Hello everyone
I was looking to build a little lighting display with LED's using the CD4017B counter but I was wondering if there is a similar one that someone could recomend that doesn't need a transistor on each output?
I looked up the datasheet and it said that the max output current for each ouput was 3.5ma at 15V supply.I only wanted to use about 9V and drive 8 LED's.

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if you're only driving LEDs, you shouldn't need transistors... or are they high power/brightness LEDs?

 

3.5mA is a little low to light a led...

 

I know,3.5ma is way too small for a normall 25ma LED.In most circuit diagrams a transistor switch is used on each output.But i was wondering if there is another decade counter that can provide up to 25ma on each output,that way I wont need so many transistors.
Im looking at cascading the 4017 so as to have about 16 outputs........... 16 transistors is abit much. :lol:

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Hi Elf,
Figure 7 in TI's datasheet for their CD4017B shows a typical short-circuit output source current of 20mA with a 10V supply. The graph also shows that it will give a current of about 18mA through a 1.8V LED. But since the chip's output transistors have an absolute power dissipation limit of 100mW, a current-limiting resistor must be used to share the power.
Using a single (the outputs are on only one at a time) 100 ohm resistor that is connected between all the LEDs cathodes and ground, and a 9V supply, the LED current should be about 14mA.
That should be plenty of current for a modern high-brightness LED such as MV8191 or HLMP-D101A from Fairchild and cheaply available at Newarkinone or maybe Farnell in the UK.
Your eyes' response to brighness is logarithmic, so if the brighness (LED current) is cut in half, then it looks only a little less bright. 1/10 of the brighness looks half as bright. Just like sound level and your ears: "My 1000 Watt amp sounds only twice as loud as your 100W amp". It does.
TI's datasheet is here:
http://www.ee.washington.edu/stores/...000/cd4017.pdf

 

i've several simple circuits lying infront of me, all of which run the LEDs directly from a 4017. They are not at full brightness, but are very close, probably for the reasons suggested by audioguru.

 

probabely also due to the fact that those parts often can handle more power then the datasheet says.
If you connect a led to it and limit the current to 20mA or so then a 4017 is very likely to just light the led...

now, if it is good for the 4017 is another question

 

Thanks a span Guys,it sure is gonna be easier than setting up 16 transistors,I didn't know that our eyesight was logarithmic and that the LED would only me a little dimmer....thanks audioguru.

Regards

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Exo,
A CD4017B (or any other regular CMOS chip) will typically supply a current-limited 14mA with a 9V supply, which is well within its ratings. Some a little more, and some a little less.
14mA through an ordinary LED is bright, and through a modern high-brightness LED is dazzling. Haven't you seen those flashing kids shoes?
The current-limiting resistor was calculated to protect the CD4017B with a generous margin, according to the max. ratings on the datasheet.

Elf,
Ignore for a moment that your sight and hearing have built-in automatic gain control. If your eyes and ears weren't logarithmic, and were setup for a sunny day and a loud jet plane nearby, then you would be blind when it is cloudy or nightime, and deaf to someone yelling or talking (or whispering?) in your ear. I'm not talking about selective blindness or deafness (Sorry wife, I didn't see your new dress, and I didn't hear you tell me to take out the garbage).
I built a linear ramp for slow, smooth LED dimming. The LED didn't dim much. Then I made the ramp logarithmic, and it worked well until offset voltage affected the dim microamp current level.

 

Instead of a long story, why don't you attach a schematic?

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Thank you audioguru for this information. That one thing about LED's and light I did not realize.

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No.
Their absolute max allowed output current is 25mA per output. They will overheat and maybe blow their output transistors with an output current more than 25mA.
The outputs go high one-at-a-time, so most outputs are low. It sources current, it is not used as a current sink.

No.
The absolute max allowed supply voltage for the IC is 7V but they recommentd a max of 6V.

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Uncle $crooge

 

If you want to be able to power it directly from a 9V battery, use the 4000 Series chip, the 4017. The 4000 series has a very low mA output, but it should be fine to drive a LED. If you want more, use a Transistor.

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Hckr69, you seem to have a a lot of self confidence, and maybe it's justified, but I wouldn't take your advice. You said This is clearly wrong, as Audioguru pointed out. Why should I trust your advice?

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Since hckr69 is talking about using a 9V power supply then he must be talking about driving series LEDs from a 4000-series ordinary Cmos logic iC. But he is wrong here too:

1) The datasheet for Texas Instruments CD4017 was linked earlier in this thread. It shows an output current into four 2V red LEDs in series of only typically 8mA with a 10V supply. The current will be less with a 9V supply. So a current-limiting resistor is not needed.

2) Into a single 2V red LED the output current is 19mA with a 10V supply and is less with a 9V supply. So a current-limiting resistor might be needed because the max allowed power dissipation per output is 100mW and with a current of 16mA and 7V across the output transistor then its power dissipation is 112mW.
With a single 3.5V white or blue LED and a 9V supply, the power dissipation in the output transistor is less than 100mW so a current-limiting resistor is not needed. The current will be about 15mA.

3) Hckr69 calculated the value of current-limiting resistors as though they are connected directly to a 9V power supply. But 4000-series Cmos ICs have a fairly high output resistance which must be included in the calculation.

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Uncle $crooge