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Decoding the extra digit

A project log for DYPLED

a thin LED display in SIP module format, much better than TIL311s to visualise your computer buses

Yann Guidon / YGDES 12/07/2015 at 08:021 Comment

The decimal display is a nice additional feature but it puts some more pressure on the design. The pressure was reduced when I found that the 74HC74 could drive the MOSFET directly (instead of using the Flash to decode one bit to 2 outputs) but the requirement remains: the display needs a fifth digit that can range from 1 to 6.

With 3 bits, it's not too hard to decode 3 bits to the required 6 segments, using a few diodes.

digitABCDEFG
1**X
2****X*
3****X*

- Segment F is not implemented.
- Segments A, D and G are on for 2 OR 3
- Segment E is on for 2 (directly)
- Segment C is on for 1 OR 3
- Segment B is on for 1 OR 2 OR 3

Total : 7 diodes only. However this limits the display range to 39999(d). Adding the 3 other digits (4, 5 and 6) is a bit more tricky but the principle is roughly the same, adding the F segment but trying to squeeze as many codes inside 4 bits. 3 bits are enough to code values from 0 to 7 but hopefully we can predecode some things and avoid using an external 74HC138...

digitAB
CD
E
FG
1**
2*****
3*****
4****
5*****
6*
*****

At this point, the codes 1 to 6 should be recoded to an intermediate 4-bits code which requires only OR logic (implemented with diodes).

The segments A-D-G are often ON at the same time so it's a first line of optimisation. Negative logic is another angle of attack (it worked well with the diode matrix). 5 and 6 differ only by the F segment so there is more hope to find an intermediary recoding.

One code (Z) is assigned to segments A-D-G, which covers codes 2-3-5-6.

Code Y controls segments B and C for the codes 1-3-4.

Sor far we have:
- 1 = Y
- 3 = Z + Y

4 codes remain with only 2 bits left. Bit X is assigned to segments C-F-G to form one part of the numbers 4-5-6. X overlaps with Y (segment C) and Z (G) but they are ORed with the diodes so there is no problem.

- 4 = X + Y
- 5 = X + Z

Digits 2 and 6 remain and get code W, which directly controls the E segment.

- 6 = W + X + Z

This pre-decoding requires a few diodes to transform the 4-bits WXYZ code (output by the Flash chip's 4 data lines) into a 7-segments digit. However a tiny extra logic is required to implement the digit 2. This is not convenient because it can't be solved with a diode, as a OR function. A AND function is required, hopefully implemented with a single transistor.

2 and 6 differ by the X code:

- 2 = W + Z (X=0)
- 6 = W + X + Z (X=1)

So the B segment's equation would be

B = Y or ( W and /X )

W is active ONLY for 6 and 2 so the electronics can be simplified, using a P-MOSFET (and a diode). I'll have to check this though...


The resulting equations are:

0 =
1 =         Y
2 = W +         Z
3 =         Y + Z
4 =     X + Y
5 =     X +     Z
6 = W + X +     Z

Converted in Hexadecimal with W=1, X=2, Y=4, Z=8,

0 = 0

1 = 4

2 = 9

3 = C

4 = 6

5 = A

6 = B

(All the codes are different so we're good).

The 4 pre-codes are :

Outside the Flash, we have :

A = Z

B = Y or (W and /X)

C = X or Y

D = Z

E = W

F = X

G = X or Z

4 segments are directly driven by the Flash output. The others are 3× OR2. That's 6 diodes and one P-FET only :-)


Update 20151214

While laying out the PCB I realise that some diodes could be saved thanks to the multiplexing... Is it possible to use only 2 output pins ? This would simplify the rest of the PCB layout and segment assignment.

ph.1ph.2
A = D = Z

*
B = Y or (W and /X)**
C = X or Y**
E = W
*
F = X
*
G = X or Z**

Some things are pretty easy :

The others are on during both phases so they are tied to 0V. It gets more complex but we can split the signals in two phases :

For G or C, it's possible that X coincides with Y or Z so there is no decoding at all, but the other needs more decoding, in the form of a MUX2 driven by the phase signal. Hey, I just got some cheap 74LVC1G157 that do that.

Segment B is a bit more tricky though. Basicly it's tied to Y/W but turned off if X is 1 during a certain phase.

So far we know that the Flash pins are tied to the segments this way :

Thus:

It's pretty simplified now ! No more diode ! There's a catch though. The light output of B/C/G can't be directly modulated. One simple solution is to tie their cathode to the /CE pin.

B must also be decoded. If needed it's possible to get the negative /CE signal, from the 2-gates oscillator. Phase can be chosen (1/2) as desired to simplify the circuit.

B = F1 and (/F2 and/or somephase) (and don't forget /CE)

This turns into a complex equation but the AND can be broken down easily because the LED must be powered from both ends. For example, B is ON when (F1 and /CE)=1 on the anode, and (/F2 and/or somephase) = 0 on the cathode.

The high side (anode) can be driven by the previous BSS84 trick (P-MOSFET) or a ANDN gate. The MOSFET is probably not electrically efficient (I'm not sure I have the right reference in stock) and it's tricky to get a cheap single-gate ANDN. But C uses a MUX2 and these gates can do a lot more than selecting one input (and I have stock :-) )

The low side can be reduced to a OR gate that can be implemented with a couple of diodes (on F2 and one phase1). The Shottky's drop requires a specific series resistor value. However the MUX2 can also implement a OR gate. In other words...

More accurately:

Ooops that's AND3...


Wait

The MUX2 on the left is useless because F1 is floating when /CE is high. So it's more like :

(phew, I feared I would have to add another MUX2 gate)


Update 20160704: I've manually checked all the equations from the latest version of the schematics, just in case, and it works.

Value:
0
1         Y
2  W         + Z
3          Y + Z
4      X + Y
5      X +     Z
6  W + X +     Z

Intermediate coding:
W = F1, ph=1 (phase 2)
X = F2, ph=1 (phase 2)
Y = F1, ph=0 (phase 1)
Z = F2, ph=0 (phase 1)

Segment decoders (high side && /low side)
A = F2 &&  ph=0
B = F1 &&  F1 & F2 & ph=1
C = (F1&ph=0)|(F2&ph=1)  && /CE
D = F2 && ph=0
E = F1 && ph=1
F = F2 && ph=1
G = F2 && /CE

In 3 cases, one segment is turned on during both phases:

But this is not critical and I can proceed to route the rest of the PCB.

Discussions

Yann Guidon / YGDES wrote 12/08/2015 at 23:49 point

Just added a little illustration to better explain my system ;-)

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