linear regulators impedance

[akash reddy]

could anyone suggest me a method on how to calculate the input impedance of a linear regulator practically.?

 

[dick_freebird]

What is input impedance? dVin/dIin. What is dIin? dIout
pretty much. dIout? That's basically your PSRR (or kVI,
in some power guys' terminology), working through dVout
into the load resistance.

So at fixed output load, figure dVout = 1+kVI, dIout =
dVout/Rload, dIin = dIout+dIgnd and Zin = dVin/dIin.

This is for DC; AC gets much messier with loop dynamics
and compensation, the output filter capacitance and so
on.

 

[chuckey]

the input impedance to DC is extremely high because any change of input voltage does not cause any extra current to flow. This is because the input current is the same as the output current which is set by Vout which is constant. As the output voltage is constant, the load is constant, the output curent is constant, so the input current is constant. However to measure it set uo your regulator with a load, and a low value resistor in series with the input lead. Measure the Vin and Iin to the regulator. Increase the Vin to Vin2, you will get another input current Iin2. So the input impedance is (Vin2 -Vin)/(Iin2 -Iin).
Frank