I have some doubts about how to select a voltage regulator. I am planning to use a Texas Instruments TPS7A05 LDO regulator to regulate my 5V (250 mA) to 3.3V (200mA) to power my STM32 microcontroller. I am using a battery as source.

My input to LDO regulator:

  • input voltage :5v
  • input current : 250 mA

Expected output:

  • output voltage : 3.3v
  • output current : 200 mA ( to power my STM32 microcontroller)

    My doubts:

Dropout voltage :

I understood that dropout voltage is the difference between input and output voltage . The difference between my input voltage and my expected output voltage is 5V-3.3V =1.75V, but in the datasheet it is given like enter image description here

So I am getting my voltage drop of 1750 mV. It is more than 230 mV so to get 200 mA of output current my voltage drop should be 230 mV or what then why they have given the maximum input voltage as 5V and maximum current as 200 mA. So based on my doubt I cannot use this regulator or ehat or my understanding is wrong and we can use it.

Ground Current : I understood that ground current is the difference between input current and output current. So the difference between my input current and my expected output current is 250 mA - 200 mA = 50 mA but in the data sheet it is given like:

enter image description here

It is given like as 6 micro amperes, but based on my input and output current difference is 50 mA it is lot more than the given.So can I use this regulator based on my input and output or not or can I use a resistor connected in series so that I am able to get the ground current as given in the data sheet.

If my understanding is wrong please correct me.

Can I use this regulator for my application or not?

Please find the attached link of Texas instruments TPS7A05 LDO regulator.


2 Answers 2


You have understood both the dropout voltage and ground current wrong.

Dropout voltage is the difference between input and output that is at least needed for the regulator to work. It means that when your load is drawing 200mA, worst case you need only 235 mV more on the input. So as the output is 3.3V, you must have 3.535V or more at the input.

Ground current is indeed the difference between input and output current. But the regulator does not take in constant 250mA, so the difference is not 50mA. The datasheet says than when the load is drawing 200mA on output, the regulator itself takes only 6uA of current to work.

Based on those, you can use this. However you have not considered power dissipation yet. 5V in, 3.3V out, 0.2A current means the device must dissipate 0.34W. Depending on which package device you use, it can heat up by 91 degrees C, which means the ambient temperature must never exceed about 34 degrees C.

  • \$\begingroup\$ thank you so much for your reply i understood perfectly what is dropout voltage , ground current and power dissipation also but i have one doubt like how to find it should not exceed 34 degree c .is it a constant value will be given in the data sheet or what. based on your explanation now i have chosen LT3063 series regulator which has max Rthja is 45 degree so for 0.34 w ithe temperature is 15 degree Celcius .so can i use that \$\endgroup\$
    – Muthu
    Commented Apr 11, 2020 at 22:23
  • \$\begingroup\$ The other answer already explained it. Chip thermal resistance can be 267 degrees per watt. Chip has maximum temperature of 125 degrees. Those above read in the datasheet. With 0.34W, temperature difference inside and outside the package can be 91 degrees. \$\endgroup\$
    – Justme
    Commented Apr 11, 2020 at 22:29
  • \$\begingroup\$ okay by 0.34 w of power dissipation it is already heating up to 91 degree out of 125 degree. the left is 34 degree so it should not exceed the rest 34 degree to exceed 125 degree. am i correct \$\endgroup\$
    – Muthu
    Commented Apr 11, 2020 at 22:40

The datasheet specifies minimum dropout voltage required for proper operation of the regulator:

enter image description here

If you hand solder it I suppose you'll use SOT-23 package (they call it DBZ) and not the tiny DSBGA package, so that's the line highlighted in yellow. That is, the regulator is guaranteed to work properly with a dropout voltage of 245mV.

Basically the question you're asking is will it work with a higher dropout voltage? Answer is Yes it will work as long as other specifications are met (maximum voltage, power, temperature, current, etc).

However it will dissipate power: (5V - 3.3V) * 200mA = 0.34W and that's a bit high for SOT-23.

enter image description here

Using RthJA of 267°C/W, at 200mA it will be 90°C above ambient so that won't work. If there are copper pours on all three pins to suck the heat away it can be cooled through the pins: in this case usinb thermal resistance to board it will be 33°C hotter than the copper pours on the board which would be okay. So it depends on your layout. If you use thin traces that won't conduct heat well enough.

I'd recommend using a SOT89 regulator instead which is small and easy to solder but has much better heat conduction to the copper on the board due to the tab.

I don't understand your question about ground current, did you measure these values?


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