Flashing LEDs - need comments/feedback

Audioguru

Joined Dec 20, 2007
11,248
On Energizer's website they show a voltage graph of their 9V alkaline battery with a low resistive current starting at 33mA. Its voltage drops to 7V in about 15 hours when its current has dropped to 26mA.

Their 9V alkaline battery has six skinny little AAAA cells. I took the graph for the high current voltage of an AAAA cell and multiplied it by six. The with a 385mA resistive load the 9V voltage drops to 7V in about 20 minutes when the current has dropped to 300mA.
 

Attachments

SgtWookie

Joined Jul 17, 2007
22,230
I thought of powering the 555 from the two batteries. But I think I've read somewhere that it doesen't like more than 15v. Thats where the 7812 regulator comes in.

OK, read you post one more time. Could I use both batteries to power the LEDs (18v), and only one to power the 555 (9v)?
Yes, that is the idea.

However, as has been said many times (and AudioGuru just repeated) that 9v batteries will not last very long at all with that large of a load on it.
 

Thread Starter

nerdegutta

Joined Dec 15, 2009
2,684
OK. Dropping the 9v battery.

Trying out one more design...

The schematic in post #35, powered with 4 1.5v AA batteries, total of 6v.

The LEDs are single, with one resistor each.

How will that be for battery-life?
 

t06afre

Joined May 11, 2009
5,934
As I simulated in post #26 the average current will be about 40mA to LEDs with your original duty cycle. About 50 mSec in on time for LEDs every second. A common AA alkaline battery have about 2700 – 2900 mAh in Typical capacity. So it will work fine. I also think you would be better of with 3 C cells and not 4 AA cells. This way ( in danger of repeating my self here). You can use a PIC an create more fancy light patterns. And save current by say not having all the LEDs on at the same time. You can split the LEDs up in cluster.
 

SgtWookie

Joined Jul 17, 2007
22,230
An alkaline 1.5v AA cell has around 2700mAh-2900mAh capacity. Let's say 2800mAh.
An alkaline 9v PP3 battery has around 560mAh capacity.
The AA cells will have five times the mAh rating as the 9v PP3 cell. However, if you are running at 6v instead of 18v, you will be drawing roughly 4x as much current.
 

iONic

Joined Nov 16, 2007
1,662
OK. Dropping the 9v battery.

Trying out one more design...

The schematic in post #35, powered with 4 1.5v AA batteries, total of 6v.

The LEDs are single, with one resistor each.

How will that be for battery-life?
Are you suggesting to run 24 - 25 LED's all in parallel? That would be 20mA x 25 = 500mA. The LED array will draw more than 3W.

With one extra battery you can place two LED's in series, paralleled by 12 more segments and consume about 2W and use only 260mA. I guess that can show just how much power loss there is in the limiting resistors.

as usual, my calculation are subject to debate and or correction!
 

Thread Starter

nerdegutta

Joined Dec 15, 2009
2,684
OK, so it boils down to 2 alternatives...

Alt 1.
Battery = 9v
2 LEDs in series, 12 strings, 24 LEDs
R=(9-(3.2 x2))/0.02 = 130 Ohm
0.02A x 12 strings = 0.240A
0.240A x 9v = 2.16 W

Battery life approximately 0.560A / 0.240A = 2.3 hrs.

Alt 2.
Battery = 6v
1 LED, 24 strings, 24 LEDs
R=(6-3.2)/0.02 = 140 Ohm
0.02 x 24 strings = 0.480A
0.480A x 6v = 2.88W

Battery life approximately 2.8A / 0.480A = 5.8 hrs

Correct or not?
 

iONic

Joined Nov 16, 2007
1,662
OK, so it boils down to 2 alternatives...

Alt 1.
Battery = 9v
2 LEDs in series, 12 strings, 24 LEDs
R=(9-(3.2 x2))/0.02 = 130 Ohm
0.02A x 12 strings = 0.240A
0.240A x 9v = 2.16 W

Battery life approximately 0.560A / 0.240A = 2.3 hrs.

Alt 2.
Battery = 6v
1 LED, 24 strings, 24 LEDs
R=(6-3.2)/0.02 = 140 Ohm
0.02 x 24 strings = 0.480A
0.480A x 6v = 2.88W

Battery life approximately 2.8A / 0.480A = 5.8 hrs

Correct or not?
Or

Alt 3.
Battery = 7.5V
2 LEDs in series, 12 strings, 24 LEDs
R = (7.5-6.4)0.02 = 68 Ohm
0.02 x 12 strings = 0.240A
0.240A x 7.5V = 1.81W

Battery life approx. 2.8A/0.240A = 11.66 hrs.

With one battery you can double the lighting duration.

In reality for LED longevity I would not feed them 3.2V each and 20mA each, but rather 3.0V and 18mA...you would never see the difference.

Thus:

Alt 4.
Battery = 7.5V
2 LEDs in series, 12 strings, 24 LEDs
R = (7.5-6.0)0.018 = 100 Ohm
0.017 x 12 strings = 0.216A
0.216A x 7.5V = 1.69W

Battery life approx. 2.8A/0.216A = 12.96 hrs.
 
Last edited:

t06afre

Joined May 11, 2009
5,934
Use 6 and place a short in the empty compartment. By the way it is not hard to make your own battery pack. By soldering the batteries together. The trick is to file down the unsolderable alloy. Till you get to the material (most often brass) that the poles are made of.
 

Audioguru

Joined Dec 20, 2007
11,248
A 9V battery has a capacity of 625mAh when its load is only 25mA. When the load is higher then the capacity is lower.
With a 240mA load a new battery is 9V then it quickly drops to 7V in about 20 minutes then drops slower to 6.5V when the LEDs will not be seen anymore after about 1 hour.
You might see the LEDs dimming.
 

Attachments

Wendy

Joined Mar 24, 2008
23,415
OK, sorry I wasn't here for the discussion, I'm back in about two hours and I'll give it my full attention. Basically this is one of the area's I've put a lot of thought into for my projects.

Even with an alternate source such as AAA or AA batteries the voltage will drop. Like I mentioned earlier in the thread you can buy battery packs for any voltage imaginable.

It is a paradox, but the more power supply voltage the less current required, because you can have more LEDs in series. At 18V the current is going to be around 120ma (20ma per chain). 24VDC is even better (I'll calculate it later).

Part of the answer is a low voltage drop out constant current source. Transistors (BJTs) do this extremely well. Don't agonize of the exact transistor needed, other than polarity it simply doesn't matter.

.. .


Some examples of LDO current regulators, I would design yours from scratch using these as concepts.

The main thing is to be willing to run your prototype dry. I can guarantee it will last longer than the predictions given for the simple reason it is a flasher. The current drain is not continuous.

Of everyone one here I probably have the most experience with this kind of circuit. If you look at my 555 projects you will find a long life LED flasher series, where I designed and tested 6 different types of circuits. I put them on top of my TV, and drained them dry. Typically the shortest time I got was 1 month off of 2 X AAA batteries, one circuit lasted 3 months.

It was why I was wanting to get involved when you proposed it.


 

Audioguru

Joined Dec 20, 2007
11,248
It is good that you have the duty-cycle of the on-time for the LEDs short. Then the battery will last for a long time.

You can reduce the on-time as low as 30ms but the LEDs will appear dimmed if you go below 30ms.
 

t06afre

Joined May 11, 2009
5,934
As I understand this is a "new" design. How do you drive the LEDs this time. Also a good design idea to have the LEDs on a separate board. This way you may change the controller circuit easy say using a MCU without a total redesign
 

iONic

Joined Nov 16, 2007
1,662
Are you sure those bolts are sufficient? I would have gone with 1/2 inch. ...just joking.
Nice video and demonstration. How are you planing on keeping the device under water, a brick? or do you plan on designing a smaller housing, like from plexiglass.
 

Thread Starter

nerdegutta

Joined Dec 15, 2009
2,684
It is good that you have the duty-cycle of the on-time for the LEDs short. Then the battery will last for a long time.

You can reduce the on-time as low as 30ms but the LEDs will appear dimmed if you go below 30ms.
I will try to get 30ms, or maybe close to it.

As I understand this is a "new" design. How do you drive the LEDs this time. Also a good design idea to have the LEDs on a separate board. This way you may change the controller circuit easy say using a MCU without a total redesign
No transistors, no IRL2703, Just the 555 driving the LEDs. I'll guess I'm pushing the IC mA limit? I needed to put the LEDs on a separate board, or it wouldn't fit in the jar.

Are you sure those bolts are sufficient? I would have gone with 1/2 inch. ...just joking.
Nice video and demonstration. How are you planing on keeping the device under water, a brick? or do you plan on designing a smaller housing, like from plexiglass.
Thanks.

I have 4Kg of lead, that will hold this under water. I will make some kind of arrangement, so the device will "float" 1m from the bottom.

I tried to take a video, but it was not that easy. Video 1 Video 2 (As you can see from the quality, I'm NOT Quentin Tarantino...)

I have some thin plywood and some plexiglass lying around so the thought has occurred to me. Then I can scale it up. To say 48 LEDs. 12 LEDs on each side, in two layers. If so, I can buy a motorcycle battery. 12v at 4.5Ah.

We went out to Tvistein Fyr, this is an old lighthouse. The flasher, lead and everything worked as planned. Only drawback, was that the water was too clear yesterday. Don't need any flashing light under these condition! Anyway we had two great dives. Both to appx 19m. :)
 
Last edited:
Top