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LED Flasher that runs forever!...

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hgg:
Hi,

I have been playing lately with several different oscillator circuits that flash an LED because
I want to use one with two or three AA batteries and install it in an outdoors alarm siren box.
I don't want to change batteries often and I was trying to find the most efficient design.

The best I could do was 316uA average current consumption with 1.2mA peak.  Today though,
I built one that has an average power consumption of 2uA (!) and a peak of 170uA for the
2.7ms pulse that lights up the LED.  The intensity of the flash is quite respectable.   The
circuit runs on 4.5V AA batteries.

I've measured the current with the UNI-T UT61E multimeter that has a sampling rate of about
twice per second.  If the numbers are correct, then with a good alkaline AA battery of 2700mA,
it will run for 154 years...   :)  (The 2.7ms pulse is not included in the calculation...)

I measured the peak value by leaving the circuit running for some time and setting the UT61E to
measure the Maximum value.  I am sure that the numbers must be wrong.  The circuit does some
strange things as well.  Sometimes at first power on, the LED lights up really bright and the period
changes.  I had to put a 22K resistor immediately after V+ to make it stable.

I've heard before that the UT61E is not very good for measuring low currents.  Do you think that
the numbers will be way off?  How can I accurately measure the power consumption of this
circuit?

Finally, if we assume that the measurements are correct, how can we calculate the total
days that the 2700mA battery will give us, taking into account the short (and high current)
pulse as well?  The whole flashing period lasts for 2125ms in which the LED flashes for
2.7ms.  Average Current = 2uA and Flashing Pulse Current = 170uA.

Thanks!



http://www.discovercircuits.com/PDF-FILES/3vledfs1.pdf
(I did not have the PN2907 and so I used the 2N3906. Not sure if that made a difference)

IanB:
If you want to measure the true average current consumption of your circuit, power it from a capacitor instead of a battery (use a reasonably large capacitor, maybe an electrolytic of 100 uF or so). Then power the capacitor from your battery, but put the ammeter in series with the charging supply, so the meter measures the current going from the battery to the capacitor. You should end up with a significantly smoothed and time averaged current measurement that will give you a much better idea of the average power consumption.

Galaxyrise:
Might I suggest a µCurrent?

hgg:
IanB if I understood correctly what you said, I connected a 220uF capacitor I had to the positive
and negative of my circuit.  Then I connected the negative side of the battery to the negative
pin of the capacitor and then connected the multimeter in series from the positive of the battery
to the positive pin of the capacitor.

The circuit makes a strange oscillation apart from the 2s flashing oscillation.  The LED starts
flashing after a while and then the power consumption is 7uA.  It then continues with the 7uA
for some time steadily, but after that, it goes down slowly to 1uA and the brightness of the LED
drops considerably.  But it does not go off.  After a surge of about 30uA it starts again and
continues normally with 7uA.  This cycle continues forever.  Not sure what is going on, but I still
don't know what the power consumption of this little circuit is...

Is there a better way to measure it accurately?  A different multimeter probably?
It looks though that it draws very little current because even without the capacitor, when I
disconnect the battery it continues to flash for 20 more seconds....

GalaxyRise will the ?Current do the job accurately?  Have you tried it?

AndyC_772:
Why not try:

- charge up a capacitor to 4.5V
- measure its voltage after 1 minute to determine the rate of self-discharge
- charge it back up again to 4.5V
- connect it to your circuit
- run the circuit for 1 minute
- measure the voltage across the capacitor

You should be able to work out the current that flowed through your circuit from Q=CV.

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