Series LED:

enter image description here

How do you make a LED circuit in series where the next LED will turn on, with equal brightness after a certain time delay?

For example, when I turn on the power, the red lights will turn on and after 10 seconds, the orange ones will light and 10 seconds after the orange LED lights up, the yellow ones will light up.

We have tried using this circuit for 1 LED but the LED did not have a delay before turning on. The following is the LED Delay circuit??

enter image description here

Where we computed the time delay as resistor value from +9V multiplied by the capacitance. (Source: http://www.ehow.com/how_7700687_add-delay-led.html)

  • \$\begingroup\$ Your links don't work and i really do think you should describe your circuit better or post a circuit diagram \$\endgroup\$
    – Andy aka
    Commented Apr 10, 2013 at 13:21
  • 5
    \$\begingroup\$ Best. Pictures. Ever. \$\endgroup\$
    – NickHalden
    Commented Apr 10, 2013 at 15:08
  • \$\begingroup\$ sorry, my reputation is just 1. \$\endgroup\$ Commented Apr 10, 2013 at 15:34
  • \$\begingroup\$ @Nick: They are understandable, so no -1, but we don't really want to encourage others to make diagrams like this either. \$\endgroup\$ Commented Apr 10, 2013 at 22:32
  • \$\begingroup\$ Should someone direct @Dan to meta.electronics.stackexchange.com/questions/963/… \$\endgroup\$ Commented Apr 11, 2013 at 15:42

2 Answers 2


That circuit is not the best way to do this, since you will a very large capacitance for any reasonable delay. As stated in the link, 0.01F (10,000uF) is a huge amount for only 3 seconds, which will make the capacitor physically large and not so cheap. For 10 seconds you will need even more than this.

A better way is to use the RC circuit together with a MOSFET for each set of series LEDs (you can use a bipolar, but you need to take the base current required into account) in a circuit such as this one. Due to the MOSFETs almost infinite input impedance, and being able to separate the RC control from the LED drive, we can use high value resistors and therefore smaller capacitors than the RC only circuit:


LED Delay


LED Delay Simulation

I have used all the same LEDs here, obviously the series resistors will need to be adjusted for the desired current in your circuit. The MOSFETs can be any general purpose N-channel capable of handling the current (if it's <100mA then almost anything will be suitable). Watch the gate-sourve voltage doesn't exceed the MOSFET specs (check datasheet) - if your supply is >10V you may need to use a divider to feed the capacitor.

The turn on time is dictated by the MOSFET threshold voltage and RC time constant. You can alter this as necessary by changing the capacitor/resistor values. The timing won't be very accurate, as threshold voltages can vary quite widely, and electrolytic capacitor value tolerances are not so good, but it should be suitable for a rough delay circuit (you can tweak to suit)
If you need more accurate timing, a timer IC or small microcontroller would be better.

  • \$\begingroup\$ Your answer is great and I can see your point telling that using the circuit i posted will result in the capacitor being large. The only problem with your answer is that I don't know MOSFETs are used. Can't a BJT be substituted? \$\endgroup\$ Commented Apr 10, 2013 at 15:38
  • \$\begingroup\$ If you know how a BJT works, then it shouldn't take long to pick up how a FET works, if anything they are easier to understand. You can follow the schematic for connections, the design should be pretty forgiving even if you are not sure how it works at first. As mentioned in the answer, BJTs can be used, but, depending on the transistor gain, you need to make sure enough base current flows for adequate LED current, which means you can't make the resistors quite as large (unless you use a darlington configuration or similar) \$\endgroup\$
    – Oli Glaser
    Commented Apr 10, 2013 at 15:49
  • \$\begingroup\$ Thank you for your help. I'll try reading into MOSFETs and would try your schematic. \$\endgroup\$ Commented Apr 10, 2013 at 16:18
  • 1
    \$\begingroup\$ +1, excellent, a solution that meets the question's spirit, rather than a cookie-cutter "go MCU" response. \$\endgroup\$ Commented Apr 10, 2013 at 19:26
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    \$\begingroup\$ @OlinLathrop "Learning" is a good reason. Telling someone who's wiring up his first LEDs, resistors and batteries, to start off with a microcontroller, is a bit detached from reality. \$\endgroup\$ Commented Apr 10, 2013 at 22:56

The obvious answer is to use a small microcontroller. That will be a lot fewer parts than trying to do this in analog. You can then also arbitrarily program the delays. You only need a micro with 4 output pins and preferably a internal oscillator to make the circuit simple. I'd use one of the PIC 12F parts. These come in 8 pin packages and have up to 5 output pins, which is one more than you need.


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