magician13134

I'm using a 555 timer with a coin cell to power very bright LEDs, but my LEDs are on for about half a second, then off for about 5 seconds. My question is, is the circuit still using lots of power when the LEDs are off, or just a tiny bit of power? Thanks

ecerfoglio

A "plain old TTL 555" will use some power - and if you use a coin cell it may be a lot of power.

Use a CMOS 555 instead - they have part numbers like "TLC555" or "7555".

__________________
E Cerfoglio
Buenos Aires
Argentina

magician13134

Ok, and can you explain what exactly the difference is between TTL and CMOS? I know CMOS doesn't like static...

And maybe could someone look over the board and schematic attached to make sure it'll work? Basically, it's just a 555 timer circuit with two LEDs. Should it have a transistor to give the LEDs higher current? If not, does the circuit look ok?
http://www.magicsoftinc.com/555/Plane%20Lights.brd
http://www.magicsoftinc.com/555/Plane%20Lights.sch




Thanks!

ecerfoglio

It's true that CMOS doesnt't like staric. ....But it uses a lot less power

The NE555 (TTL) and its CMOS versions (TLC555 or 7555) are pin for pin compatible so you may try both of them with the same circuit board design.

In fact, if you use a socket for the IC (which I advise you to do - at least in the prototype stage) you may use the same board and test it with both IC's.

I can't open those files - To share something please use some "universal" file type like .bmp or .gif (or even Word's .doc or Adobe's .pdf)

Deppending on the current you want in your leds, a transistor could be a good idea - in fact, one of the drawback's of the CMOS 555 is its low output driving current.

I don't see any connection to pins 4 (Reset - goes to pin 8 +V) and 2 (Trigger, should be connected to pin 6), but with only this file and not knowing if your board is single or double sided . If it is double sided, please disregard this.

It is a good practice to add a small capacitor from pin 5 to ground (will work without it, but it's better to put it)

The leds are in parallel and without a current limmiting resistor - again, this is not a good practice, but it works because of the coin cell's internal resistance. Just don't do it with a bigger cell or battery.

__________________
E Cerfoglio
Buenos Aires
Argentina

Krumlink

I always remember: (4 to 8) (2 to 6). Then resistor a (8 to 7) resistor b (7 to 6). Capacitor negative (1 to 2). There you go. I never put a cap (0.01uf) from 5 to ground. 7555's can operate on as low as 2V, and they work just as well, and they use MUCH less power, I think around 200ua

__________________
MechTronics

World of Warcraft is a disease

magician13134

No, it's single-sided, what you see is what you get in that GIF. I was just following a simple 555 tutorial I had found on another site. Ok, let me redesign this and I'll repost both the schematic and board in GIF format. Thanks for the information!

Ok, it's updated. This should work, right? Thanks for the advice!



Hmmm, I just put this together on breadboard and it didn't work. If pin 1 wasn't grounded the LED (I only used one LED and left out the transistors for now) glowed dimly and quickly faded off. With pin 1 grounded properly, there was nothing. Any ideas?

Attached Images

Plane Lights.gif (51.9 KB, 16 views)

Roff

Pin 4 should also be tied through your jumper to +5V (along with pin 8 and R1), instead of connecting directly.
Unless your LEDs have internal current limiting, with a 5V rating, you will need a current limiting resistor in series with each LED. Without them, you will take out the LEDs or the transistors or your battery, or perhaps all of them.
I would lay it out so that the plane is the -V (ground) node, not the positive battery node as you have it, just because that's conventionally the way it is done. It shouldn't matter in your case, unless you want to probe it with an oscilloscope. In that case, the plane is the most convenient place to connect your scope probe ground clip, and none of the voltages will make sense if you connected your scope ground to +V.

__________________
Ron

magician13134

Yeah, I know GND is the convention, but in this case it was necessary to be +5V or there would be unrouted traces. Also, the +5V is a misnomer, it's +3V, but I always just use the 5V in Eagle since I know where to find it. And I don't think there's any easy way to attach pin 4 to the switch (my jumper is actually a switch)... hmmm, that could be a problem. Are there any problems that would make this not work, though? Because it isn't working on breadboard now...

Hero999

Yes, you must use a 7555, the NE555 won't work reliably below about 4.5V.

As the battery discarged 3V will soon become 2V which is fine using the CMOS 7555.

What current are you run the LED at?

The CMOS 7555's output current reduces drastically at lower voltages, you might even be able to get away without a series resistor on the LED!

__________________
I also post at the following sites:
http://www.stop-microsoft.org http://www.heated-debates.com
Screen name: Aloone_Jonez

audioguru

I can't see the pcb layout but I see errors in the schematic.
Pin 2 and pin 6 must connect to the capacitor and pin 7 must connect between the two resistors.

I would fix your schematic but everything on it is upside down (Australia?).

__________________
Uncle $crooge

Roff

I didn't notice that. Good catch!

__________________
Ron