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Well, it is quite simple actually:

$$X_c= \frac{1}{2}\pi f c$$

That is your capacitor impedance. It changes with 60 or 50 Hz.

Your maximum current will always be:

$$\frac{V_{in}-V_{diode\;drop}-V_{zener}}{X_c//2.2M + 47}$$

And by the way, if you put any load, below the maximum current on the right side of Zener diode, your \$V_{in}\$ current will stay exactly the same... The Zener diode takes care of that.

Well, it is quite simple actually:

$$X_c= \frac{1}{2\pi f c}$$

That is your capacitor impedance. It changes with 60 or 50 Hz.

Your maximum current will always be:

$$\frac{V_{in}-V_{diode\;drop}-V_{zener}}{X_c//2.2M + 47}$$

And by the way, if you put any load, below the maximum current on the right side of Zener diode, your \$V_{in}\$ current will stay exactly the same... The Zener diode takes care of that.

Well it is quite simple actually:

$$X_c= \frac{1}{2}\pi f c$$

That is you cap impedance. It changes with 60 or 50Hz.

Your maximum current will always be:

$$\frac{V_{in}-V_{diode\;drop}-V_{zener}}{X_c//2.2M + 47}$$

And by the way, if you put any load, below the maximum current on the right side of zener, your \$V_{in}\$ current will stay exactly the same.. The zener takes care of that.

Well, it is quite simple actually:

$$X_c= \frac{1}{2}\pi f c$$

That is your capacitor impedance. It changes with 60 or 50 Hz.

Your maximum current will always be:

$$\frac{V_{in}-V_{diode\;drop}-V_{zener}}{X_c//2.2M + 47}$$

And by the way, if you put any load, below the maximum current on the right side of Zener diode, your \$V_{in}\$ current will stay exactly the same... The Zener diode takes care of that.

Well it is quite simple actually... Xc= 1/2pif*c that is you cap impedance. Changes with 60 or 50Hz

your maximum current will always be (Vin-Vdiode_drop-Vzener)/(Xc//2.2M + 47)

And by the way, if you put any load, below the maximum current on the right side of zener, your Vin current will stay exactly the same.... zener takes care of that....

Well it is quite simple actually:

$$X_c= \frac{1}{2}\pi f c$$

That is you cap impedance. It changes with 60 or 50Hz.

Your maximum current will always be:

$$\frac{V_{in}-V_{diode\;drop}-V_{zener}}{X_c//2.2M + 47}$$

And by the way, if you put any load, below the maximum current on the right side of zener, your \$V_{in}\$ current will stay exactly the same.. The zener takes care of that.