comparatoroutputtransistors
I can't seem to turn on a 2n2222a with the output of a LM339 comparator, as soon as the base is connected to the output, directly or with a resistor, even if the output is 5 volts, 9 volts, I always get 0.6 volts to the base of the transistor?? How odd…there is a pull-up resistor on the output of LM339.
I also tried a super simple comparator setup with LM324 op-amp, and same thing, I only get 0.6volts when the output is connected to base…
What am I missing? Why always 0.6 volts?
U1b can be disabled from oscillating by dragging pin 5 to +5V or by dragging pn 4 to ground. I favour taking pin 4 to ground because then the 70ohm load is not being powered at all when disabled.
To do this I would connect the output from U1a via (say 1k) to pin 4 of U1b. Because it is an open collector output it will do nothing when not active so get rid of R3 to ensure this.
Now this may mean that the pot on U1a operates in reverse but that can be fixed by swapping the circuits on the input pins of U1a.
If there is an unforseen impedance on the inactive output of U1a that disturbs the oscillations of U1b then maybe a diode connection to U1b is better (cathode to U1a o/p)
Don't connect the solenoid to ground, connect it to Vcc and use the MOSFET with its source tied to ground - drain to the other end of the solenoid. Put the protection diode across the solenoid, cathode to Vcc.
Now, if you do this you'll most likely find that it behaves much more as you would expect it to because the MOSFET will turn on properly. You need to maximize gate-source voltage from the Vcc rail to turn it on properly. What you had before was called a source follower and with your solenoid voltage required and your limited 9V supply it would never turn on effectively. Here's a snapshot of the changes: -
simulate this circuit – Schematic created using CircuitLab
Best Answer
Your circuit needs to look something like this:
simulate this circuit – Schematic created using CircuitLab
Comparator outputs are typically open-collectors, so they'll sink current in order to output a "0". You need a pull-up in order to have the output actually go high, feeding current into the base of your switching transistor. R1 needs to be high enough so you don't cram too much current into the comparator when it's trying to pull low (20 mA might be absolute max, so go with a bit less than that), but low enough to fully saturate Q2 in order to drive whatever load you're planning on using. R2 can (probably should) be zero.
As Ignacio pointed out, the base of Q2 will stay around VBE(sat), typically 0.6-1.0 V for most BJTs. If you manage to exceed that by very much, you'll destroy the device. The voltage isn't really relevant anyways; BJTs are current devices.