Counter circuit used to switch an encoder chip

[ljcox]

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?

 

[Pommie]

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

Mike.

 

[0RESET0]

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

 

[ericgibbs]

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

 

[0RESET0]

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

 

[Nigel Goodwin]

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?.

 

[Brevor]

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

 

[0RESET0]

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?

 

[ljcox]

Brevor, Thanks for the link. Hopefully that clears up some things. Yes, you were right, it is a tri state system. But the data sheet is vague. It does not show how to set the 3 states.
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.
These measurements are inconsistent with what you posted previously. You said you had measured 0.3 Ohm between 0V and the L pin. So if that is true, you should have seen about 0V on the L pin wrt the -Ve battery spring (ie. the 0 Volt line)

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?
We need to understand how the 3 stste system works before we can advise you on this.

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.

 

[ericgibbs]

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?

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.

 

[ljcox]

Since there are 12 inputs that can be used as addresses, it means that your original desire to use a 3 stage BCD counter is very simple.

However, the MC14553 is not suitable since it has multiplexed outputs.

You will need 3 decade counters such as the CD4510 which is sold by some manufacturers as the MC144510. Other manufacturers have different letters before the 4510.

This means that you can have up to 1000 combinations, or 10 000 if you add another 4510.

 

[ljcox]

I looked at the princeton web site and found a better data sheet. Attached.

There are still some gaps, but it gives info on the tri state operation.

Sean,
I still need you to make the voltage and current measurements.

 

[ljcox]

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.

You don't need relays. We can design a simple interface circuit for you now that we understand the encoder.

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. That is the point of the interface.

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. No, it is the voltage that sets the data, ie. High, Low or open. The way you are talking it sounds like something similar but we are measuring the voltage difference rather than the resistance. The resistances set the voltage. 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

Sean,
Do you have a soft copy of the data sheet? If so, please attach it to a post so we can down load it.

 

[0RESET0]

This just in...

OK, I got a new battery. Here are the new voltages. 10.1v on the L row, 1.5v on the C row and 0.07v on the H row.

 

[0RESET0]

forgot the datasheet

It looks like the display is a common anode design.

https://www.us.kingbright.com/images/catalog/SPEC/BA56-12SRWA.pdf

 

[ljcox]

OK, I got a new battery. Here are the new voltages. 10.1v on the L row, 1.5v on the C row and 0.07v on the H row.

You seem to have swapped the L & H rows.

Originally you said that L was connected to 0V as in Eric's drawing.

I was expecting more like 5 Volt on the C pins, but we don't have that detail in the IC data sheet so I'm happy ewith 1.5V.

 

[ljcox]

It looks like the display is a common anode design.

https://www.us.kingbright.com/images/catalog/SPEC/BA56-12SRWA.pdf

Yes, it is a common anode.

The 4511 decoder is designed for a common cathode display.

The 4543 (MC14543) is better since it can be configured for either CC or CA simply by making the "phase" pin high or low. See the attached data sheet.

So if you want to use this display, then you will need to buy 3 of the 4543 decoders. Alternatively, you could buy CC displays.

I post a counter circuit for you tomorrow without specifying the decoder or display.

 

[0RESET0]

Actually, the two measurements between the old and new battery are in line with each other.

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.

and

OK, I got a new battery. Here are the new voltages. 10.1v on the L row, 1.5v on the C row and 0.07v on the H row.

My original measurements were flawed because I had the jumpers installed.

 

[ljcox]

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

This is the quote I was referring to.

It appears that you saw 8.5 V on the H row and 0V on the L row.

That would also make sense. I would expect the H row to be high and the L row to be 0V.

And in your post dated 18 September, 11:43 am my time (GMT + 10) you said:-
"I measured the resistance between the - spring and the pins with the battery out. L1 = 0.2 and the other two were open"

So this shows that L1 is connected to 0V (ie. the neg battery spring)

I don't want to labour this point as it is not vital, I'm just curious why you appear to have swapped the H and L.

 

[0RESET0]

Right, that was my first attempt to measure the voltages. I measured them with the jumpers installed. So rather than measuring the voltage to the pin, I measured the voltage to a set of pins. I don't know why I did it that way the first time but after that I measured the voltage at each pin with out the jumpers. You can see in my post from 17th September 2007, 08:43 PM that I measured from the pin to the pos terminal. You said that this was incorrect so I tried again measuring the voltage to each pin WRT the neg spring. This is what the last two measurements have been, the one with the flat battery and the last with a fresh one.

In your response where you told me not to measure to the pos terminal you gave me some suggestions. After reading the posts again I realize that I didn't give you everything that you asked for. So, Here are some new measurements following your suggestions from that post.

You said "So put the +Ve probe on H1 and then Press the button, and note the voltage. it is now 9.8v. Battery voltage is 12.3, was 12.6 yesterday
Don't measure the current.

Insert a jumper H1 - C1 and repeat, ie. press the button and measure the voltage on H1 or C1. This is Also 9.8v

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."
0.08mA

In the event that you want the measurements from the L in the same way, here they are. +Ve probe on L1, No jumper - 0.05v. Jumper on L1-C1 - Also 0.05v. Current was different, it was -0.12ma with the -Ve on C1 and +Ve on L1.

This does not provide you with the resistance-continuity measurements that you previously asked for. I assume that that is because you already know what you want about that.