Here is a picture of the Tx board. This side of the board has the jumpers for the encoder chip. If you look close you can see that there are three rows of pins. The jumpers are not used in the picture, they are just sitting on one of the right pins. The right row is the High row, the center row is the common row and the left row is the Low row. Measuring the voltage with 1H jumper-ed (A) gives me 8.5v at the start and then it drops to 7.6v the amps at the jumper are 3.5mA. Doing the same with 1L jumper-ed gives me 0.04v and 0.05mA

Just for clarification, leaving 1L jumper ed and all of the positions on the receiver open results in the Rx not responding. Therefore I have three possible combinations for each of the eight positions or 3^8 combinations.

 

Even though I am now going to a computer controlled system, It would still be interesting if I could get me original idea to work. For any one interested, here are the parts that I originally gathered in order to accomplish the task. Mfg in parenthesis.

Here is what I got from Mouser Electronics
MC14511BDG 3-18V BCD/7-segment Buffer (ON semiconductor)
MC14553BCPG 3-18V 3 digit BCD counter (ON Semiconductor)
74F579PC 8-Bit Bidirectional Binary Counter (Fairchild)
(8)FX2 D3209 5V relay (Axicom)
BA56-12SRWA LED display

If I can get both ideas to work I will just buy another Tx and have two ways of firing the show, manual or automatic.

 

I'll study the photo tomorrow.

I'll also post a circuit for you.

From memory, the 3 digit BCD counter is not an up/down. It only counts up.

The 74F579PC 8-Bit Bidirectional Binary Counter is very fast and is likely to cause you problems. A slower one would be better given your level of experience.

__________________
Len

 

It is difficult to see the fine detail in the photo, but it gives me a general idea. So I've drawn the attachment to clarify.

Is there a number on the IC?

I don't understand what you mean by :-

"Measuring the voltage with 1H jumper-ed (A) gives me 8.5v at the start and then it drops to 7.6v the amps at the jumper are 3.5mA"

Do you mean that you measured the voltage with a jumper inserted between 1H & 1C?
What caused the voltage to drop to 7.6V?

You measured a current of 3.5 mA. Was this between 1H and the - battery with the jumper in or out?

You also said :- "Doing the same with 1L jumper-ed gives me 0.04v and 0.05mA"

So I assume you mean you measured the voltage and current at 1L with the jumper inserted between 1L and 1C.

How is the code set in the Rx?

I suggest that you remove the jumper from row 1 and measure the voltage on 1L, 1C & 1H.

Does your Multimeter have a continuity setting?

If so, remove the battery, connect the Negative probe to the - battery spring and see if there is a connection to 1H, 1C and

1L. If it does not have the connectivity function, then use the lowest resistance setting and measure the resistances between these points.

Can you see the tracks on the other side of the PCB? If so, take a photo of it.

Which pins are connected to the IC? ie. the L row, the C row or the H row. I suspect it will be the C row.

Attached Images

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__________________
Len

 

I hope that answeres most of the questions. Sorry about the screw up the first time.

P.S. My 12v battery is measuring 10.67v

 

We're starting to get some where.

I did not want you to measure the voltages and currents wrt (with respect to) the +ve side of the battery. You may have damaged something. Does the Tx still work?

The numbers on the IC don't mean anything to me, but don't worry.

It looks like the L pins are connected to the battery negative (this is the "ground" or 0 Volt point).

I expect that the C pins are individual and go to the adjacent pins on the IC.

I also suspect that all of the H pins are connected together and go to what you called pin 5 of the IC.

You can prove these points with your MM set in the resistance mode - and with the battery removed. {A MM with a continuity function would make this easier as it gives a buzz when there is a connection}

Note that the IC pins are numbered anticlockwise (looking at my drawing) starting at the bottom right. It looks in the photo like it has 20 pins.
So pin 1 is at the bottom right and pin 10 at the top right.
Pin 11 is at the top left etc.

It also looks like your battery is almost flat. It should read more than 12 Volt if fresh.
I expect that is why the voltage reading was dropping when you did the original measurement.

So I suggest you do the following:-

Insert jumpers in all rows (Lx to Cx) except the first, ie. no jumper between L1, C1 or H1.

Insert a fresh battery.
Connect the Neg probe to the -Ve battery spring (ie. the 0 Volt point) (I want all voltages measured wrt 0 Volt)

You said that you did not see any readings until you pressed the Tx button.

So put the +Ve probe on H1 and then Press the button, and note the voltage.
Don't measure the current.

Insert a jumper H1 - C1 and repeat, ie. press the button and measure the voltage on H1 or C1.

Remove the jumper, switch the MM to measure current and connect the probes -Ve to C1 and +Ve to H1. Press the button and note the current.

As far as I can see at this stage, there are 2^8 combinations not 3^8.
You insert a jumper between L1 & C1 if that bit 1 is the be 0 or between C1 & H1 if it is to be a 1. Ditto for the other bits.

For there to be 3^8 combinations, you would have to, in some cases, not insert a jumper. But I expect that each bit MUST have a jumper so it is set to either 1 or 0, hence 256 combinations.

