Back to inverted logic then, but it's a better alternative, I guess.

View attachment 97916​

A 2N2700 is an N-MOSFET but you are showing a P-MOSFET.
Which is it?

 

A 2N2700 is an N-MOSFET but you are showing a P-MOSFET.
Which is it?

What the.... ?

I swear I did pick an n-mosfet when I was drawing the thing!... No wonder I couldn't find the 2N7000 at first

​

Here we go again... this better?

​

 

Seems like a lot of work to drive a 3V signal at 2 mA.

 

Seems like a lot of work to drive a 3V signal at 2 mA.

I know... but it's an internal logic signal to a more sophisticated controller (see post #1). In fact, I'm thinking that using a simple opto could be simpler. But I'd also have to use a transistor and a resistor to drive the opto. So maybe I'll just stick to this design, judging from its smaller size and component count.
And now that I'm using an N-FET instead of an NPN transistor, I no longer have to worry about how much current my MCU can source or sink.

 

@Papabravo, is my post #22 more or less what you had in mind? ... just curious

 

Of course, you could replace the whole lot with one of these...

 

Of course, you could replace the whole lot with one of these...

Ahhhh... it's practically a totem-pole push-pull switch... As I mentioned before, I need to start thinking about those chips as more than just mosfet drivers. Very nice, thanks for the suggestion!

 

shame on me

That's what I call a double invert circuit. I have posted it many times.
Also, credit given to ScottWang

 

That's what I call a double invert circuit. I have posted it many times.

then DOUBLE shame on me

 

Here is an example of using an NPN in a common base configuration especially for 8051 type open drain outputs:

 

Here is an example of using an NPN in a common base configuration especially for 8051 type open drain outputs:

Thanks... I think I get the general principle now. One transistor is driving down the gate of the other one. Simple as that!
And in your circuit, the MCU is draining current through Q1's emitter and not sourcing any.

 

Here is the PNG file

 

Thanks... I think I get the general principle now. One transistor is driving down the gate of the other one. Simple as that!
And in your circuit, the MCU is draining current through Q1's emitter and not sourcing any.

Correctimundo! Because the base of Q1 is connected to the +5V supply, the collector is pulled down towards 4.3V (= +5V - 0.7V), or is allowed to float up to the +12V supply rail allowing Q2 to turn off. You can adjust the current that the microprocessor is required to sink by adjusting the values of the three resistors.