Step 4: Connecting Front-panel Items

Here, I am connecting the appropriate wires to the binding posts, power switch, fuse holder, and LED's.

In an ATX power supply, there should be a wire that is used to turn on the power supply. You can see this wire (It's green) in the second picture; it is the green wire in the middle, where it says "ON/OFF" on the PCB. I connected this to the switch, and the other pole of the switch went to ground. The +5, +12, and -12 are connected right to their wires on the PCB. The ground wire is connected through the fuse holder before the binding post.

Initially, I was going to use green LED's, but I realized I had many more red LED's than green LED's, so I switched them over to reds. In the first picture, you can see the holders I installed into the front. I connected the LED's through a common resistor to ground. The LED on the left (from the front view) is a standby LED. It is lit whenever I have the power supply plugged into the wall. It is connected to the +5V standby wire on the PCB. In my PS, it's purple. The other LED is the "Power On" LED, and it is lit when I have the power supply turned on. It's connected to the "Power OK" signal wire, which goes to +5V when the power supply detects that it has stabilized the voltages. In my PS, it's the gray wire.
<p>if i want to put more binding post do add more specific current for my 5volts and 12 volts (like 1A, 2A, 5A) what should i do? thanks..</p>
Hello, good question! If you want to limit a specific binding post to only output a certain amount of current (1A, 2A, 5A, etc) then you need to use a fuse that will blow when the current limit is exceeded. Be aware that fuses are not very precise in terms of current limits, and there is usually a range where the fuse may or may not blow.<br><br>Hope that helps, let me know if you have further questions, or if I didn't understand what you were asking.<br><br>-Matthew
<p>Matthew, have you been on ebay recently? CC/CV boards cost 5 bucks for a stated 5 amp boost or buck...reality is 2-2.5amps unless you put some metal fins or odd chunk laying around on them. .One can get a Poly fuse multi value kit for dirt cheap as well. </p><p>need more amps? If they have roughly double the budget have them search ebay for ltc3780. 10-12 bucks and a month wait to get one from china using an IC from Linear Tech in Cali. </p><p>or you can get a board from an american based ebay shop in a week for about 23 bucks.</p><p>worth noting:</p><p>its auto regulating boost buck...the chip itself can do just about any form of power converter by re configuring (sepic, cuk etc ) </p><p>I know all this because im prototyping a product and also keep a bunch of those cheap no name Dc-DC converters on hand for what ever comes up </p><p>I got my first 2 3780s for free but I didnt pay attention to the fact that the package was designed in imperial and ebay sop to dips apear to only be made in the 2 common metric pitches. I found one company that wanted like 12 bucks for the damn break out... so i paid the 23 to get the damn premade board asap and I will have 2 chips once I etch the second version at home. </p>
<p>hi Matthew<br><br>currently my ATX power supply have 12v with alot of Amps power...<br>i wonder it can charge my deep cycle battery 12v 9A or parallel connection with 12v 54Ah. <br><br>can you show us a diagrams that 12v have 1A , 2A, 5A , 10A ?<br><br>-Kokoro San</p>
<p>Hello Kokoro San. Be aware that many ATX power supplies have 12v split into 2 or more sub-supplies. Look at the sticker on the side of the ATX power supply for details. I would suggest using a battery charger to manage the charge process.</p><p>I am not sure what you're asking about for diagrams?</p>
Thx for reply Matthew Currently.. my ATX power supply on yellow color wire output is 12v 20A I wonder it able to make it 13v 3A to charge deep cycle batterie
<p>Generally it's pretty difficult to do a good job making a higher voltage from a lower voltage, especially for a high-current application like battery charging. You're probably better off looking on ebay for a cheap 14v power supply.</p>
How do you determine which load resistor to use? I'm new to all this. Learning slowly but surely. I couldn't find any 10w 10ohm resistors so I bought a 8ohm 20w &quot;non-inductive&quot; resistor and a 10w 50ohm wire wound resistor. Thanks!
