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I used LM317 and LM337 in my circuit, below is my circuit diagram. I tested it on my PCB. When their inputs are ±23.6V respectively(by DC power supply), the output of LM317 is 21.41V and the output of LM337 is -19.87V.

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According to the formula, the output should be around ±18.75 V. The resistors I chose have an error of 0.1%, so I'm wondering what's causing this problem.

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Best,

Junzhe

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  • \$\begingroup\$ Should not be R30 and R32 ... 240 Ohm ? electronics-tutorials.ws/blog/… \$\endgroup\$ Commented Oct 10 at 10:13
  • \$\begingroup\$ I have tried this, but if I make R1 and R2 both within the typical value range, I can't get the voltage I want (±19V). I think as long as I make R1 R2 meet the corresponding relationship, it should output relative values? \$\endgroup\$ Commented Oct 10 at 10:22
  • \$\begingroup\$ Reduce both "R1" and "R2" by 76%. \$\endgroup\$ Commented Oct 10 at 13:26

3 Answers 3

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The regulator output will not be within specification unless it has a proper load.

The 1000 ohm resistor does not provide a proper load.

The datasheet of the exact component you bought will specify the correct load for correct operation, but likely somewhere in the 5 to 10 mA range.

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  • \$\begingroup\$ Are you referring to the load I connected to the output? When I tested it, I didn't connect any load and measured the voltage directly at the output. My load is an op amp, and I use an LM317 and LM337 to power it. Will this situation improve after the load is connected? \$\endgroup\$ Commented Oct 10 at 10:30
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    \$\begingroup\$ @Sam Any load, including the voltage setting resistors. If your circuit does not draw the minimum required amount of current, the voltage is going to be out of regulation. Also, using 0.1% resistors is almost waste of money, for a chip with maybe 2% or more tolerance. \$\endgroup\$ Commented Oct 10 at 10:42
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    \$\begingroup\$ The regulator won't regulate properly below the minimum load current. Add some resistors to draw enough current to let the regulator work. See the datasheet for the minimum current. \$\endgroup\$ Commented Oct 10 at 10:43
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    \$\begingroup\$ I think this is probably a LM317 FAQ since everyone who has worked with it long enough will run into the problem. This here is the correct answer - pick resistors for the voltage divider that are below 1k. Saving current draw is senseless with a component which main purpose is to generate heat and as a side effect also works as a voltage regulator. \$\endgroup\$ Commented Oct 10 at 13:06
  • \$\begingroup\$ R1 should be 120Ω (guaranteed for all input-output combinations) to 240Ω (good for limited Vin-Vout). Or add enough load to the output to add up to 10mA (or 5mA) when the divider is considered. \$\endgroup\$ Commented Oct 10 at 16:17
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As others have also said, use R3, R32 240 Ohm as in the datasheet.

Simulation for a LM117 (don't have LM317 in database) with microcap v12

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And don't use a capacitor too big for C30 and C33 ... 100 nF will be ok.

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You need to carefully read the data sheet of the devices you use as there are two significant parameters you didn't consider when using these voltage regulators.
That will save you a lot of grief in any future electronic design.

The first is, the minimum current draw for the LM317 and LM337 is 10mA, so R30 and R32 should be no larger than 125Ω for a current of 10mA with no external load.
Using the 1% value of 121Ω, then R29 and R31 will be 1.69k for an 18.75V output (LTspice sim below):
The simulated voltage is slightly high due a typical ADJ pin bias current of 50µA. (next paragraph).

The second reason for using low value voltage-adjust resistors is, the maximum ADJ pin output bias current is 100µA, so even with an added 10mA external minimum load, the output could be up to 1.4V high due to that bias current going through a 14k value for R29 and R31.

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