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Timing Circuit? Counter?

Ernest George

Jul 18, 2014
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Heck it's just probably easier if I download the ORDCAD/STD3 software. That would probably be easier for both of us. It looks like they have trial-ware so that might be the easiest path (I get worried I'm taking up too much of your time!). :)
 

Ernest George

Jul 18, 2014
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Arghhh! Stupid Orcad. I downloaded and installed their trial-ware and not one of the programs will run. In each case a side-by-side error was generated and that's it. A search on the 'net shows others with the same issue and no solution from Orcad for months! :mad: So much for using their native app to view the schematic/autoroute it.
 

Ernest George

Jul 18, 2014
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Some days I'm just stupid and waste time (sorry about that!). Turns out, if I just got off my butt and stopped looking for the easy way I would have realized that laying out the exact same schematic in my own schematic editor (PCBExpress) would allow me to link it in and automagically create the PCB.

After 5 minutes this is how much I've done already....
 

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davenn

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we all have those days ;)
sounds like you are on track now

cheers
Dave
 

KrisBlueNZ

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Sure. You will need to find out which end of the PCB component is numbered 1 and which end is numbered 2.

Edit: Sorry, that was a reply to a post that was rapidly superseded!

Edit2: OrCAD/SDT III is ancient! You won't find it now, except in a computer museum. And on my laptop! I doubt whether their current stuff is worth trying.
 

Ernest George

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Dave - I need those cookies!!! ;)

Kris (and Dave) I was totally wrong. Well, I was right in what I was trying to do but when I completed the schematic what I found out was that the software does not automagically place components and traces for you in the PCB editor. Instead, what it does is just show you, after you've placed the components, which pin is supposed to go where by means of turning the mating pins blue......... you draw the lines. You place the components. In other words, the autorouter feature is not even close to what I understand a real autorouter is. This thing just turns pins blue so that you can draw a line between them.

To bring this back to the very basics. All I'm trying to do is turn Kris's schematic into a PCB layout with components placed and traces routed. So this evening I decided to approach this from a different perspective. What I'm doing now is working from the LM339N outwards. I'm hoping this is the best way to approach it :)

To that end, a simple question. At the top of the schematic, running across (left to right) is a 12V line. (see attached). My question is, does that line have to be contiguous or, can I punch down each line running up to it to the 12v plane on the circuit board?

Probably a stupid question, but at this point I'm just trying to take the schematic and turn it into an efficiently laid out PCB circuit....and it's not been easy for me :(
 

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KrisBlueNZ

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... the software does not automagically place components and traces for you in the PCB editor. Instead, what it does is just show you, after you've placed the components, which pin is supposed to go where by means of turning the mating pins blue......... you draw the lines. You place the components.
OK, so this is just assisted manual routing. It probably also has a ratsnest mode, where the connections between pins are all shown as thin, straight lines. This can be helpful when you're moving components around, so you can see where the large concentrations of tracks will be, and you can choose the orientation that results in the least track crossings.
To bring this back to the very basics. All I'm trying to do is turn Kris's schematic into a PCB layout with components placed and traces routed. So this evening I decided to approach this from a different perspective. What I'm doing now is working from the LM339N outwards. I'm hoping this is the best way to approach it :)
Yes it is. After you've placed any fixed-position components such as connectors, multiple-pin devices with lots of connections are the place to start when you're optimising placement. Join together the commoned pins, place and connect the decoupling capacitor(s), then look for components that connect only to pins on the IC and place them close by, and work outwards.
To that end, a simple question. At the top of the schematic, running across (left to right) is a 12V line. (see attached). My question is, does that line have to be contiguous or, can I punch down each line running up to it to the 12v plane on the circuit board?
You can "punch down" separately for each component, as long as any decoupling capacitor(s) are connected directly across the relevant IC(s).
 

Ernest George

Jul 18, 2014
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OK, after more/less 4 days I think may finally have it laid out :D

One more dumb question....which line is the line that is the input line from the sensor? Ie. the line going into the circuit that will be somewhere between 0.7 => 1.3V
 

KrisBlueNZ

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The circuit monitors the voltage between the two wires at the left. The top wire, connected to RIN, must be positive. The negative side of the voltage that you want to monitor must connect to the bottom wire, which is the 0V rail of the whole circuit and connects to the negative side of the circuit's power supply.
 

