here is a screenshot of the control box I am printing on a second 3D printer.
what you see is the back, all my wires from the printer will connect back here. With exception of the Main Power, it goes on front. The led’s will also be on front.
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Still using 1k resistors?
No, the 1k resistors are nothing but a holding spot, till I can get all the pieces assembled on the breadboard (or boards) and tested with power.
Just to make sure is all…
But to avoid any future confusion, I will change all the resistors to the led’s to 2.2k rating resistors.
The two resistors for J2 and J3 will probably come out to be replaced with another part.
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It’s actually pointless and meaningless having three LEDs. All three LED/resistor pairs are connected in parallel across the 24V supply. If 24V is present at the input, all three will light. If it isn’t present, none will light. So you can see that two of them are superfluous. These LEDs do not indicate the presence of the output voltages. If you want to do that, you should drive them from the output sockets.
That means, of course, replacing those 1k resistors with 12V voltage regulators, and they will need to be switched-mode regulators to keep the power dissipation down.
Nobody ever uses a resistor to drop a 24V line to a 12V line, for two reasons. Firstly, it only works if the current drawn from the 12V line never varies. If it did, then the voltage drop across the resistor would also vary - away from the desired 12V. Secondly, the resistor would need to dissipate as much power as the load, which is immensely wasteful and might need a big resistor.
Voltage regulators will keep the outputs at 12V even when the output current varies, which addresses the first point. Making those regulators switched mode will address the second point. Once you’ve done that, then you can connect the LEDs to the output rails, where they should be
Getting there 
but still a problem with the LED’s however we are in danger of “too many cooks spoil the broth” I hope you understand this 
and we aren’t realy supposed to get involved with design as we all have different processes and thoughts so it can get confusing. Therefore I suggest you look closely at the recommendations made by @SteveT at the bottom of this post as he explains a route for you perfectly that will quickly and concisely get you on your way. I must admit your enclosure looks awesome 
Good luck !
Thanks mousey for your comment, and I do understand , everyone has their own idea on what to do. In my defense though, I figured by my getting input from multiple members (each of a different skill level), it would be my responsibility to learn from the comments, then make an informed decision. As I mentioned in the beginning of this, my experience level here is - well - nill. But tackling this project has taught me a great deal.
Which is why, without ever reading SteveT’s comments, I am removing the resistors and replacing them with voltage regulators. One regulator to stabilize 24vdc for main power, and another regulator to stabliize 12vdc. This 12vdc will be my connection points for the PWM module i have coming. I am knocking out all the leds from the plan, except one, POWER. I want to visually know power is supplied to the main board.
With that said, I do hope you, and everyone else, will still contribute, point out my shortcomings, give me help. For me, the 3D printers started out with the idea of making parts for my shop. But then it turned into a MAJOR undertaking, and this 70years plus retiree is not going to let it get the best of me. I can live with what my plan is now.
And besides, once I get the pcb made, and I finally get it in place and hooked up, i’ll be coming back here for extended help in case something goes awry.
Now, off to see what SteveT commented on.
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hello SteveT, thanks for the comments. If you read the reply I made to mousey, just a short time ago, you see I am eliminating the leds cept 1 for the power. I want that visual clue its on. I’m also eliminating the Not Required resistors.
First, let me ask, your comment “If 24V is present at the input, all three will light. If it isn’t present, none will light.” Could you explain this to me? I look at those two circuits up top (to J1 and J3) and I’m thinking they won’t both light up, would it not depend on the status of the fan, if fan is OFF, no led. Maybe you could screenshot that area of the schematic, make annotations, then upload with your changes.
“Nobody ever uses a resistor to drop a 24V line to a 12V line”. Resistors OUT, regulators IN.
I’ll be adding a LM7812 regulator for theJ3 fan, a LM7806_TO220 for each D1 and D2 led. I’ll use resistors to finer tune the voltage after that.
Again, how would you wrap up the circuit J1 (J1 for power), and what would you do with J3 (for the fan).
Thanks in advance.
I’ve got to go out shortly so I’ll address just your first question for now.
I’ve disconnected all three outputs, as shown by the black marks. Follow the blue, red and green arrows I’ve drawn and you will see that all three LEDs still illuminate. Therefore they are not showing the state of the outputs. All three are showing the state of the same thing: the presence of power at the input.
Sorry for the terrible penmanship! It’s tricky drawing on glass. Anyway, this is how I would design and draw it.
I’ve left out C1 and everything to the left of it because your design is fine. Also, to stop it getting crowded, I have not drawn the 100nF capacitors I would put across the inputs and outputs of the switched-mode buck regulators.
You could neaten it up a bit further…
I’m going on about this because a well laid out schematic makes the function of the circuit as clear as possible. If you compare this latest one with your version, I think you will agree that it is easier to see what it does and how.
Also, you might want to consider fusing the 24V output (I don’t know what is downstream so it might or might not be necessary). You probably won’t need to fuse the 12V outputs because the regulators will current limit automatically, but make sure the limiting current is low enough to prevent harm downstream.
There are other protection features you might want to add, but we can talk about them another time. Don’t add protection unless the requirements call for it, otherwise you are just adding complexity.
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(Deleted). Dum de dum…, bumping the character count…
thank you Steve for the info, your drawing is spot on. I was working on some changes while you were working, and believe it or not, I was pretty close to yours. I’ll attach an update screenshot later.
Fusing, no need, the is a 20amp fuse on the primary board for the printer. Or were you thinking of a fuse on my aux board to protect it as well?
As for protection, I would be interested in your thoughts.