Hello, good question! I hope you're learning a lot and having fun! It's a bit of an experimental process to determine the size and placement of a load resistor (or resistors) for this kind of project. Some power supplies have a shutdown system where they will turn off if nothing is drawing current, and the load resistor is used to ensure that at least a little current is always being drawn, to keep the power supply from turning itself off. If you'll always have a load connected to the power supply, like if you're using it to charge RC vehicle batteries or something like that, then you probably don't need a load resistor.<br><br>From the comments on this Instructable over the past ten years, we've learned that some power supplies don't need any load resistor, some only need one on the 5v line, some only need one on the 3.3v line, and some need loads resistors on both. You'll have to experiment with it. I'd suggest trying it without any load resistor first, to see if it just works without. Then try adding it to the 5v line, then to the 3.3v line, then to the 12v line, then try adding one to both 5v and 3.3v, etc.<br><br>To calculate how much current each resistor will draw, use Ohm's Law:<br>Voltage = Current * Resistance<br><br>If you put that 8 ohm resistor on the 5v line, we can re-arrange Ohm's Law to calculate the current:<br>Current = Voltage / Resistance = 5 volts / 8 ohms = 0.625 amps of current (also called 625 milliamps).<br><br>If you put that 50 ohm resistor on the 5v line:<br>Current = Voltage / Resistance = 5 volts / 50 ohms = 0.1 amps of current = 100 milliamps of current<br><br>Once you know the voltage across the resistor, and the current through the resistor, you can calculate the power that is burned in the resistor:<br>Power = Current * Voltage<br>For 8 ohms at 5 volts, Power = 0.625 * 5 = 3.125 watts of power<br>For 50 ohms at 5 volts, Power = 0.1 * 5 = 0.5 watts of power<br><br>Since the power burned is less than the power rating for each of those resistors, you don't have to worry about melting them or having them overheat, at least on the 5v line. Repeat the calculations above to determine the resistor current and power burned for the 3.3v line and the 12v line.<br><br>I hope this helps, and good luck with your project!
That is awesome! Thanks so much for the detailed answer! That helps a lot. There is definitely a steep learning curve. I'm building a cnc/laser engraver for the final goal. Trying to get some practice under my belt and learn the basics. Do you have any suggested literature? I'll try it out when I get home. Again, thanks!
<p>I followed your instructions and I have the following voltages available to me:</p><p>12v 18A (for each of two rails)</p><p>8.7v 17A (12 to 3.3)</p><p>7v 18A (12 to 5)</p><p>5v 22A</p><p>3.3V 17A</p><p>1.7v 17A (5 to 3.3)</p><p>12v 1A (from ground to -12v)</p><p>15.3 1A (3.3 to -12)</p><p>17v 1A (5 to -12)</p><p>24v 1A (12 to -12)</p><p>My quest: </p><p>I want to use the 3.3v to turn on the transistor allowing the 24v from my ATX power supply to run a very small motor.</p><p>My problem:</p><p>3v3 is using common ground while the 12v is using -12v for ground</p><p>I think that the 3v3 will now be using the -12v making it 15.3v not 3v3. I don't know if what I'm trying to do is possible.</p><p>My question:</p><p>Can I do what I'm trying to do without getting a lot more complicated?</p>
<p>Hello. I don't think it really is a good idea to make all those intermediate voltages by connecting different power rails. Each power rail is only designed to source current, providing it to a load, which returns the current to the common ground rail. I'm not sure it will work very well to draw current from one rail, only to return the current to a different power rail, which is expecting only to source current, not sink current provided to it.</p><p>Most power rails are designed with a feedback circuit that detects when the output voltage droops too low and provides more current in that case. They don't really have any control system in place to handle the &quot;voltage too high&quot; situation, which is what you'd be doing by making it sink current from somewhere else. Imagine you have a balloon filled with regular air, so it will slowly sink but you can tap it upwards to make it higher. Your job is to keep the balloon at about 5 feet above the ground by detecting when it gets too low, and then giving it a little tap. This works great even if you start adding more weight to the balloon (equivalent drawing more current), but if you start adding an upward force (fan on the floor pointing up?) then your simple control algorithm doesn't have any way to bump it lower. I hope that strained analogy helps explain the situation.