# Pure Logic Gates Basketball Scoreboard

#### absf

##### Active Member
asbf - thanks for the effort, but you're not quite there yet. The monostables have to hold the astable 555 in reset to guarantee that it always starts with a complete cycle. In other words, the gating has to come before the astable, not after it.

ak
Thanks, and now I see the problem and how to solve it. That's why when I press the step 3 button, sometimes I got 4 pulses. I tried to sync the start of pulses of the astable 555 using its reset pin to start the pulsing from the 555 mono-stable trigger pins through AND gates but it didn't quite work as I planned.

So, I'd leave it for the OP to solve this synchronization problems.

Allen

#### Beau Schwabe

##### Active Member
I think the initial intent of the 555 timer is for a button debounce (555 setup as a one-shot) .... but as mentioned above if you have say four 555 timers, where one of the four is a base timer (sort of like a one shot but triggered from the other three 555 timers) ... the base time is the longest time interval. the other three 555's have an interval that is a multiple of the base interval .... i.e. one of the three would only be able to produce one pulse within the base interval, while a second one would be able to produce two pulses, and lastly the third 555 would be able to produce 3 pulses within the base interval.... the over all function of the pulses would be "gated" by the base interval and effectively have a debounce built in.

#### AnalogKid

##### Well-Known Member
In your circuit, the rate at which the score advances within a single gate would vary depending on the number of increments above one point. Don't know if this is better or worse than evenly-spaced pulses, which take a varying amount of time to complete an update.

As an extra-credit puzzle, generate 1, 2, or 3 evenly-spaced pulses into the 7490 clock input using only two 555's.

ak

#### absf

##### Active Member
In your circuit, the rate at which the score advances within a single gate would vary depending on the number of increments above one point. Don't know if this is better or worse than evenly-spaced pulses, which take a varying amount of time to complete an update.

As an extra-credit puzzle, generate 1, 2, or 3 evenly-spaced pulses into the 7490 clock input using only two 555's.

ak
I think it can be done by connecting the 3 buttons to the RC of the monostable 555. Will try it out later..

Allen

#### absf

##### Active Member
Here's my first trial but didnt work very well.

I've also tried 4066, 74HC238 in place of the 3 PNP transistors, but they are worst than using the BJTs.

#### Alexander Venus

##### New Member
Hello. I've seen the responses and I am thankful for the help. As I said earlier, I'm completely a newbie in this subject and I'm just absorbing infos from all of your answers so bear with me if I dont understand some terminologies :3

#### JoeJester

##### Active Member
If it is a pure love. Gates scoreboard, why is there a 555 timer in the mix.

#### Alexander Venus

##### New Member
My prof said that it must have 555 timer to generate pulses, making it possible to have 2 points and 3 points

#### Alexander Venus

##### New Member
After reading your replies, I cant make a working circuit. maybe due to my lack of knowledge about the subject :3

#### Alexander Venus

##### New Member
So i did a study last night and Ive made this circuit but i cant seem to make the 74192 counter to decrement, here is my circuit plese advice

#### absf

##### Active Member
What's the value of the resistor that pulls the MR of 74192 low?

Allen

#### Alexander Venus

##### New Member
No resistor attachwd atm

#### AnalogKid

##### Well-Known Member
Here is a circuit for the requirements in post #1. It requires that you release the 2 and 3 buttons before the circuit is finished updating the 7490, or there might be extra pulses. This can be fixed by adding a differentiator to the U2 ans U3 inputs, similar (but not identical) to the U1 input. Let me know if you want this.

ak

#### AnalogKid

##### Well-Known Member
The same functions can be achieved with two CD4093 quad Schmitt trigger NAND gate ICs and *no* 555's.

But if you're really loony, the entire circuit can can be done with only one CMOS hex inverter (and six diodes).

ak

#### AnalogKid

##### Well-Known Member
Here is the NAND gate version. It looks more complicated than the schematic in post #33, but actually has fewer parts and fewer connections. Seven, 2-input NAND gates in two IC packages. Because the 2 and 3 inputs are true monostables (unlike a 555), you can hold down the button as long as you like and the counter will be clocked only 2 or 3 times. Same for the 1 input.

U1A and U1B (and U1C-U1D) form a true monostable. A negative edge at the input produces a negative-going pulse at the output. These are OR-ed (and inverted) in U2A. (Through DeMorgan-s Theorem, a positive-true-input NAND gate equates to a negative-true-input OR gate.) So U2A pin 3 goes high when either 2 or 3 is pressed, but for different times. This enables the U2D oscillator, which starts with a negative-going edge. The 1 input also produces a negative-going pulse, and this is OR-ed with the oscillator output and inverted in U2B to produce positive-going clock pulses for the 7490.

ak

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#### AnalogKid

##### Well-Known Member
And finally, here is the circuit in post #35 reworked to use a single IC, a CD40106 hex Schmitt trigger inverter. The OR and oscillator-enable functions are done with diodes and a resistor.

ak

#### absf

##### Active Member
No resistor attachwd atm
If you want the 74LS192 step down to work, you'd need a 1KΩ resistor at the DN pin of the chip like this

If using 74192, the resistor has to be lower.

Allen

#### absf

##### Active Member
And finally, here is the circuit in post #35 reworked to use a single IC, a CD40106 hex Schmitt trigger inverter. The OR and oscillator-enable functions are done with diodes and a resistor.

ak
View attachment 111504
Using one 40106 to replace 3x LM555 was a big trade off and simplified the circuit much. I wish the OP has learned much from this thread. At least I have.

Allen

#### AnalogKid

##### Well-Known Member
Using one 40106 to replace 3x LM555 was a big trade off and simplified the circuit much.
Not as much as you might think; all of those diodes add up. Still, it was fun to work out. Not counting the three switches:

555's - 52 pin connections -- 18 components

4093's - 52 pin connections -- 14 components

40106 - 50 pin connections -- 19 components

So from a total labor point of view, the 4093 circuit easily wins the "complexity" category. The number of pc board pins to solder is almost a wash, but the number of things to insert confirms what I suspected: the 40106 circuit is cute, but the allure of a "single-chip" circuit is deceptively inefficient.

ak

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#### Alexander Venus

##### New Member
Im learning a lot in this thread (muc more than my class tbh) thank you very much for the help