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Take a look at this schematics and tell me what happens

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"Bench" tested. The 560Ω and 6.8KΩ resisters in series produce a dim glow from the LED. The transistor appears as an open switch when it's input is low. This indicates that the power supply is hooked up and the "Status" LED is functioning. When the transistors input is high, it shorts out the 6.8K resistor. That leaves only the 560Ω resistor to limit LED current, so the LED shines brightly. :)

I left the switch in the circuit and current draw is almost identical.
I think that's what can happen in "simulated" circuits. Like computer programs: "They do what you tell them to do!...exactly what you tell them to do!...and with a vengeful!" ;)

Ken
 
Another one is using an oscillator and disabling it so the LED is continuously on when the circuit is working properly and it starts flashing when something is wrong, for example, a battery low voltage detection system can be made this way using a couple of comparators.
 
I went back an tried my circuit with 100K instead of the 6.8K. Still got a dim light, but good enough to be a good, low-power indicator in room light levels. I was using a water-clear, super-bright LED. That's lit with only 90µA.

I've done the gated pulsing indicator before with one 4093 NAND gate and a PNP transistor. So combining the gated pulser and the dim/bright circuit...attached.

I think we've gone far afield from the OPs first post, and there are a hundred more possibilities...so I'm out. ;)

Ken
 

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The original posted hasn't been here in a week - I hate it when that happens. :(
 
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