</p><p>If you want different voltages than the standard voltages provided, I would strongly suggest buying or building an actual voltage regulator. It could be as simple as a LM317 regulator with a couple of resistors to set the desired current. You can even use a potentiometer (variable resistor) to make an adjustable voltage supply. The first circuit here is a good one <a href="http://www.circuitstoday.com/few-lm317-voltage-regulator-circuits" rel="nofollow">http://www.circuitstoday.com/few-lm317-voltage-reg...</a> but you can also look at the LM317 datasheet for other circuit ideas. Here's a cheap ($8) adjustable voltage regulator with voltage readout <a href="http://www.ebay.com/itm/LM317-AC-DC-Adjustable-Voltage-Regulator-Step-down-Power-Supply-Module-W-LED-TA-/111791252026?hash=item1a0747223a:g:LzcAAOSwWnFWClIJ" rel="nofollow">http://www.ebay.com/itm/LM317-AC-DC-Adjustable-Vol...</a></p><p>It makes it a lot simpler too if all your circuits have a common ground voltage.</p><p>Regarding switching 24v to a motor, don't forget that there are two ways to switch a circuit like that: high-side and low-side switching. Low-side switching is where your load (motor) is connected to the power supply voltage, through your switching device (a transistor, switch, or relay), and then to ground - The switch is below (on the low side of) the load (motor). A high-side switch is just the opposite. I attached a circuit schematic showing the difference.</p><p>Transistors are very useful for many things, but let's talk about using them as a switch. Transistors have three connections: one connection that controls the switch, and two connections that either prevent or allow current to flow between them. There are two general types of transistors, and two general technologies of transistor. The two technologies are Bipolar Junction Transistors (BJTs) and Field Effect Transistors (FETs or MOSFETs), but for simple switching applications their difference isn't really that important. The two types of transistor are N-type and P-type. N-type transistors only turn on (conduct current) when the control input is a few volts higher than the bottom connection. P-type transistors only turn on (conduct current) when the control input is a few volts lower than the upper connection.</p><p>If you are using a low-side transistor then you almost always want to use an N-type transistor, where the bottom connection is grounded and the control input just needs to be higher than a couple volts. If you are using a high-side transistor than you usually use a P-type transistor, where the top connector is connected to the power supply voltage, and the control input must be at least a few volts lower than that to turn it on.</p><p>This can be used to allow a low-voltage microcontroller to turn on a higher voltage motor or light, but using a N-type transistor. While the microcontroller can only output maybe 3.3 or 5 volts as the control signal, this will be enough to turn on a N-type transistor connected as a low-side switch, which can then allow current to flow from, say, +12v through your motor, through the transistor, to ground. You may be able to do something similar with your 24v motor even if it doesn't have a true ground.</p><p>Good luck, and I hope this reply was helpful.</p>
<p>Great idea changing the transistor type I think it will work. </p><p>Isn't the voltage for the regulators only referenced between their input and ground, not their output? The output voltage to ground would be regulated at the appropriate level but that then would be sourced to a lower voltage for the output ground causing the difference to only effect the load between the two? </p><p>I am super new to this stuff and I find that I have some misconceptions. I asked a guy I know if that would be a problem and he said that I was correct to think it would work so long as I didn't exceed the amperage rating of the lowest rated circuit. I think I trust your reservations and experience enough to question my conceptions again.</p><p>I am not doing this for any practical purpose. I am mostly just learning how all of this works. If the power supply dies I wouldn't be too upset (though finding a 1AMP -12v on a power supply is exceedingly rare so I doubt I'd ever get a 2nd chance after I blow up this one).</p>
<p>I almost done. Amph I have a few question. The ATX PSU I used have 5v with 20amp and 3v with 20amp where should I connect the 10wtt10k sand bar power resistor, in 5v or 3v? and about the fuse, is that necessary? or if yes, what is the specifcation?</p><p>Lastly, is it possible to lower the amper or 3v, from 20amp to 1amp?</p>
<p>The ampere draw depends on what you are running, to lower the ampere draw you will need components that don't draw as many amperes. If you would like to limit amperes that the power supply will allow before breaking the circuit then try a circuit breaker. You can get mini circuit breakers for about $18 from a well known online industrial automation retailer rated for as little as 0.5A.</p>
