Help with PSU (Temp control fan, load bank, & PWM circuit)

[()blivion]

Looking good ()blivion.

Thanks, I practice... like..... a lot. LOL

Did I put the capacitors in the right spot, see attachment, option 1 or 2 (maybe it doesn't matter)? Also, what about the logic circuit Op-Amps, should I put capacitors there too (I know U1 is actually a comparitor).

Nice try, but sadly you have it wired wrong.

One pin of the capacitor goes on the positive rail of the OP amp, and the other goes on the GND rail. You have it inserted in between the positive power and the positive pin for the Op-Amp, this won't work as a capacitor looks like a cut in the circuit to DC power, in series it won't let the chip get power. You want it to be in parallel with the Op-Amp IC's power, right up next to the chip.

Also, the LM358 Op-Amps are *DUAL* Op-Amp chips. This means that there are TWO(2) Op-Amp circuits (triangle things in the schematic) per 8 pin package. You only need to put one(1) capacitor per Op-Amp *PACKAGE*, for a total of five(5) packages and five(5) capacitors, but ten(10) Op-Amps. They do this to save space, and it helps quite a lot, at the cost of confusing the crap out of people some times. And to top it off.... they make packages with four(4) circuits in them too.

Look at page 9 of this datasheet and notice that there are two triangle things in the picture.

https://www.electro-tech-online.com/custompdfs/2012/08/lm158-n-1.pdf

Or just look at this picture of a different Op-Amp with about the same internal configuration...

View attachment 66661

Bottom Line
All IC's should have a capacitor on their power pins/rail, it removes power supply noise and prevents hiccups from making the chip act funny. As you guessed, you should use the extras to do this to all of the chips. So total, you should have seven(7) IC's to put capacitors on. The one(1) LM339, and six(6) LM358's. And they should all go as close as possible to the power pins of the chips. You may ignore the temperature sensor I believe. It may not actually act like it should if you add one to it, but I don't know the part very well.... So.... :shrug:

Why do we do this?
Capacitors act like little rechargeable batteries. We put them on the power wires of the IC's because they will power things for a brief moment if the power sags. And they absorb power spikes some what also. In this configuration they are called "decoupling capacitors" because they "decouple" the IC from the power supply and it's noise. Often times they are not needed and you will see no ill effects without them. But when do screw up you may never be able to figure out what or why problems are happening. Better to have them than to not.

 

[jocanon]

OK, so like this? Or is the third time a charm

 

[()blivion]

Yup! you have it 100% right now.

And Just to confirm it, though I know your quick witted enough to have caught this the first time around. But U2 and U3 (on the top left) are one chip, and have one capacitor. And U5 and U6 (on the bottom right) are also one chip, and also have only one capacitor just the same. OK? Good.

I'm going off line soon. I'll see you tomorrow though.

 

[jocanon]

Ah, I did catch U3 and U2 being one chip, but I didn't think about that U5 and U6 could also be one chip, thanks for that.

 

[jocanon]

OK, here's what I put together. I might still play around with placement, but did I at least get the connections right? Also, if anyone has suggestions on better placement those are welcome too. I used the red trace lines to represent 26 AWG jumper wire connections.

 

[()blivion]

At a quick glance, it all looks correct. But I'll need to look over the whole thing carefully before I make my judgment firm. And that will have to wait till tomorrow.

One thing, If that image is exactly how you plan on building the circuit, I would suggest that you attach the DUT ground to your dummy load ground at the point between R18 and R19 instead of on one of the extreme ends. This effectively cuts the stray resistance of your 10 AWG wire in half because the current will now get to go two directions. But it probably won't actually matter, 10 AWG has almost no resistance per foot. Plus the way the device is designed, if one FET sees less current because of large external stray resistance, the feed back will simply drop that FET's resistance to bring the current back up for it.

In short, it's a self balancing constant current system, so in theory it will automagically correct for such things when working properly.

 

[jocanon]

Thanks ()blivion. I suppose it would be just as easy to connect the DUT in the middle anyway just to be safe.

 

[jocanon]

I tested the volt meters and this is the data I got:
24.2volts: pin 1=.38, 2=.69, 3=.69, 4=.61, 5=.46, 6=.70, 7=.23, 8=.32, 9=2.1, 10=2.1, 11=1.22, 12=1.22;
12.0 volts: pin 1=.38, 2=.67, 3=.66, 4=.65, 5=.52, 6=.8, 7=.04, 8=2.1, 9=1.3, 10=2.09, 11=1.17, 12=1.2

Then I tried the same tests on the volt meter that is "factory defective" because the decimal already didn't work. I started with 12 volts and got the same readings except for pin 10, instead of 2.09 I got .78. Then, mysteriously the decimal started working somehow and the reading then jumped up to 2.09, the same as the first test...so I am pretty sure it has something to do with pin 10, probably some multiplexing going on so I don't know if I will be able to disable it easily or at all. I looked inside where that pin goes up to the LED and I could see that there was a little bit more solder on pin 10 than any of the other pins, so maybe a cold solder joint is making it work intermittently.

