I am looking to use an LDO with a negative adjustable output voltage. I have found two products sold by TI that have similar schemes for setting the output through external resistors.

I am slightly confused though, as one of the resistors in each scheme is illustrated as a potentiometer. Why is this so? Can a fixed value resistor be selected instead?

Also, the first regulator series (LM79XXAC) have fixed negative outputs but external circuitry can be configured to make it adjustable. For instance, take the -5-V fixed LDO in adjustable output configuration.

LM79L05ACZ -5-V Fixed Output LDO

If this was instead the -12 V fixed output version, would the output equation change such that 12 V is substituted for 5 V in the above?

The external circuitry for the second LDO (LM337LM) is much the same as the other LDO series as shown below. LM337LM Negative Output LDO

However, the output equation varies from that of the first LDO series, not including a current term. Is there a reason for this? LM337LM Output Voltage

Finally, both datasheets do not seem to provide an input voltage range. Is there something I am missing? I need to figure out how to select a value of Vin for a given Vout.

Here are links to the Datasheets:



Thanks, everyone

  • \$\begingroup\$ Iq should always be included, but sometimes it's so small compared to the current through the setting resistors it is ignored. Read the datasheet carefully to see what's guarranteed about Iq for any given regulator of this type. \$\endgroup\$
    – Neil_UK
    Jan 1, 2022 at 21:49
  • \$\begingroup\$ both datasheets do not seem to provide an input voltage range Look at the circuits, do these regulators (they're not LDOs, they need a somewhat high (Vout-Vin) ) "see" the input voltage directly? What if you increase the input voltage by 10 V (while maintaining a 5 V output voltage), where does the extra 10 V "go"? Maybe (Vout - Vin)? Is something mentioned about a maximum for that in the datasheet? \$\endgroup\$ Jan 1, 2022 at 21:56
  • \$\begingroup\$ They don't literally mean a potentiometer, just that you need to pick the value. Also consider something newer than the ancient 78xx regulators. The input voltage will specified, usually in the Electrical Characteristics section. \$\endgroup\$ Jan 1, 2022 at 21:58
  • \$\begingroup\$ @user1850479 Do you mean 79xx regulators? What is wrong with them? Do you have experience with the LM337LM regulator? \$\endgroup\$
    – Cole Fehr
    Jan 1, 2022 at 22:02
  • \$\begingroup\$ To be pedantic, neither the LM337 nor the 790x is an LDO. They are linear regulators, but not low drop-out regulators. An LDO is a linear series regulator with less than 1 V \$V_{do}\$. \$\endgroup\$
    – The Photon
    Jan 1, 2022 at 22:11

1 Answer 1


Of course you can select fixed resistors instead of a potentiometer to make it non-adjustable.

The 79xx regulators are intended to be used as fixed regulators, but can be made adjustable, and the actual voltage of the regulator just sets the minimum voltage, it can be adjusted higher.

The output voltage term always includes the ground or adjust pin current, the LM337 output voltage also has the pin current in the formula - the LM337 just has about 10x smaller current, and due to the LM337 already typically has an adjustment divider current of 5mA, the adjust pin current can be ignored because it does not have much effect on the output voltage.

The datasheets do contain the formula for the resistors. For 7905 the minimum voltage will be -5V and it needs at least -2 volts more than the output voltage to work properly. The LM337 can go down to the -1.25V reference voltage, and it needs -3 volts more on the input than on the output.

  • \$\begingroup\$ So a positive output LDO always requires the input voltage to be more positive than the output. Is it the reverse for negative output LDOs, with the input voltage more negative than the output? \$\endgroup\$
    – Cole Fehr
    Jan 1, 2022 at 22:07
  • \$\begingroup\$ Yes, the voltage drop over a regulator must be large enough, and in the correct direction. Yes, more negative input is required. -12V can be regulated down to -5V for example. \$\endgroup\$
    – Justme
    Jan 1, 2022 at 22:09

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