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I'm a long time lurker :)

How do I tell what is the maximum voltage drop an LDO can handle? My mentor explained adjustable LDOs like this to me (paraphrased of course):

"An LDO is essentially a high precision, low range, voltage controller that requires a slightly higher input voltage to function. For example, even though an LDO is rated at (1.24-26)V if the input is 25V, it will only be able to output (24.2-24.8)V."

How can I find these boundaries in a datasheet?

Considered components:

Sorry if my English is not good. C++ is my native language.

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5 Answers 5

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LDO means low-dropout regulator. This is understood to be a type of linear regulator. The minimum difference between input and output is called the dropout voltage. So an LDO deserves the name only if its dropout voltage is "low." Many old LDO's don't really deserve to call themselves an LDO by today's standards.

Anyway, that is not what you asked about. There are normally two constraints on the maximum input voltage. One is that there is simply a maximum recommended operating voltage. If you go above that, you may wind up with problems. Second is determined by heat dissipation. The power dissipated by a linear regulator is (Vin - Vout) * Iload. So what you normally do is figure out your maximum input voltage and maximum load current and use those to calculate the dissipation.

Then you have to check the thermal resistance of the component to figure out whether dissipating that amount of heat will cause the silicon to get too hot. I am sure you can find detailed instructions about thermal calculations on the WWW or in the search box for this forum.

In general, you will dissipate less power with lower Vin and lower load current.

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  • \$\begingroup\$ It can be frustrating at my level because I already know a good amount about electronics, but am also lacking some fundamentals. Worse off, I am bad at expressing what I mean to ask, it seems. Are you saying the only limiting factor (besides min output and max input) on the most volts the ldo can drop is determined by making sure the device doesn't over-heat and fail? \$\endgroup\$ Jul 30, 2015 at 21:28
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    \$\begingroup\$ Sorry for the late reply. Yes. That is what I am saying. \$\endgroup\$
    – user57037
    Aug 10, 2015 at 22:50
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An LDO or any linear regulator can drop any amount of voltage as long as two conditions are met. First, the maximum input voltage is not exceeded. Second, the wattage dissipated by the regulator is taken into account.

For example if using a 7805 with an input voltage of 25V with an output current of 100mA. The regulator is dropping 20V and dissipating (V*I) 2W. The wattage is quite high and would require extensive heat sinking to keep the regulator from burning up.

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    \$\begingroup\$ There is a third condition: the output voltage cannot exceed the input voltage minus the minimum dropout voltage. Thus if the minimum dropout is 2 volts and Vin is 10 volts, Vout cannot exceed 8 volts. \$\endgroup\$
    – DoxyLover
    Jul 30, 2015 at 0:18
  • \$\begingroup\$ @DoxyLover True, but besides the point, since the question was "what is the maximum voltage drop of an LDO", not "what is the minimum voltage drop of an LDO". \$\endgroup\$ Jul 30, 2015 at 10:26
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LDO stands for Low Dropout Voltage. An LDO voltage regulator is able to regulate its output voltage even if the input voltage is only slightly above the output voltage. The required minimum difference between the input and output voltages for the regulator to stay in regulation is given as the "dropout voltage" on the data sheets. It is given for different levels of output current.

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    \$\begingroup\$ So... again... what is the maximum voltage drop it can handle? \$\endgroup\$ Jul 30, 2015 at 0:07
  • \$\begingroup\$ The maximum input voltage is indicated in the datasheet, for example, the Micrel datasheet, "Absolute Maximum Ratings, Input voltage = 60V". You must also check the thermal envelope, Pdisspated = (VI-VO) * I. \$\endgroup\$
    – Grabul
    Jul 30, 2015 at 0:19
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    \$\begingroup\$ It is very bad advice to tell a newbie to look at the absolute maximum ratings. For design purposes, you should always stay within the recommended operating conditions. So, not 60V, but 26V in this case. \$\endgroup\$
    – user57037
    Jul 30, 2015 at 0:23
  • \$\begingroup\$ Indeed, in this case applying 60V for more than 100ms would destroy the part as per the datasheet (Note 1 on page 7). \$\endgroup\$ Jul 30, 2015 at 1:34
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if the input is 25V, it will only be able to output (24.2-24.8)V.

