Electronics > Beginners

LM317AHVT and Heatsink

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mike_mike:

--- Quote from: MarkF on November 22, 2018, 04:57:05 pm ---On the side, your capacitor values are way out of wack.

--- End quote ---

You mean that the capacitors have a too high or too low value ?

MarkF:

--- Quote from: AngraMelo on November 22, 2018, 05:18:44 pm ---Im sorry to go out of the main topic here but can someone please explain to me how Vin goes to the base of the transistor but Vout comes out of the collector?

--- End quote ---

The two main current paths are through the transistor and through the LM317 (green and blue).  The resistor R1 (yellow) determines how much current bypasses the LM317 by setting the emitter/base voltage as a function of current.  The  more current required the more is bypassed.

   


--- Quote from: mike_mike on November 22, 2018, 05:44:58 pm ---
--- Quote from: MarkF on November 22, 2018, 04:57:05 pm ---On the side, your capacitor values are way out of wack.

--- End quote ---

You mean that the capacitors have a too high or too low value ?

--- End quote ---

Far bigger than necessary for a DC input.  Maybe a 100uF and 1uF on the input.  And a 10uF and 0.1uF on the output.  For me, I would just have a 100uF on the input and a 10uF on the output and let whatever is being power have it's local decoupling.

mike_mike:
Thank you for your help.
I watched your video, and I understoon the following: If I need to calculate the power dissipated on the potentiometer, then I need to calculate the output voltage of the power supply, then decrease the value by 1.25V, then calculate the current through the potentiometer using Ohms Law and the using the formula P[W]=I[A]*V[V] to calculate the power ?
For example:
The output voltage is 26.25V, then the voltage on the potentiometer is 26.25-1.25V=25V, the current through pot is 25V/2000R=0.0125A, and the power dissipated on the pot is 25V*0.0125A=0.3125W.
Is that correct ?
Please consider the attached schmatic, the DC input is unfiltered.

Zero999:
Yes, you've calculated the power dissipation in the potentiometer correctly.

Note for future reference the maximum power rating of the potentiometer is only for the whole track, when current is passed through half of the track, the power dissipation will be halved. Thsi isn't a problem for this circuit since the current through the potentiometer is constant, but it's good to know.

mike_mike:
If I switch from the first schematic (with dual pots) to the second one (with 2k pot), what would change in the functionality of the power supply ?

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