Hi Jason,

>>> So it looks like the 4-16 '154 chip and the 2019 will do. But I now need to know what pins to connect to the NTE2019, and what caps I might need. If I'm already using a regulated (for sure, not a cheapy wall wart) power supply with 12 volt output, what resistor do i need for a drop to 5.4volts (that's what my 4x1.5 batteries tested in actuality, and the lights were perfect. If I need to use 6v, as it looks like the NTE2019 requires it, then the resistor value to get me from 12 to 6.)

The 74HC154 and NTE2019 will work from a 6Volt supply and the NTE2019 inputs will be compatable with the 74HC154 at 6Volts.
You cannot drop from +12V to +6V in a controlled way using a resistor, as the lamp loading changed, so would the "6V".
You require a +6V regulator that will step down the +12Vdc from your psu.
If you choose 6V lamps with the lowest wattage rating that suits the psu and +6Vreg,,, remember W=V * I, if the current was say 100mA, this would give 6*0.1=0.6Watt lamp rating
OR you could use 12V lamps and run them from the +12Vpsu and use the +6Vreg just for the logic, [this would be my choice]
Do you plan to link this system to a PIC or MCU?

>>> The other confusing point to me is the requirement of the "enable gate" in the '154. what is that, and how do I implement it? just tie it to ground?

Yes, connect the Enables G0/1 to 0Volts.

>>> Also, is my electricity running all the way through the contacts, and into the chips, or do i have separate voltage/currents going into the chips for the 4 inputs and the 2019 output voltage/current? sorta like how a relay works, small currents control large isolated ones - will I have two loops? know what i mean? Do I want two loops for better safety?

Did you have to use the metal frame of the cam housing as the common return when you did the battery lamp test??
My only cause for a little concern would be the 'quality' of the contacts on the cam housing when they are conducting the low level currents at +6V, you may have to use pullup resistors on the contacts in order to electrically 'wet' the contacts..
I good clean with meths spirit would be a good idea.

>>> Thats all - as long as it works, and now that I have the two main chips for less than $6, I can rest assured this "new" demux solution will not be more than $20 which is good.

so i really need help with the hookups and identifying any additional parts.

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hi Jason,

Attached pdf should be a start off pointer.

This drawing edited.... invertors added after 74HC154

Attached Files

WDI1.pdf (42.2 KB, 9 views)

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Nice drawing and solution Eric.

Couldn't you eliminate the 74HC14 invertor? Does it matter if the inputs to the mux' are True Lo or True Hi?

 

Hi Mike,
I suspect that the metal cam housing is the common return and that the contacts are normally open. I didn't want to have the housing at +6V.

Without actually getting my hands on the device I have had to make some assumptions.

I have a 'gut' feeling that the cams are not coded high/low in a binary pattern

Thats why I couldn't post a complete drawing.


EDIT: Added invertors on the 74HC154 outputs..... above drawing changed

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The 4 for wire output is surely just the result of the contact switches. Though one could argue that the output is, indeed, encoded (8 of 16).

Drop the inverters. There are 16 output combinations from the cam switches. The 4-16 demux will fully decode all combinations so no need for the inverters. (for example 0011 from the switches will be output 3 non-inverted and output 12 if inverted) - just pick the correct output from the '154.

I think the 2016 is a high side driver (check me on that, it's not totally clear to me from the datasheet) so you will need to use the "inverted" output of the '154. By that I mean use the output that corresponds to the inverted input. So, if NE is input as 0011, use the '154 output that corresponds to the 1100 input. The outputs of the 2016 would go to the lamp and the lamp to ground. Use moderately high resistors between the '154 and 2016. I'd say 4.7K or even 10K.

On the point about how this worked before electronics - I think there has to be an elegant solution there.

 

hi Phil,
Checking the NTE2016 data dosn't say its a high side driver.

On the point about how this worked before electronics - I think there has to be an elegant solution there.
I feel sure that the cams produced the 'code' for the 8 lamps.

Drop the inverters. There are 16 output combinations from the cam switches.
Regarding the number of switches, I'm wondering if the metal frame is the common of the 4 contact sets???

Look at my earlier EXAMP.pdf, it shows 4 switches driving the 8 lamps.

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the 2016 has a common cathode pin. perhaps I was confused by that terminology. what is common to that pin - the emitters of the darlingtons or the collectors. If it means common cathode driver then you are probably right. Also, what does the term PMOS mean in the datasheet?

which examp showed driving 8 lamps. I only saw 4.

 

the 4 wire output is a result of the N.O. contact switches closing due to raised bars on cam. there are 4 raised "bars" around the cam such that only one bar or two bars are pressing 1 or 2 active contact switches at any given time, so the 4 switches produce only 8 possibilities total due to the layout of the raised bars around the cam. the possibilities are listed in my truth table up top. the ones with <NA> are the ones that are not possible. '1's are indicating when a switch is closed by action of the raised cam bars. So it is only possible to have 1 or 2 '1's in the truth table. The ones with three, four or no 1's are <NA> - not possible.

I'm using the 2019, instead of the 2016.

And rereading Eric's post about not dropping voltage in a controlled way, if only one lamp is lit at a time, isn't this controlled? Or does having 1 switch closed produce a different voltage than having two switches closed, even if only one lamp will ever be ultimately lit? I don't think it's a big deal using a step-down regulator, just wondering.