Question? On the top leg, 24v to main power 3 3.3v for the D1 led. I thought I read from someone that using a resistor to drop voltage from 24 to 3.3 was - well, a no good practice.
Would it not be better to insert a voltage regulator, say LM 7806, then use a < 1k resistor before the led?
The advantage of using a resistor to drop the voltage is that it is extremely simple.
There are two disadvantages:
1/ The voltage drop changes if the current changes
2/ The resistor dissipates energy as heat. This has two effects: the resistor gets hot, which is OK if it’s within its specifications; and the energy is wasted, which maybe you care about, maybe not.
In the case of the dropper resistor for the LED, we don’t need to worry about 1/ because the current drawn by the LED will not change significantly.
We can calculate the power dissipated by the dropper resistor. Rather than assuming it’s a 1k resistor, I’m going to assume you want 10mA current in the LED, and you say the LED will drop around 3V. So, power (watts) in the resistor = volts x amps = 21V x 0.01A = 0.21W.
So the resistor will need to dissipate 0.21W. This is significant, but a 0.5 or 0.6W metal film resistor will be fine with that. Go with a 1W resistor if you want. You should space the resistor a few mm off the board, so air can flow around it.
Do you care about wasting 0.21W? In an industrial setting, surely not. So I would use a resistor as discussed. Inserting a voltage regulator for the sole purpose of powering an LED is, in my opinion, unwarranted cost and complexity.
Just to finish, let’s calculate the value of the resistor required to drop 21V at 10mA.
Resistance = volts / amps = 21 / 0.01 = 2.1k. The nearest convenient value is 2.2k.
So there you go: a 2.2k 0.5W, 0.6W or 1W metal film resistor will do the job.
Yeah, the heat losses add up quickly when you do the math.
There are lots of little dc-dc buck converters (from murata, oki, traco, cui…) that snuggle into a through-hole TO220 footprint – easy to use for a project such as this. Efficiencies are generally 90-ish percent, so very little heat loss. A bit of switching noise on the rail though.
Here is one that drops 24V in down to 12V out (0.5A max) and costs less than a dollar-and-a-half (plus a couple of caps):
https://www.mouser.com/ProductDetail/CUI-Inc/PXO7812-500-S?qs=Y0Uzf4wQF3kdbacxg4U5TQ%3D%3D
And this one will take the 24V down to 5V (1A max) – cost three dollars:
https://www.mouser.com/ProductDetail/CUI-Inc/VXO7805-1000?qs=HXFqYaX1Q2zTUq5iRuYtuQ%3D%3D
You can get 3.3V buck converters as well, or step down to 5V with a buck and then use an LDO down to 3.3V for a cleaner rail (eg: TC1015-3.3, LP5907_3.3…)
To: SteveT and teletypeguy,
Just so you know, I am not ignoring you. I had a power outage this morning, and for some reason KiCAD did not backup the project as it should done. No I have to recreate the entire right hand side of the schematic. Thankfully I pasted a copy to SteveT so I have something to work from. At the same time, I can look over the both you guys comments while recreating.
I respond to you both tomorrow, or at latest Saturday.
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Are you sure about that?
KiCad tends to keep a bunch of zipped up backups in the “backup” directory in your project directory, sorted by date.
Yup, I know…
Fortunately the KiCAD app failed to perform any backup’s after the 28th.
I scoured, I searched, I scanned all my internal partitions, as well as external drives. Na-Da
Quicker just to rebuild.
OK Steve, wanted to pass along some info I tested for this morning. In reading through your latest comment and trying to grasp the idea, I found a weblink that discussed some concerns with anyone upgrading their printer. I found the info valuable, and I thought I’d pass along for your info.
My power supply is rated at 120v input at 8.6amps, 24v output at 14.9amps, output calculates to 357.6 watts.
FYI. The site had this terrific little calculator.
HotBed: I measured the resistance with my meter and it is 2.1ohms, 24v. So the calculator give me amps at 11.42, and 274 watts.
HotEnd (the heating chamber where filament is melted): Resistance measured 12.4 and again 24v. So the amps = 2, and watts =46.5
Now to your comment regarding resistors (I love your comments, they are very informative, and easy to understand).
anyway, the LED’s I am going to use are metal film, I am going with the larger 5mm, they are 3.1volts, and 20mA (.02)
So using your formula (or the calculator I find that my voltage drop is : 24 - 3.1v = 20.9volts
The calculated resistor needed for my application comes to 1045ohms.
The problem is the watts. Using a .01amp led, the watts was .21watts, which you indicated was quite significant. But using my .02amp led’s at 20.9volts (24 - 3.1) the watts produced is .418, almost double your example. Does this mean I should use a higher rated resistor in the circuit?
Yes. Make it a 1W resistor. Or you could use a dc-dc converter as @teletypeguy suggests, which would reduce the heat but take up a bit more space. Only you can judge which is most appropriate for the job.
Just one thing: I’m surprised that the LED needs 20mA. I suggest that you try it first - it might be bright enough at 10mA, and too bright at 20mA.
First question?? If I use these buck converters, do I still need a capacitor in front and one behind the converter?
And what of the noise I was reading about on LDO’s? One of the goals of this upgrade was to make the printer run quieter, especially the motors that work x - y - z axis.
I will definitely look over the bucks converters.
What is meant by “cleaner rails”?
I should have thought of this long before now. I have a schematic of the printers main board, as well as a view of the physical board and all its connections. This my shed some light on the project, give a better understanding other than what I have tried to do.
Here is the board layout. The bottom left corner is where the main power connects, along with the two major hogs on power, the hotbed and hotend.
