Implementing Ripple Blanking With A CD4511

4511 blanking pin only affects that chip.
A comprehensive circuit of some kind is needed which examines every numeral, starting from the left.
If it's zero (ABDC are zero), then blank the digit (enable BL). Go on to the numeral to the right and repeat. As soon as it detects a non-zero numeral, enter a different routine that disables BL for that digit and all digits to the right.

Hi,

please post your existing circuit´s schematic.
Hand drawn is, OK, but it should be complete.

Klaus

KlausST said:

Hi,

please post your existing circuit´s schematic.
Hand drawn is, OK, but it should be complete.

Klaus

Click to expand...

Here is my ripple blanking circuit. If you need to see more of the circuit I will have to draw it up.

Misses the "ripple" (carry from digits at left).

napoleansolo said:

This did not work
as it looked like the timing was not right.

Click to expand...

What does this exactly mean? Explain "did not work"
Don´t know what you mean "the timing was not right".. please explain.

***
From my understanding this should work for the HIGHEST digit.
I even don´t know how many digits you have.

But you said "ripple" ... there is no ripple handling at all.
Not even a try.

Did you at least try it on paper?

If you want to design something ... you also need to decide how to test it. Did you think about useful test patterns?

For 4 digits it could be: 0x0000, 0x0005, 0x0010 ... and so on

The use of the OR gates is a good way ... then just use your paper and write down all the outputs of the OR gates ... no need to connect them to the BL input for this.

Then write down the expected BL states. ....

It´s always step-by-step ....

Klaus

FvM said:

Misses the "ripple" (carry from digits at left).

Click to expand...

I worked it out on paper as posted and I implemented the circuit on a breadboard.
There are currently only two digits.
How would I modify this circuit as is, to accomodate the ripple from digits at the left?

There's no blanking propagation for two digit display, in so far the thread title doesn't apply for it. For more than two digits, you have AND logic for middle digits respectivly OR logic for nBL.

nBL = nBL, left OR D[3:0]