Hello EI3KI

Why 21 IC’s ?
For The Decoder I count:
15 IC’s.
12 LED’s.
12 Resistors.

For Voltage Level Adaptor are:
8 transistors.
24 resistors.
and that's it.

Have You something else in your design ?

! ah, good thing you have a programmer for PIC's.
What PIC's can be programmed in?
You know some programming language to make the source code for the PIC?

 

I'm not quite understand all your connecting of CD4078, but from the circuit that it looks seem can be replaced by an 32KB eeprom(parallel type).

Using the eeprom as M28256 to be an encoder for all CD4078.

 

Hello ScottWang / EI3KI

Certainly, the use of a memory facilitate the development of the design.
But the originator of this topic would require more resources as I mention in one of my past posts.

I think it would require two memory chips since they are 19 LED’s that want to turn according to the figure in the 7 Segment Display.

I think it would be better to use EPROM memories.
I mean if we committed a mistake, we could erase and program again.

Keep in mind that the memory chips were not made to light LEDs.
It is necessary to use buffers in the data bus to avoid damaging the memory.

I developed the attached circuit to demonstrate how the design could be made.
Each memory chip is assigned a different HEX file to achieve the goal.
The M2732 U1, is assigned the Upr.HEX.
The M2732 U2, is assigned the Lwr.HEX.
Note, that in the address bus, the values ​​are true when they are low level.
And, on the data bus, the values ​​are true when they are high, the buffer (U3 or U4) inverts to turn on the LED.
The display called: Memory Channel Number Display. turn on those segments that are low level.
After, the signals arrive at the circuit Called: Voltage Level Adaptor. the true value, on its outputs should remain the same.

U1 to U4, chip resistors RN1, RN2, and all 12 LEDs.
Is all that is required to develop the decoder.
The other devices are to prove that the design works as expected.

The figures you see on the bottom of the image attached "With M2743.PNG", are what you should see in the "Memory Channel Number Display". otherwise the decoder will not work.

If the originator of this topic decides to make his project with memories, will require a programmer.
probably with the programmer for PIC’s, which he claims to have, can do the programming of the memory chips.

 

Hello everybody

21 IC, because I have used separate inverters for every High level in every CD4078 (my mistake). Wrong schematic attached.

17 IC, if I will use only one inverter for the same segment (like your U10 in #17).
3 pc LS14 for Voltage Level Adapter, +1 transistor and 3 resistors,
1 pc LS273,
12 pc CD4078
1 pc LS14 for Inverters (+1 separate inverter(7-th) I can use from VLA)

I have programmer and thinking to start learning, but always no time for this. Have checked Local parts shop, but there no 32kB EPROM's, 256 and more. It's possible to get on eBay, but all is used and need time for delivery.
Thanks!

 

Hello EI3KI

What is the cost of the 12 Chips CD4078.
What is the cost of those memories of 256Kb?. . Two chips are required.
For example 27C256 memory.
Besides the cost of the two ULN2803 Chips.
What would be cheaper?

But probably already have, in your hands, those 12 Chips CD4078.
If so, it would be an extra expense to acquire two Chips 27C256.
You have to add the cost for purchasing a programmer to memories.
Looks like your device programmer can not program the memories, only PIC’s.

How will continue your project ?.

 

Hello, MrCarlos

I'm so sorry for delays, it's summer holiday time...

In past few days I have designed more simple PCB for your schematic and will start to make it. All 17 IC's is 10 times cheaper and I have it already. PIC's and Memories I will leave for my future projects.

In few days I'll inform you how's going...