The question is, what is the load being driven?
How much current will the load take when given 12V?

 

I would like to change it, so that it can be controlled by a voltage from a processor like an Arduino board:
5V turns Q6 on.
0V turns Q6 off.

Yes, it does work as you asked it to be changed. On 5V input, the transistor is ON so the output is 0V.

It functions effectively as an inverter, which is sort of the opposite of how the original circuit worked, given the diode being removed.

But it's what you asked for.

 

Base of Q6 Out
-----------------------
______0V __________ 0V
______5V _________12V

Hmmm. The base of Q6 is not the output. Is your truth table for the voltage applied to the base, or it is the required output. The output of your circuit is the collector of Q6 btw.

 

Hmmm. The base of Q6 is not the output. Is your truth table for the voltage applied to the base, or it is the required output. The output of your circuit is the collector of Q6 btw.

The table doesn't look good because the website compresses the spaces.
First column is "Base of Q6"
Second column is "Out"

 

... how much current will the load take when given 12V?

With 91K at the collector, about half a microamp in practical terms. I'm sure the 91K output pullup must be a typo.

It seems the overall requirement is for a level converter, at a guess?

 

The table doesn't look good because the website compresses the spaces.
First column is "Base of Q6"
Second column is "Out"

Q6 in you circuit will invert the voltage levels, so a second transistor is needed to invert the signal again to get the voltage levels you want. See below.
Q1 inverts, then Q2 inverts again so that 0v in=0v out, 5v in=12v out.

 

With 91K at the collector, about half a microamp in practical terms. I'm sure the 91K output pullup must be a typo.

It seems the overall requirement is for a level converter, at a guess?

Hi,
The 91k resistor is correct, as the circuit continues with other parts that I didn't include because they are irrelevant to the trigger voltage being discussed..

 

Q6 in you circuit will invert the voltage levels, so a second transistor is needed to invert the signal again to get the voltage levels you want. See below.
Q1 inverts, then Q2 inverts again so that 0v in=0v out, 5v in=12v out.

View attachment 313849

That does the job. Thanks and congrats for finding the solution.