how can i increase the output current in a transformerless power supply?
Summary


ALWAYS remember that any part of this circuit MUST be considered to be live and that a user may die if they touch any part of this circuit.
If your circuit is half wave and operating at 110 VAC and 60 Hz then the capacitor needs to be about 10 uF.
If your circuit is half wave and operating at 230 VAC and 50 Hz then Iout should be about 190 mA. If it isn't then you may have the zener in the wrong position.
The zener diode should be on the AC side of the rectifier diode.
Note that capacitor tolerances can be wide - your capacitor actual value  needs to be at least as large as calculations indicate.
A comprehensive design guide can be found in
Microchip AN954 Transformerless Power Supplies ...  


An offline transformerless supply can be made using a series capacitor to provide series impedance.
 capacitor impedance is 1/(2 x Pi x frequency x C).
 For say 1 uF at say 60 Hz Xc = 1/( 2 x 3.14 x 60 x 1E-6) =~ 2650 ohms reactive

For 110 VAC if this capacitor was placed across the mains the AC current flow would be
 I = V/R = 110/2650 = 40 mA.
 To get 160 mA the capacitor would need to be 4 uF.

BUT that's for AC power.
 If a full wave bridge is used the power delivered at DC is the same but if, as is usual, half wave is used then the capacitor needs to be about double or here = 8 uF. Say 10 uF actual.

IF the 475k capacotor = 4.7 uF (as it appears) and if you are running on 110 VAC then your capacitor appears to be about half the size it should be. 

Note also the comment on zener placement below. . 

To make this into a power supply, rather than just connecting across the mains a "load" is created which uses the AC current to create a lower voltage supply. The driving voltage is now Vac - Vdc. Not much change usually. A simple version of this looks like this:

 

The resistor limits surge current. It can play a role in voltage dropping if large enough but is usually small.
 The zener diode provides its breakdown voltage in one direction and acts like a diode in the other direction. Without this diode action the capacitor would have no AC return path and the circuit would not work.
 If the zener is moved to the DC side of D2 then a reverse biased diode must be used where the zener now is 

__

The power output can be about doubled by using full wave rectification - like so:



An online calculator that allows the design of a full wave transformerless power supply can be found here .
 Using this calculator a value of 10 uF gave 218 mA at ~12 VDC for 110 VAC, 60 Hz input. 50 Hz gave 180 mA. 

230 VAC, 50 HZ, 4.75 uF full wave , 12 VDC gives 192 mA
A half wave circuit will give about half that.



ADDED

C1 MUST be mains rated.
 BR1 can be 4 x 1N400x diodes or a mains rated bridge. (Theoretically does not need to be mains rated but ...).

Caps that are rated to connect across mains (phase-neutral) are X rated and must withstand Vrms x 1.414 PLUS expected peaks and surges and transients. They are made of sterner stuff than eg just a 230 VAC x 1.414 Volt cap for 230 VAC operation. Caps that are designed to operate from mains to ground (eg phase-ground) are Y rated - a lesser but still demanding requirement.