Need help in understanding how this works.
In the attached schematic is shown the DC voltage measurements at top of C12 and C13 relative to “b” to be 11.46 volts and 5.4 volts. This means the current through R20 must be 0.0033 amps. However, this amperage through R18, the only DC source to ground suggests a voltage of 1100 DC volts which it cannot be. Just how are the voltages of 11.46 and 5.4 achieved?

 

I redrew your circuit upside down from what you drew. This circuit lives on the power line! It really does not matter if it is on the line or neutral. Do not hock a scope to it because the circuit is not connected to ground.

Left side, V1 is the power line voltage, 150Vpk, 60hz.
Power flows from the power line (ac) through C11.
--1/2 cycle, current flows C11, R17 D1 (backwards). Voltage across D1 is -0.7V.
--1/2 cycle, current flows C11, R17, D2, C12. C12 charges up. Voltage on D1 is 12V. (Zener diode limits voltage to 12v)
Voltage on C12 is about 11.4V.
Power flows from C12 through R1 to charge up C13, limited to 6V because of D3 Zener.

Not important, R18 is to discharge C11 after the power is removed. R17 limits the inrush current at power up.

Hope this helps. RonS.

 

Ron,
Thanks for your reply. May ask an additional question? As described in your first half cycle, current is flowing upward through D1 per your drawing. On the second half cycle, D1 limits the DC voltage to 12 volts, does it also limit the AC positive cycle to 12 Volts peak? And if not, why not? I like the way you redrew the circuit, but my drawing reflects what is actually on the circuit board. Does it make any difference?

 

The zener diode D1 conducts and prevents the voltage from exceeding 12V when its cathode reaches +12V. The cathode is the upper terminal (the bar on the symbol).

 

The zener diode D1 conducts and prevents the voltage from exceeding 12V when its cathode reaches +12V. The cathode is the upper terminal (the bar on the symbol).

My question is whether the zener diode in this case cuts the 150 volt AC peak to 12 volts peak? Thanks for your reply.

 

My question is whether the zener diode in this case cuts the 150 volt AC peak to 12 volts peak? Thanks for your reply.

Yes it does.

 

Need help in understanding how this works.
In the attached schematic is shown the DC voltage measurements at top of C12 and C13 relative to “b” to be 11.46 volts and 5.4 volts. This means the current through R20 must be 0.0033 amps. However, this amperage through R18, the only DC source to ground suggests a voltage of 1100 DC volts which it cannot be. Just how are the voltages of 11.46 and 5.4 achieved?

It's called a Transformer-less psu, they're use a lot in cheap electric lighting circuits like pir lights and light bulbs.

 

However, this amperage through R18, the only DC source to ground suggests a voltage of 1100 DC volts which it cannot be.

R18 is in parallel with C11. The impedance of C11 at 50Hz is 4.68k (3.9k @ 60Hz) so just 128V @33mA.

 

Thanks to all for great response.

 

C11, at 60hz, looks much like a 4000 ohm resistor. So turn C11 + R17 into a 4k resistor that feeds D1. The voltage at the top of D1 is clamped. It will be (-1V or +12V).

Here is the voltage across D1.