Hello, good questions!<br><br>1. Placement of the power resistor is somewhat of an experimental process. Some power supplies don't need one at all, some need it on 5v, some need it on 3.3v. You will probably need to experiment to see where it is needed. If you turn on the added switch and the fan doesn't turn, or the fan turns off after a few turns, then you likely need to add a power resistor somewhere. I'd try 5v first, and then 3.3v, and if it still doesn't stay turned-on, maybe try both 5v and 3.3v? Good luck, and let us know what you find!<br><br>2. The fuse is there to protect your circuit from a mistake, not to protect the power supply. I would expect most power supplies to have built-in over-current protection, either in the voltage regulators or even in a fuse inside the socket where you plug it into the wall outlet. If you make a mistake in your circuit, such as an accidental short-circuit, this kind of big power supply will be able to provide amps and amps of current through that little wire, likely melting or causing other damage to your circuit. If you know your circuit will only need a few hundred milliamps (less than one amp in total) then adding a 1 amp fuse to the +5 or +3.3 or +12 line (which ever you are using) will help protect your circuit by having the fuse blow instead of dumping tons of current through your circuit. I would estimate the largest expected current use, then use a fuse with double the current.<br><br>3. This is a very common (and very good) question, and the short answer is that you don't have to worry about the PSU having too many amps of current.<br><br>This kind of power supply is more accurately described as a &quot;voltage supply&quot;. It tries to supply a fixed voltage (like +5v or +12v) no matter the amount of current drawn. If you draw too much current (more than the rating) then the power supply will be unable to keep the voltage at the right level and the voltage supplied will droop down. If you draw less current than the rating, then everything should just work, there's no worry about it providing too much current.<br><br>By way of analogy, the municipal water supply in your town probably supplies water at, say, 100 psi. A small bathroom sink will only draw a little bit of flow, while a garden hose will draw a lot of flow. Both are tapping into that 100 psi supply, but only draws as much flow as it needs. If you turned on all the faucets and hoses and taps in your house, you might draw so much flow that you exceed the flow capacity of your water supply, and so the provided pressure would drop a bit. I hope that makes sense.
<p>Hello Matherbeckler,<br>Thank your for posting it. I like it a lot when you not only show the step, but also explain it. <br>One beginner question though, Can you add this kind switch to this project, I do see you add resisters to power up LED for indication. If possible, can you teach me how ? Thank you in advance. I just happen to have one and want to implement it into this project. </p><p>http://www.thebookyard.com/product.php?products_id=4498</p>
<p>I have connect that button with motor test to see if it &quot;works&quot;. connected as pic. however, in order to have motor running continuously. I have to hold the button firm. <br><br>On the other hand, I found another switch button. (see pic) connecting the motor as same as the Mac mini one. it works great with the motor. <br><br>My question is: Can can use this tiny switch button to connect the Power On(green) and Ground(black) ? Do we need to as some resisters?<br>Does anyone know how to implement the Mac mini one ? Thank you so much!</p>
The Power On (green) and Ground (black) wires need to be continuously connected to keep the power supply turned on. Most computers (even a mac mini) have a &quot;momentary pushbutton&quot; for the power button. The computer motherboard detects that momentary button press and then connects the green and black wires to turn on the power supply. You'll need to do the same, by holding the green and black wires connected for as long as you want the power supply turned on. I don't think the mac mini button will work very well for your power supply, unless you add some extra electronics. You could possibly use some sort of latch to turn the momentary pushbutton into a push-on/push-off sort of button, you know? https://en.wikipedia.org/wiki/Flip-flop_(electronics)<br><br>It is not possibly to know how a switch works based on a photograph. You need to experiment to determine how the connections work. Are there only two connection points for that switch? Do you push the blue part directly in or does it slide or rotate in two directions? You don't need any resistors for the power supply on connection.