As for the terminals on the cam, 4 connect to the switches, and one is voltage i.e, power most likely has to go through the cam housing - and out the switches - no way around that, right?

My original pic showed 8 lamps. re-read my lengthly previous post for the test scenerio which only hooked up 4 lamps to the four cam terminals to prove the operation of the lamps. In short, with only 4 lamps hooked up, the N lamp AND the E lamp turn on when the cam is in the 'NE' position of 45 degrees. When the cam is at 90 deg, ONLY the E lamp turns on. The inventor of the 8 light box added 4 more lamps to the original 4 so that they could (somehow) eliminate the "N" and "E" lamps lighting simultaneously and instead light up that new "NE" lamp.

I feel you guys are on the right track for sure - I can understand the solution so far, I just don't have the ability to know what is right for caps, resistors, and such for 'proper' circuit design and implementation. This I'm listening to with great enthusiasm! (and learning again)

I still don't know what you mean by "encoded". I feel my truth table explains every possible situation in a digital form quite succinctly. But If you have a moment, can you explain what you mean by that and how it is still not clear? I will further describe the device if necessary. Thanks!!!!

 

I'm working on the breadboard now. Thanks so much.

What about the 74hc14 - there's only 6 inverters, i need 8 - but can't reuse two as you show, right? Or do I get 3 chips (2 for the pre-2019, and 1 for the pre-154)

 

Hi,
The 2019 is suitable for 6V input signals.

The invertors have 6 invertors per ic. Use one 74HC14 [all 6 gates] and the other 74HC14 has 2 spare gates making 8, the number you require to drive the the 2019.

As Phil points out, you could drop the 4 invertors connected to the input of the 74HC154 and figure out which 154 output is active low for given switch contacts.

As I suspected the housing is the common 0V return for the switches, which will be at 0V for the logic, no problem.

Regarding the +6V regulation, you could, if you are confident that a resistor from the 12Vdc line will give you a constant +6V. [I wouldn't]

The HC series will operate from a +2V to +6V, so you must not exceed +6V after the dropping resistor, BUT also the 2019 expects inputs of at least +6V, so there is a close cross over point.

For the low cost of a 7806 reg I would use one to give a regulated +6V

If you had not got on the shelf the 154 and 2019 I would have suggested a simplier circuit.

Lets know how it goes.

EDIT: Would you be interested in considering a simplier circuit at this time?? Posted for reference....

Attached Files

WDI2.pdf (24.0 KB, 5 views)

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heck yea. that is simpler. if i can find the hef4514 in the stores...

Just to verify:

1) now, it's more important i stay with a regulated power supply, right? or should i put the regulator back in (yea, i agree with you, it's low cost) or did you take it out because i said i would use a regulated supply?

2) And the "1KO" is 1K ohm resistors, right? all 4 goes to 12v right?

3) what wattage for the 1K's?

4) on the '514, pin 1 and 24 go to 12v and pin 12 and 23 goes to ground?

And lastly, since I think all my current lamps are 6v (240ma was what one was drawing at 6v), and i would have to buy 8 new ones, but before i do, i want to see the brightness with one as a test. **OR** - is there now an easy way to step the voltage back down to 6v - maybe simply using a 6v power supply? but i know the 2019 need 6v...... so..... any other options easily? I forget what's the point of using the 12v lamps instead of 6v besides lower power?

 

Quote:

Originally Posted by LAJay heck yea. that is simpler. if i can find the hef4514 in the stores... Just to verify: 1) now, it's more important i stay with a regulated power supply, right? or should i put the regulator back in (yea, i agree with you, it's low cost) or did you take it out because i said i would use a regulated supply? If you use the HEF4514 it works from +12V, so you dont need the +6Vreg 2) And the "1KO" is 1K ohm resistors, right? all 4 goes to 12v right? Using the 4514 the 1KOs goto +12V 3) what wattage for the 1K's? Quarter Watt [0.25W] ,, eg: 12V/1000R= 12mA.... 12V * 12mA = 0.144W 4) on the '514, pin 1 and 24 go to 12v and pin 12 and 23 goes to ground? Thats correct And lastly, since I think all my current lamps are 6v (240ma was what one was drawing at 6v), and i would have to buy 8 new ones, but before i do, i want to see the brightness with one as a test. **OR** - is there now an easy way to step the voltage back down to 6v - maybe simply using a 6v power supply? but i know the 2019 need 6v...... so..... any other options easily? I forget what's the point of using the 12v lamps instead of 6v besides lower power? Why dont you calculate a resistor value that will drop the +12V to the lamp voltage? eg: R=V/I .... 6V/0.24A = 25R.... W= V * I = 6V * 0.24A =1.44W say 2Watt. Connect the single, 2W, 25R resistor between the +12V and the lamp supply. Remember there is only one lamp lit at any one time.

Does this help

EDIT: see pic ref lamps. The 27R is nearest pref value [ should read npv... not ppf]!

A thought, If you have a spare 6V 0.24A lamp, use that as the dropper resistor!.

Attached Files

WDI2.pdf (24.4 KB, 4 views)

__________________
Eric
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