LED Flasher that runs forever!... - Page 1
Forum section: electronics / projects

author: hgg
author_rank: Frequent Contributor
timestamp: October 23, 2013, 07:48:42 pm
content: Hi,I have been playing lately with several different oscillator circuits that flash an LED becauseI 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 the2.7ms pulse that lights up the LED. The intensity of the flash is quite respectable. Thecircuit runs on 4.5V AA batteries.I've measured the current with the UNI-T UT61E multimeter that has a sampling rate of abouttwice 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 tomeasure the Maximum value. I am sure that the numbers must be wrong. The circuit does somestrange things as well. Sometimes at first power on, the LED lights up really bright and the periodchanges. 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 thatthe numbers will be way off? How can I accurately measure the power consumption of thiscircuit?Finally, if we assume that the measurements are correct, how can we calculate the totaldays 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 for2.7ms. 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)

author: IanB
author_rank: Super Contributor
timestamp: October 23, 2013, 08:04:40 pm
content: 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.

author: Galaxyrise
author_rank: Frequent Contributor
timestamp: October 23, 2013, 08:35:55 pm
content: Might I suggest aµCurrent?

author: hgg
author_rank: Frequent Contributor
timestamp: October 23, 2013, 08:44:55 pm
content: IanB if I understood correctly what you said, I connected a 220uF capacitor I had to the positiveand negative of my circuit. Then I connected the negative side of the battery to the negativepin of the capacitor and then connected the multimeter in series from the positive of the batteryto the positive pin of the capacitor.The circuit makes a strange oscillation apart from the 2s flashing oscillation. The LED startsflashing after a while and then the power consumption is 7uA. It then continues with the 7uAfor some time steadily, but after that, it goes down slowly to 1uA and the brightness of the LEDdrops considerably. But it does not go off. After a surge of about 30uA it starts again andcontinues normally with 7uA. This cycle continues forever. Not sure what is going on, but I stilldon'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 Idisconnect the battery it continues to flash for 20 more seconds....GalaxyRise will the ?Current do the job accurately? Have you tried it?

author: AndyC_772
author_rank: Super Contributor
timestamp: October 23, 2013, 08:49:34 pm
content: 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 capacitorYou should be able to work out the current that flowed through your circuit from Q=CV.

author: IanB
author_rank: Super Contributor
timestamp: October 23, 2013, 08:53:17 pm
content: Quote from: hgg on October 23, 2013, 08:44:55 pmYes, this is what I meant.QuoteI'm not sure what is happening here, but it seems as if your circuit is not operating in a stable manner. I think you need to fix the circuit before you can measure its current consumption.You could perhaps try a larger capacitor and see if that makes it stable.QuoteI don't think there is a better way. You need an integrating time averaged current measurement, and the filter capacitor does that.QuoteI don't think the uCurrent will help. If your circuit is unstable you cannot make accurate or reliable measurements around it.

author: hgg
author_rank: Frequent Contributor
timestamp: October 23, 2013, 09:04:09 pm
content: AndyC_772 I will try your suggestion, thanks.IanB, if I don't use the capacitor and run it straight from the batteries, its stable.I have included a PDF schematic in my first post. Although I assembled the circuitexactly as in the schematic, it did some strange things and when I used a 22Kresistor in series with the positive supply everything was fine.

author: IanB
author_rank: Super Contributor
timestamp: October 23, 2013, 09:51:26 pm
content: Quote from: hgg on October 23, 2013, 09:04:09 pmBut there is already a 100 uF supply capacitor in the schematic. How can you "not use the capacitor"?QuoteIf the circuit "did some strange things" it is not stable. If you needed to put a 22k resistor in series with the supply, everything isnotfine.I recommend you to accept the evidence of your eyes and believe that the circuit you have constructed is not working properly. First get the circuit to work as intended, then measure the current consumption.

author: hgg
author_rank: Frequent Contributor
timestamp: October 23, 2013, 10:04:20 pm
content: IanB I am doing my best...AndyC_772 I think you had a nice idea. The circuit draws a lot of current initially, so 100uF were notenough. I used a 2200uF capacitor I had from a power supply. I measured the capacitor dischargeafter the circuit was fully on and stabilized.The self discharge of the capacitor in 60sec was from 4.497V to 4.479VWhen I run the circuit from the 2200uF capacitor for 60 seconds, it's voltagedrops from 4.497 to 4.4250. (The value is the average of 3 runs).So, how do you calculate the current consumption of the circuit?

