10 Year LED Flasher

12,210

185

25

About: Music: my profession for over 40 years... Electronics: my beloved hobby always.

This LED flasher will blink continuously for 10 YEARS on a single 1.5v AA alkaline cell. This is possible by achieving an average current draw of 35uA (microamperes) .

Step 1: This Is the Complete Circuit Diagram of the Flasher

Main components:

IC1: CD4001 CMOS quad NOR gate

Q1: 2N4401 NPN tansistor

T1: 1/2" toroidal bifilar transformer ( 2x25T)

Bat:: 1.5v AA alkaline cell

Led1: high efficiency led

Step 2:

IC1 CD4001 A&B are wired as astable multivibrator with frequency=1Hz . This is the flash rate.

Step 3:

IC1 CD4001 C&D are wired as monostable multivibrator with pulse width = 1mS (1 millisecond) . This is the time the LED will be ON during each second of time. IC1D biases transistor Q1 through R3 in a modified "Joule Thief"(blocking oscillator) circuit. With this configuration, IC1-D "gates" the oscillator on & off.

Step 4:

When you connect the 1.5v cell for the first time, IC1 will not operate because it needs a minimum 3V to run. At this moment Vdd is 1.5v minus 0.3v (voltage drop of D1). By momentarily connecting jumper J1, transistor Q1 will be biased through R3 and will begin to oscillate. High voltage (flyback voltage) spikes at collector of Q1 will light LED1 , and will charge capacitor C4 through diode D1. Voltage at C4 (>3v) will feed IC1 which will start running and will run until battery is exhausted ( 10 years from now ! )

Led "clamps" voltage spikes to its forward voltage V(f) . If you use a RED led, add a 1N4148 diode in series, or the voltage at C4 will not be sufficient to feed IC1 CD4001. ( red leds have a lower V(f) ).

Step 5:

Current drain of the circuit with the LED "on" is 35mA. Since the duty cycle is 1/1000 , average drain is 35uA (microamperes) . Typical AA alkaline cell has a capacity of 3000mA/H at low current drain. This is equivalent to (3000*1000)/35 , or 85,714 hours, or 3,571 days, or 9.8 years! For more details, watch the video.

Step 6: Full Video Tutorial

4 People Made This Project!

Recommendations

  • Beauty Tips Contest

    Beauty Tips Contest
  • Classroom Science Contest

    Classroom Science Contest
  • Pets Challenge

    Pets Challenge

25 Discussions

0
None
hossein saraf

1 year ago

thank you

It was really works and can be an idea for operatiing other eqipment with an 1.5 cell.

0
None
Kiểm toánP

2 years ago

what value of L1 and L2 (henry)?

thanks

0
None
lyoppi

2 years ago

Please upload circuit design too along with wiring daigram

1 reply
0
None
Yonatan24

3 years ago on Introduction

I don't know if you know this but batteries have internal resistance so the battery won't be able to survive ten years even when there is no load.

I didn't understand the part where it the battery charges itself? Perpetuum Mobile?

2 replies
0
None
michael.richey.566Yonatan24

Reply 3 years ago

Duracell seems to believe their batteries can survive 10 years. They even made a guarantee saying as much.

http://www.duracell.com/en-us/innovation/duralock-technology

0
None
Testas.T

4 years ago

I'm very interested how many current it draws, because in the begining you said that it consumes 35uA and at step 5 you said it consumes 35mA

1 reply
0
None
simpletronicTestas.T

Reply 4 years ago on Introduction

35mA is the current draw when the led is ON, which is during 1mS(one millisecond) of every 1000 milliseconds flash cycle. The time the led is off (999mS of every flash cycle) the battery current is 0, and the circuit is running only from the charge of C4, Therefore the average current from battery is 35mA/1000 or 35uA. Greetings!.

0
None
simpletronickasbah

Reply 4 years ago on Introduction

No, it's just a 1/2"frerrite core I salvaged from an old motherboard. Just wind 1meter of enameled wire folded in half , and then separate the windings by cutting end. Most cores are electrically conductive, so It'a good idea to paint it with enamel, or clear nail polish. Tutorial for transformer here:https://youtu.be/BUa3GqF4kfc

0
None
nqtronix

4 years ago on Introduction

I'm impressed.

At it smelled like bullshit, but wow, you have thought about evverything!

Yes, I did some research and this are the numbers I came up with:

The quiscent (standby) current of the CD4001 is roughly 0.25uA per gate, resumting in 1 uA total. The leakage current of a tantalum is 0.5 to 1 uA (I picked a few random parts on digikey and compared the datasheets), make sure you don't get a tantalum polymer type, their leakage current is about 10 times a big! You could also get a ceramic capacitor, they have almost no leakage at all. The value 10uF is perfectly selected, it allows a 2uA drain (1uA leakage + 1 uA due to the CD4001) for 1.4s, longer than a full refresh cycle.

Maybe the most critical part is the 1N5819 as it's reverse current can be up to 35uA @2V & @25°C (source: http://www.onsemi.com/pub_link/Collateral/1N5817-... page 5), but I'm sure you've done the homework on that as well.

If you came up with this circuit, congratulations, I haven't seen such a elegant, well engeneered in the last month on instructables. I can't wait to see with what you'll come up next.

3 replies
0
None
simpletronicnqtronix

Reply 4 years ago on Introduction

Thanks for your comment! It's really rewarding. The idea of using ceramic for C4 is really good. All the leakage currents you point are correct, but needn't be taken into account: If you look closely at circuit diagram, you will see that CD4001 is powered by capacitor C4 and NOT directly by the battery. The same applies for the diode reverse leakage... in fact we could even say that capacitor "recharges" battery during the led OFF phase through diode's reverse leakage (voltage on cap is higher than the battery's) . Battery is ONLY used during the Led ON phase (1mS) which is when C4 is recharged, and peak current is 35mA. From there on, the rest of the cycle the whole circuit operates exclusively from the C4 charge which will be refreshed on next cycle. All leakages, led current and energy dissipation come from that 35mA peak current during the 1 millisecond Led ON phase.

0
None
nqtronixsimpletronic

Reply 4 years ago on Introduction

Sure, the LED can only be on for 1ms, this is the key of the crazy battery life.

I didn't thought of the battery beeing "recharged" with the diode leakage current, I though I was missing something :)

The capacitor leakage however is still relavant, because the current is "bypassing" the battery: The faster the capacitor discharges itself during the "off-period", the more current (and thus power) it requires to be recharged in the "on-period". It does not matter which load you connect to it, everything from a simple resistor to it's own leakage counts.

0
None
simpletronicnqtronix

Reply 4 years ago on Introduction

Your reasoning is totally correct, if we use a ceramic or polyester for C4 to avoid leakage (as you suggested) we will be "treasuring" a few microJoules and that can translate into more battery life by increasing R3 (maybe 1 more year?), or brighter apparent led flash by increasing pulse width (increase R2), or higher flash rate (reducing R1). Greetings ! .