I recently remember running across a bunch of articles on web about the 555-timer, it’s my first IC.

I’m getting more familiar with it, and I want to know run it in a-stable mode, with a relay between pin-3(Output) and ground. I will be using the relay to switch between two different LED arrays.

What concerns me is whether placing a relay directly between pin-3 and ground would cause a high current situation in the 555-timer.

Would it be necessary, and or possible, for me to put a resistor in series with the coil to limit current?

I’m just now getting experience with relays to, I know that they require a certain applied voltage to magnetize the coil to a point where it can effectively move the armature. I’m just not sure how this relates to KVL?

At the high state of pin-3 12V would be applied to the coil, I’m not quite sure, but I believe without a resistor, that would probably mean a lot of current would be sinked to pin-3. If I were to put a resistor in series with the coil of the relay, At high of pin-3, would the coil maintain the required voltage across it?

  • 1
    \$\begingroup\$ it works netter with the relay between out and +12V (or discharge and +12v) as the 555 has stronger pull-down that pull-up, if you can rearrange the 555 inputs so that output=low is the on state for the relay you'll be playing to the strengths of the 555.. \$\endgroup\$ Commented Nov 5, 2017 at 19:50

3 Answers 3


To elaborate on Peter Bennett's answer, here is the standard relay driver circuit I use. It removes the requirement that the 555 can supply the full current of the relay.


simulate this circuit – Schematic created using CircuitLab


  • Relay power doesn't have to be the same power supply as the rest of the logic, but it can be if that is adequate voltage for the relay. Ensure the relay power supply voltage is adequate to close the relay.
  • Select Q1 for adequate relay drive current. For large relays, the 3904 may not be appropriate. Also ensure that the device will saturate. A darlington connection may be required for very large currents.
  • D1 is a flyback diode required to protect Q1 from voltage spikes caused by the relay coil.
  • \$\begingroup\$ Thanks a lot, but that layout requires a lot more research on my end, just started using transistors. But It does look a lot more practical, I was going to apply 12V or more Across the 555. With your setup I can just use 5V probably, and use my Radpberry-Pi to drive the 555. \$\endgroup\$
    – Iam Pyre
    Commented Nov 3, 2017 at 19:12
  • \$\begingroup\$ Does the transistor have to be a Q1-2N3904? Do I have other options as far as selection of transistor? \$\endgroup\$
    – Iam Pyre
    Commented Nov 3, 2017 at 19:16
  • \$\begingroup\$ and NPN transistor capable of passing sufficient current should work. EG: BC547, BC337, 2N3055 etc. \$\endgroup\$ Commented Nov 5, 2017 at 20:08
  • \$\begingroup\$ @IamPyre also check the power dissipation: (relay current) * (the voltage drop from Collector to Emitter), for the given Collector current and base-drive current. The hFE (current gain) must also be sufficiently high that the provided base current puts the transistor into saturation (it may burn up, otherwise). BTW the 2N3904 is a very common transistor. A 2N2222 is another option. You could substitute a MOSFET, however there may be additional design concerns such as possible high frequency oscillation and VGS limits. \$\endgroup\$ Commented Jun 19 at 20:30

If you are running the 555 on 12 volts, you should use a 12 volt relay, and ensure that it does not require more current than the 555 can supply.

If the 555 cannot supply enough current to operate the relay, or if you must operate the relay and 555 from different voltages, you may require an additional transistor to drive the relay.

You might use a resistor in series with the relay coil if the relay requires a lower voltage than the 555 output provides.


Relay coils have internal resistance so if you know this number, can measure it, or can infer it from the relay data sheet, you can then consult the 555 datasheet of the 555 variant you have chosen to determine if the current the relay needs will be available from the 555 output.

NE555 can source 100mA from the output pin, so a1 at 12V is should be capable of driving a relay with a resistance of 120 ohms or higher. this would include most 12V "sugar cube" relays and common automotive horn and lighting relays.

that said the the output can sink 200mA, so if the 555 logic can be inverted connecting the relay between 12V and output gives stronger drive.

UA555 can source only 10mA bunt can sink 100mA, so can still drive a relay connected to +12 but is too weak to directly drive a ground-connected relay

  • \$\begingroup\$ Yes, I just measured the current from the voltage source in another setup. Only 30mA flows through the coil, so it seems safe to just connect the coil directly to the output pin. \$\endgroup\$
    – Iam Pyre
    Commented Nov 5, 2017 at 20:07
  • \$\begingroup\$ Wow, I just set up the circuit where the the 555 timer sends a signal directly to the coil of the relay. What I didn’t consider is how draining the relay is. I attach about 15V to the input pin and com of the relay. The circuit will oscillate for about a minute then suddenly only one set of lights will flash. I can still hear the clicking on the relay, which means the 555 is still signaling properly. I think at that point, the voltage on my cheap batteries have dropped to the point where they aren’t providing enough voltage to “shift” the relay. \$\endgroup\$
    – Iam Pyre
    Commented Nov 6, 2017 at 0:32

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