author: lowimpedance
author_rank: Super Contributor
timestamp: October 23, 2013, 11:31:11 pm
content: I have built the cct in the pdf, for the same sort of purpose you want, except I use it in some dummy camera installations at home.I modified it a bit to improve the flash delivered by the LED to make it more visible and also added a night only sense so during the day the cct is in a quiescent state, (ie not oscillating as such). You should consider the best way to maximize the light output of the LED ie bezel used. to give the best viewing angles.I have been running this modded cct for approx 4 years on 2 * 'C size' cells no problems yet, (too lazy to have measured the average current draw!!). I suspect you would need to replace the cells more to prevent leakage damage than they were discharged to the point the cct no longer operates. I cannot remember now what the minimum supply was when the cct stopped working.

author: Dongulus
author_rank: Regular Contributor
timestamp: October 23, 2013, 11:50:22 pm
content: Quote from: hgg on October 23, 2013, 10:04:20 pmI = dQ/dt= C * dV/dtIavg = capacitance * change in voltage / time measured= 2200uF * (4.497V - 4.250V) / 60sec= 2.64uAThe self discharge of the 2200uF capacitor comes out to 0.66uA so the average current of the circuit on it's own come out to just about that 2uA figure you measured with the ammeter.

author: IanB
author_rank: Super Contributor
timestamp: October 24, 2013, 12:31:47 am
content: If you put a typical DMM across a small capacitor, the "self" discharge of the capacitor will actually be the DMM drain. A 10M input impedance is not that high when measuring a small capacitor. It would be better to use a bench meter with a 10G impedance for such measurements.

author: Dongulus
author_rank: Regular Contributor
timestamp: October 24, 2013, 04:41:51 am
content: A meter with 10Gohm impedance doesn't seem necessary. Do you have the DMM on the capacitor for the whole 60 sec duration? If you only connect the meter only long enough to get a reading, the current draw should be negligible. You can get an even better result if you increase the time before measuring (i.e. 60 sec or 90 sec).At any rate, it seems to me that you are more interested in getting the order of magnitude to validate your ammeter measurement which the capacitor voltage measurement has accomplished.

author: hgg
author_rank: Frequent Contributor
timestamp: October 24, 2013, 05:49:02 am
content: Wilfredthank you for the link. Nice info on the LM3909. When searching for a solution I came acrossthat chip as well but decided not to use it because of the big current drain. The datasheet lists anaverage 0.5mA during battery life. I wanted something that will last forever, not just for a year...(The reason is that I don't want to 'wall climb' every year to reach to the alarm and change thebatteries).lowimpedance, I too thought of using a night time sensor so I can minimize the current draw ofthe little circuit and I built a schmitt trigger (to avoid power oscillations to the flasher) with an LDR as thesensor, but that circuit was draining the battery many times more than the flasher itself! 800uA @ 3V!...3 times more than my previous circuit and tens of times more than the current one.Dongulusthank you for the calculation! I came up with a similar result except taking into account the selfdischarge voltage drop of the capacitor. It looks that the UT61E did a decent job of measuring the powerconsumption in the uA range, although a bit on the low side. It was displaying around 1.5uA and Irounded up the result to 2uA just to be sure. Not bad I think for a cheap multimeter. Wouldn't theuCurrent device be more accurate?IanB, I was aware of that small detail that the multimeter will discharge the capacitor at a faster rate,that's why I never left the multimeter on the capacitor. I made quick measurements at the end ofthe 60 second period and averaged them. Quite similar results. The first time I left the multimeter on,I got from 4.497 -> 4.463 instead of the 4.479 with the meter disconnected.If this method of measuring the current draw is correct, then this little circuit is simply amazing!With an average current draw of 2uA the only time limit is the battery quality! I have someoriginal SONY AA cells made in Japan that are in remote controls for 20 years with no signs ofleakage and working just fine. I am not sure if you can find such quality batteries any more...But with such a low current drain, a super capacitor and a small solar charging circuit, might bethe ultimate solution that will indeed last forever. What do you think?Just to make the whole circuit more extreme I would like to find an LED that draws even less current.My current LED is a bright clear white one, but its starts to be barely visible just after 2.4VDo you know if there are any very bright but low current LEDs? I saw some Kingbright LEDs thatare advertised as bright and low current drain. Has anybody tried them? More important I thinkfor my application is the angle of the LED and the one I use has 20 degree. Very narrow. Is itpossible to find an LED with at least the same output and current draw as the typical white clearone I am using but with a wider angle?Thank you for your help guys!(I still cannot believe that it draws 2uA, flash pulse included... It would be very interesting ifsomeone can explain how the circuit works and why this oscillator draws such low power.)

