I'm not understanding Proteus or something I'm doing wrong, and it's really basic, I put a 9v Battery, a Red Led, simple circuit and a 10k resistance, so the led shouldn't turn on right?

Well, is turning on even If I set a 1000k resistance...what I'm doing wrong? I've done things in Livewire and I can see when a component burns, is there a way that the simulator in Proteus do that?

Thank you!!!

enter image description here

  • \$\begingroup\$ The LED will light up, just very, very faintly. \$\endgroup\$
    – Finbarr
    Jun 30, 2021 at 19:02
  • \$\begingroup\$ @Finbarr and is there a way that this looks more? I saw a Youtube tutorial, and even with a 1k resistance, the led didn't light up, I try to put 100 ohms, or 1000k ohms and always looks the same \$\endgroup\$
    – nanokuro
    Jun 30, 2021 at 19:11
  • \$\begingroup\$ It sounds correct to me. Passing current through a LED in the forward direction will cause it to light, there is no guarantee it will be bright enough to see. Yes many RED lids with light with ~1mA of current. \$\endgroup\$
    – Gil
    Jun 30, 2021 at 19:53

1 Answer 1


I put a 9v Battery, a Red Led, simple circuit and a 10k resistance, so the led shouldn't turn on right?

The LED should 'turn on' because it has current flowing through it. Assuming the red LED drops 1.7 V, it should draw (9 V - 1.7 V) / 10 kΩ = 0.73 mA. LED light output is directly proportional to current, so 0.73 mA would produce about 7% of the light at 10 mA. However the human eye's response is logarithmic, so the perceived brightness would be a little under 50%. Whether that makes it visible or not depends on the ambient light level. In normal room light it should be quite visible.

is turning on even If I set a 1000k resistance...what I'm doing wrong?

It seems that Proteus considers even a very small current to be 'on'. For a theoretical 'perfect' LED this is correct. In practice LEDs have internal leakage which 'wastes' a small amount of current that doesn't produce light. This sets a lower limit on the amount of current needed to 'turn it on'. Leakage varies between individual LEDs, so you won't know much current is required to get any light until you test it.

I took 7 red LEDs randomly selected from the same batch and powered them from 9 V. Here's a photo of it in natural room light (on a cold winter's day). In the center are single LEDs with 1 kΩ, 10 kΩ, and 100 kΩ resistors in series. On the right are 4 LEDs all in series with a 33 kΩ resistor, chosen to draw the same current as the single LED with 100 kΩ.

enter image description here

The LED with 1 kΩ drew 7.2 mA and was very bright, the one with 10 kΩ drew 0.73 mA and was fairly bright, and the one with 100 kΩ drew 74 μA and was quite dim but still visible. Of the 4 LEDs in series (each getting the same current) 2 were dim but visible, and 2 were so dim that I couldn't tell that they were on without looking very closely at them. 100 kΩ is clearly too high to reliably 'turn on' these particular LEDs from 9 V. But neither is it high enough to ensure that they will be 'off'.

Proteus probably has a very low threshold for what current turns the simulated LED 'on' because the real value depends on the individual LED's characteristics and the ambient light it is viewed under. A more accurate simulation might show varying brightness in the LED symbol, but this might be more confusing than helpful (does dark red mean the LED is 'off', or is it just dimly lit?).

  • \$\begingroup\$ Thanks a lot for this incredible answer, after hours looking everywhere, I could understand that Proteus has two ways to show leds, Digital and Analog, and Analog will display taking into consideration the resistance, so it gets close to a "real life" like the photo you attached, and Digital will display if the LED has current or no. So thank you! I learned a lot today, and getting more excited about electronics haha \$\endgroup\$
    – nanokuro
    Jul 1, 2021 at 7:00

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