Can someone tell me what the difference between linear regulators and LDOs is?

From what I have read and understood, it is that linear regulators and LDOs don't use switching elements, but I think they are the same basically.

Is there something that I am missing to understand?

  • 10
    \$\begingroup\$ Ldo is more marketing term than anything. Your grandpoppop's LDO wouldn't be considered an LDO anymore. \$\endgroup\$
    – Passerby
    Commented Nov 19, 2021 at 6:49
  • 5
    \$\begingroup\$ The difference is the dropout voltage. \$\endgroup\$
    – user16324
    Commented Nov 19, 2021 at 15:03
  • \$\begingroup\$ A thorough explanation of LDOs can be found here: ti.com/lit/ml/slup239a/…. \$\endgroup\$ Commented Jun 27, 2023 at 14:12

4 Answers 4


They key property of the LDO type linear regulators is the P-type pass transistor. This allows controlling the pass transistor with a voltage below the input voltage \$V_{in}\$ for all possible output voltages. Therefore, the output voltage can in principle be set arbitrarily close to the input voltage, especially for MOSFETs.

Non-LDO type regulators that use N-type transistors have an inevitable dead band of output voltages close to \$V_{in}\$ in that they simply cannot reach, because those high output voltages would require a control voltage on the pass transistor above \$V_{in}\$.

  • \$\begingroup\$ The p-type pass transistor usually is working in its active region right, during the Linear regulator regulation? \$\endgroup\$
    – user220456
    Commented Nov 19, 2021 at 7:30
  • 2
    \$\begingroup\$ @tobalt The relevant mode is called forward active for BJTs, and saturation for FETs. \$\endgroup\$
    – Hearth
    Commented Nov 19, 2021 at 14:01
  • 1
    \$\begingroup\$ @Hearth thanks for adding that! I think it is a little confusing (albeit technically correct) to call the MOSFET region saturated, because when thinking in switch jargon, saturated would imply that it cannot be turned more on. But then again, I might I have a weird perception for these names because I've never really thought much about them or used them as a self-learner. \$\endgroup\$
    – tobalt
    Commented Nov 19, 2021 at 14:26
  • 3
    \$\begingroup\$ @Hearth I see where the name is coming from. If you view the gate voltage as a given, then the channel is saturated. But looking slightly further, there is of course the elephant in the room, that the gate voltage could be easily raised. For a JFET e.g. the saturation current can't be easily enhanced. But as I said, the term is technically absolutely correct, it just feels a bit weird to me from an application pov, to call a MOSFET that is barely on "saturated" (and I know it feels weird because I am weird :p) \$\endgroup\$
    – tobalt
    Commented Nov 19, 2021 at 14:34
  • 1
    \$\begingroup\$ Completely agree @Hearth Atm this answer offers a very simple explanation. I thought about adding more precision on how to control P or N type transistors, but then I thought that this would sacrifice brevity and complicate matters too much for those looking for a simple explanation. I will slightly change it hoping to achieve both. \$\endgroup\$
    – tobalt
    Commented Nov 21, 2021 at 13:55

An LDO is a type of linear regulator.

All linear regulators have what's called a dropout voltage, a minimum input-to-output differential that they can't work without. The original linear regulators (including the still-popular 78xx and 79xx series) had dropout voltages on the order of 2~3 volts.

Since the input-to-output differential is directly related to power loss (among other considerations), it's desirable to operate with as low a differential as possible. So the low-dropout (or LDO) linear regulator was invented. Initially, these had dropout voltages in the range of 1~2 volts (such as the LM1117 series), but that would hardly be considered "low" dropout today; modern LDOs can achieve dropout voltages of only a few hundred millivolts.

It's not recommended to use an LDO when you don't need one, though, as they tend to have poorer stability characteristics than standard linear regulators. A 78xx is typically stable no matter what as long as you give it adequate capacitance on the output, but a fancy LDO with only 100 mV of dropout may require the output capacitance be in a narrow range, and that capacitance needs to have high enough (but not too high) ESR to avoid oscillation. See this question for some more details on this.


A low drop out regulator (LDO) is a linear regulator.

Linear regulators control the output voltage by adjusting the current through an output transistor.

In all linear regulators, there is a minimum difference between the input voltage and the output voltage that you have to maintain.

Take the ancient 7805 linear regulator. It needs the input voltage to be at least 2V higher than the output voltage. You need at least 7V in to get 5V out of a 7805.

enter image description here

By comparison, the TLS850C2TEV50 only needs a difference of about 0.1V between input and output. You only need 5.1V in to get 5V out. That's "low drop out."

enter image description here

Both are linear regulators.

The TLS850C2TEV50 is a low dropout (LDO) regulator.

The 7805 is a linear regulator but not a low dropout regulator.


A low dropout regulator is a type of linear regulator where the drop out voltage is relatively low. For example, an LDO might require an input voltage only a few hundred millivolts above the output voltage, while a classic linear regulator like the 7805 requires at least 2 volts.

For more information, see: https://en.wikipedia.org/wiki/Low-dropout_regulator


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