author: IanB
author_rank: Super Contributor
timestamp: October 24, 2013, 06:30:12 am
content: LEDs don't "drain" current, they pass however much current you supply to them. What varies is how much light they emit for a given current.In the circuit linked from the first post, the LED current is limited by the 47 ohm resistor. If we assume a 2.2 V drop over the LED and a 3 V battery then the LED current will be 0.8/47 = 0.017 A = 17 mA.With a 2 ms pulse every 2 seconds the average current used by the LED would be approximately 17 mA / 1000 = 17 uA. Since the circuit description says the average current is 30 uA there are evidently some other significant current paths in the circuit.You could make the circuit use less current by increasing the value of the current limiting resistor in series with the LED. This would cause the brightness to decrease, but an extra bright LED might compensate for this.I doubt your measurement of 2 uA given the documented value of 30 uA, but it is hard to know what current to expect without seeing the exact circuit you have built.How the circuit works:When first switched on the 0.68 uF capacitor starts out discharged. The PNP transistor is turned off because the base is pulled high and the 0.68 uF capacitor slowly charges up through the 4.7M resistor. When the voltage at the base of the PNP gets low enough the transistor starts to turn on, pulling up the base of the NPN transistor and making that start to turn on. As soon as that happens there is a strong feedback effect causing both transistors to turn hard on and the LED lights up. However the 0.68 uF capacitor now discharges through the NPN transistor in a matter of a millisecond or two, causing the PNP to turn off again. Once the PNP turns off the NPN is turned off and the system quickly resets ready for the next cycle.[Edit: It seems this explanation is not quite correct. The charging and discharging of the 0.68 uF capacitor seem to be the opposite of my description.]

author: AndyC_772
author_rank: Super Contributor
timestamp: October 24, 2013, 06:31:35 am
content: Have you tried simulating the circuit in LTSpice to see how it works and what really governs its power consumption?Here's the circuit... you'll need to change the extension from .txt to .asc.

author: kayvee
author_rank: Regular Contributor
timestamp: October 24, 2013, 07:00:01 am
content: Back in the day (early '80's) I knocked a hundred or so LM3909 based flashers epoxy encapsulated into a length of 50mm plastic water pipe, powered from a D cell, that ran for almost 3 years.Two mounted on each windowsill of the family beach cottage did a good job of preventing break-ins, which was/is common in the area. Every other December holidays it was my job to do the battery refresh.It didn't stop the baddies from removing roof tiles and entering that way, however!Ah, such is life in darkest Africa

author: AndyC_772
author_rank: Super Contributor
timestamp: October 24, 2013, 07:25:12 am
content: An LED flasher circuit powered from a small coin cell, encapsulated in epoxy and attached to the collar of a large dog seems to work quite well in that respect too. You only need one, and it seems to remain oddly effective even when the battery has run down.Must run a simulation one day... maybe using a coin cell instead of AA's has something to do with it?

