switching power supply

[3dOptics]

See image. What wattage rating would I need for 50K potentiometer and the 1.21k resistor.

 

[Hero999]

Why not work it out using Ohm's law?

P = V²/R

 

[crutschow]

The regulator feedback maintains 1.23V nominal across the 1.21kΩ resistor R1, thus for 35V output, R2 will be approximately 33.8kΩ. From that you can calculate maximum power in both resistors.

 

[3dOptics]

Why not work it out using Ohm's law?

P = V²/R

For the potentiometer: P = 40²/50,000 = 0.032watts. So a 1/8watt or higher would work, correct?
For the resistor: P = 40²/1,210 = 1.32watts. So a 2watts would work, correct?


For the inductors I would need 33uh and 20uh, minimum of 3amps, and a frequency of 150k, correct?
https://www.electro-tech-online.com/custompdfs/2010/02/LM2596-DPDF.pdf

 

[3dOptics]

Do the inductors have to match the frequency of the switching regulator(150Khz)?

 

[crutschow]

For the potentiometer: P = 40²/50,000 = 0.032watts. So a 1/8watt or higher would work, correct?
For the resistor: P = 40²/1,210 = 1.32watts. So a 2watts would work, correct?


For the inductors I would need 33uh and 20uh, minimum of 3amps, and a frequency of 150k, correct?
https://www.electro-tech-online.com/custompdfs/2010/02/LM2596-DPDF-1.pdf

The resistor has 1.2V across it, never 40V

Those requirements are ok for the inductors. They should be rated for a switching power supply but may not necessarily be spec'd for a particular frequency.

 

[Hero999]

Just use 1/4W if it's through hole because it's actually cheaper than 1/8W which should only be used if you're short of space or if it's SMT.

If you're not going to adjust the pot. then why not replace it with a 34R fixed resistor?