By the way, which diagram are you using? Are you using the single NPN transistor diagram (which I think I posted somewhere)? I lost the context somewhere.OK, this time I got a base resistor value of just over 3K.
The PNP needs to sink around 40mA so the base current is roughly 4mA. Added to the current coming from the diagnostic pin makes roughly 14.3mA, so the current at the base of the NPN will be around 1.43mA. So the base resistor needs to drop 4.3 volts and pass a current of 1.43mA which makes it just over 3K.
How does that sound?
You would never need a pull up resistor. The PNP has a tendency to pull its base up by the means of the emmiter-base diode (unlike MOSFETs).The ECU diagnostic pin would be connected between the collector of the NPN and the base of the PNP, which should hold the base of the PNP at 12v without needing a pullup resistor.
In that case, a NPN darlington transistor would be better.The only concern I have now is the amount of current being drawn from the Q output. It is almost 1.3mA and, although the datasheet says the outputs can source up to 20mA, at 5v VDD the 4013 can only source around 1mA so it is potentially over by 0.3mA. Maybe I should find a darlington NPN to use for Q3 instead of a 2N3904?
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