Seeking Advice for LED Circuit

glad to help

OK, so I did a little testing and found some interesting results. I hooked up my prototype circuit (schematic attached) to 8 AA batteries and measured the supply voltage. It was around 11.5 V. The circuit was working fine. To simulate the extra loading that the extra LEDs would place on the batteries I placed a 10 Ω power resistor in between supply and ground. So that should have drawn about 1.15 A (about twice what the extra LEDs will draw). When I measured the supply volatge after hooking up the power resistor it was about 8.5 V So the extra current draw dropped the supply volatge from 11.5 to 8.5 V. I figured that if the 555 and 4017s could still do thier jobs with this much of a voltage drop then the actual voltage drop (which should be smaller) from the extra LEDs should be OK. Well, it didn't work. When I hooked up the power resistor the pulsing stopped and whatever LEDs were lit at the time just flickered. So I disconnected the batteries and reconnected the prototype to my bench power supply. I then varied the voltage on my power supply to see at what voltage the circuit would stop working. From 12 - 9 volts it worked fine. Then from around 9 - 5 volts the pulsing stopped and I got that wierd flickering. Then from 5 - 2 volts it worked fine again. There was obviously some wired resonance thing going on here so I decided to try the circuit with out the .068 µF capacitor (shown as C2 in the schematic) and the circuit worked fine at all voltages. I then hooked the prototype circuit back up to the 8AAs and tested it with and without the power resistor. It worked perfectly. The 555 and 4017s still did their thing even when the supply voltage dropped from 11.5 to 8.5 and then went back up to 11.5. So it appears I will not need a voltage regulator or a capacitor to smooth out the supply voltage but what is the purpose of C2? The circuit seems to work fine without it. Could I just leave it out or should I replace it with a different value that will work between 11.5 and 8.5 V. Thanks for everyone's input so far. I'm almost there.

Oh and BTW Valence, the LEDs are not orange. The ones that will be pulsing are warm white and the ones that I will occasionally turn on via a switch are bright blue. Why do you ask? Maybe you know what I'm making?

Willbe said:

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.

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In reading this thread, I've noticed that a twist has crept in regarding multiplexed display and how they supposedly save all this power. Multiplexing LEDs was devised to save cost in driver hardware. Power was never the concern. However, it was empirically noted after the fact, that a side benefit of multiplexed displays was, that because of the how the human eye perceives brightness, a *bit* of power was indeed saved. Lets keep in mind that multiplexing saves in the count of driver chips, transistors, and/or pins on a micro. If a bit of power is saved, wonderful.

Multiplexing is first and foremost a parts cost savings measure, NOT some power savings scheme...

Saturn,

After reading the application note posted Ubergeek it was obvious that the theme throughout the document was that multiplexing was geared towards saving cost not power. That was why I decided to just forget that idea because the extra work it would have created for me wouldn't have been worth the power savings. Nevertheless, thanks for the clarification. Do you have and ideas about my last questions concerning the capacitor? Thanks.

Valence_4 said:

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.

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Placing LEDs in series is indeed a power savings methodology, especially if one has sufficient voltage to begin with. This avoids the losses of numerous limiting resistors. But, be careful in quoting power savings in terms of mA, as above, because although your stipulated drive currents above looks really attractive, you're not really addressing the *power* it takes to drive them, i.e. the larger voltage required for a given series string.

I'm not trying to be nitpicky here, really, but sheesh, some considerable twisting seems to be going on here (the total thread, not just this particular message), if you ask me.

I think I'm done now

Look at the datasheet for the 74HC4017. Its absolute max supply voltage is only 7.0V but 6V is recommended. You have 12V or more? Don't they smoke?

Use a CD4017 instead. Its absolute max supply voltage is 18V to 22V and it might be able to supply enough current to your transistors.

vne147 said:

After reading the application note posted Ubergeek it was obvious that the theme throughout the document was that multiplexing was geared towards saving cost not power. That was why I decided to just forget that idea because the extra work it would have created for me wouldn't have been worth the power savings.

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Right.

