Actions
Forum Thread Details
  • Replies 8 replies
  • Subscribers 461 subscribers
  • Views 615 views
  • Users 0 members are here
Related

Overcurrent protection, what is a cheap and simple way

Catwell

I hope this message gets read....

 

 

 

In a circuit I am working on, I am powering two pins at the same time. These pins are powering a single LED.

When I power both pins I am doubling the current powering an LED.

 

What I need to do is protect the LED from excess amperage. What suggestions do you have? I know there are many ways of doing the same thing, I just want to collect them all.

 

Any help would be appreciated.

 

Cabe

  • Hi Cabe, please explain better, anyone reading your message is hit by several ambiguities, perhaps start by explaining what two pins you mean, if you mean the LED pins; then that doesnt make any sense?

  • High Cabe,

     

    If you want to have a controlled current source to drive an LED, which accommodates supply voltage variation and could be driven by more than one signal, that's easy.  You could do this with two transistors and 2 to 4 resistors (and 2 diodes if you need to control with two signals).  Unsure on the requirements.  It's also possible to use only one transistor, given the control signal is referenced to a regulated power supply.

  • in reply to Former Member

    Eclipze,

     

    I like where your head is at on this obstacle.

     

    Could you post a possible circuit of your idea?

     

    Cabe

  • in reply to Catwell

    Hi Cabe,

     

    I don't have enough information about the circuit you have.  You have 2 pins powering a single LED.  But I don't know what type of signals these pins have. It's better if you can define what you need to acheive, along with the systems contraints.  What's the power supply, is it regulated, what current are you targeting, what are the two controlling signals... 5V, 3V, a push button or switch etc...

  • in reply to Former Member

    Eclipze,

     

    Input voltages on both pins from an unregulated supply (no clue what it is, to be honest):

    Pin 1 :14.5V

    Pin 2 :14.5V

     

    Current on the LED should be 20mA.

     

    Normally, one one of the pins will be on, and the system will be fine. But my customers are not quite sure what they are doing, and might power both pins at the same time.

     

    I have limited space on my circuit board, and need a small, simple solution. Possibly close to the LED itself after the point where both inputs combine.

     

    Thoughts?

     

    Cabe

  • You really should have a specification for what the input voltage is.  Making assumptions is best kept to where you have experience with the application, otherwise it's the road to failure.

     

    There are sooo... many ways this could be accomplished, however it's also easy to pick the wrong approach without knowing enough information.  If you could find out what the input voltage was, you might be able to get away with two diodes (cathodes wired together) and a 620ohm 1/2W resistor.  Giving a solution with a nominal current of 19mA.  If the supply goes to 16.5V, the current would be 22.2mA, or 15.7mA at 11.5V.  Assuming you using a red LED.  That sort of variation isn't going to effect the brightness on a typical LED much.

     

    That's the lowest component count and footprint size.  Least complicated and least expensive.  You could even use a SOT23 with dual diode, common cathode.

     

    If the LED current needs to be more accurate, or the voltage swing on the inputs is higher, then we would need more circuitry.  This could be in the form of a zener diode to clamp the voltage.  Or it could be with a controlled current source.  But depending on the voltage, you may or may not need an additional resistor and you may also need something bigger than a SOT23 to dissipate the power.  If your circuit had a regulator somewhere, perhaps this could even be used.

     

    It really comes down to needing more information.  It would be a waste of time to go through and cover all bases, or direct you with a circuit that is otherwise overkill and may not fit etc... Otherwise I'd be here for an hour hahaa...

  • in reply to Former Member

    The easiest thing to do is add a Transient Suppression DiodeTransient Suppression Diode on each input. I plan on insisting to do so.

     

    I believe I can get these is small packages. Heat is an issue, but the whole board will be attached to a heatsink. I also have a few other components in the design that get incredibly hot, dissipating the excess power.

     

    Cabe

  • The easiest way is to put a resistor in series with the led. 14.5 Volt / 20 mA = 0.725 kiloohm.

    A 750 ohm resistor will do. If the outputs have a current control for the LED, the current now only will be limited by the resistor. If the output currents are limited with an internal resistor, the total resistance will be higher, resulting in a different brightness when two outputs are driving the LED. The maximum current however will not be exceeded. In that case a lower resistor value can be used, so in case of two outputs are driving the LED, the maximum current will not be exceeded.