|
Shame Boy posted:If pressing on the front bezel bit causes it to start working it's probably the flat flex board that connects the drivers to the actual LCD panel itself, those are generally a massive pain in the rear end to solder (they're made of plastic) so I wouldn't get your hopes up too much. I'd still say open the thing up and find the precise bit that causes it to work when you press on it (be careful of wall voltages while the case is off!) and post a picture here, it might be something easier to fix than I'm thinking, but yeah. Good luck! I see. That’s probably what it is. I’ll take a look once I get the new one in place. If nothing else maybe I could shim it back into place somehow. But maybe long term that would make it worse?
|
#
?
Oct 14, 2018 05:52
|
|
|
|
| # ? May 28, 2024 05:52 |
|
|
I guess this is gonna be an odd bench power supply. A wooden one. I got the idea from Matthias Wandel  Stuff to do yet, top and front / back panels are not glued together so I can take it apart. I like the way it's turning out looks wie and in the future I might want to replace the innards with a more powerful PSU, a DC voltage booster circuit might also be useful, seen them on ebay. And yeah my worksurface is a mess and looks like a fire hazard but I promise I clean it up and I solder over a metal sheet.
|
|
#
?
Oct 15, 2018 06:00
|
|
|
Looks gorgeous to me. Mine's a wooden holder for an ATX supply with a cover made of hardboard, so you're doing better than me. I assume those holes are for binding posts? Does the switch light up or do you have another on/off indicator? Only change I might suggest is to have more than one ground post - can make things a little easier if you're doing a multi-voltage project.
|
|
#
?
Oct 15, 2018 15:15
|
|
|
Thanks I'm still not done with the exterior, needs a good sanding and some varnish yet. Good idea about a second ground. The switch does not light up so I want some way to see it's running, personally I was thinking if I could get a red light or something inside the box to light up, and make red light shine out through the slots on the front and the back grille. Just something different than an indicator light.
|
|
#
?
Oct 15, 2018 16:59
|
|
|
If you look at case mod stuff, you can use any of that -- it all eats 5V or 12V. LED strips, code cathode tubes, light-up fans, etc. For my simpler build, I just ran LEDs from the standby ATX line (indicating AC is connected) and the 3.3V line (it's on). At home I have a fancier one with a light-up switch, and I had to go a bit nutty and wire it for active high from the standby line so the light would work, then run that through a NOT gate made out of a transistor so I could hook it to the enable line, which is active low.
|
|
#
?
Oct 15, 2018 18:04
|
|
|
Notes: - Potentiometers do, in fact, contain magic smoke. It can only be released by morons. - Spinning 140mm case fans can remove skin, and they should not be caught when they fall. - Breadboarding doesn't work if you don't line things up right. - Discretion is the better part of valor. Soldering can wait. Normal Barbarian fucked around with this message at 04:12 on Oct 16, 2018 |
|
#
?
Oct 16, 2018 01:52
|
|
|
scandoslav posted:Notes: oh dear this is quite a compilation
|
|
#
?
Oct 16, 2018 10:51
|
|
|
I once had a breadboard set up to program a PIC micro that belonged to a board that I didn't put an in-circuit debug port on. So I'd pull it out, put it in the breadboard, program it, and put it back. One time I did this, the IDE couldn't detect the device. Then I smelled something burning. Then I touched the chip and it was burning hot to the touch and kind of hurt. I pulled everything out and let it cool down. When I pulled the chip out the breadboard was melted under it. It turned out I had socketed it in one set of pins over so it was grounded through an I/O pin and powered via the ground pin, so it was probably sinking a lot of current through a protection diode on that one I/O pin. The pin didn't work after that but the rest of the micro worked just fine. It kind of is true that PICs will survive anything.
|
|
#
?
Oct 16, 2018 11:15
|
|
|
Splode posted:oh dear this is quite a compilation All projects require a blood sacrifice. Hail Satan. Blood for the Blood God.
|
|
#
?
Oct 16, 2018 19:35
|
|
|
I was absolutely, 100% certain I had wired it this way: I even used color coded wires. Turns out I had somehow done this:  Smoke happened. I rewired it like this, then wondered why I wasn't getting any readings.  Life is beautiful. Learn through blood. Learn through lungfuls of magic smoke.
|
|
#
?
Oct 16, 2018 19:45
|
|
|
scandoslav posted:Learn through lungfuls of magic smoke. TCC is thataways.
|
|
#
?
