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I'm pretty new into electronics, But I do this mostly for educational purposes. Even if there are cheaper/more effective ways to build a power supply, I want it to be based of the LM317.

This project isn't finished yet, I just wanted to ask for any insights or tips that might make this better/safer. This is my circuit atm:

Diagram From Multisim (Copy Image link for a higher resolution)

I was also thinking of using two extra pots for fine tuning the current + voltage.

My main goal is to have a simple power supply based of LM317 with both Current and Voltage adjustment. Thank you!

EDIT: If i use this design and what to add a transistor for increased load like this: https://embed.gyazo.com/e8ff3fd621ad4c0f7444903f880a4804.png

Where does the collector go?

EDIT2: One guy here suggested that i might use a wire-wound potmeter to control the current. Is there any other solution? Maybe controlling the current using an op-amp/transistor? If so can anyone show me an example circuit? I have checked out most of the datasheets but i don't think that they explain it that well. Cheers!

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  • \$\begingroup\$ You might want to add clipping diods at the output (to make sure the output can not be lower then GND or higher then your DC-supply voltage). \$\endgroup\$
    – NeinDochOah
    Commented Jan 20, 2016 at 10:03
  • \$\begingroup\$ Thanks for the tip! Never heard about clipping diodes before tho. Could you explain why u would need these? \$\endgroup\$
    – Xane
    Commented Jan 20, 2016 at 10:12
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    \$\begingroup\$ This is a general measure against overvoltage on the output, since someone could wrongfully apply a voltage higher then what your circuit delivers at the output (or one with wrong polarity). Your freewheeling diode may counteract some of those scenarios to some extend, but in general its a good idea. \$\endgroup\$
    – NeinDochOah
    Commented Jan 20, 2016 at 10:19
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    \$\begingroup\$ I have a remark about the current limiting part. Note how the output current has to go through those resistors. This will make these resistors heat up. For R3 it will be easy to calculate what it can manage. For potmeter R4 it is more complex. Maybe you will need a wire-wound potmeter so that it can handle the current ! \$\endgroup\$
    – Bimpelrekkie
    Commented Jan 20, 2016 at 12:42
  • \$\begingroup\$ Thank you! I didn't think of that. What is a wire-wound pot? I know what is looks like, but can they handle more current? I was thinking of adding a 2N3055 power transistor so that i don't need to take the load through the resistors. For example like this: i.gyazo.com/e8ff3fd621ad4c0f7444903f880a4804.png I don't think that circuit will work tho. Just an example. Any tips are welcome! (I'm still learning about transistors) \$\endgroup\$
    – Xane
    Commented Jan 20, 2016 at 16:27

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If you want an adjustable current limited linear power supply, I suggest you build one on 5-pin ICs instead of two 3-pin ICs. That way you don't pay the voltage drop price [over the pass element] twice. Look at L200 etc. The difference is that these 5-pin regs have a single pass element with two sources of control for it (error amp for voltage reg and comparator for current limit).

enter image description here

If you want to stick with the twin LM317 solution, look at LM317 based power supply with current limiting where this topic has been discussed before.

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  • \$\begingroup\$ Thanks for your answer. Yes i have checked that one and even tried to simulate it. The problem is that i don't understand the Q2 part that requires -10v, how do i get a negative 10 volt? Thanks! \$\endgroup\$
    – Xane
    Commented Jan 25, 2016 at 0:08
  • \$\begingroup\$ @Xane: that negative voltage is necessary if you need the output to be adjustable all the way to 0V (and/or 0A). The same guy who posted that question also asked for solutions for that... electronics.stackexchange.com/questions/102096/… \$\endgroup\$
    – got trolled too much this week
    Commented Jan 25, 2016 at 0:21
  • \$\begingroup\$ Yes i saw that too, But i do not have a center tap transformer. Will it work anyway? Or do i need to buy a "negative converter" or similar? \$\endgroup\$
    – Xane
    Commented Jan 25, 2016 at 13:14

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