trouble with logic counters

Rather use a 4000 series counter. This can run directly from 9V. Using a resitor divider to provide power for the 74HCT163 is not a good idea. 1/2 of a 4520 IC should do the job. Connect the clock from the 555 to the clock input of the chip. Tie the enable pin high and connect bit 4 of the counter output to the reset pin. The counter will then reset to 0 every time it reaches 8, hence only count from 0 - 7. Also remember to provide some decoupling capacitance.

Regards Anton Erasmus

... snip ...

... snip ...

That logic series, IIRC, uses 5 volt supplies. You are effectively supplying 4.5 volts through a 5 M resistor, which won't run it. 1 uA draw will result in zero volts at Vcc. Get a 5 volt regulator chip.

Also, where does the voltage to the LEDs come from? If the 9 volt supply, you have probably already destroyed the chip. It the alleged 4.5V thru 5 megs there is no power to run them, but you probably haven't destroyed the chip.

Read up on Ohms Law. It's not hard.

Yes you were correct, that was the problem, I put in a regulator and the thing started running, thanks

The voltage to the LEDs is coming from the outputs of the counter, they light up and I can see the thing count.

No, the chip was never connected directly to the 9 V, but through that voltage divder I mentioned, so the chip lived.

Right

ok,

Here's one thing I don't understand. I thought that as when you put something in parallel with something, the voltage across the 2 things stay the same and only the current changes. When I put LED's at the outputs of the counter, they light up fine, the counter is ouputting 5 Volts and everything's great. But then when I take this output and attach it to the controls of an 8 to 1 MUX, the voltage drops and I can't drive the MUX unless I disconnect my LEDs. Then the 5 Volts output from the counter go straight to the mux, but sharing an output form the counter with an LED and the MUX control input..... doesn't work. Isn't that a parallel connection? And shouldn't the voltage stay the same with a load connected in parallel with another load?

thanks

happy new year

... snip ...

Most logic has greater current drive ability in the low state than in the high state. Thats why you normally connect led anodes to the supply through a current limiting resistor, and drive them by the low signal. A led with 5k to +5 needs about 1 mA drive, 1k will need 5mA drive (roughly, actually less because of the drop across the LED, which will be in range 0.7 to 3 volts, depending on brand and color.) Again, apply Ohms law.

With the LEDs connected to the counter's output, the counter's HI and LOW voltage levels may no longer be correct for the MUX chip. For

-Dave Pollum