I have an input signal that is either 2.7V (low) or 9.9V (high). Is there any way to use passive components (excluding op-amps) to turn an LED off when the signal is low and on when the signal is high? This seems straightforward, but nothing is coming to mind.

  • \$\begingroup\$ What power supply rails do you have available, as well? \$\endgroup\$ – jonk Jun 5 '19 at 20:07
  • \$\begingroup\$ 24V. I considered adding just a logic level shifter and a couple of linear regulators but I'm not sure if a logic level shifting IC could handle the current draw. \$\endgroup\$ – ccolton Jun 5 '19 at 20:12

Zener diodes do not conduct in the reverse direction until their breakdown voltage is exceeded.


simulate this circuit – Schematic created using CircuitLab

Figure 1. A simple solution.

How it works:

  • A 3.3 V Zener requires a voltage > 3.3 V before it will experience reverse breakdown. At 2.7 V input the Zener will not conduct and D2 will be off.
  • At 9 V the Zener will breakdown and 3.3 V will be dropped across it. You didn't specify the LED colour so we'll assume 2.2 V is dropped across it which leaves 9 - 3.3 - 2.2 = 3.5 V to be dropped across R1.
  • Setting R1 to 470 Ω will give a current of \$ \frac {V}{R} = \frac {3.5}{470} = 7.5 \ \text {mA} \$.

Your input signal needs to be able to source that much current or else you can't do this passively.

  • \$\begingroup\$ Much appreciated, this is exactly what I was looking for but couldn't think of. \$\endgroup\$ – ccolton Jun 5 '19 at 20:19

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