I looked at the data sheets of the MC14553 and the MC14511.

I don't have a data sheet for the LED display. Do you have one? Is it common anode or common cathode?

__________________
Len

 

Ignore this, I didn't read the second page.

Mike.

 

I know this may sound like I am beating a dead horse but it may be integral to figuring this out. The encoding jumpers do not need to be in place for the Tx to function. I have three possible states for each "bit" Open - no jumper, H - jumper connected between H and C, and L - jumper connected between L and C. At this point the number of combinations is moot because the computer only has 8 output bits on the parallel port. I just want to make sure that it is understood that I am considering 0 to be all bits open, no jumpers, and 255 to be all bits closed, jumpers between all C and L pins.

I am beginning to think that it may be easier to just use the eight data bits from the computer to activate relays and have them in the place of the jumpers set to NO. I will then need one more relay connected to one of the control bits of the port that I can connect to the contact points of the fire button.

I know you said that I may not need the relays but I just can't get my head wrapped around how the other way is going to work. The data pins on the port give 5v when set to 1, how would applying a voltage to a pin work if that pin is normally jumper-ed to another pin. If the C pins are neutral and the H and L pins give a voltage that might make sense but then I would have to find a way of making the 5v from the port into what ever voltage the pin is expecting.

You guys have to be getting frustrated with me. This is like doing surgery by e-mail. I hope you will bear with me long enough for me to get a grasp of how this stuff works. I thought some one told me once that the encoding worked by creating a resistance at the pins of the IC. The way you are talking it sounds like something similar but we are measuring the voltage difference rather than the resistance. I don't know, you guys are the experts.

As for the display data sheet, I have one printed off at home, I will check it out and let you know CC or CA.

Thanks
Sean

 

hi Sean,
Looking at your photo of the pcb.

The row marked 'L' are all connected to 0V [common]

The row marked 'H' all seem to have a pullup resistor, probably to +V..

The centre row of pins is connected to the chip ic pins.

So when a jumper is connected from the centre pin to a 'L' pin the ic sees a logic low on that pin.
When the jumper is connected from the centre pin to a 'H' pin then the ic sees a logic high on that pin.

So you have 8 combinations 0 thru 255.

If the logic levels on the ic are +5V and 0v [high/low] then if the jumpers were removed and the centre row connected to the PC's parallel port, then the ic pins would see high or low depending upon the voltage on the port pin.

Does that help.

Look at the routing of the pins on the TX pcb to confirm this.

I won't get involved in this part, as Len is doing some logic drawings for you.
I'll help with the PC program.

EDIT: Dwg of 2 pins

Attached Images

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__________________
Eric
"Good enough is Perfect"

PIC tutorials:
Gramo's: www.digital-diy.net/
Bill's: www.blueroomelectronics.com/

 

Eric, I still don't think you guys understand about the possible coding combinations. I will try to explain it another way. I have 8 pins in each row, each of those eight pins can be either open, high or low. thus three combinations. If I were to type it out, a possible coding could be

HOOOLOOH

O being open, I do not have to pull it high or low. For instance, the system I used last year was used with no encoding, all pins were open, and it is designed to work that way.

Now, since I can only control eight bits with the printer port I an going to ignore the possibility of setting any of the pins to high so my coding would be

LOOOLOOL or 10001001

HTH
Sean

 

Assuming this is a binary system, and it almost certainly is?, you can only have HIGH or LOW, there's no third option - if you leave the input OPEN, it will either be pulled LOW or HIGH by internal resistors, or drift HIGH or LOW randomly.

It's certainly possible to design a three level logic system, which in your case would mean an OPEN input would need to be biased at 50%, using two ressitors - but it's probably a useless idea?.

__________________
PIC programmer software, and PIC Tutorials at:
http://www.winpicprog.co.uk

 

http://www.halaxion.com/PT2262.pdf
Here is the link to the remote control IC.

 

Brevor, Thanks for the link. Hopefully that clears up some things.

I have measured the voltages across the pins with out the jumpers and with the + lead on the pin, - lead on the spring. Here is what I got.
8.5v on the L, 1.5v on the C and 0.25v on the H. This is with the old battery, I will get a new battery tomorrow and get some new readings.

So, are we going to apply a voltage from the printer port to the pins to create the encoding?

 

I suggest that you do an internet search and see if you can find a better data sheet for the IC.

The manufacturer may provide an Application Note on the IC.

__________________
Len

 

Hi Sean,
I have looked thru the ic's datasheet, that Brevor posted, it does use and recognise tristated inputs. Total combinations 12^3.

Of the 8 jumpers on the address A0 thru A7 lines you have 8^3 combinations, ie: 6561.

The other 4 high order address lines A8 thru A11 are shared with Data, see the coding D0 h D3 on the otherside of the ic.

There are ic's available with tristate ouputs so its still feasible to drive from the port.

Can you send or direct me to the datasheet for the completed TX module, so that I can checkout what the module manufacturer claims.

__________________
Eric
"Good enough is Perfect"

PIC tutorials:
Gramo's: www.digital-diy.net/
Bill's: www.blueroomelectronics.com/