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. board By the way I didn't make the mess with all the solder, this is the way it looked when I cracked the case.
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
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)