 

[()blivion]

Now that I have thought about it more, we are most likely not going to get enough information from just probing the wires with a multimeter. I think we would need a scope or logic analyzer on each pin to unravel the mystery. Sorry for having you do an unnecessary test.

From what we know so far, I can tell you there is definitely multiplexing going on. The normal way to do it is to have one wire for each of the letters + decimal place, and then one wire for each individual part of the number. So 7 number segments plus the one decimal makes 8 wires, then 3 more for each whole number. That is a total of 11 wires to control the whole screen. And if you look, there are 12 solder spots on the bottom of the device, but one is not soldered. So that makes 11 connections which fits well with that theory. What we would expect is one wire that controls all the decimal places, which would be safe to unsolder as we need none.

Thing is, there are other ways to multiplex than the above that are even more wiring efficient and more complex. A designer can take advantage of the fact that LED's only conduct electricity in one direction, then make each number selection wire be able to be three different states. float, ground, or V+. This would be able to cut the number of wires needed per LED roughly in half and most any driver chips could do this easily. It would also mean that one of the decimal points would share a common drive wire with some other LED. SO if you disconnect it you would be shutting down some other part of the screen.

In the end, I don't really know which it is. I would need to think about it more or have it in my possession. You can assume that the LED display is the most commonly found type and say it has 8 wires, one per LED, and 3 digit wires. For the mod, You could maybe just try and unsolder pin 10. If that doesn't work, resolder it and try a different wire. If at any time you notice a bunch of segments go out, but all the numbers still have at least one segment on, then it is NOT the most commonly found type of screen, and I would think it can't be done.

You could try a drill bit to the little dot until it stops making light, Or cover it with black tape. (⌒▽⌒)

 

[jocanon]

I was thinking about the drill bit idea too.

 

[()blivion]

OK, Looked at the circuit a bit more. Got a few things to point out so far. Hopefully this will prove to be a complete list of "problems", but I am still inspecting it. And neither of these things are your fault or big show stoppers for that matter. In fact... they are quite minor considering this is your first time.

1) You have it hooked up right, and I'm sure ronv knows what he is doing, but I don't think that your alarm will function as is. In simulation, with the values shown, it just buzzes constantly no matter what U1's output is. If you hook it up and it screams at you right away, disconnect it. We will figure it out later.

2) We didn't tell you, and I don't think it actually matters with this PARTICULAR part, but generally when one is working with basic IC's and one is not using all the circuits inside them, you have to tie the unused inputs either high or low manually to avoid over heating. The LM339(U1) has 3 unused circuits inside it. I'm 98.5% certain it does not actually matter with comparators because their inputs are made to take analog voltages. But with other logic chips leaving the inputs unattached can be destructive and is not optimal. Better learn this lesson now and learn how to fix it.

You want to tie pins 4,6, & 10 together, then to Ground through any resistor. After that do the same for pins 5, 7, & 11 but this time tie to V+ through a resistor. The two resistor values don't really matter. Should be anything greater than 100 Ohms... so any spare you have left over will work fine. Then you can leave the three unused outputs open circuit and unattached to anything. What this did is put a logic 0 on the unused (-) inputs, and a logic 1 on the unused (+) inputs. This will keep each comparator from seeing it's own internal voltages on it's inputs, then thinking that is signal and changing it's output, which would of course rearrange it's internal voltages and change what it sees all over again.

That's all for now. Again, I'm still looking, but there is no reason you can't build it as you have it that I can see. You have the Op-Amps all attached the right way, all the sense resistors, and looks like you have where all the FET's are going to go figured out too. You may want to optimize the spacing of parts and lengths of some wires more, but other than that and the things outlined above, it looks 100%. Well done.

 

[jocanon]

Thank you ()blivion, being my first time I was sure that I would need some help, that's why I wanted to get it all put together on "paper" first. As far as spacing, is it OK to have things as close as I do? Particularly the current sense resistors, or would it be better to have a little more space between them for better heat management?

 

[jocanon]

Will this work as far as tieing the unused inputs?

 

[()blivion]

Will this work as far as tieing the unused inputs?

Yeah, that's fine.

is it OK to have things as close as I do?

Also OK. I was thinking more about optimizing the arrangement of parts just right to make all the connections as short as possible. But you most likely have it as close to as good as your going to get it.

No complaints here. Build it.

 

[jocanon]

Cool. Thanks again!

 

[jocanon]

Two things, first, I just realized I forgot to add the LED so I put that in there at D1 now. Did I wire the LED correctly (see attached picture)? Second, is it safe to run it without the MOV idea we were tossing around the other day if I just make sure I don't unplug the DUT with the current turned up?

 

[ronv]

Hi guys,
The - side of the alarm needs to go to ground. The diode cathode goes to the ground side as well.

 

[jocanon]

Hi ronv, do you mean like this:

edit:
note, the red line from the LED goes to ground (it represents a jumper wire even though it goes over the +12v line I am not connecting it to +12v, I am going over that and soldering it to ground, sorry if my drawing is confusing )

 

[ronv]

Yep. The flat side of the diode and the long lead is usally the cathode on the led. They are polarity sensitive.

 

[jocanon]

Ah, gotchya, so the long lead on the LED goes to ground...I was wondering about that, thanks.