Improve battery life in LED Chaser

Discussion in 'LEDs and Optoelectronics' started by batkin, Aug 22, 2013.

1. batkin

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0
Jul 23, 2013
Yesterday I made my first LED Chaser from the schematics below with the addition of a 400ohm resistor between the LEDs and ground. It worked well but the newish 9v alcaline battery only lasted about 20 minutes or so. For the sake of learning energy efficiency what changes can be made to the circuit to improve operation time?

http://www.eleccircuit.com/led-chaser-by-ic-4017-ic-555/

2. Harald KappModeratorModerator

11,310
2,588
Nov 17, 2011
1) for the timer 555 use the CMOS version, it will require ~10 times less current than the bipolar version. It is not clear, which version you use.

2) Instead of 1.5k resistors and a 1µF capacitor, use e.g. 15k resistors and a 100nF cpacaitor. All that charge that is pumped into and out of the capacitor is wasted energy.

3)The 4017 delivers almost 9V in the high state. A red LED has a voltage drop of approx. 1.6V (give or take a few millivolts). That leaves the reistor you added (a good idea, btw.) with approx. 7.4V voltage drop. That makes for a current of ~18mA. You can reduce the current down to e.g. 5mA - 10mA and use high brightness LEDs to increase efficiency.

A 9V alkaline block has a capacity of 500-600mAh WIthout any other current consumption a fresh block can last max. 600mAh/18mA= 33h. In reality, the current for the timer plus the less than ideal discharge characteristic of a 9V block will lead to a much earlier death of the 9V block.
20 Minutes, however, sounds like the block you used either wasn't that "newish" as you thought or of very low quality. Or both.
Calculating backwards and assuming a total current of 20mA, the capacity was nearer to 20mA*1/3h= 7mAh

3. BobK

7,682
1,688
Jan 5, 2010
I would expect the battery to last maybe 10 hours with that circuit. Perhaps there is a (near) short somewhere.

Bob

4. batkin

9
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Jul 23, 2013
Excellent feedback Harald. Thanks!

Bob, I took a second look and I must have grabbed the wrong resistor. It was a 100ohm and not 400 which probably explains the super short life. Should I be shocked the LEDs didn't pop? Should the person who made the original schematic be ashamed for not putting any on at all or is that sort of thing traditionally implied?

The 555 I used is is a NE555 http://www.taydaelectronics.com/datasheets/A-249.pdf

I'll try it again with the different resistor/capacitor values on the timer and get the right resistor for the LEDs. The LEDs I have are MV5774C with a forward voltage of 1.5 @ 20mA so the 400 should be a good fit if I'm doing everything correctly.

5. BobK

7,682
1,688
Jan 5, 2010
That would give you maybe 70 or 80mA. The battery still should have lasted longer than that. At 100mA, an Energizer Alkaline 9V is rated for 5 Hours discharging to 4.8V.

Bob

6. (*steve*)¡sǝpodᴉʇuɐ ǝɥʇ ɹɐǝɥdModerator

25,482
2,830
Jan 21, 2010
Not really.

Their life may have been reduced, but considering that each LED was only on for 10% of the time, the extent of the damage may have been slight to non-existant.

If your frequency was high enough that all LEDs appeared lit, you could increase the current significantly -- this is how multiplexing works.

Yes (to ashamed).

No (to undrawn components being assumed).

About the only thing you can assume is a connectin to the power supply pins in cases like op-amps being drawn with only 3 leads (2 inputs and the output). You *know* they're connected to power as well. If there is more than one possible option though, you would be very well advised to note what was used.

7. KrisBlueNZSadly passed away in 2015

8,393
1,271
Nov 28, 2011
CMOS devices, especially the old 4000 series, have relatively high output resistances (around 300 ohms, I think - somewhat dependent on the supply voltage). So there is normally no need for current limiting resistors. So the original design is acceptable.

There is a problem using a resistor from the common rail of the LEDs to ground to reduce the LED current. The higher this resistor, the more reverse voltage the OFF LEDs will see. Reverse voltage above around 5V will damage a standard LED. This is another reason why no resistors were used in the original diagram.

If you want to reduce the LED currents (which is a good idea, especially if you use high-efficiency LEDs), you should use a resistor in series with EACH LED.

You could also save power by using a lower supply voltage - say 4.5V from three AA or AAA cells. If your supply voltage is comfortably less than 5V, you can go back to using a single current limiting resistor without damaging the LEDs.

You can replace the 555 with a CD40106 (either a whole IC or a single-gate device) to operate as the oscillator; this saves components, space, and supply current.