Input power and output power of LDO

[engr_joni_ee]

I am wondering how we can find the input power of the LDO linear regulator.

Consider the following load current and supply voltage to the load. This is also called output current and output voltage of the LDO linear regulator.

Output current: 1 A
Output voltage: 3.3 V

If the input voltage to the LDO linear regulator is 5 V, then how we can find the input current to the LDO linear regulator ?

 

[betwixt]

Current in = current out + whatever the regulator itself needs.

Brian.

 

[KlausST]

Hi,

We learned this in basic physics class in regular school. (no special electronics class)
Every electronics related person should know this .. without the need of research.
Please try an internet search for "most basic electronic formulas" ... The formula comes up directly after Ohm´s Law without the need for clicking a link.

--> Every basic electronics tutorial (wikipedia, documents, videos...) will tell you: P = V x I

And as simple as it sounds:
P_out = V_out x I_out
P_in = V_in x I_in

.. or P_LED = V_LED x I_LED
.. or P_diode = V_diode x I_diode
.. or P_resistor = V_resistor x I_resistor

it´s always the same ..

Klaus

--- Updated Yesterday at 9:45 AM ---

and without much surprise (since there is law of conservation of energy) ..

The difference between P_in and P_out .. is the power of the heat dissipated by the part.

P_heat = P_in - P_out

If the input voltage to the LDO linear regulator is 5 V, then how we can find the input current to the LDO linear regulator ?

the current" whatever the regulator itself needs" (as Brian describes it) .. can be found in the according datasheet. As so often.
Sometimes it is called "no load current" or "bias current" or "GND leg current" or similar. (the current will deend on output current)

Similar: the sum of all currents into a device is the same as the sum of the output currents.
Thus if you have I_in ... and a different I_out ... then the difference current needs to flow in the GND leg.
I_GND = I_in - I_out

Every designer needs to read the datasheets.

Klaus

 

[engr_joni_ee]

Current in = current out + whatever the regulator itself needs.

Brian.

Is this the "Quiescent Current" which is used as current the regulator itself needs ?

Part number: TPS7A4533KTTR

TPS7A45 | Buy TI Parts | TI.com

TPS7A4533KTTR - 1.5-A, 20-V, ultra-low-dropout voltage regulator with reverse current protection & enable in a TO-263 (KTT) package with 5 pins

www.ti.com

I found Quiescent Current for TPS7A4533KTTR is 1 mA.

Then in my case.

regulator input current = regulator out current + the current regulator itself needs
regulator input current = 1 A + 1 mA


The required input power of the regulator is = regulator input current x regulator voltage = 5 V x (1 A + 1 mA). Is that correct ?

 

[danadakk]

The required input power of the regulator is = regulator input current x regulator voltage = 5 V x (1 A + 1 mA). Is that correct ?

Short aswer is yes.

 

[betwixt]

confirmed!

 

[KlausST]

Hi,

what about Figure#12, on page#9, red line, when V_in = 5V?
which almost tells the same as Fig.#13 @ 1A

Sometimes it is called "no load current" or "bias current" or "GND leg current" or similar. (the current will deend on output current)

Klaus

 

[danadakk]

The sum of all currents entering and exiting a N port network = 0

Kirchhoff's circuit laws - Wikipedia

en.wikipedia.org

Regards, Dana.

 

[Easy peasy]

Output current: 1 A
Output voltage: 3.3 V

If the input voltage to the LDO linear regulator is 5 V,

If the above is true the power dissipated in the LDO will be 1.7 watts - so at least 1square inch of pcb copper to dissipate this