Connect with us

Flashing LED problem

Discussion in 'LEDs and Optoelectronics' started by GM Creations, Oct 29, 2018.

Scroll to continue with content
  1. GM Creations

    GM Creations

    Oct 29, 2018
    I have been building a Gainclone amplifier over the past few months.

    On one of the many websites that I visited for advice, it was suggested to connect a capacitor across the contacts of the power switch, to avoid arcing. It is an LED illuminated single pole switch. The LED is powered by an old USB charger, that I am also using to power a Bluetooth module. The charger only receives current when the power switch is on. The issue is that when the switch is off, the LED flashes every 3/4 seconds. I presume that it is the capacitor that is allowing a small amount of current through. I have checked, and it is definitely the line, not neutral, cable that is switched.

    Could someone please advise me as to how I can prevent the flashing?
  2. duke37


    Jan 9, 2011
    What is a Gainclone amplifier?
    Draw a circuit diagram of what you have and the specification of the capacitor.
    Is the power switch in the mains or low voltage supply?
  3. (*steve*)

    (*steve*) ¡sǝpodᴉʇuɐ ǝɥʇ ɹɐǝɥd Moderator

    Jan 21, 2010
    It's an amplifier based on the LM3886 (nor 386!)
  4. dave9


    Mar 5, 2017
    Wire the LED with a series resistor to the amp positive power rail and ground. If the switch LED has a built-in resistor to get the appropriate current from a 5V supply, approximate the forward drop of the LED based on its color, measure it's current draw from 5V, to approximate the resistor value in the switch (or if you can open the switch and measure it, do that).

    Once you know that resistor value, merely subtract it from the calculated resistor value you'd need for that LED if it were powered through an external series resistor.

    Perhaps it is easier to show with an equation (based on ohm's law).
    5V supply
    "IF" Red LED, approx forward voltage 1.8V
    Measured Current (10mA = 0.01A in this example)

    (5V - 1.8V) / 0.01A = 320 ohms resistor estimated to be in the switch if the above #'s are true.

    Now if you had a 20V positive to ground power rail for the amp, you use the 320 ohms number above like this:

    (20V - 1.8V) / 0.01A = 1820 ohms

    You already have 320 ohms in series in the switch, so 1820 - 320 = 1500 ohms resistor in series from positive power rail to switch. Remember this depends on the variables above, which LED color and that forward drop voltage, and what current you measure the LED consuming off 5V.

    I would also think about getting rid of the USB supply to the bluetooth module using an LM317 (configured to 5V) instead of the USB supply for powering the bluetooth module, assuming that module is a modest current consumer around 20mA or less? Cleaner power might also help it perform a little better.

    I mention LM317 instead of LM7805 because it has a higher input-output voltage difference capacity. If the rail is no higher than 25V you can use LM7805 instead and not have to use a couple resistors that would've set the LM317 voltage but aren't needed for LM7805.
    Last edited: Oct 30, 2018
Ask a Question
Want to reply to this thread or ask your own question?
You'll need to choose a username for the site, which only take a couple of moments (here). After that, you can post your question and our members will help you out.
Electronics Point Logo
Continue to site
Quote of the day