<p>I have to turn it on and of 4 times and quikly to make it run, but hey, it is working!</p>
<p>nice dmm, u an eev blog fan? if you havent heard of it check it out. I ask because your choice in meter is one of his recomended ones</p>
<p>Just an idea, but if you need a load to draw a small current on each voltage rail, could you not use an LED and an appropriate resistor to drop the voltage instead of these power resistors? It would act as a status light to prove each voltage rail is powered up. </p><p>Or would it not work due to the current being too low?</p>
A &quot;power good&quot; status LED for each voltage rail is an excellent idea in and of itself, and they might even draw enough current. The whole &quot;small minimum load&quot; stuff is really black magic, nobody really knows anything except &quot;this worked for my power supply&quot;, and really takes a bit of experimentation with your specific power supply in your specific use case. Some people use their PSU as a battery charger, so there is always a large load and no power resistor is needed. Most LEDs only draw a max of a few dozen milliamps, so that might not be be enough current draw to keep the mystery &quot;minimum load&quot; circuits happy :-) Good luck!
<p>I would bet one could take the &quot;black magic&quot; out of it if they read the markings on the DC-DC converters and used google to find the data sheets. </p>
<p>Thanks for the instructable it helped me a lot with my PSU. One thing you forgot to mention, or it didn't apply to you, was that you have to connect the 3.3V brown sense wire to the 3.3V rail or else the unit wont turn on. Idk if this apply's to everyone but my unit would only turn on for a second then turn off when it wasn't connected. Also added a LM2596 buck converter in my unit so I can have a adjustable voltage output. I also used 5x 15Ohm 1 Watt resistors in parrell to get a comparable 5W resistor then I used another bundle of those 15Ohm 1Watt resistors and added it in series to have a total of around 10W at about 6Ohms. </p>
Nice work, I like the blue light and the 7-segment displays. My PSU was so old it didn't have a 3.3v sense wire, so you make a good point that newer PSU will need that.
Lol my power supply is anything but new, I think. When I bought it the fan grill was completely covered in dust and I mean like a thick layer of it. But yeah I guess it varys with PSU. Also those aren't 7 segment displays its just a cheap 3 digit voltage meter, you can buy them in ebay.
<p>have you ever attempted to count the little lines that make up the digits in your &quot;cheap 3 digit voltage meter&quot;... u might find that there are 7, matt wasnt wrong calling it a segment display. Next time try to understand some people know more than you...trust me when I say it takes a lot of effort to find something to like about your build but he did and yet, you give additude in return</p>
have soldered a 25ohm 10watt resistor on my 12 volt rail since it has the most power.<br><br>but the resistor is getting too hot that i can't touch for long... is that normal? do i need to use a heatsink? or something else? please help..
