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I'm relative new to this, I have a question about designing the power for a circuit.

What does efficiency mean for a voltage regulator? Here's an example

Let's say I have an LDO that create 5V at 2A with an input of 10V at 1A with an efficiency of 50%

Does that mean that the LDO can only output a maximum of 5V at 1A and the other 1 A is lost as heat?

Similarly, if the efficiency is at 60%, does the LDO can output a maximum of 5V at 1.2A?

Thank you for the help

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    \$\begingroup\$ Let's say I have an LDO that create 5V at 2A with an input of 10V at 1A with an efficiency of 50% a regular LDO regulator could never do this. A buck regulator can do this (nearly). You need to fix your question so that you don't make statements about impossibilities. \$\endgroup\$
    – Andy aka
    Mar 13, 2021 at 16:37
  • \$\begingroup\$ Thank you for the feedback Andy. I just placed a hypothetical example to make sure if I understood the efficiency. I've done some research on the different type of voltage regulators with their own pros and cons, but I'm still trying to understand the efficiency portion \$\endgroup\$
    – Leotwin
    Mar 13, 2021 at 16:51
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    \$\begingroup\$ Always use SMPS if efficiency counts , linear loss is simply voltage drop x I \$\endgroup\$ Mar 13, 2021 at 17:04
  • \$\begingroup\$ Another thing wrong is the claim of LDO: if you're reducing 10V to 5V, that is not "low drop-out"; that's a pretty high, poor drop-out. A linear regulator could do that, but LDO refers to linear regulators that are optimized for dropping a small voltage. \$\endgroup\$
    – Kaz
    Mar 15, 2021 at 3:26

5 Answers 5

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Let's say I have an LDO that create 5V at 2A with an input of 10V at 1A...

You have 10 W input, and 10 W output, therefore you have 100% efficiency.

...with an efficiency of 50%

This is a contradiction. In the first part of your sentence you gave numbers that show the efficiency is 100%.

But, a linear regulator (the LDO being a sub-type of linear regulators whose capabilities aren't important in the scenario you presented) could not produce the numbers you stated. A linear regulator requires the same input current as output current (plus a few microamps or milliamps to run the regulator circuit itself).

So realistic numbers for a linear regulator to get 5 V, 2 A output would be 10 V, 2.001 A input. And those numbers do give an efficiency close to 50% (20 W input, 10 W output).

If you can reduce the input voltage, it won't affect the input current significantly and the efficiency will improve.

If you want efficiency with a large voltage difference between input and output, consider using a buck switching converter instead of a linear regulator.

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Efficiency in a linear regulator is quite easy to find.

You put some power in and the regulator:

  • Uses some for itself (usually the quiescent current is stated); it also need some extra voltage to work correctly but this is a limitation on the regulation, not an efficiency thing;

  • Let the needed amount to the output (the regulation itself)

  • Dissipates the excess as heat

Quiescent/working current (going from input to ground) is usually trascurable (unless doing micropower work), so you have a simple current path: the same current that enter will leave from the output.

Example: you want 5V from 12V; and you sink 1A from the regulator (easy numbers).

You are using 5V*1A=5W of power. Ignoring the quiescent current the regulator will take 1A from the input, so you have 12W of input power.

Of these 12W, 5W will power the load, the remaining 7W will be dissipated as heat. So you'll have something like less than 50% efficiency and a very hot regulator.

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    \$\begingroup\$ Is "trascurable" from Italian "trascurabile"? "negligible" or "insignificant" would be the usual English words for something that's so minor that it can be neglected or ignored. \$\endgroup\$
    – ilkkachu
    Mar 14, 2021 at 12:35
  • \$\begingroup\$ yes, sorry, negligible is the correct word. Modern linear regulator can reach microamps off Iq so unless you are doing micropower it is insignificant compared to the load. It is there however and it lower the efficiency even more \$\endgroup\$ Mar 15, 2021 at 10:11
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Efficiency in any system is 100% * (useful power out / total power in)

So for an LDO the power out depends on the output current * output voltage
Pout = Io * Vo

And for any electrical system, power in is input voltage * input current
Pin = Ii * Vi

For an LDO the input current is almost the same as the output current (we'll call that I) but in addition LDOs take a small current to work, called Iq.

So bringing this all together, efficiency:
Ef = 100% * ( (I * Vo) / ((I + Iq) * Vi))

Or to put it another way, if there is a large difference bettween Vi and Vo then they are very inefficient. But also if Iq is anywhere near I then they are also inefficient.

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The efficiency for any regulator is:

Eff = POUT / PIN

Where Eff is efficiency, POUT is output power and PIN is input power.

For linear regulators, the input current is always the same or greater than the output current. It is not possible for the input current to be less than the output current for a linear regulator. The LDO is a type of linear regulator. So the scenario in your question is not possible.

Linear regulators are typically used when the output power needed is very low, or when the input voltage is only slightly higher than the regulated voltage.

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The efficiency is in terms of power, not current.

Let's say I have an LDO that create 5V at 2A with an input of 10V at 1A with an efficiency of 50%

It's not a valid example, for two reasons:

  1. The efficiency is just a name for the ratio Pout/Pin, where Pout=Iout*Vout and Pin=Iin*Vin. So you can't just make the efficiency up, you have to calculate it.

  2. A linear regulator always maintains Iout >= Iin - that's a necessary condition maintained by linear regulators. It's not sufficient by itself to know that it's a linear regulator, since e.g. switching buck converters will maintain that relationship too, but all linear regulators behave that way.

    An LDO (a linear low drop out regulator) will never deliver 2A of current while consuming just 1A (on average). To deliver 2A it will have to consume at least 2A, and usually it'll be 2A + a few mA (at most).

Does that mean that the LDO can only output a maximum of 5V at 1A and the other 1 A is lost as heat?

An efficiency of an LDO is bound by the Iout >= Iin equation and is mostly determined by how you operate it in your circuit. I.e., in case of the relatively large currents you speak of, pretty much all linear regulators will have the same efficiency, and it'll be very close to the Iout = Iin criterion.

From this point onwards, it's just a basic high school algebraic exercise. Suppose you select Iout = Iin and set Iout = 1A, Vin = 10V, Vout = 5V. We immediately get Iin = 1A. Now for the efficiency:

Pin = Iin*Vin = 10V*1A = 10VA = 10W
Pout = Iout*Vout = 5V*1A = 5VA = 5W
Eff = Pout/Pin = 5W/10W = 50%

In this particular circuit, real LDOs will have an efficiency a tiny bit worse (lower) than 50%, but most modern LDOs have fairly low ground pin currents, so the actual efficiency may be something like 49.95% if the LDO's ground current is 1mA (i.e. Iin = 1A + 1mA).

Do note that since out of 10W going in, only 5W leaves as the electric current, another 5W is dissipated as heat. Typically you need an LDO with a high-power package well heat-sunk into a 4+ layer PC board, or you need a discrete heatsink for the LDO. Also, since the LDO is dropping out 5V, you don't really need the low-dropout aspect of an LSO, but that's mostly academic since there's no cost benefit to choosing a "non-LDO" regulator over an LDO these days, at least as far as low volume applications go.

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