author: hgg
author_rank: Frequent Contributor
timestamp: October 24, 2013, 07:27:52 am
content: QuoteI've built the circuit exactly as in the diagram I have attached and as I've noted the only thing I've replacedwas the PNP transistor from the PN2907 to a 2N3906 I had. I am not sure how that affects the circuit.I also run the circuit from 4.5VdcDon't you agree that the capacitor discharge measurement (suggested by AndyC) I've made is correct?How else can I measure it? I will leave it today with the multimeter again measuring max current, but fora long time just in case there is a current drain surge after many 2sec oscillations.Thank you for the explanation of the circuit! I think I understood what you described except thepoint of the strong feedback effect. I also feed the positive side of the circuit through a 22k resistorwhich was not included in the initial circuit diagram. Do you know how this will affect the circuit atleast in theory?AndyC_772I will indeed try to simulate it in LSpice although theory and practice do not alwaysgo together...For example from what I see capacitors are very inaccurate parts. At least I will try tosimulate it with the actual values I measure with my multimeter.kayvee3 years are indeed quite a lot but I want to avoid that "every other December" batterychange because in order to do it I have to climb a very steep ladder...

author: IanB
author_rank: Super Contributor
timestamp: October 24, 2013, 07:32:15 am
content: Apparently my evaluation of how the circuit works was not quite right. I should have predicted that from the polarity of the 0.68u capacitor. However...Can you please solve this puzzle for me?On the right is the simulation I built before seeing Andy's post. It didn't work until I turned Q1 upside down (reversed collector and emitter terminals).On the left is the simulation that Andy uploaded. This has Q1 in the orientation originally shown on the schematic.Both circuits simulate identically. So...huh?Did I find a bug in LTspice?

author: IanB
author_rank: Super Contributor
timestamp: October 24, 2013, 07:35:59 am
content: Quote from: hgg on October 24, 2013, 07:27:52 amNote that my explanation of the circuit was not exactly correct in its details, although the broad principle was right.The simulated current consumption can be determined from the LTspice model, and it is less than 30 uA but more than 2 uA.

author: hgg
author_rank: Frequent Contributor
timestamp: October 24, 2013, 07:50:20 am
content: Quote from: IanB on October 24, 2013, 07:35:59 amTo tell you the truth the polarity of the capacitor was bothering me as well but I though that I did notfully understood what was going on. I get the principle though. Did you try to simulate it withthe 22k resistor in series with the 4.5Vdc power? I get some strange measurements with and withoutit! I will try the simulations myself as well but I have to go now for a while.

author: IanB
author_rank: Super Contributor
timestamp: October 24, 2013, 07:51:59 am
content: Attached is my LTspice file with the oddly inverted transistor (see post above).Any thoughts from LTspice experts would be welcomed...

author: BravoV
author_rank: Super Contributor
timestamp: October 24, 2013, 08:04:35 am
content: Not sure if this is related, using a cmos 555 chip, 2 resistors and one electrolytic cap, you can blink a high brightness "red" led using just a single alkaline battery until 0.6 volt, at this voltage is basically more than dead for an alkaline cell.With a D cell sized battery, I guess it will last for a very-very long time.Actually its still blinking at 0.5V but it was too faint to see with bare eyes, only knew it when using a scope to see the blinking "signal".Remember, this works only using "cmos" version of 555, not the ordinary 555 since the bipolar won't work properly at these voltage, and its usually more expensive.Photo below is the simple circuit that I built in few minutes as an emergency test rig when I was testing out a newly bought batch of cmos 555 chips.ge

author: hgg
author_rank: Frequent Contributor
timestamp: October 24, 2013, 11:28:06 am
content: BravoV Hello, I am aware of the 555 flasher as well but it still chews the battery faster.I researched all the available flasher circuits and the one I ended up building was the mostefficient. I did not expect it though to draw so little current. Do you know what is the currentdraw of the circuit you just built?I have not yet run the simulation but the PSpice result is nothing like the one I am getting.Below is the oscilloscope output while probing the LED pins. Maybe that is why is different.Ian where did you probe the circuit?Light PulsesSingle PulseCan someone explain the following?I measure the current with the UT61E in two ways.a) By putting it in series with the positive 4.5Vdc of the power supply. (no resistor)b) By putting it in series before the 22k resistor that is placed immediately after the 4.5Vdca) average of 2uA and a peak of 170uAb) average of 7uA and a peak of 14uA even after two hours.I also measured several times the consumption with the Capacitor Discharge method and got thefollowing:a) average over 30 seconds of 50.5uF (no resistor)b) average over 30 seconds of 7.7uFTwo conclusions. First the measurements of the Capacitor Discharge and the UT61E match!It looks like that the UT61E can measure uA very accurately. Second the 22K resistor I putin the V+ supply made all the difference! Although the initial circuit was rated for a 30uAmean, using that resistor it went down to7.7uAwith the same flashing brightness!Maybe the explanation is that the resistor limits the current to the whole circuit no matterif at some point the circuit needs to draw an excessive amount of current that creates the170uA peak.Using 3 AA 2700mA batteries in series the light will flash for 40 years!Now, I have to find batteries that will last so long...

