LDO Dropout Voltage vs Input Voltage Range

[Vitor Przedzmirski]

Hello guys, I appreciate very much If you could help me. I can't understand the following:

I know that the dropout voltage of an LDO is related to the minimum input voltage necessary to maintain regulation. So, If a datasheet says the dropout for a specific regulator is 100mV, and the output voltage is 3.3V, I should get 3.3V with a minimum 3.4V , right?

The specific point that I cannot understand yet is the relation between the minimum input voltage and the dropout voltage. Example:

The TI LM3940 LDO has an input voltage range of 4.5 - 5.5V, and there is a note saying:

"Because the LM3940 is a true low dropout regulator, it can hold its 3.3-V output in regulation with input voltages as low as 4.5 V."

Based on this information, I can conclude that I should have at minimal 4.5V in my input to get 3.3V at the output.

But, looking at the Electrical Characteristics Table, one can see the defined Dropout Voltage, that is typically 0.5V with maximum load. In this case, I understand that with 3.8V I could still regulate the output.

So, how come the two values can be so different? I'm trying to design an LDO and I'm stuck with this concepts.

For the dropout voltage, I am considering the differential voltage Vin-Vo when the regulated output drops 100mV (commonly used condition test in TI datasheets).

So, at last, based in what information I may decide my input voltage range?

Thanks in advance

Link for the LM3940 datasheet: https://www.ti.com/lit/ds/symlink/lm3940.pdf

 

[Audioguru]

It looks like there is an error on their datasheet about the dropout voltage.

 

[ads-ee]

This part was originally from 1999 "SNVS114G –MAY 1999–REVISED FEBRUARY 2015".

This was a low-dropout regulator back then.

The maximum dropout is from -40c to 125c which is 1V not the typical 0.5V, you should always look at maximum values, typical is not a guaranteed value, it's just likely where the top of the bell curve lies.

 

[KlausST]

Hi,

oh dear!

Usually I thought: Simple answer: Dropout_voltage = V_in -V_out.

But it seems there are different definitions for the term "dropout_voltage"

Texas Instruments uses the folowing definition: (From LM3940 datasheet)

Dropout voltage is defined as the input-output differential voltage where the regulator output drops to a value that is 100 mV below the
value that is measured at VIN = 5 V.

***
In other word this means: (According Texas Instruments)
* connect 5.00V to the LM3940, connect a load of 1A (or other test conditions. See datasheet)
* measure the output voltage. Let´s say we measure 3.33V.
* calculate the output voltage at dropout condition: = 3.33V - 0.10V = 3.23V
* slowly drop the input voltage until the output voltage is 3.23V.
* measure the input voltage (@ output voltage = 3.23V). Lets say it is 4.52V
* calculate dropout voltage: V_dropout = 5.00V - 4.52V = 0.48V
--> this 0.48V is the dropout voltage according Texas Instruments.

***
I´m sure I´ve seen other definitons from other manufacturers.
I´m not able to say one definition is correct or wrong. It´s just another definition.

***

I assume you/we are not the only one that has/have problems with TI´s definition of "dropout voltage".
Therefore they have a document for clarification:
www.ti.com/lit/an/slva207/slva207.pdf

Klaus

 

[Vitor Przedzmirski]

Thanks for your answers!

ads-ee:

I understand when you say we have to account for the maximum values, but even in this case, with 1V dropout, the minimum voltage should be 4.3V... Maybe I could consider that the minimum input voltage is selected as 4.5V for safety, or what?

KlausST:

I totally agree with what you said! And like I said, I'm satisfied with this test condition where you consider 100mV drop at the output, and measure Vi-Vo. The only thing I cannot understand yet is the relation between the minimum specified input voltage and the minimum voltage needed to regulate the output.

Another example:
AS1353, from AMS (first in list): https://ams.com/eng/Products/Power-Management/LDOs

The VIN range = 2.5 to 5.5V.

The dropout voltage @150mA, typical, is stated as 60mV. There is a note saying that, in this case, the values are valid only for output voltages >= 2.5V.

Ok, in this case, if i wanted 2.5V at the output with 150mA load, I understand that I could use 2.56V at the input. But in the majority of cases (based on datasheet informations) it seems that, ---------> Output Voltage + Dropout Voltage < Vin(MIN) <--------- THAT's what is confusing me. =D

 

[KlausST]

Hi,

AMS clearely has a different (from Ti) definition of "dropout voltage".
They say: V_in - V_out.
and as remark:

Dropout voltage = VIN - VOUT when VOUT is 100mV < VOUT for VIN = VOUT +0.5V (applies only to output voltages ≥ 2.5V).

****
On the other hand TI also talks about "nput-output differential voltage".

I´m a bit confused now. I need to go through the documents again....


Klaus