I am playing with a power supply that went bad, to expand my knowledge and just see if I can repair the circuit level board. Here is my problem. I have a resistor that burned up. I have no idea of it's value as it is a char-coaled crispy critter and it is open so I cannot measure any resistance. I have already replaced the following parts 2 caps-1000MFD 16v, the fuse, a 22pf 1kv ceramic cap, 1 pnp S9015 trans, and 1 npn 9014 trans. Output voltage is 12v-1.5A and 5v-1.5A. I have tried to use the forumla I learned along time ago to calculate the resistance required but it doesn't seem right. V/A=R. I have even looked it up on line and they tell me that if I have a 12-volt circuit requiring 0.015 amperes of current, I will need a resistor with 333 ohms (12V / 0.015A = 0.333). That doesn't seem correct so I did the math and got a completely different answer. Can you help me calculate what I need and show me where I am screwing up please? I can send you pics of the board and where I have replaced the parts so far. The resistor is physically located between the 9014 trans. and the ceramic cap. boardboard2board3board4 Thanks for the tip on uploading the pics. By the way I didn't make the mess with all the solder, this is the way it looked when I cracked the case.

  • \$\begingroup\$ Ohm's Law (V = IR) only applies when you want to know the amps through the resistor in question or the volts across it. In your case, if the 1.5 A provided by the power supply were going through a resistor that's part of the power supply, there wouldn't be any current left for whatever you want to power with the supply. So Ohm's law isn't what's relevant here. \$\endgroup\$ – The Photon Jan 14 '12 at 19:59
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    \$\begingroup\$ To actually understand what resistor you need, we'd probably need a full schematic of the power supply circuit --- which you probably don't have. If you can at least give us a photo of the circuit, though, there is at least a slight chance we can help. \$\endgroup\$ – The Photon Jan 14 '12 at 20:00
  • \$\begingroup\$ I'll also point out that 12 / 0.015 is 800, not 0.333. Also, where did the number 0.015 come from. First you were talking about 1.5 Amps, then all of a sudden 15 milliamps. \$\endgroup\$ – The Photon Jan 14 '12 at 20:02
  • \$\begingroup\$ @ThePhoton - near as I can tell, the 0.333 came from dividing 5V by 15A; where he gets 15A, or 0.015A from, I don't know. The decimal points are all over the place here. \$\endgroup\$ – JustJeff Jan 14 '12 at 20:59
  • \$\begingroup\$ 1000MFD? One thousand Megafarads?! That's a Gigafarad! That must be about the size of one city block! \$\endgroup\$ – Majenko Jan 14 '12 at 22:40

If you are dividing a supply's rated voltage by it's rated current, you're off track. First, understand that the current rating is just a maximum that should be drawn from the supply. Think about it; with no load connected, it will actually supply no current, right?

Chances are nearly 100% that the resistor you are looking at is part of a regulation circuit, and any similarity of its ideal value, and rated V over rated I would be coincidental. Maybe if you could trace out the schematic and post it, someone here could offer good estimation of what the value should actually be

  • \$\begingroup\$ Thanks for your help folks and here is the explanation for the number variances. I goggled calculating a resistor value and found a webpage that showed the example I posted above 12/.015=.333. I knew this wasn't correct but I just wanted to run it by you guys before I emailed the guy that wrote the article. Now I do have pictures of the board and will gladly email them to you or post them if someone will tell me how. I liked to have never got it to allow me to post my question to begin with since I am a newbie to here. \$\endgroup\$ – Ralph Waldrop Jan 14 '12 at 21:48
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    \$\begingroup\$ @RalphWaldrop You can post an image by editing your question and clicking on the add a picture button. Or edit your question and hit Ctrl+G \$\endgroup\$ – Kellenjb Jan 14 '12 at 22:18

This is a switched mode PSU.

This means it takes the mains, converts it to DC then chops it up at high frequency and feeds it into a transformer, then the supply is smoothed and fed out to your load.

If used on 240V supply the voltage across the input capacitor will be around 350V - THIS IS VERY DANGEROUS - PLEASE TREAT THE BOARD WITH CAUTION... (even on 120V it can be LETHAL)

The resistor is part of the control circuit for the input and as such it's value is unknown, it seems to be part of feedback from the main switching transistor - it will be used to limit the transistor to prevent it overloading...

I have no Idea of its value, but it is an important part of the circuit and without it the main switching transistor will blow...

As far as I can see this resistor (R6) is part of the over current feedback in the control circuit (this is added to the over voltage feedback from the opto isolator) this is used to make sure the output voltage does not rise too high and if the output is overloaded the switching transistor won't blow.

Without an other PSU to compare the value to, or the value from a circuit diagram, I don't know of any way to work out its value - Sorry

Your attempt to work out its value as 12V/1.5A = 8R - is meaningless for this part in this circuit. (If you wanted to load a 12V Supply so that you took 1.5A, you would need a 8R resistor - but it would need to be rated at 18W)

  • \$\begingroup\$ Wow that is a mouthful Wonko and I appreciate it. Seeing as how this was the power supply for my external SATA Dock I guess it is useless to me unless I can come up with another power supply for it. I have been thinking about trying to take an extra one from an old computer and modifying the leads so it would power it. Got any suggestions anyone? Doesn't seem like it would be that hard to do. \$\endgroup\$ – Ralph Waldrop Jan 15 '12 at 0:11
  • \$\begingroup\$ The two power lines on that brick seem to be 12V and 5v rated at 1.5A. This should be easy to source and connect to your SATA dock. \$\endgroup\$ – R.Joshi Jul 19 '17 at 10:37

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