Electronics > Beginners
LM317AHVT and Heatsink
MarkF:
--- Quote from: mike_mike on October 07, 2018, 10:51:15 am ---Dear Forum,
I made a layout for a LM317 power supply, using the attached schematic.
I don't know if I correctly made the layout. Please have a look at the layout and tell me if it is good.
I want to know only if the layout is correct. The schematic that I draw in KiCAD is correct because I checked it. I also added a LED and a current limiting resistor (R4 and J3).
The refdes of the components does not correspond from the schematic to the layout.
I was a little bit confused by the copper pours, because I don't know if I correctly made the copper pours.
--- End quote ---
I did a similar PCB which will accept either AC or DC input.
Zero999:
--- Quote from: mike_mike on November 15, 2018, 01:48:15 pm ---Thank you for the reply David Hess.
I tested the power supply and the results are (the schematic is the one from reply #27, with C4=2200uF/63V instead of 1000uF/63V)
without load V=19.80V, with 6.17A load V=19.79V
without load V=26.0V, with 7.24A load V=26.0V
I don't know if the wire that come from the pin 1 or another pin of LM317 to the PCB was making good connection with the screw terminal located on the PCB.
If there was a bad connection, could the power supply work good and if I repair the bad connection by tightening the screw could the power supply to work bad ?
--- End quote ---
The is no issue with those measurements. A 10mV change in output between no load and 6.17A of load is equivalent to a resistance of 10m/6.17 = 1.62mOIhms, which is very good. If there's a problem with a mechanical connection then it isn't affecting the circuit operation but it should still be resolved to improve the reliability.
mike_mike:
1. Can someone explain how to calculate the voltage on the potentiometer ?
It is the output voltage minus the voltage on the resistor (1.25V) ?
2. If I remove the 3 8.2K resistors and the 2 potentiometers and I use instead of them only a single 2K/2W multiturn potentiometer, what can change in the power supply operation ? excluding the output voltage ?
MarkF:
--- Quote from: mike_mike on November 22, 2018, 03:21:09 pm ---1. Can someone explain how to calculate the voltage on the potentiometer ?
It is the output voltage minus the voltage on the resistor (1.25V) ?
2. If I remove the 3 8.2K resistors and the 2 potentiometers and I use instead of them only a single 2K/2W multiturn potentiometer, what can change in the power supply operation ? excluding the output voltage ?
--- End quote ---
First, get rid of all that resistor nonsense you have.
Keep the two potentiometers if you want a coarse and fine adjustment (with the fine adj. being about 10% the value of the coarse).
On the side, your capacitor values are way out of wack.
Then watch this video:
AngraMelo:
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?
Navigation
[0] Message Index
[#] Next page
[*] Previous page
Go to full version