LM338 Resistor Wattage

dsime42

Member
2010-05-24 3:19 am
Dumb question but please send me link or answer. :-(

I find calculators to find R2 for a given R1 but I don't find anything telling me how to calc the current involved. Probably saw it and just didn't recognize it.

I see that the LM338 has a max adj current of 100uA but also see that current flows from output to ground. (real shocker there)
So is current through R2 100uA > that the current through R1?
Is the current through R1 = Vreg/(R1+R2) ?
 
Your answers are in the datasheet, using Ohm's law and some understanding of how the regulator functions (which is easy enough even I understand it). The LM338 maintains a typical 1.24V potential from its output to the adjust pin. So R1 will have 1.24V across it. Ohm's law then determines the current flow through R1. That same current then must flow through R2. So the R2 value can again use Ohm's law to determine the voltage desired, V=I*R or R=V/I. So far this has ignored that current flow out of the adjust terminal, 45μA typ and 100μA max. This "represents an error term" in the device, and normally the current through R1 is much greater than that and so it can be ignored. It is included in the standard formula,
Vout = Vref (1 + R2/R1) + Iadj*R2.
The max 100μA will have even smaller consequence in determining the R2 power rating. Calculate the current through R1, add 100μA, and use that sum to multiply by the voltage across R2 (which is Vout - 1.24V). That gives the wattage dissipated in R2.
 
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