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A Low Dropout Regulator (LDO) is a Linear Voltage Regulator with a very low minimum input-to-output differential voltage of a few hundred millivolts. It means that the output voltage can be maintained at the set value even if the input voltage is at a level very close to the output voltage level.
/This circuit has been modified. It now uses MOSFETs everywhere and not a combo of MOSFETs and BJTs/
This circuit uses the IRF4905 P-MOSFET as the series pass transistor and the BD522 N-MOSFET and the BD512 P-MOSFET as auxiliary transistors.
The circuit is configured to output 5V, and features a Dropout Voltage of 50mV*. It has a quiescent current of 786uA**. The Line Regulation is 0.006%***.
To achieve low dropout, a current sense resistor is omitted. The obvious disadvantage of this is that there is no current limiting.
*Measured at Io = 500mA. But the circuit has not been tested in real life.
**Measured at Vi = 10V and Io = 0A.
***Measured at Io = 500mA and 20V<Vi<170V, which is obviously not something that can be done IRL as the FET would blow up, but the scope has a low resolution, leaving me with no other choice. I am not able to measure the load regulation for the same reason.
Derivation of the output voltage:
Applying Voltage Divider Theorem at the output, we get
Vref = Vo∙(R2/(R1+R2))
Where R1 is the upper resistor in the voltage divider network and R2 is the lower resistor.
We need to solve for Vo. Rearranging the equation and putting Vref = Vgs, we get
Vo = Vgs∙(1+R1/R2)
Here, Vgs is the Vgs(min) value of the auxiliary MOSFET. For the BD522, it is 2V****.
Selecting R1 for a given Vo and R2:
Putting Vo = 5V, R2 = 9.9K and Vgs = 2V, we get R1 = 14.6K.
****This value could change with temperature and affect the output voltage.
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