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Step 3: Build the LED circuit (with tilt switch)

Picture of Build the LED circuit (with tilt switch)
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A string of LEDs... (pictures 1, 2, 3)
The six leds in the tube are connected parallel (all the led's plus-sides are connected, and all the minus-sides too). Use stranded (supple) wire to connect the leds, so they fit into the tube nicely later on. To connect the leds to eachother, I soldered 6 strings of wire pigtail, and then soldered the pigtails onto the led's leads. See picture 3 on how to do that. If it's not clear to you then, have a look here for a more detailed explanation...

Blinking LED circuit (pics 4 through 8)
The circuit that makes the leds blink (picture 4) is a very basic circuit around a 555 timer. This is not the most efficient way to do the job, because the 555 timer consumes a fair amount of energy just to stay alive. I used this scheme just because I had the soldered pcb lying around (it is a remnant from a previous I'ble I made). On the pcb a switch was soldered already. I replaced this switch with a roller ball switch.

This circuit should be turned on when the Sound Tube is set vertically, with the speaker-side up. A roller ball switch (or a tilt switch) takes care of this. 
  • Solder the components onto a pre-etched pcb. Cut the copper strips where marked with an X in the drawing (picture 5).
  • Solder pieces of supple wire of about 8 cm (3") to the roller ball switch' leads.
  • Eventually, solder the string of six leds to the pcb and test the circuit.
I tried to make a roller ball switch myself, but the attempt failed miserably. See pictures 10 through 15. None of these switches work :-s

Powering the circuits
The voicerecorder and the led-blinking circuit are powered by the same 9 Volt battery.
I made a seperate, small pcb on which the battery's leads and a switch are mounted. The powerleads to and from the voicerecorder and led-blinking circuit are all connected to this pcb.
The switch is used only to shut down the Sound Tube completely.

Optional (picture 9): Leds fading in and out
What I really wanted was an array of leds that fades in and out. I spend hours trying different kinds of circuits, but none of them worked the way I wanted to. Making a led fade in was not so hard, but letting a led fade in and out without a microcontroller is truly a tricky task to accomplish...
Eventually, I got help from Jan Leisink, probably the best science teacher in The Netherlands. He came up with a circuit that almost worked the way I was looking for. After some tweaking, I got it right. Unfortunately, that was after the Epilog Challenge deadline, so that's why it's not in the Sound Tube I made.

Now, if you build the circuit in pic 9, chances are that it will not work just as you expected: The way the leds fade in and out depends strongly on the values / specs of four components in the lower left side of the circuit: The leds, the transistor, the capacitor and resistors. To make tweaking more easy, I put a variable resistor in the circuit. Adjusting it's value influences the way the leds fade in and out.

A big advantage of this fade-in fade-out circuit is that it is much more efficient in energy than the 555-timer circuit. So using this circuit will make the 9V battery last a fair bit longer. Give it a try, if you like.
 
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yoh-there1 month ago

An alternative circuit is a straightforward a-stable using the 7555 (actually, you gave me that idea, used it for a blfnar). A very nice setup is to put 6 LED's anti-parallel and put a decent sized capacitor in series. if you then also put a resistor parallel *) to the LED's it ensures the full (dis)charge of the cap. Whenever the 7555 flips, 3 LED's will flash on, fade out. Very nice effect.

*) you best size the resistor that multiplied by the value of the series capacitor is roughly the same as the R*C of the 7555.

ynze (author)  yoh-there1 month ago

Sounds interesting... What do you mean by "anti-parallel"? Could you draw a circuit to explain?

yoh-there ynze1 month ago

See image, sorry for the quality, draw in yoh-draw ;-)

The idea is if 3 goes high, current rushes through the top 3 LEDs. As the C charges, the current will fade away. The LEDs will be off under about 1.6 V at which point the resistor will ensure the capacitor is fully charged. When the 3 goes low, the capacitor discharges over the bottom 3 LEDs and again the resistor will ensure a full discharge.

The resistor is needed if that last 1.5 V is really needed. It is when powering from a 3V source. If you stick with say a 9 V battery, no need at all as the C will charge to 7.5V (battery minus LED forward voltage).

I would recommend a small current limiting resistor in series with the C, though the 7555 can be mishandled somewhat.

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ynze (author)  yoh-there1 month ago

Cool! All clear, thanks!