Forums

trouble with logic counters

Started by panfilero January 2, 2007
Hello, I'm trying to hook up a 74HCT163 to count from 0 to 7 then loop
and keep doing it.  I'm having a lot of trouble making sense of the
data sheet on how to connect this thing.  I have a 555 chip creating my
clock, it's lighting an LED right now, I took that from the positive
end of the LED and am running it into the counter.  I was worried about
going straight from the 555 chip to the counter cause i've heard these
type of chips are fragile, real sensitive to too much current/voltage.
I'm powering the circuit with a 9 Volt battery, I'm splitting that in
half using a couple resistors (of equal value 10 Mohm) in series and
taking the voltage from one resisttor and using it as my high for the
counter.  I've connected the pins in this way:

1   MR - high
2   CP -  555 output
3   P0 -  high
4   P1 - high
5   P2 - high
6   P3 - high
7   PE - high
8   GND - ground
9   SPE - high
10  TE - high
11   Q3 - not used
12   Q2 - to a 1K resistor in series with an LED
13   Q1 - to a 1K resistor in series with an LED
14   Q0 - to a 1K resistor in series with an LED
15   TC - not used
16    Vcc - high

I can't get anything to happen, and can't tell what some of these pins
are even for.  Any help with this would be greatly appreciated, thanks

On 1 Jan 2007 21:06:23 -0800, "panfilero" <panfilero@gmail.com> wrote:

>Hello, I'm trying to hook up a 74HCT163 to count from 0 to 7 then loop >and keep doing it. I'm having a lot of trouble making sense of the >data sheet on how to connect this thing. I have a 555 chip creating my >clock, it's lighting an LED right now, I took that from the positive >end of the LED and am running it into the counter. I was worried about >going straight from the 555 chip to the counter cause i've heard these >type of chips are fragile, real sensitive to too much current/voltage. >I'm powering the circuit with a 9 Volt battery, I'm splitting that in >half using a couple resistors (of equal value 10 Mohm) in series and >taking the voltage from one resisttor and using it as my high for the >counter. I've connected the pins in this way: > >1 MR - high >2 CP - 555 output >3 P0 - high >4 P1 - high >5 P2 - high >6 P3 - high >7 PE - high >8 GND - ground >9 SPE - high >10 TE - high >11 Q3 - not used >12 Q2 - to a 1K resistor in series with an LED >13 Q1 - to a 1K resistor in series with an LED >14 Q0 - to a 1K resistor in series with an LED >15 TC - not used >16 Vcc - high > >I can't get anything to happen, and can't tell what some of these pins >are even for. Any help with this would be greatly appreciated, thanks
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
panfilero wrote:
>
... snip ...
> I'm powering the circuit with a 9 Volt battery, I'm splitting that in > half using a couple resistors (of equal value 10 Mohm) in series and > taking the voltage from one resisttor and using it as my high for the > counter. I've connected the pins in this way: >
... snip ...
> 12 Q2 - to a 1K resistor in series with an LED > 13 Q1 - to a 1K resistor in series with an LED > 14 Q0 - to a 1K resistor in series with an LED
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. -- Merry Christmas, Happy Hanukah, Happy New Year Joyeux Noel, Bonne Annee. Chuck F (cbfalconer at maineline dot net) <http://cbfalconer.home.att.net>
CBFalconer wrote:
> panfilero wrote: > > > ... snip ... > > I'm powering the circuit with a 9 Volt battery, I'm splitting that in > > half using a couple resistors (of equal value 10 Mohm) in series and > > taking the voltage from one resisttor and using it as my high for the > > counter. I've connected the pins in this way: > > > ... snip ... > > 12 Q2 - to a 1K resistor in series with an LED > > 13 Q1 - to a 1K resistor in series with an LED > > 14 Q0 - to a 1K resistor in series with an LED > > 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.
Yes you were correct, that was the problem, I put in a regulator and the thing started running, thanks
> Also, where does the voltage to the LEDs come from?
The voltage to the LEDs is coming from the outputs of the counter, they light up and I can see the thing count. > If the 9 volt
> supply, you have probably already destroyed the chip.
No, the chip was never connected directly to the 9 V, but through that voltage divder I mentioned, so the chip lived.
>It the > alleged 4.5V thru 5 megs there is no power to run them, but you > probably haven't destroyed the chip.
Right
> Read up on Ohms Law. It's not hard.
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
panfilero wrote:
>
... snip ...
> > 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?
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. -- Chuck F (cbfalconer at maineline dot net) Available for consulting/temporary embedded and systems. <http://cbfalconer.home.att.net>
panfilero wrote:
> CBFalconer wrote: > > panfilero wrote: > > > > > ... snip ... > > > I'm powering the circuit with a 9 Volt battery, I'm splitting that in > > > half using a couple resistors (of equal value 10 Mohm) in series and > > > taking the voltage from one resisttor and using it as my high for the > > > counter. I've connected the pins in this way: > > > > > ... snip ... > > > 12 Q2 - to a 1K resistor in series with an LED > > > 13 Q1 - to a 1K resistor in series with an LED > > > 14 Q0 - to a 1K resistor in series with an LED > > > > 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. > > Yes you were correct, that was the problem, I put in a regulator and > the thing started running, thanks > > > Also, where does the voltage to the LEDs come from? > > The voltage to the LEDs is coming from the outputs of the counter, they > light up and I can see the thing count. > > > If the 9 volt > > supply, you have probably already destroyed the chip. > > No, the chip was never connected directly to the 9 V, but through that > voltage divder I mentioned, so the chip lived. > > >It the > > alleged 4.5V thru 5 megs there is no power to run them, but you > > probably haven't destroyed the chip. > > Right > > > > Read up on Ohms Law. It's not hard. > > 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
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 74HCT chips a LOW input must be between 0 and 0.8 volts, and a HI must be between 2 and 5 volts. And, as others have suggested, use a 5 volt regulator to supply 5 volts to the power pins of your chips. -Dave Pollum