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I have an LDO TLV733 that output 3.3 volts. I need that the rail would always be 3.3v because anything below that the slaves data pins might get damaged because the host signal is 3.3v. The input voltage can range from 3.3 volts to 5 volts. Anything beyond 3.75volts is guaranteed to output 3.3volts on this regulator but i am not so confident voltages below it.

How do i compute the voltage output of the LDO compared to its input. In the datasheet i could not find a graph of its input output relationship.

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    \$\begingroup\$ Datasheet will cover that; it helps if you know the load current. Right there on the first page as it happens : just subtract the dropout voltage. \$\endgroup\$
    – user16324
    Commented May 23, 2020 at 15:35
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    \$\begingroup\$ Check the data sheet. It will tell you what the drop-out voltage is. Do note this voltage is dependent on load current. Some Low-Dropout regulators are really quite good - dropout voltage is often less than 100 mV. \$\endgroup\$
    – Dwayne Reid
    Commented May 23, 2020 at 15:38

2 Answers 2

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The parameter you are looking for is known as the Dropout Voltage - that is the minimum required difference between the input and desired output voltage in order for the regulator to maintain the correct output voltage.

This parameter is highly dependent on the load current, so you will often find a graph showing dropout voltage vs current. You simply need to add on the nominal output voltage to this value to get the corresponding minimum input voltage.

For your part, an example curve is shown on the very first page of the datasheet you linked to.

It should also be noted however the curve will also be dependent on temperature as well, so you will need to check the datasheet for further information regarding temperature effects. There are more complete graphs later in the datasheet showing dropout vs current for different temperatures for different output voltage settings. For example:

Dropout Voltage Curves

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  • \$\begingroup\$ Looks like i will have to compute my current but my estimate is its lower than 100mA whhich is good that means about less than 50mV drop out \$\endgroup\$
    – DrakeJest
    Commented May 23, 2020 at 15:48
  • \$\begingroup\$ @DrakeJest given the regulator will warm up and current might be spikey, I'd air on the side of caution and design for 100mV dropout for <=100mA load \$\endgroup\$
    – Tom Carpenter
    Commented May 23, 2020 at 15:53
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Elaborating on Tom's answer, there is a bit of a subtlety about dropout voltage.

enter image description here

The pass element is a PMOS, so when it is fully ON, it will behave as a resistor RdsON whose value depends on temperature and Vgs. This is why dropout voltage graphs in Tom's answer are all straight lines.

In dropout mode, when Vin is too low, the "opamp" shown in the internal schematic will be clipped, its output will be the lowest possible voltage to get the FET to turn on as much as possible. If we assume TI engineers are smart, then the opamp output should be able to reach ground, so the FET's Vgs is equal to the input supply voltage.

So the subtlety about dropout voltage in PMOS based LDOs is that it depends on the input voltage since that's what determines the available FET Vgs. So whatever LDO you use, make sure you look for a dropout voltage graph at Vout=3.3V which means they set the input voltage to just enough to get 3.3V on the output, and not for any other voltage which would result in a different FET Vgs.

Especially if you pick another LDO, if the only dropout voltage graph in the datasheet is measured at 5V output, it won't be the correct dropout for 3.3V output at close to 3.3V input.

Since max dropout voltage at 3.3V is 220mV for 300mA, the PMOS' maximum RdsON is about 733 mOhms, at 85°C Tj. In dropout mode, the FET is basically a resistor of at most 733 mOhms which means there is no PSRR and the output impedance is higher than when the LDO regulates properly.

So you have to be careful that the input voltage isn't too noisy and does not have voltage dips, because that will end up directly in the output. And if the load draws pulsed current, output voltage may dip too. You may need a larger output cap, so it's a good idea to place a footprint for it, even if you don't use it. This could be important if the load is sensitive to noise, and you use a LDO to give it a clean supply.

Also your slave's data pins should not be damaged until the ESD diodes conduct, which should allow for signals up to about 0.5V above its supply.

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