Re: 40khz 555 IR transmitter overheating?!

Aug 28, 2003 4 Replies
A


-=-=-=-=-=-


>
>- Due to the very small pull-up resistor the 555 sinks about 109mA. The specs
> say the 555 can sink up to 225mA so it's well below the danger level.

Ah, but the way you have it connected, the 555 is SOURCING not sinking.


Move the LED up so it is connected between Vcc and OUT and then the


555 will be sinking current.

You don't need either a pullup or a pulldown on OUT.


Give it a try and let us know.


(I know, I know. It doesn't make sense to me either. I learned electron motion, and the electrical guys learn hole-motion)

Vote
Twitter Facebook LinkedIn
M

The continuous current of the IR transmitter LED is 150ma. Can the 555 safely sink this without a resistor? Should I add a 100ohm resistor between the led and OUT?

Thanks everyone,

-tom

Vote
Twitter Facebook LinkedIn
A

A 100 Ohm resistor between OUT and the LED will limit the current to a maximum of 5 (volts) divided by 100 (ohms) equals 0.02 Amps (20 milliAmps)

So if you want to drive allow the LED to draw a maximum of

0.15 Amps, use a

5 (volts) divided by .15 (Amps) = 33 Ohm resistor. And remember to make it larger than 1/4 Watt so it doesn't burn up.

Vote
Twitter Facebook LinkedIn
E
150ma is about 0.75 watts, I think the 555 will still overheat, since your running it continuously 100 - 220 ohms would be about right. The resistor should goto Vcc, the other end to the LED and then other end of the LED goes to the output on the 555, if you want to sink it. Reverse it to source it. If your using a MCU of some type, and then are intermittnetly sending data out the IR LED, you could have the I/O pin on the MCU drive a driver that drives the LED at 150ma pulsed and intermittently, but not continuously. If I remember right the Radio Shack LED's I used last had a max rating of something like 50ma.
Vote
Twitter Facebook LinkedIn
D

With a 10 ohm resistor you're putting about 300ma through the circuit IF the LED was connected correctly, your circuit isn't one of those. If you want to pump that kind of current through the LED, I strongly recommend you get an external power transistor (TIP120, whatever) and use that with a heatsink to power the LED.

DLC

:> Move the LED up so it is connected between Vcc and OUT and then the :> 555 will be sinking current. :>

:> You don't need either a pullup or a pulldown on OUT.

: The continuous current of the IR transmitter LED is 150ma. Can the 555 : safely sink this without a resistor? Should I add a 100ohm resistor between : the led and OUT?

: Thanks everyone, : -tom

Vote
Twitter Facebook LinkedIn

Join the Discussion

Have something to add? Share your thoughts — no account required.

Didn't find your answer?

Ask the community — no account required