Switching Led

[willeng]

<Code471#>

 

[willeng]

Thought I would post the modified sketch I have been playing around with for the BCD switch.
Not sure what to do from here though at this moment?

The output is still inverted!

/*Willeng
20/9/2014
Modified Sketch
BCD TEST?
*/
const byte BCDpin1 = 2;
const byte BCDpin2 = 3;
const byte BCDpin4 = 4;
const byte BCDpin8 = 5;
void setup()
{ Serial.begin(9600);
pinMode(BCDpin1, INPUT_PULLUP);
pinMode(BCDpin2, INPUT_PULLUP);
pinMode(BCDpin4, INPUT_PULLUP);
pinMode(BCDpin8, INPUT_PULLUP);

}
void loop() {
// read
unsigned int val_BCDpin1 = digitalRead(BCDpin1);
unsigned int val_BCDpin2 = digitalRead(BCDpin2);
unsigned int val_BCDpin4 = digitalRead(BCDpin4);
unsigned int val_BCDpin8 = digitalRead(BCDpin8);

int switchNo = 0; //reset

if ( val_BCDpin1 == 0) { switchNo = switchNo + 1; }
if ( val_BCDpin2 == 0) { switchNo = switchNo + 2; }
if ( val_BCDpin4 == 0) { switchNo = switchNo + 4; }
if ( val_BCDpin8 == 0) { switchNo = switchNo + 8; }

{
Serial.print("[BIT8: ");
Serial.print(val_BCDpin8);
Serial.print("] ");
Serial.print("[BIT4: ");
Serial.print(val_BCDpin4);
Serial.print("] ");
Serial.print("[BIT2: ");
Serial.print(val_BCDpin2);
Serial.print("] ");
Serial.print("[BIT1: ");
Serial.print(val_BCDpin1);
Serial.print("] ");
Serial.print("[switchNo: ");
Serial.print(switchNo);
Serial.println("] ");}

delay(3000);
}

 

[KeepItSimpleStupid]

Willeng:

You should be able to replace all of these switchNo = switchNo + 1 type expressions with corresponding lines similar to: switchNo = switchNo && 1; etc,

After this if ( val_BCDpin8 == 0) { switchNo = switchNo + 8; } line add
switchNo = !switchNo or this line switchNo = !switchNo; wwhatever works. Not sure if the space is required.

Take the ULn2003, shown above and connect pin 8 to ground. Leave #9 disconnected.
Take pin 10 and connect it to common of the BCD switch your currently using and remove the ground that's currently connected there.

With pin 7 open, you should read all 1's with your program and 15 with the modified program.
Connect pin #7 to +5 and you should get the same BCD results that you had. 0-9 with the modified code.

With this chip, you can float the inputs with no damage and the floated inputs will be interpreted as a LOGIC 0.

==

The next step is to tie pin 7 to an output port.

==

Then think about if you need to "conserve bits".


==
Notes:
DS 2560, #219
Pressure sensor datasheet: MPX2010DP

 

[willeng]

Thanks Again,
I will test things out now & get back to you later.

Cheers

 

[KeepItSimpleStupid]

To save bits, you would have to find a chip like the 74HC251 or similar https://www.google.com/url?sa=t&rct=j&q=&esrc=s&source=web&cd=1&ved=0CCMQFjAA&url=https://www.onsemi.com/pub/Collateral/MC74HC251A-D.PDF&ei=ZckcVI2wIc-iyASL-YLQAg&usg=AFQjCNEiUyYnIP1CzzoJrTVs18uCGSXjjQ&sig2=BtvoBrAsNz3cma8Kf6dn9A&bvm=bv.75775273,d.aWw&cad=rja

That one has both inverted and non-inverted outputs.

==

If you get past the previous exercise then you should understand what's going on. Conserving bits is a design decision. Then, I think your looking at a minimum of building a 3 digit xx.x for setpoint and a 1 digit BCD switch, each with endcaps. Lots of diodes.

==
Notes:
DS 2560, #219
Pressure sensor datasheet: MPX2010DP

 

[willeng]

Ok, I finished the exercise & the results are as follows:

ULN2003 (Pin 7 Open Original Code) = [Bit8: 1] [Bit4: 1] [Bit2: 1] [Bit1: 1] [switchNo: 0] on
all 0-9 settings on Thumbwweel switch.
-----------------------------------------------------------------------------

ULN2003 (Pin 7 Open Modified Code) = [Bit8: 1] [Bit4: 1] [Bit2: 1] [Bit1: 1] [switchNo: 1] on
all 0-9 settings on Thumbwweel switch.
---------------------------------------------------------------
So they both read 15 in Binary or 1111!

