timer circuit

Discussion in 'Electronic Basics' started by mike, Jan 26, 2004.

1. mikeGuest

Howdy,

I have looked through several websites but I could not find specifics on
this so I was wondering if someone more experienced could give me a hand.

I want to use a 555 to drive 20ish LEDs. All the example circuits I have
seen have used 2 or so LEDs. The LED I was thinking about using is 1.9 V @
36 mA. I was wondering how I could figure out the ouput voltage and current
of the 555.

I was thinking the output voltage would be close to the input voltage. For
the current, I saw something about a 200 mA output current. So, I was
thinking that using a 9V battery would allow 4 diodes in series across 5
parallels. eg:
-------- 4 LEDs ---
- -
-----4 LEDs----
... etc...

I am not sure if this would work. Any advice on this or a better way to do
it would be much appreciated.

Thanks,
Mike

2. John PopelishGuest

I think you have the right general idea, but you have to take into
account the slight variations in forward drop between each of your
parallel strings. You will need to add a ballast resistor to swamp
out these variations and to stabilize the current in spite of small
variations in supply voltage and output swing of the 555. \$ 1.9 volt
drop diodes in series will need a total of 7.6 volts. The 555 will
not be a perfect switch but will drop about 1 volt at 100 ma, all the
higher the chart goes on page 5 of:

http://www.national.com/ds/LM/LM555.pdf

So you have used up 8.6 volts and you are at only 100 ma and have left
only .4 volts for both the ballast resistors and the battery decay.

At this point it should be pretty obvious that the combination of 9
volts and a bare 555 are not going to drive 5 parallel strings of 36
ma (180 ma total) that each need at least 7.6 volts.

If you could come up with more voltage, say, two 9 volt batteries in
series, you are right at the absolute maximum rated supply voltage for
the 555, but you could drive 3 strings of 7 leds at 36 ma each with a
volts. So even when you lose a volt through the output, you still
have about 3.7 volts to waste across 3 resistors (one in series with
each string of 7 LEDs) to stabilize their currents. 3.7 volts across
a resistor that carries .036 amps implies a resistance of 102 ohms.
Since the batteries will not hold up very well at this current, you
are safe in lowering this to 100 ohms. And you get one extra LED
lit. But you are working the 555 very hard.

3. Robert C MonsenGuest

[a question about driving LEDs with a 555]
Another simple scheme that wouldn't stress the 555 would be to use a current
mirror, controlled by the 555:

<http://home.comcast.net/~rcmonsen/misc/20leds.GIF>

The trimmer R1 allows one to adjust the brightness of the LEDs by adjusting
the current through them. It can be omitted if you don't want to limit
current below 36mA.

Also, nothing should get hot, and the scheme can be extended to more LEDs
without worrying about pulling too much current out of the 555. (For those
wondering, the .IC=3V is required for the circuitmaker model to converge.
Its not a part.)

Assuming there are N strings of LEDs, the controlling resistance of R =
R7+R1 should be something like

R = (Vcc-0.6)/(((N+1)/ß + 1)*I)

where I is the current you want to pass through each of the LEDs, and ß is
the beta of the 2N4401 at that current.

Regards,
Bob Monsen