Hello, good question!<br><br>Using Voltage = Current * Resistance, we know that 12 volts across a 25 ohm resistor results in 12/25=0.48 amps of current flowing through the resistor. The other important formula to use is Power = Voltage * Current, so 12 volts * 0.48 amps = 5.76 watts of power, that becomes heat in the resistor. Since the resistor is rated for 10 watts, you're not going to melt the resistor but it may get warm over time.<br><br>I wouldn't expect it to get too hot to touch, so perhaps the resistor has less resistance, and is therefore drawing more current? If I remember correctly, the wirewound resistors that can handle many watts usually have a large variance in the amount of resistance it actually has, compared to the &quot;official&quot; resistance. If you have a multimeter you can measure the resistance of the resistor by itself, or you can put the meter in current measuring mode, and insert it in series with the resistor and measure the actual current. Let me know if you need more info about how to use a multimeter for this kind of investigation.<br><br>If you could find a 100 ohm power resistor, that might work better. It would result in 0.12 amps of current instead of 0.48 amps. I really don't know how much current is needed for the minimum load, but I would guess that 0.1 amps would be enough. This would reduce the power burned as heat in the resistor down to 1.44 watts, so you could use a 5 watt power resistor instead of 10 watt.
<p>Thanks alot, here's my version. Converted from an old 250W Hipro unit from my mother's ancient Pentium 4 machine that just finally gave up last month.</p>
<p>I'm having a problem with one. I have the brown connected to the orange as it was in the plug, and the resistor on the red and grnd as it should be. But when I try to fire it up it will only spin the fan for a second and then quit. It won't stay running. Any suggestions? Much appreciated.</p>
What do you mean by &quot;the brown connected to the orange&quot;?<br><br>Perhaps try moving the power resistor to a different voltage rail, such as 12v or 3.3v? Maybe your particular power supply needs the minimum load on a different voltage line? Or it needs a minimum load on two or more of the voltage lines? I've never seen any specifications for this sort of thing, so you might need to experiment with your power supply. Let us know if you figure out what's going on.
<p>There was a brown wire piggybacked onto an orange in the 20 pin plug so I after removing the plug I tied the brown wire to an orange. Also, I've tried the resistor on the 3.3v, the 12v, and the 5 volt. Also tried two resistors. One on 3.3v and one on the 5v at the same time. I've set this one aside for now pending a solution. If I can figure it out I'll post my results.</p><p>Thanx!</p>
<p>TheLonesometoad, I was having the very same problem you were, but I was able to resolve it using two 10&Omega; resistors in parallel, on the 5v line. This gives me an equivalent resistance of 5&Omega;, and it seems that is enough for my power supply. I got the idea for this from matthewbeckler's reply to harshesh below. Hope this helps!</p>
<p>Why the extra fuses? I am sure I read that the reason converting psu's into desktop lab psu's is because of the built in protection they already have. I Am trying to convert a modern 535 watt into a desktop psu. I don't know how many wires I should join together being that my psu is much larger then all the examples I can find. I think if I joined them all together it would be far too much power possibly far to much wattage but my memory on the subject is very cloudly. Can anyone tell me how many 12 v wires I can safely join together?</p>
<p>Also, at least for me, the reason people convert PSUs into a bench PSU is because there are many old and cheap (usually free) PSUs that aren't powerful enough for a new computer, but still work perfectly fine and provide more than enough power for hobbyist needs.</p>
Hello, sorry if this results in double-posting, but my first reply seems to have disappeared. The extra fuses are added to protect your load from over-current situations, and are not intended to protect the power supply itself (the built-in protection is to protect the power supply, not to protect your circuit).<br><br>A power supply like this is known as a &quot;voltage supply&quot;. It creates the rated voltage (5v or 12v for example) and tries to provide as much current as needed by the load circuits. It can provide up to the rated load current (listed on the label on the side of the PSU). There's no danger of providing &quot;too much&quot; current, as the load will only draw as much current as it needs. If you join too-few wires, then you draw a lot of current, the individual wires might start to heat up, so I would connect as many wires as you can. It's sort of like how in your city's water supply, they provide water at a certain pressure, no matter if you use a little bit of water, or use a ton of water.<br><br>One thing that complicates the answer is that modern power supplies seem to have split their 12v rails into two or more &quot;domains&quot;. Presumably this is because it's easier and cheaper to include several smaller 12v regulators than to include one giant 12v regulator. For example, take a look at the attached image, showing this power supply has three 12v regulators each capable of providing 18 amps of current. I'm not sure if the PSU would like having all three 12v domains connected together (they might fight each other). I'm not sure how to tell which domain a certain yellow wire belongs to, but you might be able to look at the circuit board in the PSU, and maybe the yellow wires are grouped together into two or three groups? Of so, I would feel somewhat confident in combining <br><br>Hope that helps, let me know if you still have questions.
