With a circuit of a charged capacitor, a resistor, and an LED, I can create a lighten up LED that gradually dims until the capacitor run out of charge.

With a combination of a battery, capacitors, resistors, and LED, is it possible to create the opposite effect, that is an LED that gradually lighten up? If not, what kind of components that I might need to produce this effect?

  • \$\begingroup\$ you can put the capacitor in parallel with the LED, the resistor in series with both of them.. its an RC filter.. its not very efficient but you will create the effect you want \$\endgroup\$ – Wesley Lee Oct 8 '15 at 23:38
  • \$\begingroup\$ @WesleyLee - probably not. The LED current is so large that the cap will need to be enormous. And the problem is made worse by the exponential V-I curve of the LED, which produces a large change in current (and brightness) for a small voltage change near the turnon threshold. \$\endgroup\$ – WhatRoughBeast Oct 8 '15 at 23:58
  • \$\begingroup\$ True! But considering the components stated I think it would be the only possible configuration.. although in practice it would be useless \$\endgroup\$ – Wesley Lee Oct 9 '15 at 0:23

The other component you need is a transistor. Try something like this, and it should take several seconds to turn on.


simulate this circuit – Schematic created using CircuitLab

  • \$\begingroup\$ Hey it works, but I'm yet to make sense of what's going on. So I think the purpose of the big R1 here is to slow down the capacitor's charging process? Does it serve any other purposes? \$\endgroup\$ – Andree Oct 10 '15 at 2:36
  • 1
    \$\begingroup\$ @Andree - It does two things. First it charges the capacitor. For a large capacitor/resistor combination charging is slow enough to be seen. If you replaced C1 with a 0.1 uF the LED will seem to turn on instantaneously. At the same time, it provides a very small current into the transistor base. The transistor amplifies this current to a level high enough to turn on the LED. You'll have noticed that the LED does start to show light for a second or two. That's because the capacitor voltage is not yet high enough to provide the LED voltage needed. \$\endgroup\$ – WhatRoughBeast Oct 10 '15 at 14:00

An LED is very non-linear. Most of the time folks use PWM to control the brightness. (Full on or full off for varying fractions of time and our eyes do the averaging).

  • \$\begingroup\$ This is not an answer to the question. \$\endgroup\$ – Transistor Oct 14 '15 at 19:27

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.