Input power and output power of LDO

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

Brian.

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

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

engr_joni_ee said:

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 ?

Click to expand...

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

betwixt said:

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

Brian.

Click to expand...

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

Part number: TPS7A4533KTTR


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 ?

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

Click to expand...

Short aswer is yes.

confirmed!

Hi,

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

KlausST said:

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

Click to expand...

Klaus

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




Regards, Dana.

engr_joni_ee said:

Output current: 1 A
Output voltage: 3.3 V

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

Click to expand...

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