I have a few questions regarding the following schematics: imga![/imga]


Initially I was testing the first schematic to see if it works good. The result of the tests were good, the power supply voltage drop under 7A load was 0.01V. I have a few questions regarding the 2 attached schematics:

  1. The first modification was that I removed R5, R7, R8 and the 2 potentiometers (1k and 10k) and I used instead of the mentioned components a single 2K/2W multiturn potentiometer. What modifications will appear in the power supply functionality after this modification besides the modification of the maximum output voltage ?

  2. How to calculate the power dissipated in the 2k/2W multiturn potentiometer ? It is sufficient a 2W power rated potentiometer ?

  3. How to calculate the voltage on the 2k/2W potentiometer ? it is equal to Vout-1.25V ?

Probably there will be more further question.

  • \$\begingroup\$ It's a way to alter the linearity to get a better adjustment at high voltage ranges. With 2k pot you will have a linear voltage variation from 1.25V to 26.25V \$\endgroup\$ – Dorian Nov 23 '18 at 16:28
  1. Theoretically nothing. R8 || R7 || R5 || (RV1 + RV2) (where || means "in parallel" and + means "in series") is equivalent to Rp || Rv where Rp is R8 || R7 || R5 = 2.73kΩ and Rv is a single pot from to 11kΩ. Rp || Rv is equivalent to a pot from to 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.

  2. 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 Adj is 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 , near 2kΩ and near the middle (1kΩ) to ensure you've found the worst case scenario. Let's consider near , say , first - this is often the worst case because the variable resistor's power handling capability is worst at low resistances.

So at R = 1Ω, current through R is 1.2625/101 = 12.5mA which is <1mW.

At R=1kΩ, current through R is 13.75/1100 = 12.5mA which is 156mW.

At R=2kΩ, current through R is 26.25/2100 = 12.5mA which is 313mW.

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.

  1. The LM317AHV datasheet is terrible, but yes, provided the regulator is regulating, Vadj will be Vout - 1.25V.

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.