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?

  • 3
    \$\begingroup\$ ... Because that's the Vbe of a silicon transistor. \$\endgroup\$ Jun 8, 2013 at 6:12
  • \$\begingroup\$ Wow, I'm retarded. Back to the drawing boards \$\endgroup\$ Jun 8, 2013 at 6:14
  • \$\begingroup\$ Can you add a circuit please. however i guess your missing a pull up resistor for the comparator output. \$\endgroup\$
    – Spoon
    Jun 8, 2013 at 6:14
  • \$\begingroup\$ Nay! There is a 4.6k pull-up resistor. It's the most basic comparator circuit, I'll figure it out after I get some sleep...If I have 5 volts at the output (with a 4.6k pullup), as soon as I connect that to the base of 2n2222, the voltage drops to 0.6volts \$\endgroup\$ Jun 8, 2013 at 6:28
  • \$\begingroup\$ electronics.stackexchange.com/questions/70477/… \$\endgroup\$
    – Phil Frost
    Jun 8, 2013 at 12:03

1 Answer 1


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.


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