I'm trying to build my own Bench Power Supply but I'm having a hard time to find a good design (circuit/project) for it.

Here are my requirements:

  • Input: main using a 18V 2A transformer
  • Output: 0-18V
  • Current: 0-1.5A
  • Current Limiting controlled by a potentiometer
  • Voltage output controlled by a potentiometer
  • And I'd love it to be based on the LM317

I know there are some designs with two LM317, one limiting current and the other controlling the voltage output, but I couldn't find any good reference for those circuits.

  • 2
    \$\begingroup\$ LM317 is a bad choice. 18VAC in the secondary is ~25.4V RMS, if you regulate down to 1V at 1.5A that's 36.7W of dissipation which the device won't be able to handle without a few cubic meters (ish) of heatsink. \$\endgroup\$ Feb 23, 2014 at 1:06
  • \$\begingroup\$ LM317 won't go below 1.2V either... \$\endgroup\$
    – Matt Young
    Feb 23, 2014 at 6:38
  • \$\begingroup\$ You'd have to chain multiple TO220 LM317 and give a heatsink to each. Not a good idea. \$\endgroup\$
    – Lundin
    Aug 17, 2018 at 7:51

3 Answers 3


There is a schematic in the ON-Semi datasheet

enter image description here

It needs a negative voltage that feeds the two depletion mode JFETs, they operate as constant current sources and in conjunction with the two diodes connected to the output provide about -1.4V to the pot.

  • \$\begingroup\$ Thanks @alexan_e . I saw that design on the datasheet, but I wasn't sure about a few things: a) I was not able to find the Q1 and Q2 components. Can you suggest me any alternatives to them? They seem to be obsolete. b) Do I need to provide exactly -10 for Q1 & Q2? Can it be something around -10? How well needs this -10 be in terms of stabilization? c) What is the value for Rsc? \$\endgroup\$
    – mlemos
    Feb 23, 2014 at 13:48
  • \$\begingroup\$ @mlemos The -10 voltage doesn't have to be exact , it can even be -8 or -12 or lower. The Jfet work as constant current source so the current will change slightly (IDSS parameter) but this is not a problem. For calculation of the resistor you can refer to figure 23 page 9 of the datasheet. You can also refer to a similar post electronics.stackexchange.com/q/41549/33841 , it suggests an alternative Jfet. \$\endgroup\$
    – alexan_e
    Feb 23, 2014 at 14:15
  • \$\begingroup\$ The voltage rating of the 2N3822 JFET is only 50V, so -10 is about as low as you would want to go with 40V in. One of them sees Vin + 10V -2V (or so). \$\endgroup\$ Feb 23, 2014 at 15:51
  • \$\begingroup\$ @SpehroPefhany His input is not 40v but about 25v from a 18v transformer. \$\endgroup\$
    – alexan_e
    Feb 23, 2014 at 16:07
  • 1
    \$\begingroup\$ Sure. Should be okay for him even with bad transformer regulation etc. Just in case someone else tries to use the circuit as drawn I thought I'd mention it. \$\endgroup\$ Feb 23, 2014 at 16:12

You can use the circuit alexan_e presented and replace the obsolete JFETs with current regulator diodes. Compared to the loose tolerance 2N3822 (\$2mA \lt I_{DSS} \lt 10mA\$), they are much more tightly specified.

Power dissipation rating is a little close on the NSI50010YT1G, though, so give it lots of copper and don't allow ambient temperature \$T_A\$ to get too high.

Or you could use the IXYS TO-220 450V IXCP10M45S, which should be bulletproof (but more expensive).

  • \$\begingroup\$ Thanks @SpehroPefhany . How should I connect this component to the circuit? \$\endgroup\$
    – mlemos
    Mar 7, 2014 at 18:05
  • \$\begingroup\$ @mlemos They're connected just the same as the JFET in your schematic. Gate connected to source. \$\endgroup\$ Mar 7, 2014 at 19:04
  • \$\begingroup\$ and with the same negative sourcing, right? Does this negative sourcing can be really well regulated and filtered or I can solve it with just a bunch of diodes and caps straight from the 'negative' output of a transformer? \$\endgroup\$
    – mlemos
    Mar 7, 2014 at 19:12
  • \$\begingroup\$ Yes, you could use a 50/60Hz charge pump provided the caps are big enough. \$\endgroup\$ Mar 7, 2014 at 19:21

You should use switching regulators. TI has some grate products to help, including modules (cost more but are simpler. just add a pot and filter for instant power supply. we used one to power a robot through an umbilical and found the accuracy was very good [much better than we needed]) but regardless switching regulator will allow high current with low power dissipation. also many of the parts out there have a signal line separate from power output allowing the regulator to even compensate for any voltage drops (from ammeter for instance) making the Output what you want.

  • \$\begingroup\$ OP requested LM317. Namely. Linear regulators has some disadvantages but also some advantages over switching regulators. OP probably has some reason for LM317. \$\endgroup\$ Apr 29, 2017 at 21:55
  • \$\begingroup\$ @Chupacabras You can still use a switch reg as first regulator, then put a LDO after it. That way you get the advantage of the LDO (less noise) without massive heat. \$\endgroup\$
    – Lundin
    Aug 17, 2018 at 8:04

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