author: hgg
author_rank: Frequent Contributor
timestamp: October 24, 2013, 04:13:26 pm
content: mojo-chanHi,Actually I was looking right now for a solution using a supercap and solar panel combinationbut your RF power harvest sounds really interesting! Do you have any references to get anidea?Thanks!

author: IanB
author_rank: Super Contributor
timestamp: October 24, 2013, 04:16:44 pm
content: Quote from: hgg on October 24, 2013, 11:28:06 amI probed the base of Q1 (observing the charging and discharging of C1).Plots of the LED current are below. I think I used a white LED with a higher Vf which is why the current is quite small.

author: hgg
author_rank: Frequent Contributor
timestamp: October 24, 2013, 04:31:08 pm
content: QuoteCorrect! That's exactly what I am getting if I probe the base of Q1 and ground, althoughthe flashing frequency doubles when I connect the oscilloscope probes.

author: IanB
author_rank: Super Contributor
timestamp: October 24, 2013, 04:51:42 pm
content: If you attach your oscilloscope probe across the 4.7Meg resistor you will change its resistance and thus the time constant of the circuit. A basic x10 scope probe has something like a 10Meg input impedance.

author: hgg
author_rank: Frequent Contributor
timestamp: October 24, 2013, 08:16:41 pm
content: Quote from: IanB on October 24, 2013, 04:51:42 pmIndeed, it makes sense.I think that instead of a battery, I will use a 5F/5.4V/0.07Ohms super capacitor and a small5.5V/130mA solar cell to power the little circuit. Instead of using a shunt regulator I chosethe parts so that they will be "self regulated". I will connect the solar panel with diodes tothe supercapacitor so that it will not be discharged back to the solar cell during the night.The supercapacitor will be connected directly to the flasher circuit.Super Capacitor :http://gr.rsdelivers.com/product/cooper-bussmann/phv-5r4v505-r/supercapacitor-phv-5f-54v-007-ohm/7637848.aspxSupercapSolar Cell:http://dx.com/p/lx-b580-0-8w-pet-laminated-solar-monocrystalline-silicon-cell-panel-black-blue-184629Solar Cell 8.5x8cmI found some online calculators that will give you the discharge time of a super capacitor. In mycase the 5F will provide the circuit enough power to run for at least 11 days. What I cannot findis how to calculate the charging time. How long it will take to charge the super capacitor.Any ideas?

author: IanB
author_rank: Super Contributor
timestamp: October 24, 2013, 09:30:22 pm
content: OK after reading the comments others have made and doing a few more simulations, I now understand the circuit better.As presented in the original schematic it did not start up reliably and could easily lock up into a non-oscillating state.Apparently the cause was the gain of the 2N2907 PNP transistor and the value of the 4.7 Meg base resistor. If this resistance is too small the transistor can be permanently biased into a partially on condition and the circuit will stop oscillating.Things got much better when I increased this resistor to 10 Meg and reduced the timing capacitor proportionately.See the picture of the modified circuit and simulation attached. The average current consumption is measured at about 4.5 uA, so it is pretty low.Note that increasing the supply voltage to 5 or 6 V might require some values to be changed. The original circuit was designed to be run from a 3 V coin cell.

author: hgg
author_rank: Frequent Contributor
timestamp: October 25, 2013, 01:56:32 pm
content: QuoteIan, that is exactly what it was doing in the beginning!I replaced the PN2907 with an S9012 and its works fine now! 2uA heavier thought...I am going to order today the supercapacitor and the solar panels to do some tests.mojo-chanthank you very much for the info! I will try that method as well.