Our well intentioned friends sometimes forget that although many ideas make sense in the world of serial production, i.e. with thousands or millions of units, that for a one-off unit, "good enough is perfect," as Eric would say...

vne147 said:

Do you have and ideas about my last questions concerning the capacitor?

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I'm not sure. Where did this schematic come from?

Audioguru,

In my protoype circuit I have one HCF4017BCE and one NTE4017B. They don't smoke. They don't even get warm. Thanks for the input though. I'll double check all my compnent specs before ordering the project from Mouser.

Saturn,

I made the schematic on Eagle but the idea came from here:

**broken link removed**

I copied it almost exactley. The above attached schematic is just of my prototype circuit. It does not show all the extra LEDs that will be part of the final project. I can attach a schematic of my entire project if you're interested. I just thought the prototype circuit was enough for people to get the idea of what I'm trying to do. Thanks.

Please do not call an ordinary CD4017 a 74HC4017 that is very different.
Your schematic shows 744017 ICs that do not exist.

The bottom CD4017 is directly driving the bases of transistors that have their emitters grounded. Then the current from the CD4017 is too high and it might burn out.
A CD4017 can directly drive the base of a transistor without a series current-limiting resistor if its supply voltage is 7V or less.
The circuit you copied has 10k series base resistors to limit the current.

Multiplexing LEDs reduces their brightness. Then a higher current is needed. The LEDs are less efficient at the higher current so even more current is needed.

Adioguru,

Thanks for pointing that out for me the isse with the 74HC4017. I was going to order the wrong chip. i don't know why the chips I have in my prototype circuit seem to work fine without even getting warm. I checked out the data sheet for the CD4017 and it appears that it has different pin assignments than the 74HC4017. My board is all laid out and don't want redo all that work. Do you know of an IC that has the same pinout as the 74HC4017 but can handle a 12V supply? If not I'll just put a 5V voltage regulator in the circuit. I think that would be easiest. Thanks.

I think I found it. The HCF4017M013TR is good between -.5 to 22V. It looked like the pinout was the same. Still needing some help on the capacitor question though. Anyone?

An HCF4017 is an ordinary Cmos CD4017. Its supply voltage is from 3V to 18V or 22V. It has the same pins as a 74HC4017 that is high speed Cmos with a supply voltage from 2V to 6V.

Each manufacturer has their own prefix letters for a CD4017. HCF is used by ST Micro. Motorola (now called ON Semi) uses MC14017. Philips (now called NXP) uses NTE4017.

OK, so I just wanted to thank everyone again for all the help and to show off the results. The project was a Halloween costume of the flux capacitor from the movie Back to the Future. Yes, I know I'm a geek. I attached a bunch of pictures of the boards I made and below is a link to a youtube video of the project in action. Everything except the LEDs are SMT components.

YouTube - Flux Capacitor Back to the Future Halloween Costum

Hero999 said:

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.

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not that, it means turning one row on for a short time, switching it off and turning on the next one, switching that off and turning on the next one, all really quickly. then you only have one row on at once instead of all 40, which means you wont need as much current at one time. switch between rows fast enough and your eye cant see that they're all not on.

solis365 said:

not that, it means turning one row on for a short time, switching it off and turning on the next one, switching that off and turning on the next one, all really quickly. then you only have one row on at once instead of all 40, which means you wont need as much current at one time. switch between rows fast enough and your eye cant see that they're all not on.

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I think that was understood by Hero. The problem was that if I operate the LED at a 50% duty cycle, it would only be on for half the time. Even though you could switch it off and on quicker than the eye could see, it would appear less bright because it was not continuosly illuminated. So, to make an LED on a 50% duty cycle seem as bright as one that was continuosly lit, you would have to make it shine brighter when it was on by allowing twice as much current to flow through it. That was what was being said I think. However, that turned out not to be the case because of how the human eye percieves brightness (refer to the ops note posted by Ubergeek). A continuosly lit LED drawing 10mA would appear as bright as an LED at a 50% duty cycle drawing ~15mA. The net result would be some power savings. The 15mA is just an estimation based on the fact that the percieved brightness is somewhere between peak brightness and average brightness. It could be slightly more or slightly less but it would not be as much as 20mA.