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Is it possible to use an analog circuit to turn an LED on for 10 seconds after 5v power is applied, then turn it off until power is removed and reapplied? This would be useful in small battery-powered designs; if a "power status" LED turns on for a short time after you've applied power, you are sure that the circuit is working...no need to constantly keep it on and drain more of the battery.

So, this is what the process would look like:

  1. Plug power in
  2. LED turns on
  3. 10 second delay
  4. LED turns off
  5. infinite delay

Is there a simple analog circuit that can do this?

Update: Thanks for the help on this one. I agree that a single big cap and resistor solution would be too expensive/big to be practical. I think I'll use a MOSFET or some transistor array using Olin Lanthrop's schematic below.

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    \$\begingroup\$ This is identical to question asked recently but the delay wanted was 30 seconds. Find that question and look at the answers. Essentially, in your case the answer is use a '555' timer configured as a monostable and connect the LED via a resistor to pin 3 = output. \$\endgroup\$
    – Russell McMahon
    Feb 20 '12 at 22:52
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    \$\begingroup\$ Is there any way to do this with just a capacitor/resistor combination instead of an IC? \$\endgroup\$
    – mr_schlomo
    Feb 20 '12 at 23:13
  • \$\begingroup\$ @mr_schlomo You would need to charge a cap that would sustain the desired amount of power to light the led. I think you would have to use diodes (maybe even a transistor/ or gate) in a clever way to cut out the charging after the cap is full.. but this could potentially cost more than a 555 timer...(bulk 5k+ = USD0.09, single USD 0.40) \$\endgroup\$
    – Piotr Kula
    Feb 21 '12 at 15:54
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Here is a circuit that should work (untested):

These are all cheap jellybean parts and will be together still significantly cheaper and smaller than a large capacitor to provide the LED current directly. The LED will also be on at full brightness for most of the time with a short, if any, soft off near the end.

D2 ensures that C1 is discharged when power is off. C1 therefore starts discharged when power is applied. C1 slowly charges thru R2 to provide the timing. Adjust R2 to get the desired LED on time. The value shown is a rough guess. Experimentation is required at these low currents.

Q1 is on as long as C1 is charged to below 4 to 4.5 Volts or so. Q1 turns on Q3, which turns on Q2, which turns on Q4, which turns on the LED. When C1 charges up high enough, Q1 turns off, turning off all the rest, including the LED. The LED current is set to about 10 mA in this example. Adjust R3 according to the desired LED current.

Leakage when off should be low enough to be ignorable. If not, add 1 MΩ resistors accross B-E of Q2 and Q4.

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    \$\begingroup\$ A whisper of hysteresis would allow a rapid turnoff. Maybe Q3b - Q2c or similar. || A single FET would allow a 10 second delay with ease. \$\endgroup\$
    – Russell McMahon
    Feb 21 '12 at 15:03
  • \$\begingroup\$ Just out of curiosity, what does the four bjt configuration buy us here? \$\endgroup\$
    – drxzcl
    Feb 24 '12 at 18:11
  • \$\begingroup\$ @drxzcl: The 4 transistors are part of the circuit and are required for it to function as intended. If you mean to contrast this scheme to another, you have say what the other scheme is. In other words, buy you compared to what? \$\endgroup\$ Feb 24 '12 at 18:56
  • \$\begingroup\$ Fair point, I'll study the schematic a bit longer and come up with a more precise question. Intuitively I would have thought it could be done with two (turn on/turn off) in a flip flop-like arrangement, but I don't have a concrete proposal. \$\endgroup\$
    – drxzcl
    Feb 24 '12 at 21:39
  • \$\begingroup\$ @OlinLathrop: I think drxzcl is asking why the second stage of BJTs, instead of running the LED (with limiting resistor) directly into Q3's collector. \$\endgroup\$ May 2 '12 at 19:59
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Is there any way to do this with just a capacitor/resistor combination instead of an IC?

A: Capacitor from +5V to resistor to LED to ground.
Reverse diode across LED.

When 5V steps o -> +5 LED will light.

I_LED initial for red LED ~= (Vcc-Vled)/R = (5-2)/R = 3/R for red LED
Or R = 3/I

LED will dim exponentially.

Time constant ~~~= RC.
So C =~ t/R

For say 20 MA initial R = 3/I = 3/0.020 = 150 ohm.
C = t/R = 10/150 =~ 0.06 Farad = 60,000 uF.
That's doable with normal caps but large.
An appropriately rated supercap would be easier to do.

Doing this with a single super cheap transistor and a few Rs and cs is so easy and so superior in result that the cap and R solution makes zero sense in almost any context.


Supercap vs Standard Al electrolytic - not as big a difference as many would assume:

The phantom commentless downvoter is back.
IF this is for the statement that

  • "that's doable with normal caps but large"

then it shows a lack of understanding of what was being said
AND a lack of appreciation of what is available.

A supercap would probably be the best choice but the point was (as the words say) that you COULD do this with conventional caps BUT it is at the very high end of the range where you would.

You can buy an eg a 47 mF 5V5 supercap here for $1.58/1
and you can buy a 47 mF 10V standard aluminium electrolytic cap here for $3.75/1. In this case the supercap is about 42% of the price of the Al cap BUT the Al cap is 10V rated so can have = 3.3 x as much energy at full charge so in terms of energy storage per $ the std cap is cheaper. ie a bit of looking around would probably find a lower voltage Al std cap that is cheaper for the supercap wity the same desired spec. BUT the differebnces are close enough that it's not really important in most cases. Other attributes would make the difference.
eg the std cap is much much much less susceptible to over voltage damage,
BUT it's far larger - in many cases the supercap would be the preferred choice. But not all.

"Horses for courses", and be wary of forming opinions about what "old tech" is capable of just because "new tech" seems to be so much better. Often it may be. Sometimes it isn't.

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    \$\begingroup\$ A supercap is an absolute "must" here not only because 60 mF/ 60,000 uF is a really big aluminum electrolytic, but also because it's far cheaper. That said, you can buy several dozen transistors or several 555s for the price of one of those big caps... \$\endgroup\$ Feb 21 '12 at 5:50
  • \$\begingroup\$ The phantom commentless downvoter is back. IF this is for the statement that "that's doable with normal caps but large" then it shows a lack of understanding of what was being said AND a lack of appreciation of what is available. A supercap would probably be the best choice but the point was (as the words say) that you COULD do this with conventional caps BUT it is at the very high end of the range where you would. See addition to answer \$\endgroup\$
    – Russell McMahon
    Feb 21 '12 at 13:49

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