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need schematic.

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sparky3489

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I've been searching for a sound activated LED circuit/schematic with little luck.

I need one that'll drive 8 high output blue LED's with the following specs:

Luminosity: 5000mcd @ 20mA
Viewing Angles: 25°
DC Forward Current: 30mA max
DC Forward Voltage: 3.3V typical
Reverse Voltage: 5V max

I've calculated I'll need an 82 ohm 5 watt resistor to drive them all. So I'll also need a pretty hefty transistor as well.

I don't want any electrical connection to my source (must use a mic), has to be non-latching and will run off of cars electrical system.

I really don't have time to bread-board at the moment, so an "out-of-the-box" answer would be great.

I've seen kits that resemble what I need, but I'd rather build it myself.

Any ideas? Please e-mail me at sparky3489@yahoo.com
 
hmmm, how did you get 5 W? I come up with .034W for an If of 20 mA

Your transistor will need to handle 160 mA (8 * 20 mA) which is pretty small - any small signal NPN can handle it.

look here for sound activated switch circuits.
 
I think Sparky used 30 mA*8 for total current, and I2R for wattage. John
 
jpanhalt said:
I think Sparky used 30 mA*8 for total current, and I2R for wattage. John
I wouldn't use max current.

1.7V/.02A = 85 ohms. .02*.02*85 = .035W
1.7V/.03A = 57 ohms. .03*.03*57 = .051W

If he paralleled them with a single resistor then it would be 1.7V/.16A = 10 ohms and .272W (1/2 Watt resistor). However, you should never use a single resistor for 8 LEDs but one per LED. 1/8W would do fine.
 
sparky3489 said:
So if I use a resistor for each, then a 560 ohms resistor would do. Is this right?

sigh. ohms law is V = I*R. rearrange R = V/I = (5V-3.3V)/.02 = 85 next higher standard 5% value is 91. W = V * I = 1.7*.02 = .034 1/8 watt will do fine. However, you could probably use an 82 ohm resistor with no ill effects. This presumes you are using a 5V supply. If your supply is different, you will need to recalculate the numbers. Note that the LED "consumes" 3.3v so you only need the resistor for the remaining "unconsumed" voltage in order to set the current. (yes, I know this is imprecise but it makes it easier to explain)
 
sigh. ohms law is V = I*R. rearrange R = V/I = (5V-3.3V)/.02 = 85 next higher standard 5% value is 91. W = V * I = 1.7*.02 = .034 1/8 watt will do fine. However, you could probably use an 82 ohm resistor with no ill effects. This presumes you are using a 5V supply. If your supply is different, you will need to recalculate the numbers. Note that the LED "consumes" 3.3v so you only need the resistor for the remaining "unconsumed" voltage in order to set the current. (yes, I know this is imprecise but it makes it easier to explain)

...will run off of cars electrical system.

This is what I have so far.
**broken link removed**

The original circuit called for a 6 volt supply. Will the 14.4 cause any foreseeable problems, or do I need to use an LM7806.
 
I just figured out that this won't work as the LED's are set to see the 14 volts.

So I thought about separating the emitter of the drive transistor from the circuit and connecting it (on its own) to the 14.4V. Then I can use the 7806 like I mentioned.

Would I need to to lower the base resistance a little to compensate for the lower feed voltage to reference voltage ratio? or is that just for current limiting?
 
I suppose that would work. If I were doing this, I would use a VReg for the LEDs since the auto voltage isn't super stable. Also, don't forget that people sometimes put batteries in backwards - protect against that. You may be very unhappy if you don't.

I don't see why that circuit won't work for 12V though I'd use 3 LEDs in series (2 strands of 3, one of 2).
 
I shouldn't have to worry about "reversed battery syndrome" since this circuit will be swithed. Besides, I change my own battery.

This circuit is going to be used for my speaker box. The LED's will be in the box with these subs.

**broken link removed**

My only limit is I'm going to use two banana jacks on the box to connect the LED's with. So the resistors and LED's will be inside, while the rest of the circuit is outside the box.

Change in design:

**broken link removed**
 
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sparky3489 said:
Does anyone forsee any problems with the previous diagram?

try it. build it in stages and test each stage before moving to the next.

I shouldn't have to worry about "reversed battery syndrome" since this circuit will be swithed. Besides, I change my own battery.
heh heh, famous last words.
 
heh heh, famous last words.

Seriously, a car battery. Who puts one in backwards?!?!?

Actually, in my car it's impossible, the wires are intentionally short so it will only go in one way. Whether front to back or side to side, the cables will only reach so far.

Besides, Ive changed designs again.

**broken link removed**
 
reversed car batteries happen all the time. it's one of the things on automotive electronics designers checklists to protect against. wrong jumper cable connection and a truely dead battery can do it. but hey, you build it however you want, it's just a diode, though.
 
True. The .7 volt drop isn't critcal either.

Alright, you talked me into it. It won't be very hard to add it to my PCB layout.

Did you notice design #5. I think this is it.

**broken link removed**
 
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Well crap!!

I found what I was looking for with no guess work involved. I'll still have to tweak it a bit, but I think I've got a handle on that.

**broken link removed**

**broken link removed**

So I would like to thank everyone who helped me out in my quest.

peace.
 
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The last two circuits are missing the capacitor that holds the peak voltage for a moment to allow you to see it. The circuits without the capacitor will flicker very dimly on the peaks of the sound.
 
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