Resistor watt requirement calculation?

[bitman]

I've got a few problems getting a 1085-5.0 (http://www.ti.com/lit/ds/symlink/lm1085.pdf) working at a constant 5 voltage.  I'm using a 110 and 320 (Approx values) which should give me around 5.1 volts if my math is right. I'm not seeing that (16v actually - only feeding the curcuit 12 volts so well odd). I'll figure that side out, but I'm not sure how to figure out how the wattage required for the resistors between ADJ/OUT as laid out in the datasheet. I have small basic 0.5watt resistors right now and I think that may be on the low side since I'm pushing about 1.5 amps through the voltage regulator.

I get how I can calculate the wattage on the output - but how do I calculate between the output and the adjustment pin? Or between adjustment and ground?

[Andy Watson]

I've got a few problems getting a 1085-5.0

The datasheet indicates that the -5.0 version is a fixed voltage regulator - it will be trying to develop 5V between output and gnd.

For the wattage of the resistors, use \$\frac{V^2}{R}\$ where V is the voltage across the resistor.

[Audioguru]

Maybe you do not understand that the regulator you have has a 5V output when it does not use any resistors, connect the ground/ADJ pin to ground (0V). The resistors increase the output voltage to the 16V that you get.

[bitman]

I've got a few problems getting a 1085-5.0

The datasheet indicates that the -5.0 version is a fixed voltage regulator - it will be trying to develop 5V between output and gnd.

For the wattage of the resistors, use \$\frac{V^2}{R}\$ where V is the voltage across the resistor.

Put in that way I have to say *duh*! 
Not sure why I didn't realize that myself. I was wondering why I needed resistors ... but it never clicked.  Now, I have a heating issue but that I expected. Thanks - in a simple answer you fixed 2 days of hitting my head on the top of the table not understanding what was going on.

[jaycee]

As you've now realised, you dont need resistors for the fixed 5V version

However lets suppose you were using the Adjustable version, and you still wanted to know about resistor wattage. The important thing to realise is that the load current doesnt flow through R1 and R2, they are simply feedback resistors.. so it's irrelevant how much current you are drawing from the regulator.

The regulator works by trying to maintain the 1.25V reference voltage between the Out and Adjust pins. For good regulation, you want a certain amount of current between Out and Adjust.. with the example 120 ohm resistor shown in the datasheet for R1, 10mA will flow into the reference. Using P=I^2R you can see that this is 12mW dissipated in R1. As R2 is typically a much larger value, you can see that the dissipation in it will be even smaller still. So your 0.5W resistors are more than capable

The reason adding resistors has an effect on the fixed 5V version is because you are effectively shunting the internal resistor network, and changing the voltage. If you look at Figure 7.2 in the TI datasheet you can see why that is.