HCF4017BE problem

Can any1 help me with this IC . I'm trying to make a traffic light using IC 555 and 4017 .
I use multisim to create a design like the file i attached .
There not much problem when i run it in multisim .
Then I bought 2 IC : NE555P and HCF4017BE.
A problem is that when i do everything like in design, only 4th and 12th leg of IC 4017 light up my led . But it light up just like if i connect my led directly to the output of IC 555 . It feel like my 4017 IC only connect between leg 4, 12 and 14 and that all .
So any1 know what the problem here ?
Thank for your help and sorry because im bad at english .

you have to connect pin 8 to ground (negative) and pin 16 to supply voltage. you can refer the data sheet..
CD4017 Datasheet pdf - Hex D-Type Flip-Flop Quad D-Type Flip-Flop - Fairchild Semiconductor

dont know why . seem like it work now . if there a problem i will really need your help thank very much

you are welcome, also read the bellow thread until the end. its for 4 street system.
https://www.electro-tech-online.com/threads/traffic-light-help-sos.99450/

The 4017 will source only enough current to make the LEDs quite dim. You will likely need a transistor to switch the LEDs...

mbarazeen said:

you are welcome, also read the bellow thread until the end. its for 4 street system.
https://www.electro-tech-online.com/threads/traffic-light-help-sos.99450/

Click to expand...

I done everything with my design but still, thank for the link

MikeMl said:

The 4017 will source only enough current to make the LEDs quite dim. You will likely need a transistor to switch the LEDs...

Click to expand...

I use 9V input and put a 680Ω before each Led, they work well, not dim at all . But thank for advise .

Here's a circuit:


**broken link removed**

If you have a low impedance clock driving a CMOS circuit, the circuit may operate even with no power applied since the V+ will charge up through the protection diode between the input and the Vdd line. This diode acts like a rectifier to provide power to the circuit from the clock. But that's not a desired way to operate the circuit.