Oct 17, 2018 00:05
|
|
|
scandoslav posted:Learn through lungfuls of 22 Eargesplitten posted:TCC is thataways. Don't feel bad, scandoslav. Any of us who work with electronics regularly have done something... exciting like that at one point or another. My most recent in memory was completely forgetting to check the voltage rating when replacing an electrolytic capacitor on the power supply for a friend's DJ lights. Pretty sure that little bastard entered low Earth orbit after it popped. The cats still look at me a little sideways when I turn on the soldering iron.
|
|
#
?
Oct 17, 2018 01:28
|
|
|
I recently hooked up a comparator wrong and when I kicked on my power supply it drew a couple amps. Yeah that one was toast. Also got some magic smoke early in the year when I was mucking around with a mosfet driver and forgot its max VCC rating. On triggering it made a pretty good pop!
|
|
#
?
Oct 17, 2018 01:33
|
|
|
We had an issue where turning on a device caused a 1A supply to brown out. When we finally debugged it using a current probe, our little 200 mW device was spiking to >25W for a few milliseconds (not capacitive inrush). It turned out to be an issue with ramp rates on internal power supplies. I calculated it out and it was something like 10 mW*s, but a stupid amount of power flowing through some pin somewhere for a short amount of time.
|
|
#
?
Oct 17, 2018 03:25
|
|
|
I had a battery charging circuit with a part wrong. When the board tried to switch from wall power to charging the battery, it shorted the battery terminals together. A decent lead acid battery can surge a few hundred amps. All the soldermask above the traces the current went through boiled off and left shiny copper behind. Battery didn't explode though! (The board actually still worked after replacing the FET that melted. The battery did not survive)
|
|
#
?
Oct 17, 2018 05:00
|
|
|
Tinkering some more with the PSU. I wanted a light to indicate when the power is on because the power switch didn't have a built in light, so I added a separate one. The PSU also has a small green led on the back side that lights up, but I had decided to put that on the back, it was so small anyway and this big red light I bought from the local radioshack equivalent looks much nicer on the front.  The green led was flickering though and not lighting up properly which indicated to me that the system needs a dummy load to function properly and most likely on the 5V line from what I've read. If I were to believe some of the stuff I've read, i.e. the stuff about modern PSUs wanting a dummy load 10% of the total capacity or 30W and adding in a safety margin I'd need a 50-60W resistor at 0.8 ohms IIRC. That seemed excessive draw to me, just a gut feeling. I figured a draw of 10 watts would be fine so I bought a 20 watt 2.7 ohm resistor. If I understand the theory behind resistors after some excessive googling that means it should draw (or produce? what's the diff) 10 watts on a 5V line. Well it worked and the LED lights up properly now. Pretty drat big resistor. I wonder if I should mount it to a piece of aluminum with thermal paste just to be sure. Maybe soon it will be assembled, I only have worked a few minutes on it since last time, busy busy. 
|
|
#
?
Oct 17, 2018 07:09
|
|
|
You have a +5V voltage supply. The circuit in it will maintain a 5V difference between the +5 and its ground, except that it was designed to always have some amount of current flowing and its regulation becomes unstable without it. Normally this load would come from the computer, so it wouldn't be wasted. On your system, you're going to stick in a resistor instead and turn it into heat. You think that that minimum required load is 10W. 10W / 5V is 2A. So you're trying to find a resistor that will allow 2A to flow through it when exposed to 5V and not melt in the process. The less resistance the path has, the more current will flow. A wire would let a lot of current flow. The air (high resistance) won't let very much at all flow. The relationship that describes this is Ohm's law: V (voltage drop across a thing) = I (current through the thing) * R (resistance of the thing) Filling in your numbers: 5V = (2A)*(R). A R of 2.5ohms will have 2A go through it If you want a feel for how hot it will get, a LED bulb that you would put in a desk lamp dissipates about 10W. The resistor itself is unlikely to get hot enough to damage anything as long as the inside of your case has airflow. Also, carefully unplug your supply, then tape up or otherwise shroud the exposed wall power pins on the other side of the entry plug. You're going to go to unplug the thing and zap yourself. It also should really have a fuse in the power entry if there isn't one in the PSU.
|
|
#
?
