Iso 4017 Pdf

[Rell Jette]

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I'd like to power my LEDs with 20mA each, so the plan is to connect the 4017 to the Arduino's pin via a current limiting resistor, then connect two LEDs in serial, directly to each of seven pins, and then to ground... the idea being to make full use of that 5v coming off the pin to power both leds with 20mA without exceeing the 4017's output capability, and to reduce the number of resistors and arduino pins needed to just one.

PS:

I plan to wire the eigth pin in the 4017 sequence to the 4017's reset pin, which I assume will trigger the first led in the sequence to illuminate the moment I hit the 8th clock in my sequence.

PPS:

The current limiting resistor I've calculated I'll need for the 4017 is 47 ohms, based on a 5v source, a 2.1v drop, and 20mA of current, powering two leds per pin in serial. Also, according to the LED wizard, the leds in this configuration will dissipate 84mW, which is well below the 500mW which another CMOS 4017 I found said it could handle. (The TI one doesn't list the power dissipation from what I could tell, but I assume its similar.)

I know there is an attitude that says "well it works so it's OK" but that does not auger well for a long life of these components. There is also package power dissipation issues that are difficult to address as this chip is not designed for driving LED loads.

You will probably find that at a high current you don't actually get 5V out, the voltage drops with current. You will see that the output can drop as low as 3.7V under some loads at some temperatures. This means the resistor value has to be small and with small resistors they act less like a constant current source and as so less effective at regulating current.

Yes, and I was referring to the absolute maximum. That's why I said 'max'. I'm aware I should not actually run that much current though it for any period of time. I quoted that figure because that was my starting point for determining what a safe current to run through it is. I'm not sure how else I'm supposed to calculate what a safe current is.

And the thing is, I have to drive a display with 7 leds on it as well. So I gotta use a 4017 for that too. I don't want to have to use four chips in this prop just for the sake of making the leds a little brighter.

Besides, I just tested the leds at 10mA, and you're right. I cannot see a difference. That extra 10mA made a big difference in my led array, but those leds were kinda dim and being multiplexed. These are super bright ones.

Let's say I connect the 4017 to the Arduino pin via a 100ohm resistor, and connect two LEDs with a forward voltage of 2.05 in serial to each of the 4017's outputs, proving each pair with around 5v at 10mA.

I've never seen a datasheet include a 'recommended operating level'. The only thing I've seen that might be considered a reccomendation is when the manufacturer lists a 'test coniditon' for a part, like an LED where they usually list how it performs at 20mA.

And besides, why would you run at 20% below the operating level which is reccomended? That doesn't make sense to me. Why reccomend an operating level if it's not actually reccomended you run at that level?

The point is you asked if this chip was up to it. My answer is NO. It was not designed as a driver and was not meant to be used in the way you are planing to use it. If you don't like the advice then pay no attention to me, I have only been in electronics 40 years what do I know.

You know, normally I'm in favor of treating data sheet "max" ratings with some respect, but the 4017 has been used in SO many hobbyist LED projects that I think it has proven itself capable in that role, and your original idea should work just fine. A carefully calculated resistor value will give you less current than you think, since the 4017 internals are NOT ideal switches; it has some internal resistance as well.

Mike, don't get me wrong. I really appreciate the advice, and I'm taking what you say seriously. I just want more information. I need to understand the 'why' of things. Answering 'Can the 4017 handle this?' with 'No.' isn't useful to me. If it can't handle 20mA, I need to know what it can handle, so I know when I can use it and when I can't.

Also, why does it matter what it was designed to do, as long as I run it within safe parameters? If it can handle 5-10mA and I'm fine running my LEDs with that level of current, then what's wrong with using it in that instance? Current is current, right?

I'd still like more info on the 4017's capabilities, but I thought I should let ya know that I'm now looking at using the ULN2803A (or its 7 input cousin whose part number I forget) in conjunction with the 4017.

The reason for the change is I realized that if I was gonna have to use two 4017 chips anyway if I wanted to run two displays with a total of three LEDs, that I might as well make one of those chips a darlington array and shift the load onto that.

Hm... I may be able to spare a couple more pins. Might be a bit cheaper than buying two chips and two ice sockets to go with them. Would definitely take up less space and space is at a preminum on this board.

I haven't read the detail of the data sheet but what can go wrong is that logic chips don't guarantee that the 'high' output voltage is a full 5v. It only needs to be high enough for another logic chip to recognise the input as high, and that can be significantly lower, maybe 4v or 3v. Especially when the output is under load and the chip has a bit of internal resistance the output voltage may drop. So to get 20mA out you may need to lower your series resistor. You might even find that with 2 LEDs in series you can't get 20mA out of a pin even with no series resistor.

It can happen that you get different voltages under heavy load from each different pin, leading to uneven brightness (I've had this when trying to make a 8x8 LED display driver using only 74hc595s without buffers; the resulting uneven brightness is particularly annoying). Different chips can give different output too, in the worst case you could find that you swap the chip and one of the outputs is overcurrent.

Ultimately if it's your own personal project it doesn't really matter, you may as well do it anyway and if it works out too dim or uneven it isn't really a problem. But if it was commercial or on display then I'd recommend using a buffer.

I have noticed that the if you assume the same current flow, the illuminousity will change between the different colour leds. Most of this is due to the difference in forward voltage, and some in the human eye.

This one relates to when the wall supply delivers a voltage greater than 5v, and you use a 5v regulator on the board.

You can opt to use a 7805 which is physically large, and drop the voltage to everything, including LEDs, etc.

OR use something like a 78L05, and run the essentials.

The latter method also removes the current switching spikes, which in digital ICs sometimes comes back to haunt the designer. (I've seen many a circuit work using a bench supply, with lots of filtering, to fail when the final unit is made)... :-?

No. Power is the product of voltage and current. Suppose you have an LED with 2V across it and 20mA through it. The power dissipation in the LED would be 2 * 20 = 40mW.

Now suppose you switched that with a transistor that had a saturated collector emitter voltage of 1V then the power dissipated in transistor switch would be 1 * 20 = 20mA. Then suppose you switch that LED with a FET that had an on resistance of 0.5 ohm. Then with 20mA through it it would have a voltage of 0.5 * 20 = 10mA therefore the FET would dissipate 10 * 20 = 30uW.

So you see the power dissipated in the load is not the same thing as the power dissipated in the switch.

One thing about chips in general is that the maximum current rating is nearly always higher than the total power dissipation, so it is power dissipation that becomes the limiting factor. see my notes on the Darlington drivers:-

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