4060 question

[quantumplasmathingie]

Howdy to all you wire-wrapping, solder-smelling, electro-buddhas out there. I humbly lay my bewilderment at your feet and beg for your guidance.

I'm lighting 6 blues and 15 whites. The blues and 8 of the whites will be constantly lit. The other 7 are going to be 4060 driven.

Question is: what's the maximum current the 4060 can handle? I dug around the intarwebs for a while looking at datasheets, but I only succeeded in confusing myself...which isn't difficult.

My concern revolves around the divide-by frequency synchronizing between multiple LEDs. If two or more pins pulse simultaneously, I would like the current to stay high enough for the LEDs to light, and not skip a beat due to lack of current, or worse, have two or more LEDs flash weakly. I decided I'd provide ~72 mA to the IC, but before I go burning out any other components, I'm doing the smartest thing I've done so far, which is turn to you guys for help.

[en'til Zog]

Hi there! May your plasmathingie never loose quantum singularity.

The 4060 chip is wonderful for driving LEDs. Mainly 'cause it doesn't deliver more current than any known LED can stand. EACH output from the 4060 can SOURCE or SINK only about 11 milliAmps - can only deliver 11 mA in either a POSITIVE or a NEGATIVE state.

Now, you can drive several LEDs in SERIES from one output as long as the total voltage the series string needs doesn't exceed the voltage you're supplying to the 4060.

EXAMPLE: you drive the 4060 with 12 volts. You can drive a series string of 6 RED LEDs rated at 1.9 volts @ quite nicely, but not at full brightness (11.4 volts for the string). OR you can drive a series string of 3 BLUE or WHITE LEDs rated at 3.6 volts @ (10.8 volts total). BUT - each string will have only 11 mA flowing through it.

If all the 4060's outputs are lighting LEDs there may be some dimming if you haven't given the 4060 enough milli-Amperage to begin with. Say, 12 LEDs at 11 mA will need about 150 mA supplied to the 4060. Since I usually drive my circuits with a 9 volt battery with no current limiting to the 4060, I don't have a lot of trouble with dimming or power starvation. One circuit uses 3@ 4060 chips, and 60 or so LEDs - there is no noticeable dimming using one (rather stressed) 9 volt battery.

First suggestion? Get a cheap "Experimenter's block" or "Breadboard" and layout your 4060 circuit along with the LEDs you're trying to drive. The old engineering "Cut and Try" method. ALL and JAMECO have these boards for about 4 buck US.

[quantumplasmathingie]

Ahhhhhso! I believe I'm sassing your hoopy en-'light'-enment. It's a matter of volts available, not current. By george, I think I've got it.

I was under the misunderstandingment that one varies the current to the 4060 depending on the quantity of leds that might light in sync. I've got the volts available (12V wall-wart!).

My heartfelt gratitude, sir, for imparting your wisdom upon this lowly student.

I'm gonna' go get a breadboard.

[Balok]

In each two lit Tyderiums I built, I lit 4 Superbrite blues in two series off one pin and 8 superbrite whites each in series of two off another pin all the 4060. I used two standard 9V batteries wired in parallel , still 9 volts just more current available. Yeah test em on a board, you'll know you've got too many when they dim. Remember to protect your LEDs with a resistor for each individual and each series string.

[Sparky]

The point Zog is making is that the 'protection' resistor is not needed.
One is already under powering all the LEDs.
They can try to draw more power than they should but the 4060 won't pass it to them.

All resistors will consume power from the system, if you don't need them to drop the power to the LED don't install them. On a battery powered system don't waste power if you don't need to.

BTW a resistor in no way helps if the power supply starts changing voltage. it protects only if the LED begins to experiences either thermal runaway or for some reason draws less current than rated. As it tries to draw more current, the resistor will pass more current but uses (drops) more voltage, so the net power (voltage times current) left for the LED is still within the LEDs tolerances., and vice a versa for the state when the LED draws less current.

If the power supply voltage changes, the resistor cannot change its value, and the value is set by formula that assumes power supply voltage is constant. Try it on paper. pick a resistor, using the java tools or the mathematical formula (that the java tools use). Then change increase the voltage of the power supply in the mathematical formula, and see what is delivered to the LED, it will be the same current but higher voltage! At that point the LED will getting too much Power, and the resistor hasn't helped.

Ok so the nitty gritty is that, if you choose a power supply voltage, say two 9 volts in series, you have an 18 volt power supply. 18 divided by 3.6 is 5. That means the best configuration is 5 blue or white LEDs (that want 3.6 volts). you won't need a resistor, putting one in will under power the LEDs with fresh batteries, and as the batteries run down, you will reach the cutoff point (when the LEDs won't even light dimmly) sooner.

[quantumplasmathingie]

Aaaaarrrrgh!

Yep. Just burned out 8 of my whites. I was holding the V+ wire in some tweezers, it slipped and fell right onto the inside bit of the circuit. They all lit up really brightly and then emitted a funny smell.


They don't light no more

Curse these sausage-fingers of mine!

