds my quick answer is I don’t care about that. My point is this seems much much simpler than the method you described in your answer nobugz. And it wouldn’t be as ugly if the .Net SerialPort class gave me more control over the pin signals. Are there other serial port APIs that do?

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1 Answer

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  1. Editorial Team
    Editorial Team
    2026-05-14T01:30:02+00:00Added an answer on May 14, 2026 at 1:30 am

    As far as I can tell, the approach is similar to the I2C approach described here.

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