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?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
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
Now, with every press, each digit should increment or decrement.
You can preset them to different values
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)
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.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.
The 74xx series chips pins float high when unconnected. This series doesn't require inputs to be connected. 74yyyxx could be a different story.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 "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.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.
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by Luke James
by Luke James