my question is about the IC 4017 decade counter. how much is the maxumium current that the output can produce withoput the chip being destroyed. i have search he net and connot seem to find anything that is of use. plz let me kno

 

Here is a link to the datasheet for TI's version of 4017B.

According to the datasheet, with +5V powering the device, the output can sink 1mA or source 1mA.

hgmjr

 

If you wish to control devices that require more than about 0.35 mA, I suggest you look at using ULN2004/ULN2804 ICs as drivers. These are, respectively, Seven/Eight Darlington Driver IC's with inputs suitable for control via CMOS. These driver ICs are configured as open-collector; they cannot source current, but they can sink 500mA per Darlington pair, and they can handle up to 50V on the outputs. They also have built-in diodes to take care of reverse EMF if you're driving inductive loads like motors or relays.

 

http://www.doctronics.co.uk/pdf_files/4017.pdf This one gives a complex formula for it. They never come out and state it in a useful fashion. I know why but they don't.

 

If you wish to control devices that require more than about 0.35 mA, I suggest you look at using ULN2004/ULN2804 ICs as drivers. These are, respectively, Seven/Eight Darlington Driver IC's with inputs suitable for control via CMOS. These driver ICs are configured as open-collector; they cannot source current, but they can sink 500mA per Darlington pair, and they can handle up to 50V on the outputs. They also have built-in diodes to take care of reverse EMF if you're driving inductive loads like motors or relays.

this is meant to be 35 mA am i correct?

 

No, I meant 0.35mA. If you look at the tables for the outputs, at 5v operating at 125 degrees, that's what it'll output.

Sure, it'll output 1mA at 5v Vcc and 25 degrees, but that's operating in optimal conditions. One should plan on things going more towards the worst-case scenario; that way your circuit is more likely to stay functioning under such conditions.

 

Let me preface this by saying that you won't get valid logic levels under heavy loads, but if you are trying to drive an LED, or the base of a transistor, this may not be a problem.
Looking at the TI datasheet (Fig. 7), the short circuit output sourcing current is about 20mA (typ) when Vcc=10V. This is the most current an output can source, so the answer is, something less than 20mA. 20mA will cause the chip to dissipate 200mW. This should not hurt the chip, because it is rated at 500mW. Of course, the chip may not be typical, but I doubt it would source 50mA.
Sinking is another issue, since 9 outputs are always low. Sinking loads can easily exceed the 500mW limit if you have multiple loads connected.

 

ok thanx for the specs and the advice, i am greatful

 

ok thanx for the specs and the advice, i am greatful

I should have mentioned that the output current will vary with the supply voltage, as you can see in the datasheet. Power dissipation will be a function of voltage, as you probably know. I just used 10V as a convenient example.