Hallo Russell,

unfortunately, I do not totally agree with your statements.

You wrote:

Note that on port0 you will still need the pullup resistor of around 100k. How to come to this resistor value? It comes down to how much current the input of the load device is - in this case it is a 74hc04 which needs very little current and because we expect the frequency to be low (as we're driving relays) we don't have to worry too much about capacitance.

Only because a line is labelled to carry 'DC signals only' does not automatically make them immune against high frequency EMI. Keeping in mind, that turning-on a relay can produce lots of interference, especially if no adequate power supply decoupling was used or if a solid ground plane was omitted, an impedance of 100kOhm at input of 74HC04 seems a bit high. If it's not just a battery powered application, I would always use 10kOhm pull-ups. This would decrease impedance at involved points by a factor of ten and result in a dramatically decreased susceptibility against EMI.

I do not want to tell, that 100kOhm will the circuit make fail and a 10kOhm succeed, but for me it seems to be better to use 10kOhm pull-ups.

You wrote:

Your BC547 has a current gain of around 300 (see transistor spec 'hfe'). So ,for example, your relay consumes 300mA, you will need to supply 1mA into the base of the transistor...

Unfortunately, this high current gain can only be achieved, when Uce is 5V. A situation which rather differs from that, when a relay is switched-on. Then, we expect an Uce in the range of 0.1V. For so little collector emitter voltages usable current gain heavily decreases!

When designing a transistor stage to switch-on a load, normally Ic / Ib = 10...20 is choosen. Only by this it can be guaranteed, that collector saturation voltage becomes minimal.


Regards,
Kai.