Ernest George

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Oh-oh.... There's only one input from the sensor that I'm monitoring :eek:

Minor heart-attack.....I bet it makes a big difference if the input voltage is 5V and not 12v :confused:

Oh no...."The circuit must be powered by a clean regulated 12v supply..."

I think I just screwed the pooch - royally
 
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KrisBlueNZ

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A voltage can't exist at one point. A voltage is a DIFFERENCE in potential between two points. When you measure the voltage, which is 0.7~1.3V with a multimeter, you connect your red probe to that wire, right? And what do you connect your black probe to? That's what you need to connect to the circuit's 0V rail.

Yes, you said you have a 12V supply. But it doesn't need to be accurately regulated, or clean, now that there's a reference voltage generator in the circuit.
 

Ernest George

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I really screwed it up Kris :mad:

For some reason I was under the impression (at the outset) that the power feed to that sensor was 12V. It's not, it's 5V - groan.....the ground for it comes off the sensor ground circuit - groan again. A third wire coming from it is the output (the line with the voltage to be measured). I don't know how I could have screwed that up (other then that I was working on 20 different things at the same time).

I'll go draw a bath and hold my head under the water... o_O
 

KrisBlueNZ

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LOL :)

Don't be so hard on yourself.

I think the existing circuit will work OK with a few component value changes.

269993.003.GIF

RA changes from 4k7 to 1k5
R1 changes from 30k to 10k
R2 changes from 120k to 15k
RIN changes from 30k to 15k (this change is just to avoid an extra line on the BOM)
CT changes from 4.7 µF to 6.8 µF.
 

Ernest George

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Bear with me for a sec and always keep in mind my limitations ;) I'm painfully aware that this is a mess of my making and confusion in really messing with me on all fronts when it comes to this circuit - I don't know why which makes it harder. Backing up for a second... :eek:

OK, in really simple terms the sensor I am measuring is like this;

Fed shared 5V VCC with other sensors (power)
Has a Shared Sensor Ground with other 5V sensors (ground)
Has one output line that varies from 3.6 to 4.4 (all OK) and drops down to .7 to 1.3 when the circuit should activate. (let's call it sensor output)

So, with respect to the new schematic above, the output line from the sensor (typically .7v to 4.4v) should be connected to the top VIN line?
The ground for this circuit should be connected to the lower VIN line? (the shared sensor ground)
The rail across the top (marked 5V) should be connected to the 5V VCC feed (ie, always has 5V VCC).
Line #2 marked out is the line that latches low (goes to ground) when the circuit is activated.

If my understanding is correct, then this should work. The specific point of concern for me is the top VIN line on the left. That would be the sensor output right? ie. Would typically be 3.6 to 4.4 or .7 to 1.3 when the sensor activates (so to speak).

Lastly, would this circuit draw much power from the 5V VCC line.

Fire when ready! :D
 

KrisBlueNZ

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Yes, that's all exactly right. The circuit will draw a few milliamps. Allow for 10 mA worst case.
 

Ernest George

Jul 18, 2014
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Awesome! Thanks Kris :) I'm glad I finally got my head around some aspects of the circuit! Burn-out has been setting in lately, I'm almost done for :confused: and could use an extended break as I've been at this non-stop for 2 months now (you can probably tell!).

Unfortunately, I'm going to have to implement this specific circuit on revision 2.0 of the board. I'm running pretty far behind schedule on the whole and need to test out all the other circuits I've built up (as well as test my measuring/placement skills - there's a LOT of components that have to fit just right in order for the boards to fit in their little box.) I've been over the board so many times my head is spinning BUT...I'm at the point where I'm as sure as can be and only actually seeing the initial proto's and soldering in the components will tell me what I've missed or didn't think of.

I factored in doing two prototypes from the outset as I'm sure I won't nail it 100% the first time through and probably will want to add some more stuff to it after actually using it :D I'm about to pull the trigger on the first batch (4) so I'll see what this step brings and how far I've come...or if it even works :D

I will have some more questions about this circuit though and also it would be worthwhile for me to shed some more light on what the output is latching down when active. For some reason I get the feeling that my design might not be right in that area. But I need to take a brief detour first to see how the rest of the project is shaping up :) I just wanted to let you know so you don't think I've disappeared on you!
 
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