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What do I need to turn off an LED 10 seconds after it powers on?

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GT3

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Hello everybody!

I'm looking to power off an LED around 10 seconds after it gets powered on. I would need it to have it as SIMPLE as possible and as CHEAP as possible as it'll be used in mass production. What will I need?

Any help would be appreciated - thanks! :)
 
plz mention the duty cycle, how long it would be on before turning off for 10sec?

When the device is turned on, the battery assembly gets continuity which then powers a low-power LED. As soon as it turns on, I'd like to have it turn off around 10-15 seconds later (the time isn't critical - I'm using this only as a simple power indicator).

I hope it is DC voltage you supplying.An RC network feeding to the combined two pins of a NAND gate will work.

Yes. My plan is to use a CR2016 coin cell battery to light the low-power LED. Do you have any examples you can link me to? Electronics is not my area of expertise so I'm completely lost!! ;) Are there any kind of IC's that has everything in one unit so it'll be quicker to manufacture (i.e. not having to solder multiple components)?

Thanks for the help! :D
 
If the LED is to be turned off never to function again until the next power up cycle, you could use a 555 timer in monostable configuration that would do the job. While this would involve some additional cost and labor, the end result would be exact and lasting.
 
You don't need big value capacitors, but big value resistors. :)

The attached circuit disables the LED after approximately eight seconds.

Boncuk
 

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use a single transisitor circuit with RC low curent trigger at the base. LED can be between collector and source in series with a current limiting resistor.
when you power on transistor is on, thus you get LED on, after a short time capacitor connected between base and source would be fully charged and transsistor would turn off.

hope it might work,

Razeen
 
I'm not sure on transistor idea that it can't delay longer time periods.Because 0.6 will fill on capacitors more quickly.

The OP has already selected the battery before the design.Surely he needs some low power design.
 
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GT3,

Does the battery power anything else, once something switches the LED off?

Ken
 
Wow, thanks for all the replies, guys! I'm completely new to this so I'm trying my absolute best to understand what you're all talking about.. :) The battery was chosen as it's very low cost ($0.13 or under in quantities of 5000+) and the battery will only be used to power this temporary LED power indicator light.

Once the LED is turned off, the light will never power up again until the next time the device is powered on. Efficiency of the design is not crucial, I'm all for a "quick, dirty and crude" way to make it simple.

What mbarazeen posted (with a capacitor, resistor and transistor) sounds like the simplest so far, but I'd like to get it even more simpler (like a all-in-one integrated circuit that I can buy, but maybe I'm dreaming!) :) I'm just trying to get this down to as little parts as possible to reduce the labor of making these, as soldering five parts in thousands of boards will suck! :p

Thanks for the help and any ideas!
 
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Does the switch that connects the battery to the LED (and timer) parallel another switch that turns on another part of your device? Most of the circuits mentioned will continue to draw some current from the battery after the LED goes out.

Can you post even a rough schematic of your device?

Ken
 
It's a 3v battery. You could use an LMC555 - CMOS Timer, as suggested. Standby current might be 0.16 ma. I saw this figure for the 7555.
 
Here is a circuit that take no power once the LED is extinguished:
**broken link removed**
This LED illuminates for a few seconds when the power is turned on. The circuit relies on the 47u discharging into the rest of the circuit so that it is uncharged when the circuit is turned on again.
 
Here is a circuit that take no power once the LED is extinguished:
**broken link removed**
This LED illuminates for a few seconds when the power is turned on. The circuit relies on the 47u discharging into the rest of the circuit so that it is uncharged when the circuit is turned on again.

It seems to me it's a perfect power saving circuit. With a base resistor of 47K (at 3 to 5V Vs) the transistor will never turn on.
 
If you use one a Single gate schmitt trigger you can replace the two smaller resistors and the transistor with one component.

The input current of the gate is zero (for all practical purposes) so the 47k isn't needed. The output is current limited so the 100 Ω isn't needed either. It also has the advantage that the input clamp diodes of the gate will discharge the capacitor when the power is turned off.

However, any circuit that has a large capacitor and take low current, can take a long time to discharge when the power is removed.
 
colin55 has posted the circuit for the same idea i explained, hope it would work much better.
as Diver300 told a FET can be replaced instead of BJT, removing two other resistors but it may increase the cost. also draining the capacitor when power is removed can be another problem, so may need another resistor across the capacitor.

gayan zoyza missed one point, transsitor will remain on until the voltage to the base drops bellow 0.6. can be set to get 10sec "on" without any problem.
 
It seems to me it's a perfect power saving circuit. With a base resistor of 47K (at 3 to 5V Vs) the transistor will never turn on.

It'll turn on the first time, however, there is no discharge path for the capacitor, so it won't turn on the second and subsequent times...
 
Hi all,

I guess bipolar transistors are the wrong choice. After the LED has extinguished the circuit should require minimum current.

Using an N-channel MosFet (BS170 would be perfect) the power on current flow decreases to 3µA after a short period. Using bipolar transistors the current stays well above 100µA for a considerable time and reduces to 30µA finally.

With the given dimensions the LED stays on for approximately 8 to 9 seconds.

Concerning discharging of the electrolytic cap a compromise has been met comparing the value of the cap with the discharge time. For a short discharge time the ON-time will be affected.

R4 must not omitted. Otherwise the circuit won't work.

Boncuk
 

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Wow, you guys rock! Thank you to everybody for the replies. :D I took a trip to RadioShack to get a few supplies in order to try out Boncuk's circuit as it sounded like it used the least amount of power so far. I bought the only N-channel MOSFET that they offered, specifically the IRF510. I built it according to the schematic and.. IT WORKED! :)

In practicality, I found that 10 seconds was a bit too short and that 20 seconds would be better, so I raised R1 to about 1.5mΩ (up from 470kΩ) which kept the LED lit for around 25 seconds. The capacitor discharge time takes around 30-60 seconds, which isn't a big issue.

I have a few questions:

1. The circuit would not light the 2v LED with a single 3v coin cell battery. Using two batteries (for a total of 6v), it worked fine. I am assuming that this is due to having a 2v LED and a 2v Transistor in the circuit, thus requiring a minimum of 4v. What alternatives can I use to get it down, so that I can power this circuit with a single 3v battery?

2. Why is R1 needed? Wouldn't it be the same thing if I combine R1 and R4?

3. Can I scrap R2 as R1 and R4 is already limiting the input voltage?

Now, I love Diver300's idea of using a Single gate schmitt trigger as it would eliminate two components to solder. Is anybody able to supply a Digi-Key or Mouser part number for the correct component in a through-hole design? I'm not quite sure what I'm looking for here. :p And, last question: If something like this is used, what kind of "standby" current will it use after the LED has extinguished? I'm not quite sure what this would be referred to in the data sheet.

Again, thank you to everybody for all of the help - I couldn't be doing this without all of you and it's highly appreciated! :)
 
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