Oct 17, 2018 07:40
|
|
|
I used this calculator to figure out what kind of resistor I needed. I entered voltage and desired wattage, then I got a resistor with double the wattage but same resistance: http://www.bowdenshobbycircuits.info/ohmslaw.htm As far as I've seen resistors are rated in watts, not amps so I never took that into close consideration what amperage the resistor can handle. I assumed, perhaps incorrectly, that whatever combination of voltage, amps and resistance resulted in a lower wattage than the rated wattage of the resistor, then whatever the amperage is, should within the specified limits of the resistor? Is this logic right or did I miss something obvious? As for taping and shrouding, everything now is not how it will go together. It's just mocked up very experimentally because I was only testing the setup. I don't even touch the power hookup there but instead plug it in and out at the wall instead. When it's done it will be mounted inside the box and everything will be shrouded and protected from electroboom style mishaps.
|
|
#
?
Oct 17, 2018 08:38
|
|
|
His Divine Shadow posted:As far as I've seen resistors are rated in watts, not amps so I never took that into close consideration what amperage the resistor can handle. I assumed, perhaps incorrectly, that whatever combination of voltage, amps and resistance resulted in a lower wattage than the rated wattage of the resistor, then whatever the amperage is, should within the specified limits of the resistor? Is this logic right or did I miss something obvious? Nah that's right. If you think of the power supply like an "ideal" 5V source, it will pump out as much current as it needs to to keep the circuit at 5V. Your low-value resistor allows quite a lot of current to flow, so you can sorta imagine it's trying to "pull down" the 5V - imagine a pump that you attach to a huge pipe, it's going to need to work way harder and pump way more water to keep the same pressure than if it were attached to a narrow pipe. The power supply will respond the same way, by outputting more and more current until it's back at 5V. Since this type of "constant voltage" power supply will always try to output however much current it needs to bring the voltage to the correct value, by picking the right resistor you can get it to output a specific current. Now here's the weirdest bit (in my opinion, anyway): the reason resistors are always specified in watts is because they actually don't care about volts. If you look at the formula for calculating watts from volts and amps: P = V * I (where P is power in watts, V is voltage in volts and I is current in amps) ...and then look at ohm's law: V = I * R (where R is resistance in ohms) You might notice that you can substitute one into the other: P = (I * R) * I ... and simplify: P = I2R So the amount of watts the resistor dissipates as heat due to resistance is directly proportional to only the resistance and the square of the current. By telling you the resistance and the wattage rating, it's implicitly telling you the maximum current as well. It's a bit easier to work with watts for resistors though, because the maximum wattage of the resistor also happens to be the amount of heat it's going to need to deal with  Edit: Note that what I said isn't 100% universally true, resistors do care about high voltages, so if you're thinking about shoving like 1000V through a tiny resistor get ready to enjoy a nice fireworks show Shame Boy fucked around with this message at 14:22 on Oct 17, 2018 |
|
#
?
Oct 17, 2018 14:18
|
|
|
His Divine Shadow posted:amperage current
|
|
#
?
Oct 17, 2018 15:55
|
|
|
FYI, you can get 50W wire wound resistors with integrated heatsinks for fairly cheap -- they're commonly used in retrofitting signal lights of older cars to LED.
|
|
#
?
Oct 17, 2018 19:40
|
|
|
Acid Reflux posted:Don't feel bad, scandoslav. Any of us who work with electronics regularly have done something... exciting like that at one point or another. My most recent in memory was completely forgetting to check the voltage rating when replacing an electrolytic capacitor on the power supply for a friend's DJ lights. Pretty sure that little bastard entered low Earth orbit after it popped. The cats still look at me a little sideways when I turn on the soldering iron. Oh, I don't feel bad! This could be how I break out of my perfectionist/paralysist shell. The magic smoke did not kill me. Now I'm curious; why did my potentionmeter smoke? It's just a resistor, right? Was the wiper close enough to a pole to allow smoke-level current? Foxfire_ posted:All the soldermask above the traces the current went through boiled off and left shiny copper behind. 
Normal Barbarian fucked around with this message at 21:26 on Oct 17, 2018 |
|
#
?
Oct 17, 2018 21:24
|
|
|
Any type of resistor can smoke, burn, pop, etc. if you put too much current through it and it dissipates too much power (P=VI and V=IR, so P=I2R=V2/R). Pots tend to not have very high power ratings which can make it easy to destroy them if you aren't careful. How did you hook up the pot? If it only smoked if you turned it too far in one direction, then you hooked it up in a way such that when the resistance from wiper to one terminal got too low, too much current went through it and it dissipated too much power. Note how in the equations above, if you halve the resistance but the voltage across it stays the same, twice the power is dissipated. So even small turns near the end of the pot's range can cause exponential power increases. edit: to avoid this, consider adding fixed resistors so that there's a minimum safe resistance even if the pot is turned to extremes BattleMaster fucked around with this message at 21:49 on Oct 17, 2018 |
|
#
?
