On Mon, 13 Aug 2007 12:56:47 +0000,
MrJogo@gmail.com wrote:
> I'm an absolute beginner to electronics with basic understanding of
> the principles (resistance, voltage, Ohm's law, etc). I need to make a
> battery-powered circuit that has two separate groups of ~20 LEDs (each
> will eventually be arranged into an image) that blink so that when one
> group is off, the other is on. Some online research leads me to
> believe that I want a 555 timer (mainly these pages:
>
http://www.instructables.com/id/E6GMNH7 ... /?ALLSTEPS and
>
http://ourworld.compuserve.com/homepage ... 55leds.gif),
> but I'm not totally sure how the circuit should be built since I don't
> completely understand circuit schematics. To ground the circuit, do I
> just connect the wire to the negative terminal of the battery?
Yes. Every point which is marked with a ground symbol should be connected
to the negative terminal.
> Is the
> 555 suitable for what I want to do?
Yes.
> If I want each LED group to be on
> for about 2 seconds, how do I calculate the values for R1, R2, and C
> (all the calculators only go the other way)?
The interval for which pin 3 is high is proportional to (R1+R2)*C and the
interval for which pin 3 is low is proportional to R2*C. The cited values
of R1=4k7, R2=150K, C=1uF result in on/off times of ~100ms (R1 is much
less than R2, so R1+R2 is approximately equal to R2).
So, increasing the product R2*C by a factor of 20 will result in 2 second
on/off times (i.e. a total period of 4 seconds). Pin 6 will draw up
to 0.25uA, which (along with the leakage current of C) determines the
upper limit of R1+R2.
Simply multiplying R2 by 20 would give 3MOhms; without any capacitor
leakage, that would result in a constant 0.75V drop due to the current
drawn by pin 6. The capacitor voltage oscillates between 1/3 and 2/3 of
Vcc, i.e. 2-4V for a 6V supply. A 0.75 drop would be enough to make the
output noticably asymmetric, i.e. one group of LEDs would be on for
noticably longer than the other.
Also, at such long periods, capacitor leakage current is likely to be a
major factor with an electrolytic capacitor. My instinct says change
C1 to a 10uF tantalum electrolytic and R2 to 330K. That gives ~18uA
charge/discharge current, which is significantly above the currents
sourced/sunk by pins 2 and 6 and typical capacitor leakage (note that a
normal aluminium electrolytic could easily leak the entire 18uA, so that
isn't an option).
If you want a higher resistance and a smaller capacitance, you'll probably
need to abandon electrolytics in favour of e.g. polyester, which will have
a significantly lower leakage at the expense of a larger package and
higher cost (Farnell quotes £0.17 for a 10uF tantalum vs £1.20 for a
10uF polyester; the fact that the poly can handle 100V vs 6.3V for the
tantalum doesn't help you here).
OTOH, for a one-off, you may as well ignore the cost and just go with
C = 10uF poly, R2 = 330K. That will allow you to treat both the capacitor
and 555 as "ideal" (i.e. you can ignore currents other than the actual
charge/discharge current).
> If I wanted to be able to vary the period after it was built (say, with
> a knob) how would I do that?
Use a variable resistor for R2.