Hi everyone.
First post so please excuse me if I am posting incorrect. Takes me a while to catch on to a new forum. I am a retired CBMET (Certified Bio-Medical Electronics Technician) with more then thirty years in electronics. Still get a big kick out of it.
Anyhow, I have a question, of course. Working with the CD4017BE CMOS chip recently on two different circuits. A four way stop light for toddlers to help them learn traffic rules while still on their trikes. A second circuit for manual counting up to fifty items.
No problem with the circuits, real simple but it brought to mind a question I have pondered for a long time.
The datasheet for the 4017 gives max source and sink current for the outputs as somewhere around 8 mA. Depends on manufacturer exactly what. Okay, the only thing is, I have used them over the years to drive LED and this last project powered the 15,000 MCD LEDs at 25mA. What gives, or rather what am I missing here? I have read the datasheet over and over but I must be missing something. Anybody have a clue.
Thanks!
Roland

 

When you say you powered an LED at 25ma, was that the actual current or the rating of the LED?

 

You should post a schematic of the circuit you used to do that, if you drive it through a transistor, you can multiply your available max current by 100 or something like that.
And chips can usually be pushed past their datasheet limits, just not forever.

 

When you say you powered an LED at 25ma, was that the actual current or the rating of the LED?

Thanks Digg, but the question is not about the circuit, just the rating on the datasheet. Thanks for your reply.
Roland

 

This has been debated in the past about the drive
capabilities of 4017. Audioguru once posted an explanation
but I cannot find it right now. The gist of the explanation
involved the graphs on page 3-53 Figs. 5-8 of the datasheet
I have attached. I remember the post was very informative
including the explanation of how the graph was used to arrive
at the conclusion. I wish I could explain it as well as he did.

 

This has been debated in the past about the drive
capabilities of 4017. Audioguru once posted an explanation
but I cannot find it right now. The gist of the explanation
involved the graphs on page 3-53 Figs. 5-8 of the datasheet
I have attached. I remember the post was very informative
including the explanation of how the graph was used to arrive
at the conclusion. I wish I could explain it as well as he did.

Hi.
That is the same datasheet I am using or one of them. I did study the graphs but even there it seems to be less then I have experienced. My thoughts were/are it may have to do with the Vin as it is listed as 0.15V for -6.8 mA.
Well, thanks and if you run across that writeup I would appreciate reading it.

I am having a time trying to get to the right person, I mixed up some respondents

Roland

 

When you say you powered an LED at 25ma, was that the actual current or the rating of the LED?

No, the LED is rated up at to 30 mA, that was the measured current through the LED. This is not the first time I have wondered about this chip and I realize you can push components beyond the datasheet limits. In this case I have run different circuits hundreds of hours without failure just to be sure. I am still wondering??. In case you ask, I am using this chip because I have forty or so remaining from when I closed my business, old but good, just like me, well the old part at least.

 

A 4017 can drive a LED at 25 mA, but the output is no longer a ligit logic level as V willbe down to about half of Vdd at output. Figure that the output impedance is around 300 Ω.

 

You should post a schematic of the circuit you used to do that, if you drive it through a transistor, you can multiply your available max current by 100 or something like that.
And chips can usually be pushed past their datasheet limits, just not forever.

Maybe this time I will get to the right person. I posted earlier but to another answer apparently.

Anyhow, I know you can boost the current with the addition of a transistor/s but this was a case of the fellow wanting to drive the LEDs direct, no BCD or display and the question has been around, with me at least ever since the 4017. So a circuit would not help much, it is the data sheet I am questioning.

I do know you can push components beyond the limits set by the datasheet but this has come up with me so often in the past I have run the 4017 at 20 to 25 mA for hundreds of hours with no ill affect. That is why I decided to ask just to see if I am alone or if someone has a better idea then me.
Thank very much for your fast response, everyone if I miss someone, you have at least assured me I have found a live forum.
Take care.
Roland

 

The datasheet of the CD4017 from Texas Instruments has very detailed graphs showing typical and minimum output current at different supply voltages and with different output voltages. It doesn't rate the max allowed output current, instead it lists 100mW as the max allowed output transistor power dissipation.

For example, with a 10V supply and a 3.5V blue LED with no current-limiting resistor the output transistor has 6.5V across it and the graph shows a typical current of 17mA. The power dissipated in the output transistor is 6.5V x 17mA= 111mW which is too much.
If a 9V supply or less is used then no current-limiting resistor is needed.

Use two 3.5V blue LEDs in series, a 10V supply and no current-limiting resistor. The typical graph shows a current of 11mA and a power dissipation of only 77mW so it is fine.