[Sparky]

were they wired in series to take the V+ volts? It sounds like you had them in parallel so that they wanted 3.6 volts each. Must be in some parallel configuration, all in series they would need 28.8 volts. (assuming 3.6 votl white LEDs).

Had that happen, the wiring points weren't marked and I forgot which point was supposed to be the feed, bypassed the resistor for just a moment, during final power test. They didn't die completely just got real dim.

[Balok]

They all lit up really brightly and then emitted a funny smell.

I felt the disturbance in the Force. You have just taken your first step into a larger world.

Don't let it deter you for long. You learned something important. Something...

[quantumplasmathingie]

sparky, yeah, this segment of the circuits had the 8 in parallel. poof!

Oh well, I can keep going on other parts of the circuit until the replacements arrive.

Hee hee Balok, it was as if 8 voices cried out in terror and were suddenly silenced...

If I've learned anything, it's don't power up a lead until it's wired to the post on the breadboard...or something along those lines.

"Okay Chewie, that's it. Try it...OFF! TURN IT OFF! TURN IT OFF! OFF!"

[en'til Zog]

"..dropped the +V lead and it shorted..."

Oh, the memories that brought back.

The horrible memories....





At least they didn't literally explode like mine did. Teeeeeny plastic shrapnel....

[quantumplasmathingie]

Well squeeze my feet and call me a jelly donut.

It works!

This is so ubercool, I can't believe that I haven't played with a 4060 before!
The ultrabright whites are so dazzling, I can't even look at it for very long! It's like watching an episode of "Japanese Siezure Robots". I'm not worried about the brilliance though, I need it. The flashing LEDs are going to be individually housed in light boxes made of styrene tube, and will backlight gels of dials and indicators that I printed out on transparency film. The effect looks awesome! They're all going into the head of the "Analyzer" robot from Yamato.
I can't thank you fellows enough. Really tip-top karma is going your ways, merely for the spirit of support and confidence you instill in others to just go for it and play with these things.

But I gotta 'nother question:
I've got 12 LEDs (more than I need, actually) all blinkin' away happily at their various rates, but I still think I'm not understanding something. I'm using a 12V (1000mA) wall wart for juice. I tested the mA between the ground and the cathode of one of the LEDs (when it lights), and I'm reading ~110 mA.
I didn't want to shorten the life of my LEDs, so I damped the voltage down with a 470 ohm resistor in series before the 4060 "+" pin. That dropped the mA between the LED cathode and the ground to a nice 20 mA.
I thought that the 4060 limits the current to 11 mA from the output pins. I was worried that was going to be too weak for my LEDs, but they're pretty darned bright.
I'm using a Fairchild semiconductor 4060BCN. Do you guys know if this 4060 is different than other brands?

[jwrjr]

No, the 4060 does not have any kind of current limiting. The 11ma rating is the maximum that you can drive from the output without overheating it and eventually damaging it. You should consider removing those 470 ohm resistors and using 1k resistors instead. The leds will be a little dimmer but the 4060 will last much longer.

[quantumplasmathingie]

Ah, I get it now.

I bought a tube of 10 of these little wonders, so I'm probably going to use a fresh one in the Analyzer. When I first hooked it all up and got it working, I was running it with only a single 470 ohm in series with the V+. I was so pleased that the little lights were blinking that I left it on for a few days to admire it. I felt the 4060 and the other components, and nothing was getting hot, but I'm guessing that I probably shortened the life of that component by running it with so much current for so long.

I'll throw this one into another giveaway project. I've been slowly filling my nephew's walls with "Tech Panels". Every time I come across a dead VCR, monitor or laptop, I take 'em apart, and hot glue greeblies to the plastic outer casings. A quick coat with the airbrush and it's good to hang on the wall. Eventually, he's going to live in a sci-fi movie set.

I've been lighting them up with mini mag lights and plastruct fiber optics. Now (*rubs hands together eagerly*) I can start adding the flashing lights.

[Sparky]

Hooking a resistor up on the chips power lead is not a good idea, since the resistor only works properly for steady current draw, the chips, internal clocking and driving inherently means that the current requirement will be spiking and changing.

Remember most devices only draw the current they need, the safety circuit breakers are there only to keep the wiring from burning when something goes crazy and draws to much current.

The resistor should be on the LED and you need one for each LED, since you have some turning on and running off:

If the voltage*current delivered to the LEDs is greater than its optimal running point you either shorten the life of the LED or (under powering it) lessen the brightness of the LED.

In the case of the LED/resistor combo the resistor works since you pick the resistor for some steady state current. The LED is either all the way on or all the way off, and you don't worry about what happens during the ramp up to power and the ramp down.

If you try to regulate power with a resistor in a circuit doing anything more than on or off you won't get the desired effect and may be shortening the life of the components.

Power = Current (Amps or milliAmps) * Voltage (volts)
Resistance = Voltage / Current

A resistor absorbs the extra power in a circuit like this:
Voltage of the power supply - Voltage the LED wants = Voltage extra

Current is the current the LED wants/is rated for, you assume the LED wants that current all the time its on and will get it all the time its on, aka steady state.

