Can someone explain how this circuit works?

I believe the green led is supposed to be turned on when the input is positive however I can't seem to figure out a situation when the red led would be turned on.

From the direction of the Red LED it would look like it would be turned on if the input was negative. However if that's the case I can't seem to see where it could source current since the QLED1 would be blocking any current to it.

LED Circuit

  • \$\begingroup\$ What's down stream of the red LED (to the right)? \$\endgroup\$
    – Andy aka
    Commented Feb 1, 2014 at 21:24
  • \$\begingroup\$ Transistors will sort-of work in reverse (swap E and C) but with a lousy gain and low breakdown voltage. Without a lot of analysis, I think it will conduct but not regulate the current if the input is negative. \$\endgroup\$ Commented Feb 1, 2014 at 21:31
  • \$\begingroup\$ It wouldn't have to regulate well if the red LED is just supposed to indicate incorrect polarity power! \$\endgroup\$
    – user16324
    Commented Feb 1, 2014 at 21:51
  • \$\begingroup\$ @Andyaka it's the rest of the demo board. The complete schematic can be found if you search digikey for DC2027A-A-ND it's in the data sheet. \$\endgroup\$
    – user16105
    Commented Feb 1, 2014 at 23:15
  • 1
    \$\begingroup\$ So you missed that part of my comment that mentioned a link huh? \$\endgroup\$
    – Andy aka
    Commented Feb 1, 2014 at 23:44

1 Answer 1


Keep in mind that the B-C junction of a transistor will function like an ordinary diode when forward-biased. In normal transistor operation, this junction is usually reverse-biased.

When the Input is positive, the two transistors will regulate the current through the green LED such that the voltage drop across RLED2 equals the VBE of QLED2. 0.6V / 200Ω gives a current of about 3 mA, and this isn't particularly sensitive to the input voltage.

When the Input is negative, DLED3 is reverse-biased, so the current flows through RLED3, RLED2, the B-C junction of QLED2 and the B-C junction of QLED1 and then through the red LED. The current will depend on the voltage; an input voltage of about -12V will give the same 3 mA in the red LED.

  • \$\begingroup\$ I didn't realize that when reverse biased the B-C junction functions like a diode. That make sense. \$\endgroup\$
    – user16105
    Commented Feb 1, 2014 at 23:19
  • \$\begingroup\$ So then why wouldn't they just use two simple circuits each with a Diode, LED and resistor all in series together? that would be a lower component count and function the same. What's the benefit of them doing it this way? \$\endgroup\$
    – user16105
    Commented Feb 1, 2014 at 23:28
  • \$\begingroup\$ Good question. I would have just connected the two LEDs back-to-back (in antiparallel) and then used a single 3300-ohm resistor in series with that. It would seem that the designers wanted the green LED to light to full brightness over a wider range of input voltages. \$\endgroup\$
    – Dave Tweed
    Commented Feb 1, 2014 at 23:40
  • \$\begingroup\$ It allows them to use either discrete red and green LEDs or a single two-lead bicolor LED depending on package requirements without having to change the board. \$\endgroup\$ Commented Feb 1, 2014 at 23:53
  • \$\begingroup\$ @IgnacioVazquez-Abrams: I don't understand your point. Both the original circuit and my proposed simple circuit would allow that. \$\endgroup\$
    – Dave Tweed
    Commented Feb 1, 2014 at 23:59

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