ULN2003 (Pin 7 +5V Original Code) reads the same as the original code I had before 0-9 in
Binary & shows the switchNo like before.
--------------------------------------------------------------------

ULN2003 (Pin 7 +5V Modified Code) reads the same as the original code I had before 0-9 in
Binary except for the actual Thumbwheel switchNo because it now has && in the code & the serial
won't read that.

If you get past the previous exercise then you should understand what's going on.

Could you explain what I should be understanding.

Then, I think your looking at a minimum of building a 3 digit xx.x for setpoint and a 1 digit BCD switch, each with endcaps. Lots of diodes.

How do I go about doing & testing this?

Cheers

==
Notes:
DS 2560, #219
Pressure sensor datasheet: MPX2010DP [/QUOTE]

 

[willeng]

I have the SN74LS251N which look similar?

 

[KeepItSimpleStupid]

Somehow you got the inputs and outputs mixed up I believe and your also making little sense. Explain:

No because it now has && in the code & the serial

Pin 7 is an input and pin #10 is an output.

You had a ground connected to the common of the BCD switch, right? Remove that ground.

Connect the Common to (pin #10), an output of the ULN2003. Pin #10 is labeled O7 for output #7. Pin #7 is the corresponding INPUT.

Putting a high (+5) on pin #7 will put pin #10 to "ground", thus it's now the same as when the common of the BCD switch was grounded. This will read the BCD value or complement depending on the code.

Putting a LOW (ground or OPEN) just opens the common connection and all bits should be a 1. It should read 15 or 0 depending on the code.

==

==
Notes:
DS 2560, #219
Pressure sensor datasheet: MPX2010DP

 

[KeepItSimpleStupid]

I picked the wrong part. Back to the drawing board. We need the reverse of that chip. One input is multiplexed to n outputs where n is like 7 or 8 with a 3 bit address.
Sorry. My head is behaving badly. I'm outside at night with flood light re-enforceing a brand new snow shovel for use in a few months.

 

[KeepItSimpleStupid]

For the "conserve bits mode" you would need a 74x138 part https://www.google.com/url?sa=t&rct=j&q=&esrc=s&source=web&cd=1&ved=0CCAQFjAA&url=https://ecee.colorado.edu/~mcclurel/sn74ls138rev5.pdf&ei=y_gcVMfMIvL_sASTm4HADg&usg=AFQjCNHeqCshjQI-em-T1KQjb-m4ZocISA&sig2=sQDr5_yT2PTlVrabJIX_cA&bvm=bv.75775273,d.cWc&cad=rja

It would have a consequence of having to add an inverter for each wafer. There are usually 6 on a chip.

UNLESS you happen to have a 74xx07 buffer with open collector or open drain outputs. That would eliminate the ULN2003 and the inverter.

==
Notes:
DS 2560, #219
Pressure sensor datasheet: MPX2010DP

 

[willeng]

Somehow you got the inputs and outputs mixed up I believe and your also making little sense. Explain:

Pin 7 is an input and pin #10 is an output.

You had a ground connected to the common of the BCD switch, right? Remove that ground.

Connect the Common to (pin #10), an output of the ULN2003. Pin #10 is labeled O7 for output #7. Pin #7 is the corresponding INPUT.

Putting a high (+5) on pin #7 will put pin #10 to "ground", thus it's now the same as when the common of the BCD switch was grounded. This will read the BCD value or complement depending on the code.

Putting a LOW (ground or OPEN) just opens the common connection and all bits should be a 1. It should read 15 or 0 depending on the code.

No, I have it connected exactly as you mentioned previously, not sure why it's not making any sense.
Only pin 8 to ground connection.

Cheers

 

[willeng]

Somehow you got the inputs and outputs mixed up I believe and your also making little sense. Explain:

You should be able to replace all of these switchNo = switchNo + 1 type expressions with corresponding lines similar to: switchNo = switchNo && 1; etc,

After this if ( val_BCDpin8 == 0) { switchNo = switchNo + 8; } line add
switchNo = !switchNo or this line switchNo = !switchNo; wwhatever works. Not sure if the space is required.

