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I posted a more vague version of this question a few days ago here and this question was linked as being similar. