Seeking Advice for LED Circuit

[vne147]

Hello everyone. I am seeking advice on the design of a circuit that will go inside of a Halloween costume I’m making. The circuit basically blinks 75 LEDs, 9 at a time in sequence to give the illusion of traveling light. I am using a 555 timer that sends pulses to two 74HC4017 decade counters. The 4017 outputs turn transistors on and off which turn the LEDs on and off. I am planning on powering the circuit off of 8 AA batteries wired in series so I have an approximate supply voltage of 12 V. All the ICs, transistors, and LEDs draw about 200 mA at 12V. I prototyped the circuit already and it works great off of the 8AAs with no voltage regulator or caps at the V+ side. Pretty straight forward so far however, I am also planning on occasionally powering an addition 40 LEDs all at once that I will turn on and off through a switch. The additional 40 LEDs will be powered off of the same 8 AA batteries. When the 40 LEDs are lit they will draw about 1 A. OK so here is my question. When I turn on the 40 additional LEDs will the sudden increase in current draw adversely affect the rest of the circuit? I want the 555 timer and 4017s to keep working the whole time. Should I add a voltage regulator to the circuit for the logic portion? Could I get away with just adding a few caps to the supply voltage side of the logic circuit? If so, how would I go about selecting the proper size caps? Should I add any diodes? Any advice or suggestions are greatly appreciated. Please let me know if you need more information and I can post a schematic or the .sch file. Thanks.

 

[Mikebits]

Since all your LED's will not be lit at the same time, do you really need that much juice? ( leds 20ma * 9led)= .18 A at one time. According to you.

This assumes all 40 not lit at once.

 

[vne147]

There are a total of 115 LEDs in the circuit. 75 of them will be blinking in sequence with only 9 lit at any one time. The 250 mA for this part of the circuit also accounts for the transistors and TTL chips. 250 mA is what I measured from the prototype circuit. On top of that I will occasionally turn on an additional 40 LEDs that will all be lit at once. They draw ~25 mA each so (.025*40) = 1A. Does this clarify the problem? Thanks.

 

[Mikebits]

Spare yourself the burden of packing that much bat power. Multiplex the leds. Doing this will give the illusion of all being lit, but save a lot of power.

 

[vne147]

Mikebits,

Thanks for the input. I hadn't thought of multiplexing them but unfortunatley I don't think that will work for me. The board is already pretty much laid out and I don't have the space for the extra traces and complexity multiplexing would add. I also don't have the time to redesign the circuit and board at this point. I'm not too worried about making this thing awfully efficirnt as it will only be used for a few hours once a year and I don't mind lugging around the 8 AAs for the same reason. Does anyone have any advice about whether caps or a voltage regulator would be required? Thanks again.

 

[Andy1845c]

Does anyone have any advice about whether caps or a voltage regulator would be required? Thanks again.

Not an expert here, but I can't see why you'ed need a regulator. As long as all your components are spec'ed for 12v or higher, and your led current limiting resistors are sized for 12v, then you should be fine. With batterys, the voltage will sag over time as they drain, but can't go over 12v unless you add more batterys. A regulator needs a few more volts in then out, so using one at 12 volts would require you to lug around even more batterys, plus they do waste some power when operating.

I can't see a need for caps. Battery power should be pretty smooth and clean, so no filter caps should be needed.

You may want to add a diode to protect the circuit should all the batterys get put in backwards. I've killed more then onr 555 by hooking it up backwards.

 

[vne147]

Andy,

Thanks for the help. All my components are spec'ed for operation between 3 and 18V. So I should be good from that standpoint. Also all the resistors for the LEDs are sized for 12V. If I did use a regulator I would probably go with 8V, that way as the battery output decreased as it was drained, 8 AAs even close to death should probably still give enough voltage for an 8V regulator to work. If I went with the 8V reg I would either resize the LED resistors or just take the hit in brightness. I know the battery output should be relatively smooth under normal operation, I just wasn't sure about the transient effect on battery voltage the extra current draw from turning on the 40 LEDs all at once would be. Would the supply voltage dip momentarily? Would that screw up the logic ICs? From your post, it sounds like I shouldn't worry about it too much. I was planning on testing the circuit before I made the boards but I haven't received the LEDs in the mail yet and I kind of want to get moving on it. As for as the diode to protect the 555. Should I just place it between ground and the 555? I am exploring Mikebits' suggestion of multiplexing the LEDs. I hadn't though of it before so my board really wasn't laid out to add that functionality. It is creating quite a bit of backwards work for me but it truly is the best solution in my opinion and in Mikebits' too. Anyway, please keep the thoughts coming everyone. Thanks!

 

[Willbe]

For how long will the 40 LEDs be on? How long off?

 

[vne147]

Well I'm not really sure. That will be controlled by me through a switch. So basically however long I feel like it. If I had to guess I'd say something like, on for 10 seconds once every couple of minutes. Why does that matter? Just curious.

 

[Hero999]

Spare yourself the burden of packing that much bat power. Multiplex the leds. Doing this will give the illusion of all being lit, but save a lot of power.

I don't see how multiplexing saves power. For example if an LED is powered at 20mA 50% duty cycle it will appear as bright as if it were powered continuously at 10mA.

 

[Willbe]

10 seconds once every couple of minutes.

For 10 sec at 1A with a voltage droop of 2v you'd need an energy storage C of 5F so a smoothing capacitor for this circuit doesn't seem practical.

I've also heard that the eye responds to peak brightness. so muxing the LEDs with some kind of current feed may save a lot of mA.

 

[Andy1845c]

Is there a reason that the 40 that will all come on at once need to be wired alone? Could you string 4 together? Assuming a voltage drop of 2v as Willbe did, if you wired 4 in series, you'ed have 10 strings, alot less wires and resistors and you'ed only use about 250mA if i am thinking correctly, or am I missing somthing?

My referance shows a reverse polarity diode in series with the supply or across the power rails with a fuse. I personally don't see why it couldn't go between the 555 and ground. Maybe one of the resident pros can chime in.

 

[Hero999]

I've also heard that the eye responds to peak brightness. so muxing the LEDs with some kind of current feed may save a lot of mA.

That's not true, the eye acts as an integrator so average brightness is important, not peak brightness.

 

[vne147]

So Hero, let me see if I understand what you are saying. Basically you telling me that if I use larger resistors that this would reduce the current draw and result in a less brightly shining LED and that this would be equivalent in terms of percieved brightness and power consumption as the multiplexing arrangement. Do I understand? Is there a linear relation of LED current to brigtness? For example, the LEDs I ordered are 20000 mcd and will draw 20 mA. So a 50% duty cycle, would be the same as 10 mA of continuos current draw and would be percieved as 10000 mcd. If I just sized the resistors to draw 10 ma continuosly, would that result in a 10000 mcd continuos output? Let me know if I'm unclear in my question. If I do understand it seems I would either have to take the hit in brightness to reduce power consumption, or keep the brightness but have to deal with the extra current draw. In either case it sounds like I should just stick with the original design. Thanks.

 

[vne147]

Willbe,

I'm not trying to use a capacitor to maintain the 12V the whole time the 40 LEDs are on. And I'm definetly not walking around with a 5F capacitor strapped to my back. My concern stemed from any transient effects that might occur when I suddenly turn on a switch that will draw 1A from the batteries. There will be a transient voltage drop until the circuit settles out. What I don't know is how much of a drop will it be and for how long. Any transient would probably be on the order of milliseconds if that but would that voltage dip over that period of time be enough to make the ICs I have malfunction. I don't have the knowledge or tools to predict this. Am I out in left field on this one? I'm not a EE but it seems like that would happen. Or if it would happen is it so quick that I don't have to worry about it? Like I said I would prefer to just test a prototype but don't have all the parts yet. Anyway, I hope I'm being clear with my question or maybe I'm just misunderstanding the responses. Thanks for the help.

 

[Willbe]

My concern stemed from any transient effects that might occur when I suddenly turn on a switch that will draw 1A from the batteries. There will be a transient voltage drop until the circuit settles out. What I don't know is how much of a drop will it be and for how long.

What is the internal resistance of your battery?

Info on a battery's transient response might be hard to come by. Being electrochemical it may be slow. This response might be improved by the use of a small cap.

 

[Willbe]

That's not true, the eye acts as an integrator so average brightness is important, not peak brightness.

The early handheld battery calculators that used red LEDs had some kind of peak current feed with muxing that fooled the eye as to actual display brightness.

If I come across it I'll post it. Maybe it only works at certain wavelengths. . .?

 

[Ubergeek63]

The early handheld battery calculators that used red LEDs had some kind of peak current feed with muxing that fooled the eye as to actual display brightness.

If I come across it I'll post it. Maybe it only works at certain wavelengths. . .?

Here is an old Siemens app note... the last page tells about why... This effect is offset some in new tech LEDs that are more efficient at low current densities, you need to look at the spec sheet to know for sure.

For absolute max light it apparently does no good, only for under say 20% of max. If you are muxing or just PWMing for efficiency it makes sense-not for torches.

 

[Valence_4]

Since your 75 LED portion is almost laid out, it's hard to change it. However, your 40 LED portion doesn't seem to be advanced. I'd suggest you to multiplex them (either in ten consecutive 4-LED sets or in five 2 x 4 LED sets, depending of the desired effect.) This way, your 40 LED will draw (@ 20mA per LED) either 20 or 40 mA.

BTW, let me guess... The LED's are ORANGE ?

 

[Willbe]

Here is an old Siemens app note... the last page tells about why... This effect is offset some in new tech LEDs that are more efficient at low current densities, you need to look at the spec sheet to know for sure.

For absolute max light it apparently does no good, only for under say 20% of max. If you are muxing or just PWMing for efficiency it makes sense-not for torches.

"The eye does not behave as an integrating photometer,
but as a partially integrating and partially peak reading photometer.
As a result, the eye perceives a brightness that is
somewhere between the peak and the average brightness
The net result is that a low duty cycle high intensity pulse of
light looks brighter than a DC signal equal to the average of the
pulsed signal."

Thank you, Mr. U - I rest my case!