I don't understand why the green portion of LED3 is able to conduct. Since the path marked in green represents the negative half-cycle of the 24VAC isn't the green LED reverse biased therefore no current flow. Isn't the red portion of LED3 FW bised- shouldn't it be lit? Or since this is AC does the "-" or "negative" have to do with current direction as seen here?

Circuit in Question

  • \$\begingroup\$ If door number 1 is closed, then there's a path -- LED3, D8, LED2, D5, LED1, & finally D2. All the LEDs will light both red and green, appearing amber or yellow. Open any one door, and they should all show red. Which is fine if your customer base doesn't include the 3% of the male population who can't distinguish between green and red, and thus can only tell when one of those @#$% tricolor LEDs is on, but not whether it's saying "OK" or "DANGER!!!". \$\endgroup\$ – TimWescott Oct 18 '18 at 0:52
  • \$\begingroup\$ @TimWescott I must not be understanding here, but how will any of these LEDs light both red and green at the same time? They're antiparallel, so I'm pretty sure there's no way they could both be on at once... \$\endgroup\$ – Hearth Oct 18 '18 at 2:49
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    \$\begingroup\$ @Felthry: that sound you just heard was a light coming on over my head. They don't light at the same time. Each lights for one half cycle of the AC. Your eyes see the lights as solid on because of persistence of vision. You can sometimes detect that an LED is blinking past your ability to see it by moving your eyes to the side and observing dots, or if it's a handheld thing, waving it from side to side. I built a box that blinked developer error codes on the LED that you could see by waving the box, but looked like solid on otherwise. \$\endgroup\$ – TimWescott Oct 18 '18 at 16:48
  • \$\begingroup\$ @TimWescott Ah, of course, I didn't think of the fact that it's AC. \$\endgroup\$ – Hearth Oct 18 '18 at 17:03

Your schematic shows two basic topologies: one with and one without a resistor in parallel with the switch. For each of these topologies, there are four possible states: two with the switch closed and two with the switch open. So, altogether, there are eight situations to examine.

Let's start with the cases on the left, without the resistor in parallel with the switch:


simulate this circuit – Schematic created using CircuitLab

At the far left, you can see that there is only one path, through the RED LED, to complete the circuit. So for this half-cycle, the RED LED is activated and showing. But just to the right, with the AC reversed, there is NO path available. So both LEDs are off. So for the left two cases, with the switch open, only the RED LED can be lit. And only then, for one half-cycle. Given the 50-60 Hz frequency, this is sufficient for you to perceive the RED LED as "ON" and the GREEN LED as "OFF."

Now with the switch closed, you can see on the left side of this right-pair of cases, that the current flows through \$D_3\$ rather than through \$LED_{1_\text{R}}\$. This is simply because a "normal" diode has far less of a voltage drop than does an LED. So the current does NOT flow through the RED LED, despite the fact that it is oriented with the right polarity. \$D_3\$ effectively bypasses the RED LED here. So the RED LED is still "OFF" in this half-cycle. On the far right side, with the AC reversed again, the current cannot go through \$D_3\$ or the RED LED. It can only flow through the GREEN LED. Luckily, there is also \$D_2\$ to complete this circuit. So for the right two cases, with the switch closed, only the GREEN LED can be lit. And only then, for one half-cycle. Given the 50-60 Hz frequency, this is again sufficient for you to perceive the GREEN LED as "ON" and the RED LED as "OFF."

In your last case, with the added resistor in parallel to the switch, this only affects things for the case where the switch is open:


simulate this circuit

(I've removed the inactive diodes above.)

This will appear as a YELLOW/ORANGE color to your eye, since both colors are being emitted; one in each half-cycle.

In the case where the switch is closed, the new resistor is bypassed and the behavior will be as before (a GREEN LED lit.)

So here, you will have YELLOW or GREEN. But never RED, exactly.

The exact currents in this last case are a little more tricky. But it's not terribly important. The value of \$R_2\$ can be adjusted according to need, easily.

I've not addressed the details with respect to how many of these are series-chained. But that's another detail regarding the selection of \$24\:\text{VAC}\$ and isn't quite as important. Fewer sections will mean brighter LED emissions. More sections will mean dimmer. But otherwise, the logic remains similar.

This isn't a well-managed system in the sense of the voltage drops for each section stacking up always the same. But it's probably good enough for this use. They could add more circuitry, I suppose. But they felt it wasn't necessary.

  • \$\begingroup\$ thanks for your thorough answer. I have one question; when the switch is closed w/o the resistor in parallel (the right most circuit on the first diagram) why doesn't any current pass through D3? Is this because of repulsive forces upon like charges or voltage potential must travel from high to low ( from the bottom of the circuit to the top)? \$\endgroup\$ – Niko_Jako Oct 18 '18 at 4:05
  • \$\begingroup\$ @Niko_Jako Isn't \$D_3\$ arranged in opposition to the applied voltage polarity there? How exactly would current flow through that pathway? \$\endgroup\$ – jonk Oct 18 '18 at 8:05
  • \$\begingroup\$ I think you confirmed my suspicion in my last comment-I struggle to put the reason into words but I understand why. It wouldn't flow through D3 because the polarity is "pushing" electron flow from + to -. Is that correct verbage? \$\endgroup\$ – Niko_Jako Oct 18 '18 at 16:56
  • \$\begingroup\$ @Niko_Jako There is no "correct verbage" I can suggest. Others use English to think with. I don't. I didn't even speak until I was 7 and I don't use language for thinking. I use images as best I can convey the idea (psychologists tell me they call them "gestalts".) You should use what language works for you. But here is how I'd write it: D3 is reverse-biased (off) because forward-biased D2 places a "minus" on the top/north end of D3 and R1 places a "plus" on the bottom/south end of D3. That's sufficient to know. \$\endgroup\$ – jonk Oct 18 '18 at 21:32

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