Oct 17, 2018 21:38
|
|
|
BattleMaster posted:Any type of resistor can smoke, burn, pop, etc. if you put too much current through it and it dissipates too much power (P=VI and V=IR, so P=I2R=V2/R). Pots tend to not have very high power ratings which can make it easy to destroy them if you aren't careful. Oh right, since it was connected between ground and 5v, there was no "only draws as much current as it needs" thing. BattleMaster posted:How did you hook up the pot? If it only smoked if you turned it too far in one direction, then you hooked it up in a way such that when the resistance from wiper to one terminal got too low, too much current went through it and it dissipated too much power. I put 5v on the wiper, ground on one of the lugs, and the Arduino's analog input on the other lug. It started smoking as soon as I plugged in the Arduino. I didn't do any troubleshooting. I wish I had made note of where the wiper was when this all happened, but I didn't. I had INTENDED to put 5v on one lug, ground on the other, and analog input on the wiper. God knows how it ended up like it did. BattleMaster posted:edit: to avoid this, consider adding fixed resistors so that there's a minimum safe resistance even if the pot is turned to extremes
|
|
#
?
Oct 17, 2018 21:52
|
|
|
BattleMaster posted:Any type of resistor can smoke, burn, pop, etc. if you put too much current through it and it dissipates too much power (P=VI and V=IR, so P=I2R=V2/R). Pots tend to not have very high power ratings which can make it easy to destroy them if you aren't careful. They posted the wiring diagram a page or so ago: scandoslav posted:I was absolutely, 100% certain I had wired it this way:
|
|
#
?
Oct 17, 2018 21:52
|
|
|
My eyes slide off of schematics that try to look like the real thing so I managed to block it out, sorry
|
|
#
?
Oct 17, 2018 21:53
|
|
|
BattleMaster posted:My eyes slide off of schematics that try to look like the real thing so I managed to block it out, sorry Noted for future
|
|
#
?
Oct 17, 2018 21:54
|
|
|
scandoslav posted:Oh right, since it was connected between ground and 5v, there was no "only draws as much current as it needs" thing. Well I mean there was, it's just the amount of current it "needed" was like, multiple amps
|
|
#
?
Oct 17, 2018 21:56
|
|
|
scandoslav posted:Noted for future I mean do whatever you want, don't just change because some rear end in a top hat on the Internet doesn't like those kind of diagrams scandoslav posted:Word. How do I choose resistance for the fixed resistor? In this case you shouldn't need one if it's wired correctly because modern micro inputs shouldn't draw very much current (theoretically 0 current, but in reality there's a bit) and therefore there will be no power dissipation no matter how you turn it. If you want protection against a misconfigured micro input, such as having it set as a logic output which will sink current when outputting logic 0 (killing the pot when you turn it all the way "up") or source current when outputting logic 1 (killing the pot when you turn it all the way "down") then add a resistor of like 1K between the wiper and micro input. Note that not all analog to digital converters enjoy having a high resistance on their input, but if you're using a pot as you are then another 1K shouldn't be too bad for it.
|
|
#
?
Oct 17, 2018 21:59
|
|
|
scandoslav posted:Now I'm curious; why did my potentionmeter smoke? It's just a resistor, right? Was the wiper close enough to a pole to allow smoke-level current? You're supposed to hook up a potentiometer with ground and Vcc on the two outside pins, which act as a 10k (or whatever) resistor no matter how you turn the knob. The pot forms a voltage divider and you read its value by measuring the voltage at the wiper. If you connect ground to an outside pin and Vcc to the wiper, or vice versa, then the resistance in that circuit will vary as you turn the knob. Turn it all the way to one side and you'll see the full 10k; turn it to the other side and it's effectively zero ohms. Your power supply will push as much current as it can through the wiper and the carbon trace, and the inside will burn up. Every semester probably a third of my students destroy small potentiometers in this manner. e: scandoslav posted:I put 5v on the wiper, ground on one of the lugs, and the Arduino's analog input on the other lug. It started smoking as soon as I plugged in the Arduino. I didn't do any troubleshooting. I wish I had made note of where the wiper was when this all happened, but I didn't. it was close to one end of the travel. Sagebrush fucked around with this message at 03:01 on Oct 18, 2018 |
|
#
?
Oct 18, 2018 02:59
|
|
|
And not to harp on it, but honestly, if you learn how to draw a schematic, it makes it really clear exactly how to hook them up and what you did wrong
|
|
#
?
