I wanted to ask what the purpose of the high pass filter does in the below circuit. Is it necessary?

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  • 1
    \$\begingroup\$ Is it from a LED lamp by any chance, because that is a common technique to ballast them? \$\endgroup\$ – winny Feb 27 at 7:05
  • \$\begingroup\$ As winny mentioned, it is used in LED circuits. Check out "transformerless power supply" \$\endgroup\$ – C K Feb 27 at 8:26
  • \$\begingroup\$ I hate to state the bleedin' obvious, but why would they fit it if it wasn't? \$\endgroup\$ – Finbarr Feb 27 at 9:30
  • \$\begingroup\$ Could I call it a capacitive dropper without discharge resistor but with a current limiting resistor? \$\endgroup\$ – Max Ried Feb 27 at 13:21

Is it necessary?

If it isn't there you have full mains voltage on the bridge rectifier.

The capacitor has an impedance to AC given by \$ Z = \frac {1}{\omega C} = \frac {1}{2 \pi f C} \$ where f is the mains frequency. The result is that it will drop the voltage to the load and, since it is a capacitor, there will be very little power dissipated in it and this eliminates the heating problem associated with a resistor voltage dropper.

Please be aware that this circuit provides no isolation from the mains and that the low voltage side must be considered live.

  • \$\begingroup\$ Thanks. Since it drops the remaining voltage efficiently, this circuit is a current limiting circuit - not a high pass filter - right? Also, can I get rid of the 27 ohm resistor since it doesn't really affect the overall impedance? \$\endgroup\$ – Alex Feng Feb 28 at 1:48
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    \$\begingroup\$ Yes, it is current limiting. The 27 ohm resistor may be required to limit the current to a safe value in the case when the mains is at a high voltage at the moment of switch on. I advise you to study the theory behind these very carefully before building one. \$\endgroup\$ – Transistor Feb 28 at 7:09

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