If the supply is 13.5V and the output drives a 2V red LED without a current-limiting resistor then the current would be 25mA because you are throwing away 11.5V and the LED is shorting the output. The power dissipation in the output transistor is 288mW which might be OK if the CD4017 cycles through its 10 outputs quickly then one output dissipates an average of only 29mW. But how quickly is OK?

I use a 20mA load on 74HCxxx Cmos logic ICs because their max allowed output current is 25mA.

 

I would think it was about the degree by which the device voltage output deviates from VDD (in source mode) or VSS (in Sink mode) as (source or sink) current increases.

Consider the device in current source mode. The graphs mentioned by Pencil show that, given sufficient current drain in source mode, the entire VDD supply value can apparently be absorbed across the internal on-state CMOS device (Drain to Source) with nothing left at the output pin.

The static electrical characteristics table states that for the source mode with VDD=15V, Vout can fall to 13.5V with a typical current drain of 6.8mA. It can be "worse" with the variability in devices meaning a drain of only 3.4mA might produce the same internal drop. For a typical case of 20mA drain at a supply VDD=10V, it is conceivable the output would have fallen to 0V.

Similar arguments would apply to the current sinking mode.

 

A 4017 can drive a LED at 25 mA, but the output is no longer a ligit logic level as V willbe down to about half of Vdd at output. Figure that the output impedance is around 300 Ω.

Hi Bernard.
Now that makes sense. I am and was not concerned with maintaining true logic level, just a progressive sequence of LED and as the fellow wanted, or did not want to use transistors I set it up, this last circuit, and breadboarded it to be sure. It worked and he was happy so what can I say. I did recommend transistors in the first "design" I gave him but,,, You know how that goes, the customer is always right even though this was gratis work.
So that scratches an itch that has been around for some time.
I am quoting you message because when I tried to respond before I was not getting to the correct person. Got to figure this forum out.
Thanks again and take care.
Roland

 

I would think it was about the degree by which the device voltage output deviates from VDD (in source mode) or VSS (in Sink mode) as (source or sink) current increases.

Consider the device in current source mode. The graphs mentioned by Pencil show that, given sufficient current drain in source mode, the entire VDD supply value can apparently be absorbed across the internal on-state CMOS device (Drain to Source) with nothing left at the output pin.

The static electrical characteristics table states that for the source mode with VDD=15V, Vout can fall to 13.5V with a typical current drain of 6.8mA. It can be "worse" with the variability in devices meaning a drain of only 3.4mA might produce the same internal drop. For a typical case of 20mA drain at a supply VDD=10V, it is conceivable the output would have fallen to 0V.
Similar arguments would apply to the current sinking mode.

Hi.

Another good answer and sort of along the lines I was thinking although stated much clearer then I could have.

Thanks for the fast response and take care.
Roland

 

The datasheet of the CD4017 from Texas Instruments has very detailed graphs showing typical and minimum output current at different supply voltages and with different output voltages. It doesn't rate the max allowed output current, instead it lists 100mW as the max allowed output transistor power dissipation.

For example, with a 10V supply and a 3.5V blue LED with no current-limiting resistor the output transistor has 6.5V across it and the graph shows a typical current of 17mA. The power dissipated in the output transistor is 6.5V x 17mA= 111mW which is too much.
If a 9V supply or less is used then no current-limiting resistor is needed.

Use two 3.5V blue LEDs in series, a 10V supply and no current-limiting resistor. The typical graph shows a current of 11mA and a power dissipation of only 77mW so it is fine.

If the supply is 13.5V and the output drives a 2V red LED without a current-limiting resistor then the current would be 25mA because you are throwing away 11.5V and the LED is shorting the output. The power dissipation in the output transistor is 288mW which might be OK if the CD4017 cycles through its 10 outputs quickly then one output dissipates an average of only 29mW. But how quickly is OK?

I use a 20mA load on 74HCxxx Cmos logic ICs because their max allowed output current is 25mA.

Hi.

Okay, now I am beginning to understand. Going by that I am dissipating right at 100 mW. Still everything worked after breadboarding it an testing for an extended time. The only LED I used was the green and the current on that one as I remember was a bit lower then the stated 25 mA, I think around 20. I had checked the temperature on the 4017, many a time using it over the years and it was never hot or even warm. Also this time the circuit was advanced manually using a one shot and I ran it for hours on at each stage, well at least overnight for a couple outputs. I did use a current limiting resistor for each output.

How long is fine? Good question and I would rather not learn by letting all the magic smoke out.

Thanks for the information and fast response. Everyone has been really great.

Take care.

Roland