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
trouble with logic counters
Started by ●January 2, 2007
Reply by ●January 2, 20072007-01-02
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, thanksRather 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
Reply by ●January 2, 20072007-01-02
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 LEDThat 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>
Reply by ●January 2, 20072007-01-02
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
Reply by ●January 2, 20072007-01-02
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>
Reply by ●January 2, 20072007-01-02
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 yearWith 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