You mentioned to change part of the code to switchNo = switchNo && 1; etc, when I do this the Serial will not read the switchNo but the original code will?

Cheers

 

[willeng]

I still have the Diodes connected from the BCD switch to the Arduino Ports as mentioned in post #224, is this an issue now?

Cheers

 

[KeepItSimpleStupid]

You mentioned to change part of the code to switchNo = switchNo && 1; etc, when I do this the Serial will not read the switchNo but the original code will?

Cheers

What's wrong with me today. https://www.tutorialspoint.com/cprogramming/c_logical_operators.htm

The && is the AND operator. The || is the or operator. You want the OR operator.

The add that you did was perfectly fine. Replaceing with OR should just make it faster.

This switchNo = !switchNo or this line switchNo = !switchNo should just turn 000 to 1111 etc.

 

[willeng]

No problem, I will give it a go & let you know.

I couldn't figure out what the heck was going on with anything & got totally confused.

All Good!

Cheers

 

[willeng]

Just did a test using the Or operator as suggested from 0-9, +5V to Pin 7 ULN2003, see attached:

Cheers

 

[willeng]

I have to admit trying to understand this is beyond my capability at this stage.

It may be better if I just return to using voltage references as the set points & see what error exists & then try to compensate for it?

I don't know?

Cheers

 

[KeepItSimpleStupid]

Post your code again. Thanks. See next post.

It should look like:

if ( val_BCDpin1 == 0) { switchNo = switchNo + 1; }
if ( val_BCDpin2 == 0) { switchNo = switchNo + 2; }
if ( val_BCDpin4 == 0) { switchNo = switchNo + 4; }
if ( val_BCDpin8 == 0) { switchNo = switchNo + 8; }

or

if ( val_BCDpin1 == 0) { switchNo = switchNo | 1; }
if ( val_BCDpin2 == 0) { switchNo = switchNo | 2; }
if ( val_BCDpin4 == 0) { switchNo = switchNo | 4; }
if ( val_BCDpin8 == 0) { switchNo = switchNo | 8; }

They should both give the identical result.

switchNo is an integer - Not good! This means that NOT will give very wierd results. In any event, the bitwise NOT operator is apparently "~" and not the "!" Logical OR wasn't right either.

It has to be an "unsigned int" for NOT to work correctly.

Don't get too upset yet. Please?

 

[KeepItSimpleStupid]

Better yet, try this. I put the code within code tags. [code]code goes here[/code]

/*Willeng
20/9/2014
Modified Sketch
BCD TEST?
*/
const byte BCDpin1 = 2;
const byte BCDpin2 = 3;
const byte BCDpin4 = 4;
const byte BCDpin8 = 5;
void setup()
{ Serial.begin(9600);
pinMode(BCDpin1, INPUT_PULLUP);
pinMode(BCDpin2, INPUT_PULLUP);
pinMode(BCDpin4, INPUT_PULLUP);
pinMode(BCDpin8, INPUT_PULLUP);

}
void loop() {
// read
unsigned int val_BCDpin1 = digitalRead(BCDpin1);
unsigned int val_BCDpin2 = digitalRead(BCDpin2);
unsigned int val_BCDpin4 = digitalRead(BCDpin4);
unsigned int val_BCDpin8 = digitalRead(BCDpin8);

unsigned int switchNo = 0; //reset

if ( val_BCDpin1 == 0) { switchNo = (switchNo |  1); }
if ( val_BCDpin2 == 0) { switchNo =  (switchNo | 2); }
if ( val_BCDpin4 == 0) { switchNo = (switchNo | 4); }
if ( val_BCDpin8 == 0) { switchNo =  (switchNo | 8); }

{
Serial.print("[BIT8: ");
Serial.print(val_BCDpin8);
Serial.print("] ");
Serial.print("[BIT4: ");
Serial.print(val_BCDpin4);
Serial.print("] ");
Serial.print("[BIT2: ");
Serial.print(val_BCDpin2);
Serial.print("] ");
Serial.print("[BIT1: ");
Serial.print(val_BCDpin1);
Serial.print("] ");
Serial.print("[switchNo: ");
Serial.print(switchNo);
Serial.println("] ");}

delay(3000);
}

Yep, lots of problems. My fault. Your almost there. I edited the last two posts.

 

[willeng]

Hi Kiss,
Thought I would leave you alone today & spent some time studying, more on that later.

I"ll have a look at what you have posted then get back to you.

Thanks Again

Cheers