# Convert Leds to 7 Segment Display

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#### Almazick

##### New Member
I would like to convert 6 leds to 7 segment display. What is the best way to do it? The way it works right now that LED’s lit up in a continuous style (for example: 1st will light up the 1st LED, 2nd will light up the 1st and 2nd LED, 3rd will light up the 1st 2nd and 3rd LED’s, etc. etc) I got 7 negative inputs and 1 positive signal as you can see from the picture that every led get grounded to lit up.

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#### kinjalgp

##### Active Member
How can you construct 7-segment display using 6 LEDs?
Please be more clear.

#### Almazick

##### New Member
I want to replace 6 Leds with 7 Segment display so it will display number from 1-6 instead of leds to lit up

#### bogdanfirst

##### New Member
simple solution: use a 8 to 3 priority encoder and then a converter from 4 bits to 7 segment display. the priority encoder must be active low.
i'll look for some IC's for this.

#### kinjalgp

##### Active Member
CD4532 is a 8:3 priority encoder chip and 4511 is a BCD to 7-segment display driver.

#### Sebi

##### Active Member
Or You can use diode array... and don't forget the serial resistors for segments.

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#### mechie

##### New Member
Woops !

Sebi
If input 3 is active (as I understand it) inputs 1 and 2 will also be active, so the display will show all 7 segments lit (displays '8' ).
Some priority encoding is required ?
Or is the diode matrix intended to still use the CD4532 8-to-3 encoder (and a decoder).

#### bogdanfirst

##### New Member
i think that the 8 to 3 and decoder is the only way.
the diodes would have worked better if you had only one led on for the specific number

#### Sebi

##### Active Member
Mechie,

maybe with my bad english cannot understand right this sentence:
" I want to replace 6 Leds with 7 Segment display so it will display number from 1-6 instead of leds to lit up."
I think, only one input active (not two or more) and need to feed to 7segment display. In this case the diode-matrix can work.

#### bogdanfirst

##### New Member
Almazick said:
1st will light up the 1st LED, 2nd will light up the 1st and 2nd LED, 3rd will light up the 1st 2nd and 3rd LED’s, etc. etc)
so whatever number you have, that number of leds you have lit.
1: LED1
2: LED1 & LED2
3: LED1 & LED2 & LED3
.
.
.
.

#### Sebi

##### Active Member
O.K.

......i thinking....

#### Sebi

##### Active Member
Here is a possible solution, but this is only a small exercise for my brain...

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#### Almazick

##### New Member
Hey Sebi! It actually looks like it might work. One more question. I get 6 negative inputs and 1 positive input to 7 segment display and as I can see from your picture that all 6 inputs are positive. Is it possible to put HEX Inverter to make it work? Thanks a lot

#### e

##### New Member
couldnt you do this...?

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#### Sebi

##### Active Member
Yes, ofcourse.In this case no need the pull-down resistors, but the inverter inputs need pull-up resistors.

#### mechie

##### New Member
Almost...

Sebi
Sorry if I seem to be picking on you but -
that last circuit of yours with the NAND gates has a flaw...
If input '3' is high (active) then the NAND will block input '2' - Great! but...
As input '2' is now blocked there is nothing to stop input '1' from sneaking through ...
so the 7-seg display still shows an '8' lit up ('1' & '3' superimposed).

If I steal your circuit and add some diode 'AND' gates and add the input inverters aleady mentioned ...

I still think the Bogdanfirst's circuit using two ICs (plus input inverters) is a little more compact even if it is nowhere near as much fun :wink:

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#### laroche73

##### New Member
Converting leds to 7 segment display

bogdanfirst had a good solution (e said the same thing in schematic form), and it's probably the quickest to implement when design time is considered. The package count is 3 IC's, with the inverters. Sebi's idea also does the job, with 6 NAND gates, 6 inverters, and 41(!) diodes (as mechie pointed out). Note that the priority encoder w/BCD to 7segment decoder will display a "0" when none of the comparators are on, while the "diode ROM" will blank the display.

Here's another way to approach the problem. It doesn't require any diodes or MSI parts, just eight 2-input gates, for a total package count of two ICs. (btw, this would make a good test question for an introductory course in digital electronics)

Partial decoding of the inputs can produce the required outputs, since there are many "don't care" states (57 out of 64). In general, the more "don't care" states there are, the better the decoding logic can be minimized (where the "don't care states" are located in the karnaugh map is also important).