That's not right.

The maximum input voltage is clear enough, the input cannot be higher than that.

The drop-out voltage is the minimum amount the input voltage has to be higher than the desired output. Be aware that the drop-out voltage depends on the current you draw from it.

So if the drop-out is 0.5V at the highest current you will draw and you want 6V out, you need to always at least supply 6.5V.

Then, the maximum voltage differential between in and out may be given for one or two types out there, but it's not likely to happen, because that's what the maximum input voltage and maximum power dissipation are for.

A linear regulator (LDO regulator is a subclass of linear regulators) is not magic, any power drained at the input, but not supplied at the output will be turned into heat. As opposed to a switching regulator, which will - at the cost of extra components and more output ripple in most cases - try to be as efficient about it as possible in the situation.

If you put 20V in and take 2V out, but you want 100mA, that means the LDO has to "absorb" 18V at those 100mA, which makes 1.8W of energy it turns right into heat. So if the regulator is not equipped for that, or the way you mount it doesn't support getting rid of that, it will be destroyed.

However, if you take 2mA out of it, that's only 36mW and that can be handled by any linear regulator, even when it's not mounted at all. Unless, of course, it's a high precision regulator with a maximum of 1mA output, but then it's the limited current that does it, not the power. 36mW is absolutely negligible.

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  • \$\begingroup\$ So you are saying that in a datasheet, there would be a "maximum power dissipation" part, and with that, I could derive how many volts can be safely dropped without a heat-sink? \$\endgroup\$ Jul 30, 2015 at 21:34
  • \$\begingroup\$ @JamesOldiges Depending on the way it is mounted. It has a note saying something about footprint required for that rating. With smaller or larger areas you need to calculate that with the thermal conductivity given in the datasheet combined with that of your solution. For TO220 and similar it often has a free-air value that doesn't depend on PCB mounting. \$\endgroup\$
    – Asmyldof
    Jul 31, 2015 at 8:01
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Do you mean the maximum voltage drop or the maximum output voltage?

The maximum output voltage would be the maximum input voltage minus the dropout voltage.

For the mic29150 the dropout voltages are listed on page 7 for the LT3015 on page 5 for a variety of current.

You have to allow for ripple and variations of the input voltage. The lowest rectifier output voltage must be above the LDO output voltage plus dropout voltage plus ripple plus many variation of the input voltage.

As an example if you wish 20 volts output at 100mA with the mic29150 and there can be 10% variation of the input voltage. The normal input voltage would need to be 20 + 2 + 200mV (dropout) = 22.2V.

At high voltages you may not need an LDO as the other variations may dominate.

Also be careful about the difference between the absolute maximum voltage and maximum operating voltage - for the mic29150 the absolute max voltage is 60V but max operating voltage is only 26V. The circuit is not guaranteed to operate at the absolute max, only not be destroyed.

At high input voltages it is probable that you may be limited by device dissipation.

Also with bipolar LDOs be aware that the quiescent current can increase greatly when the input voltage drops close to or below the set output voltage - for example see the graphs on page 9 for the Lt3015.

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  • \$\begingroup\$ "At high input voltages it is probable that you may be limited by device dissipation." Which means that there is only so many volts it is able to drop? \$\endgroup\$ Jul 30, 2015 at 21:20
  • \$\begingroup\$ That is correct - in the example I gave you have a minimum of 2.2V across the device so at 100mA there will be 220mA dissipation. \$\endgroup\$ Jul 30, 2015 at 21:37
  • \$\begingroup\$ So I guess my mentor is wrong..? That isn't very encouraging. Thank you very much. \$\endgroup\$ Jul 30, 2015 at 22:08
  • \$\begingroup\$ What he said was correct but not sufficient when applied to a real system. If you had an accurate 25V supply with no ripple, you wouldn't need to use an LDO to create a supply only slightly lower! You would use the 25v rail directly. (my previous comment should have said 220mW not 220mA for the dissipation) \$\endgroup\$ Jul 30, 2015 at 22:36
  • \$\begingroup\$ So the LDO isn't able to drop a large amount of volts unless we were to add a giant heat sink, making it effectively lower range? \$\endgroup\$ Jul 30, 2015 at 22:47

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