Electronic – Resistor failed open

First off, I'm going to presume that the TV is out of warranty. If it is still covered, your best bet is to call the warranty line and get it fixed that way.

It's hard to tell from your popped resistor photo, but is there a heat sink to the left of it in the image that is black? There might be something under there that blew out too.

As for DIYing the repair, you could try swapping out that through hole resistor that blew. However, while it's common for low quality capacitor to give out, resistors don't usually pop unless there is an over current situation, so there might be a larger issue at hand here. It also looks like there are 0 ohm and a 47K (? hard to read) ohm surface mount resistors in the damaged area on the back of the PCB that might need replacing.

If it's currently broken, there's no harm in attempting a repair. Find a friend who does electronics to help you source the parts, especially if you think you'll need to replace those surface mount resistors.

As for getting a replacement board, rather than having it replicated from scratch, you should be able to get a replacement board. The various boards in TVs are usually modular (power, tuner, etc), and that appears to be the power supply board. Try searching for "TVMAKE TVMODEL Replacement Power Supply Board" and see what comes up. I know the Vizio 50" plasmas have a nasty habit of blowing power supplies and thus they're pretty easy to find replacements for online.

Good luck and stay away from your boyfriend's books lest they burst into flames. ;)

If you draw 1mA from the resistor divider circuit you mentioned, it will output one volt (the upper resistor will have 1.1mA flowing through it, thus dropping 11 volts; of that 1.1mA, 0.1mA will go through the bottom resistor while the remaining 1mA will go into your load). The 6K resistor would drop 6 volts, thus feeding 6 volts into a 100mA load.

If either the load current or the load resistance is a known constant value, one can calculate a series resistance which will convert a known input voltage into any desired known, lower, load voltage. If the load current or resistance isn't known precisely, however, deviations from the ideal will cause the load voltage to vary from what is intended. The greater the difference between the input voltage and the load voltage, the greater the variation in load voltage.

Adding a load resistor will effectively add a known fixed load in addition to the potentially-variable one. Suppose one had a 12-volt source and the intended load were 10uA +/- 5uA at 6 volts. If one just used a series resistor sized for the 10uA case (600K), it would drop only 3V at 5uA (feeding 9 volts to the load) and 9V at 15uA (feeding 3 volts to the load). Adding a 6.06K resistor in parallel with the load would cause the total current draw to be about 1.000mA+/-0.005mA, requiring the upper resistor be changed to 6K; since changes in the load current would only affect the total current by about 0.5%, they would only affect the voltage drop of the upper resistor by about 0.5%.

If the source voltage is stable, and the output current is small, a voltage divider may be a practical means of generating a stable voltage. Unfortunately, for the voltage divider to generate a stable voltage, the amount of current fed through the lower resistor (and thus wasted) must be large relative to the possible absolute variation in load current. This is usually no problem when the output current is on the order of picoamps, is sometimes acceptable when the output current is on the order of microamps, and generally becomes unacceptable when the output current is on the order of amps.