Circuit to light up LED only once

Question

I have a battery-powered product with an ON/OFF switch with two leads. I am looking for a circuit to shine a LED for a short duration each time the switch goes from OFF to ON. Power consumption and cost are of great importance, so any leakage current or similar has to be absolutely minimal; 0.1 μA is too much. The LED pulse needs to be big enough to be noticeable when looking directly at it. The device does not turn on and off often.

I was thinking about charging a capacitor when the switch is open and then dumping that energy through the LED (plus disconnecting the cap) when the switch is closed. What would that look like? What about considerations for leakage? Other ideas are welcome too!

Answer 1

^{simulate this circuit – Schematic created using CircuitLab}

Dare I suggest something as crude as the circuit above? The only quiescent current is the leakage through the capacitor. D2 is there to help discharge the capacitor when the switch is off. Or it could be be removed, and a very large (>10M) resistor added across the capacitor.

Answer 2

The 74C14 (unbuffered) hex schmitt inverter has a static whole package (6 inverters) current of typically 50 nA at 15V supply and maximum 15 uA across the temperature range. At 25C it will usually be close to the typocal value. At 5V supply it will be lower and at the maximum alllowed supply of 3 V lower again.

The CD40106 hex schmitt trigger {buffered} has a 50 nA typical quiescent package current at 25 degree C and 1 uA maximum. At 3V it will be lower again.

It will be easy to built a monostable based on schmitt inverters that meet your requirements.

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Gates must be schmitt triggered.

3V, 25 C, 74C14 - Well under 50 nA quiescent typical.
5V, 25 C, CD40106 - 50 nA typical

RC sets pulse period.
R2 not needed if Vin is always high or low. D1 not essential - aids recovery time after Vin falls.

^{simulate this circuit – Schematic created using CircuitLab}

Answer 3

I am not sure this will work. I didn't simulate it. But some variation should work.

^{simulate this circuit – Schematic created using CircuitLab}

Answer 4

Check out this circuit below

The circuit above is done by 555 timer. When the circuit is in off state it consumes about 353ua current. When the circuit is in off state the capacitor charges through pin 6 and 7 connected to positive voltage source. when the switch is been pressed, it connects pin 2 to ground which causes an astable state that makes the capacitor shut circuit and is totally discharged. It then charges through pin 6 and 7, in the charging state of the capacitor, the circuit is on which means current flows out through pin 3 and it can be connected to load. When the capacitor reached the minimum voltage needed by the timer it turns of the timer no current flow through pin 3 waiting for another trigger from pin 2.

Note that by modifying the resistor values on pin 7 and the capacitor value, the charging time of the capacitor chenges too, you can also do that with the help of the pot to change the charging time. And modifying the value of the (pot, resistor on pin 7 and the capacitor) , the current consumed in off state changes too.

I might not be good in explanation but i think when you simulate it yourself, you will understand better.

I decide to use an npn transistor for driving the load but you can also connect it directly to the load. You can remove the transistor like below

Answer 5

Here's the circuit I mentioned in my comment last night.

In my case, the output fed an I/O to a microprocessor. You won't be able to light an LED directly with this, instead you'd probably sent 'output' to a MOSFET gate.

The pulse duration is controlled by C1. Clearly the low quiescent current is achieved by the very high value resistors used.

To trigger the circuit, as shown you would close the switch. In your circuit, you might AC couple Q1's base to your load. Basically it just needs to be 'tickled' to get it to trigger.

If you wanna try pursuing this, write back and we can work through your implementation.

^{simulate this circuit – Schematic created using CircuitLab}

Answer 6

A reset / power monitor device can do this, and consume very little power in the process.

0.1uA may be difficult to attain however. This one’s less than 2uA: https://www.maximintegrated.com/en/products/power/supervisors-voltage-monitors-sequencers/MAX835.html

And this one claims 125nA: https://www.maximintegrated.com/en/products/power/supervisors-voltage-monitors-sequencers/MAX16056.html. Close enough?

Answer 7

Here is another circuit, it consumes about 100 fa current when not used

Push the button to turn on the circuit, you can adjust the on time by tuning the potentiometer or changing the value of the capacitor along with the resistor and the potentiometer.