Oct 18, 2018 03:39
|
|
|
Do flatscreen monitors have potentially fatal capacitors in them like the CRTs did? I don’t think so, but I want to make sure.
|
|
#
?
Oct 18, 2018 03:50
|
|
|
Not fatal to you. I mean, some of the power supply caps can still pack a punch, but not outright deadly.
|
|
#
?
Oct 18, 2018 03:51
|
|
|
22 Eargesplitten posted:Do flatscreen monitors have potentially fatal capacitors in them like the CRTs did? I don’t think so, but I want to make sure. Nothing anywhere near as bad as CRT's, but if it has an internal power supply (as in, the wall cord goes right into the monitor instead of into an external power supply brick first) there's probably dangerous voltages on the power supply capacitors. e: In the past when I've taken flat screen monitors apart the power supply was a completely separate circuit board from the rest of it so it was pretty easy to just avoid. They're usually easy to identify too, since it's... the part with a bunch of big capacitors that the power socket is attached to
|
|
#
?
Oct 18, 2018 03:53
|
|
|
I'm starting to see a lot of these three-color e-ink displays showing up on the usual parts sites: I want to pick some up for prototyping because that's pretty dang cool. How do they work, though? I thought e-ink was just black microparticles in white capsules that float up and down. Where does the color come from?
|
|
#
?
Oct 19, 2018 18:14
|
|
|
There are black particles and red particles. Not three colour, unless they've brought out some new ones I haven't seen yet. Waveshare sells a bunch of different kinds, along with driver boards that have open schematics and open driver code so you can dive in if you like.
|
|
#
?
Oct 19, 2018 18:28
|
|
|
white is a color, hth There's a few articles from like 2010-15 describing one method, essentially sub-pixel e-ink blocks and clever colored filters above them.
|
|
#
?
Oct 19, 2018 18:33
|
|
|
The one I've heard is that the red and the black are different sizes. It works normally when both particles are near the screen, one color is higher contrast so you only see it. When the field is changed the smaller particles move away faster, leaving other color there.
|
|
#
?
Oct 19, 2018 19:19
|
|
|
I'm working on what I thought would be a simple project and keep running into stumbling blocks. Basically I'm trying to make a circuit that lights up a bunch of LEDs when triggered by an audio input, specifically a synthesized kickdrum. Like an LED strobe but light tied to an audio pulse instead of a steady beat. Power is coming from a standard 9V/1A wall-wart. The LEDs I scavenged from dollar-store work lights, basically four arrays of 24 little white LEDs each. They're all wired up in series, going only through the resistor built into the first array (very low resistance, I think 6-10ohms which surprised me), not sure that makes sense so I drew a little diagram. I'm using a 5V relay and a 386 audio amplifier IC. I have a 100uF capacitor between +9V and ground. Simplified this schematic: I wired the 386 correctly and the audio ground is attached, etc.  What I have right now works, like it does turn on the lights when I hit the kick drum, it's just kinda weak. The LEDs are way less bright than when I've just hooked them up to the power supply. It was way better when I had it on a breadboard, and for a few minutes after I soldered it. I'm confused because my understanding is that the relay is just activating a physical switch, so when it makes contact shouldn't the LEDs be exactly as bright as if I just had it wired straight up to a 9V supply? I'm thinking maybe the audio amplification is robbing power from the LEDs. Or I somehow damaged the LEDs, maybe they need a higher resistance (I have lots of experience building circuits from schematics I find online, but current and voltage and power calculations are terribly confusing when I have to play with them myself). Or I damaged the relay somehow? By hitting it with a signal that's larger than 5V? Any ideas? Thanks in advance.
|
|
#
?
Oct 19, 2018 20:16
|
|
|
|
| # ? May 28, 2024 05:52 |
|
|
Radio du Cambodge posted:I'm working on what I thought would be a simple project and keep running into stumbling blocks. Basically I'm trying to make a circuit that lights up a bunch of LEDs when triggered by an audio input, specifically a synthesized kickdrum. Like an LED strobe but light tied to an audio pulse instead of a steady beat. What is the current draw of your relay coil? I would guess that alone would be about 250mA, your 9V 1A supply is probably not going to cut it. Also your diagram shows 10 Ohm resistor on the relay, is that additional resistance(and if so, why?), or is that meant to represent the ones built into the LEDs (in which case it would make more sense to draw it on the LED side of the circuit)? peepsalot fucked around with this message at 21:06 on Oct 19, 2018 |
|
#
?
Oct 19, 2018 21:03
|
|
