I tried to make a simple 555 timer circuit which turns on one LED when its output is HIGH and turns on another LED when its output is LOW. I tried connecting two LEDs with single 1k ohm resistor (connection 1). The top LED turns on and the bottom one turns off during LOW state of 555 timer. During HIGH state, top LED just reduces its brightness and bottom LED turns on. Moreover, it spoiled the IC. I initially thought there is some issue with IC and tried another IC. The results were same. Later I realised, both of them were spoiled because of this connection (when I used these ICs for some other project). I searched in the internet and found connection 2 in every search results.

I measured the voltage at pin 3. It was like 2 v lesser than VCC during its on time. My questions are

  1. Why it spoiled the IC?
  2. What difference it makes when there are two resistors which make the circuit to work as intended?

Note: R1 and R2 values were high enough to produce visual observation of LED. This image I just took from random site for this question

enter image description here

  • \$\begingroup\$ I already wrote something on this topic. It's not an exact fit. But the BJT drivers shown there will do what you want done. You do not want the two LEDs forward-biased with respect to the supply rail (connection 1) and ground! Better not even if you place resistors in series (connection 2), though that starts to get more argumentative. I'd just add the BJTs and be done with it. That said, connection 1 is not acceptable under any conditions. Ditch it. \$\endgroup\$ Commented Sep 2, 2023 at 6:38
  • \$\begingroup\$ Thank you or your reply. I went through your post. Connection 2 is working. I completed what I wanted. So, could you explain why Connection 2 is also not preferable. My main concern with Connection 1 is, it destroyed the IC. Why did that happen? \$\endgroup\$ Commented Sep 2, 2023 at 9:25
  • \$\begingroup\$ The NE555 is a very old design that uses an old TTL output circuit that has an output high voltage that is about 2V less than the supply voltage. \$\endgroup\$
    – Audioguru
    Commented Sep 2, 2023 at 14:48
  • \$\begingroup\$ The voltage/current curve of an LED is very steep. For example, with a blue or white LED, no current will flow until the forward voltage exceeds ~3.00 V. Without a limiting resistor, the LED might draw 10 mA at ~3.1 V, 50 mA at 3.2 V, and 250 mA at ~3.3 V, at which point it emits the "magic smoke" and ceases to work. (All numbers are very approximate, and vary with LED chemistry!) \$\endgroup\$ Commented Apr 15 at 17:22

2 Answers 2


Connection 1 should not damage the IC, rather it would kill or damage one or both LEDs unless you had a much lower supply voltage than shown.

I can only assume that you had a connection error in the first case or something else happened to kill the ICs. Or perhaps the ICs are not dead but don't work because the other project was not well designed (and perhaps some other kind of 555 did). Nothing we will be able to figure out, I fear.

Bipolar 555 chips (eg. NE555) do not have a "rail to rail" output, even when loaded with just a few mA. At (say) 5mA sink/5mA source the output will typically go from about 0.1V to about 10.5V with a 12V supply. The problem is that 1.5V (12V - 10.5V) is enough to get some light out of many red LEDs, so it may not turn off completely with circuit 2.

CMOS 555s would typically work at a few mA but they tend to have weak drive capability and can suffer from the same problem if the LED current is even 5mA (worst-case).

Here is a way to use the NE555 that requires a couple more parts. R2 is made a bit lower than R4 to account for the typical 10.5 output high voltage (rather than 12V) with 5mA load and 12V supply. It wastes about 5mA though R1 when D1 is off, so average waste is 2.5mA. If you're concerned about 2.5mA you should not be using a bipolar 555 which wastes 5 or 10mA just sitting there.


simulate this circuit – Schematic created using CircuitLab

  • \$\begingroup\$ Thank you. Yes, after providing such poor connection (1), I realized it should have burned the LEDs. But it didn't, instead I faced issue with IC. That's what puzzling. Anyways, I leaner that even the second connection wouldn't be good. Thanks once again for educating me. \$\endgroup\$ Commented Sep 2, 2023 at 10:18

I know this is an old thread, but here is an alternate theory of the crime.

When the 555 output is high, D1 is reverse-biased and off; D2 is powered through R2.

When the 555 output is low, D1 steals the current away from D2. This is because the D1 forward voltage (Vf) is lower than the combined Vf's of D2 and D3. D1 effectively "shorts out" D2.

Also, this schematic shows an alternate scheme for the timing components. With this arrangement, the 555 output duty cycle is almost exactly 50/50, using one fewer component.

NOTE: This circuit works only for low values of Vcc. At 12 V, the reverse voltage across D1 in the off state is almost 9 V. Most LEDs are not rated to withstand this.

enter image description here


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