<p>Is a 10 Ohm 10W Resistor really needed ? </p><p>And what fuse have u used and why ?</p>
Some power supplies won't run if there is nothing connected, and need a minimum small load to keep it from doing an auto-shutdown. It really depends on your specific power supply, so you must do some experiments to determine what your power supply needs. If you will always have some load connected, then you don't need the power resistor.<br><br>This original design used a single fuse on the ground wire, but that is a really bad idea. The better way is to put no fuse on the ground wire, and put a fuse on each non-ground wire (like 5v and 12v). I would use a fuse that can handle just a bit more current than you expect to use in your project. This way if something goes wrong and your project starts to draw much more current, the fuse will blow and prevent a fire or other dangerous situation.
<p>maybe a stupid question, but wouldn't it be sufficient to use the cooler fan for the required minimum load? </p><p>Nice project though!</p>
Thanks gungajin, that's a good suggestion. There are a million different power supplies out there, and many require a minimum load separate from the case fan, and many don't require an additional load. Your mileage may vary, and these things require a little experimentation to see what it needs. Good luck!
<p>this guide has been really helpful to me as I work on my own PSU. I found this handy wire color and purpose chart, hope it helps. Heres the link to it https://www.instructables.com/id/A-Makers-Guide-to-ATX-Power-Supplies/?ALLSTEPS</p><p></p>
<p>I love this project. I have everything connected but I can only get it running if I turn on and off the switch quickly many times. It is the only way I can get it running. Any Idea why this happens?</p>
<p>Hook the load up and then turn it on. Mine stays on just fine with a load on it, but shuts off after a few minutes if I remove it. </p>
<p>Great idea. I'm building mine now. Going to one thing a bit different though since I also work on computers. I'm leaving the computer wires on the supply intact so I can use it on a motherboard or use the terminals installed on it for running my ham radios, testing things, etc. Be even more useful that way. The one I'm working with puts out 12 volts at about 17 amps. That such power an Icom IC 735 ham rig just great!</p>
<p>Thank u for this =)</p><p>One question...</p><p>Can I add a regulator on the 12v? and connect a voltmeter? So I can change between 1-12v? Please let me know :) And maybe give me a link to a regulator =)</p>
You can definitely use a regulator to convert the 12v down to a lower voltage.<br><br>The LM317 is a popular adjustable voltage regulator, but the max output voltage is about 2 volts less than the input voltage, so you'd be limited to 0-10 volts. There maybe other types of regulators that have a smaller &quot;dropout&quot; voltage. This is a &quot;linear&quot; regulator, meaning it burns excess power as heat, equal to the current * (Vin - Vout), which can waste a lot of power. Maybe you do or do not care about that, but you might need a heatsink.<br><br>There are also switch-mode buck converters that don't waste as much power, since they are more efficient.<br><br>Good luck!
<p>I connected PSU's green wire with black (ground) wire, but not with the gray wire. Should that make any problems or is it OK?</p>
The gray wire is only used to indicate if the power supply is turned on, so you don't need to do anything to it if you don't want to have an LED indicate when the power supply is turned on.

About This Instructable




Bio: If you need to get in touch, please email me instead of sending an instructables message. matthew dot beckler at gmail dot com
More by matthewbeckler:Laser cut desk name plate Convert old electric toothbrush to plug-in Ye Olde Chain Maille Rings 
Add instructable to: