R8 || R7 || R5 || (RV1 + RV2) (where
|| means "in parallel" and
+ means "in series") is equivalent to
Rp || Rv where
R8 || R7 || R5 = 2.73kΩ and
Rv is a single pot from
Rp || Rv is equivalent to a pot from
2.73kΩ || 11kΩ = 2.19kΩ. So a single
2kΩ pot will achieve much the same thing.
Practically all those extra resistors might be there for a good reason. Power dissipation seems unlikely because they won't share much current. Maybe the reason is layout, part availability, allowing circuit variations, or just to test you. Who knows.
You need to know the current flowing in the pot. The most the LM317 will provide is 100µA and can be safely ignored. The most the VO pin will provide is the output voltage divided by
100Ω + R (where
R is the variable resistor resistance). Noting that the voltage at
Vo * R / (R+100) and re-arranging, I get an output voltage of
0.0125R + 1.25. Please double check my derivation below.
Vo * R / (R+100) = Vo - 1.25
Vo * ((100)/(R+100)) = 1.25
Vo = 1.25 * (R+100) / 100
Vo = 0.0125R + 1.25
Calculating power in a pot is tricky because you have to consider the entire operating range. It usually enough to consider a setting near
2kΩ and near the middle (
1kΩ) to ensure you've found the worst case scenario. Let's consider near
1Ω, first - this is often the worst case because the variable resistor's power handling capability is worst at low resistances.
R = 1Ω, current through
1.2625/101 = 12.5mA which is
R=1kΩ, current through
13.75/1100 = 12.5mA which is
R=2kΩ, current through
26.25/2100 = 12.5mA which is
Given that a pot's power handling capabilities scale roughly with the resistance setting, and in this circuit the power in the pot scales in roughly the same way, and that the maximum power is well under the 2W rating, it looks like the pot will be fine.
- The LM317AHV datasheet is terrible, but yes, provided the regulator is regulating,
Vadj will be
Vout - 1.25V.