Starting with the truth table:

Code:
       LED                    Segment
---                    -------

L1 L2 L3 L4 L5 L6         A B C D E F G    Display
-- -- -- -- -- --         - - - - - - -    -------
1  1  1  1  1  1          0 0 0 0 0 0 1      "0"
0  1  1  1  1  1          1 0 0 1 1 1 1      "1"
0  0  1  1  1  1          0 0 1 0 0 1 0      "2"
0  0  0  1  1  1          0 0 0 0 1 1 0      "3"
0  0  0  0  1  1          1 0 0 1 1 0 0      "4"
0  0  0  0  0  1          0 1 0 0 1 0 0      "5"
0  0  0  0  0  0          0 1 0 0 0 0 0      "6"
The simplest case is segment "G". From inspection, "G" is identical to L2, so no logic is required. Likewise, segment "B" is the inverse of L5. Decoding the other segments is only slightly more complicated...

Example: "F" is the same as L4, except at digit "0". A high on L1 or a low on L4 should drive the segment low.

So, segment "F" = (L1 + L4')' => (L1' * L4), using demorgan's theorem.

A similar approach to decoding the other segments yields:

Segment "G" = L2

Segment "F" = (L1' * L4)

Segment "E" = (L1' * L2) + (L3' * L6)

Segment "D", "A" = (L1' * L2) + (L4' * L5)

Segment "C" = (L2' * L3)

Segment "B" = L5'

This looks pretty simple, and as it stands would take 9 gates to implement; 8 two-input gates and 1 inverter ("D" and "E" can share the L1' * L2 term, and "G" doesn't need any logic). The only problem is that most of the gates are of type "AND with one inverted input (AND2N1)". OK if you're implementing the decode logic within a PLD or FPGA, but standard TTL and CMOS families don't offer this type of gate as a primitive. To build the logic using standard SSI parts would require 5 AND gates, 2 OR gates, and 6 inverters.

Even using three SSI packages (74xx08, 74xx02, 74xx14, for example) and the fact that a NOR gate is equivalent to a negative logic AND, you'd still be one inverter short.

We can minimize the logic for a better fit with standard ICs by noting that a XOR gate implements the function (L1' * L2) + (L1 * L2'). Since the second term always falls under one of the "don't care" states in this truth table, XOR gates can substitute for the AND2N1 gates throughout the decoding logic. XOR gates can also replace the two OR gates, for the same reason. Finally, if a XOR gate is used to implement the inverter (by tying one of the inputs to VCC), we end up with a grand total of 8 XOR gates, or two SSI packages (74xx86 or CD4070). Not bad. 8)

The new equations using XOR gates are:

Segment "G" = L2

Segment "F" = L1 @ L4

Segment "E" = L1 @ L2 @ L3 @ L6

Segment "D", "A" = L1 @ L2 @ L4 @ L5

Segment "C" = L2 @ L3

Segment "B" = L5 @ Vcc

(Lacking a simulator at the moment, I tested this circuit on a breadboard to check my logic and it works as advertised)

The logic assumes a common-anode seven segment display, similar to the original LED setup. To make the display brightness more consistent across digits, a current limiting resistor should be used for each
segment, rather than having just one at the display's anode.

Of course, a simple PAL like the 16V8 could do the job in one package, but where's the fun in that?

- CAL

http://www.electro-tech-online.com/custompdfs/2003/08/elesimp.pdf
--------------

Note 1:

In the original design no LEDs are on when all of the comparators are off. If you allow the same for the 7-segment display (blank instead of "0") one more gate can be eliminated, but the IC package count remains the same.

The equations become:

Segment "G" = L2

Segment "F" = L4

Segment "E" = (L2 @ L3) @ L6

Segment "D","A" = L4 @ L5 @ L2

Segment "C" = L1 @ (L2 @ L3)

Segment "B" = L1 @ (L5 @ Vcc) -- (XNOR function)

Segments "C" and "E" share the (L2 @ L3) term.
As before, XOR gates are used to minimize the package count.

Note 2:

Another minor point to keep in mind; the SN7445 and CD4511 BCD-seven segment decoder ICs do not output a "tail" on digits 6 (seg "A") and 9 (seg "D"). Most modern displays do (check your VCR or digital clock) and people seem to prefer this. It takes additional logic to add on the "tails" when using these decoder ICs.

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#### laroche73

##### New Member
truth table hard to read

Sorry about the truth table display in the preceeding post, the spaces got squished out. It's the same as the one on the schematic, and it's more readable there.

#### czxjuren

##### New Member
i think that should add one led

i think that should add one led

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