Help with 2 digit BCD counter

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Deleted member 289596

Joined Dec 31, 1969
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Did you look at page 3-254 of the 4510 datasheet posted earlier?
Yea I basically did the first picture that says "parallel clocking", and the clock goes into the 1s place to carry over the 10s to the next 4510
 

Thread Starter

Deleted member 289596

Joined Dec 31, 1969
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Now, the project really makes sense to me.

1. Totally ignore the multiplexor. I mean totally. leave room for it, but IGNORE IT.

2. Start with one counter and connect the outputs tohe single LED.s. The count push buttons are preconditioned.

3. So, you should be able to create a single BCD digit binary coounter
e.g.
0000
0001
0010
0011
0100
etc

You should be able to get preset and load to work.

Now, duplicate the most significant digit. Make it a single digit. For the time being connect the clocks together. Add the other LEd's

0000 0000
0001 0001
0010 0010

Now, with every press, each digit should increment or decrement.

You can preset them to different values
1001 0111
1000 0110 count down

Now remove the clock to the most significant digit. and connect carry in//out.

Now it should count like
0000 0001 ---> 01 decimal
0000 0010 ---> 02 decimal
...
1001 1001 --- > 99

You can still leave the single LEDs in place

Now, you can try to take a big step and multiplex them. Initially don't use the function generator. Use a switch or a wire.
Be warned that a miswire on the multiplexor wlll give you wierd numbers.

So, start small and grow. Leave room to grow.

I can't expect you to wire something and expect it to work. Designing, simulating and building works.

Seeing the breadboard made all of the difference in the world.

There are still breadboard issues. I'm not crazy with you using CD4xxx parts.

REMEMBER that all unused inputs have to be connected to something. (GND or VCC whatever is convenient)
To use one 4510 by itself, do you just ground all inputs, connect outputs to 4 LEDs and connect clock to a pulse switch or regular switch?
 

Thread Starter

Deleted member 289596

Joined Dec 31, 1969
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Power pins. They are usually not shown, but make sure they are connected.
Tie all unused inputs to something
Preset wasn;t wired.

What you can do is, pull the inputs to the 74157 and hard wire a 1 and a 7.
See if a 17 appears.
that will check you mux. Not 100%, but something predictable should happen.

remove the outputs from the counters and connect to the 8 individual LEDs.

There is a parallel clock mode, http://people.ece.cornell.edu/land/courses/ece4760/FinalProjects/s2011/kj83_twl46/docs/CD4510B.pdf (Pdf page 6) but make sure CI on the LSD (Least significant digit) is set to zero. In the figures the LSD is shown first.

Note, it say if CI is held low, the counter advances on a positive going pulse. So make sure your using the right pulser.

You missed the boat, or got eleuded with the counters. You kinda have the 1s and 10s not interconnected properly.
The way TI drew it, it's 1's, 10s, 100's
You drew it as 10's, 1's . CI has to be low on the 1's digit.
the 1's digit CO, goes to the 10's digit CI.

If you use the TI method, the drawing would be neater.
but the chips are layed out as 02 for displaying 20.

Make sure the pulse signal goes below 0.5V and above 4.5 V and goes to >4.5 when pushed assuming your using a 5V supply.. This is critical.

Actually using a voltmeter to check the power pins can be helpful.

Kinda sorry, I wasn't quite paying attention. I do suffer from migraines and I'm suffering with one as I write.
So yea I left out the VCC and GND pins cause I feel like that's common sense. I did VCC all 16 pins and GND all 8 pins.

BTW, on the actual breadboard that contains my chips, I have the 74157, 4510, 4510LSD and 7404 chips in that order. I use a 9v battery with a resistor. The board also has my extra switches. The other "digital lab" board is just to connect the clock, D1, D2 and the 4 mux outputs.

I fixed the LSD out of order thing. Now the 1s are first, then 10s. I also put a switch going into clock of the 1s 4510, so everything else should be good now.

With grounding unused pins, in past labs, people have said that it isn't necessary. Are there certain times where it's required or something? I saw a diagram somewhere and none of the unused pins were grounded.
 

Thread Starter

Deleted member 289596

Joined Dec 31, 1969
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GOOD NEWS GUYS! I've finally done it! It's working! I'm gonna get it graded today!

Basically from my last circuitry I showed you all a pic of, my problem was that C in wasn't grounded, and the inputs to the 4510s should've been connected.

Now reset works, up down works, I can parallel load and blah blah.

I am extremely happy guys. Thank you so much for basically spoon feeding me all this info. I came in here not knowing anything and now, not only do I have the project done, I also understand how it works perfectly!

Thank you all for your help. I really appreciate it!
 
Wow! Congratulations!

I think you learned something instead of someone designing it for you. Sorry about the mis tracks.

With grounding unused pins, in past labs, people have said that it isn't necessary. Are there certain times where it's required or something? I saw a diagram somewhere and none of the unused pins were grounded.
The 74xx series chips pins float high when unconnected. This series doesn't require inputs to be connected. 74yyyxx could be a different story.

The CD4xxx chips do. If you don't they may oscillate, get hot and blow up.

Learning about the logic family helps figure this out. See: https://en.wikipedia.org/wiki/4000_series

Wikipedia said:
Some care needs to be taken with the design of circuits using CMOS chips. Many parts offer multiple logic gates in a single package and it is common to not need all of them. An engineer who forgets to 'tie off' (connect the unused gate inputs to VSS or VDD) may find the chip draws excessive current. The problem is caused by biasing in each gate. With the inputs disconnected, the gates may be biased into a mode where the outputs are partially conducting; this leaves the output buffer drawing a great deal of current since it is not fully on or off, creating a low resistance current path between the power supply rails.
The "short rule" is to connect unused inputs. The longer rule is to connect the unused input to the logic state that dissipates the least power.
Layout might dictate which one you use. It's better to use pull ups or pull downs to a logic state. They cost more. You really should do the same to OP amps as well.

Now, there is an interface type of chip that holds it's state when the inputs are disconnected. So, sometimes it can be 1 or a zero.

FWIW: I once repaired an Auger Spectrophotometer where the interface boards were wire wrapped. This machine worked for the longest time. One day, it lost a color It turned out that the power pins were soldered and the signal pins were wire-wrapped. they missed soldering a power pin.

In this project the CD4xxxx (CMOS) inputs had to be connected. With the CMOS and it's low power, sometimes the inputs can actually power the chip.
 
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Bypass capacitors are missing. These caps are usually ceramic, around 0.1 uF, and located close to the power pins of the IC, They should exist in digital and analog designs.
 
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