I'm working on my term project which is designing a dual DC regulated voltage that gives a DC voltage between +or-ve 3 to +or-ve 15. I have completed the testing for positive DC output part, but I have few inquires. (refer to design image below first please)

First for my choice of resistors values in the regulator part R3,R4, and R5. For variable resistor R4, since I'm given a potentiometer of 10K I have choose that value, but the other two resistor I used this formula taken from LM317 regulator data sheet

v0 = 1.25v(1+Radj/R1)

My question is does the values of R3 =1.1k and R5=1.5K practical for the real implementing of the circuit? is it logical?

Also, when I tried to test the negative DC out voltage it's not correct although I have used the same design done for the positive DC?

enter image description here

  • \$\begingroup\$ Looking at your schematic I wonder why r u using a full-bridge rectifier instead of a half-bridge, since you have a ground at the half of turns. I know that this is not what u r asking, but your design have a incredible low efficiency. See, you are using a 1:10 turn ratio. So you have a 22V AC. If you calculated the capacitors to give you nearly 0 ripple, you will have 22*sqr(2) = 31.1V (DC). At your worst output of 3V you will use only (3/31,1) 9.6% of your energy! At your best (15/31.1) 48.2%! Can't you design a switch regulator or, at least, change your transformer ratio? \$\endgroup\$ Apr 17, 2015 at 10:48
  • \$\begingroup\$ How much current are you expecting to drawn? What is the purpose of your design? It seems you are trying to design a symmetrical DC source, but why? \$\endgroup\$ Apr 17, 2015 at 10:53
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    \$\begingroup\$ @PedroQuadros Unfortunately using this transformer with full bridge rectifier is what our instructor has requested us to do in the design frame. When I observed the ratio between vin 31 volt and vout 15 maximum I noticed that this will cause heat dissipating through the regulator, I will ask him to clarify more about this. \$\endgroup\$
    – AMH9
    Apr 17, 2015 at 10:57
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    \$\begingroup\$ @PedroQuadros Our instructor said something about rotating the potentiometer in clockwise or counterclockwise to reduce or increase the voltage, so I assume that means having one adjustment for both. About last point, I think I will try to change the circuit in the way you have mentioned because it seems better to do it that way. \$\endgroup\$
    – AMH9
    Apr 17, 2015 at 11:22
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    \$\begingroup\$ You need to use a negative regulator (LM337 or similar) for the negative side of the supply. \$\endgroup\$ Apr 17, 2015 at 15:46

1 Answer 1


First, you need to go back to the datasheet for the regulator. There is a specification for minimum output load current. My recollection is that value is 5 mA.

That current determines the value of the resistor connected from the output of the regulator to the Adj pin. R3 & R9 in your schematic diagrams.

1.2V / 5 mA = 240 Ohms. Note that 240 Ohms is used in all of the app notes for the regulators.

Note that you can now treat the regulator Adj pin (with that resistor installed) as a current source. The output voltage is the voltage across the resistor from the Adj pin to ground PLUS 1.2V. Because there is a constant 5 mA current from R3, it's easy to calculate the value of the resistor to ground.

Note that the value of R3 & R9 can be changed to make things easier for you. 240 Ohms is the highest value you should use but you can safely go lower. This may make it easier for you to pick a pot value that is more readily available.

Also note that your negative regulator can't be the same as the positive regulator unless you use separate unregulated supplies. The negative regulator would need to be a LM137 (or LM337), not a 117 or 337.

  • \$\begingroup\$ Why value of R3 should not exceed 240? I have read in other website that suitable value for R3 would be between 100 to 470 oh'm. See first I'm restricted to use one of the following potentiometer: 1k,5k,10k,50k, or 100k. And my Instructor wants to have 3v to 15v using the whole precision of potentiometer - from o to 100%-. So say I used the 240 R3, In order to have 15v output, value of R4+R5 must be 240*((15/1.25))+1)= 2640 which cannot be applied using any of the potentiometer that I have. I hope you got my point. \$\endgroup\$
    – AMH9
    Apr 18, 2015 at 8:45
  • \$\begingroup\$ So pick a current that allows you to use the pot that you have. You could use a 1k pot and adjust the value of R3 to give you the voltage range that you want. \$\endgroup\$ Apr 18, 2015 at 11:50
  • \$\begingroup\$ You could also use a 2-gang (stereo) pot with value of 5K and simply wire the pot sections in parallel. That is very close to your desired value. \$\endgroup\$ Apr 18, 2015 at 11:51
  • \$\begingroup\$ I'm not familiar with the 2-gang pot, however I added resistor in parallel with my pot and it worked on multisim, thank you. \$\endgroup\$
    – AMH9
    Apr 18, 2015 at 13:59

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