Resistor needed = Voltage extra / Current

If the voltage extra is negative then you need a resistor which has negative ohms, also known as a battery, thats right your circuit is under powered from the get go and no resistor is needed.

If you have a dirty power supply that may change its voltage over time (its not steady state any more) or is providing higher voltage then you use in the above formulas, the resistor won't protect your LED. You can't use the resistor to regulate current, you must get a better power supply or build a regulator into the system.

[quantumplasmathingie]

heh heh, just when I think I've sassed this hoopy thing, I get served a grilled ignorance sandwich with a side of humble pie .

Okay, so...no resistors between the 12V wallwart and the 4060 + input. Resistors go between the ground and the led(s). The light that burns twice as bright, burns half as long, so limit the current for each output to 11 mA.

Since juice isn't going to run out, I can go hog-wild with the resistors. Individual White LEDs from an output @ 3.7V, 11mA get a ~910 ohm resistor, 2 in series get 470 ohms, 3 in series get 82 ohms. Not as bright, but the LEDS will live long and prosper.

Each white LED of the same rating in parallel from a single output gets a resistor of ~910 ohms.

uhhhhhh....right?

[Sparky]

Those numbers look good, accept the first one, 920 ohms. While that might be the next nearest value to the actual value calculated:
12-3.7 volts => 8.3/ 0.011 => 754.54 ohms since you are already well under a regular white LED's current draw you don't need to round up so high.
For the LED not to get burned we should really use:
12-3.7 => 8.3/ 0.020 => 415 ohms. So anything at the value or higher protects the LED given the constant 12 volt supply.

Now the standard 4060 might not be able to give you that much current, I don't think it will burn out if you setup the circuit to try and use more current, basically there's a current limiting resistor in the chip's output transistor. (You can get versions of some chips with "open collectors" that let you drive more power by adding the appropriate resistor outside).

We should double check your Chip and see what it can source as far as power. I have noticed that some of my 4017's can provide a lot of current accept on one particular output, very annoying to find that the current limit they put on all the outputs is really caused by one of the outputs. (we made an 8 Blue LED ring with the off LED walking and one of the outputs was always dimmer than the rest just that one).

[Sparky]

Yep, min current output from this chips is 8.8 mA there's no max listed, I'm betting its 8 mA

Look at page 5 of the pdf doc, big nasty table, look at the IOL and IOH
http://www.fairchildsemi.com/ds/CD/CD4060BC.pdf

Note these are at the high voltage supply (15 volts at the chip, and the output voltage on the pin(s) is not that high, 13.5 there's a limit in the chip's ability to swing the output all the way up to the power supply and still deliver the current).

The effect is that the chip is current limiting.

This really means you're under powering the LEDs, we should look at the curve of your LED, as current changes the acceptable voltage changes too. maybe the LED can take 12 volts at 8 mA and be happy, in the end it's about power (heat). Sure there are limits to the voltage at which the little junction in there starts to fall apart (molecularly speaking).

[Sparky]

Quick search on digikey and then a browse through the data sheets (where available). shows that most of the White LEDs they have a very limited on their max forward voltage, i.e. 3.6 volts running and a max of 3.8 volts. One did go upto 4.0v.

Now there still maybe a an actual range that the LED could operate safely in but without the current/voltage graph there's no way to know if the LED can take it higher.

What does this all mean?

Well you have to either count on only 12 volts @ 8 mA of current from the chip, or assume it can give you more current assume 12 volts @ 20mA and get your resistor from that. Remember that the chip also has some loss between its supply voltage and the max output voltage anywhere between 1/2 and 1 volt difference).

One experiment would be to go with a resistor based on getting 12 volts @ 20 mA and then measure the voltage across the LED and the current going through it. You might find that the chip can supply this but not on every output or worse, you loose some current when you hook up more outputs (total current through the chip starts getting divided up as more outputs get used).

I think either way the LEd will be starved for power, the voltage might be to hot, so the resistor will be needed if the chip is really keeping the voltage on the output, when loaded with LEDs, at the power supply rail.

I have some junk LEDs from some Christmas light sets, I can try to breadboard a 4017 and take some measurements.

By And By:
http://www.kc6sye.com/techmages_11_10_06.html
This 4060 ran a single white LED without resistor, however it was supplied with 3.3 volts. The LED wasn't putting out a lot of light but it was good enough.

[quantumplasmathingie]

Thanks a million Sparky, I really appreciate all that you're doing to clear up my muddlement.

I think that the 8mA should still be fine for these whites. I just accidentally burned the fuse in my multimeter. I wanted to see how steady my wall wart was, but forgot to change the setting from 150 mA to 15VDC.
Until I can drop by radio shack for a new fuse, I just kept adding 470 ohm resistors in series between the led and ground. I'm up to 2350 ohms and the thing is still bright as all get-out. I'm sure I'm below the 11 mA draw, but I won't know for sure until I get the multimeter back up.

[jwrjr]

The nice thing about using ultra-bright leds (especially white) is that you can run them on reduced current and still have good brightness. This is especially useful if the project is to be battery powered (not that this one is), or you want to use a lot of leds without having a monster power supply. My refit is lit that way.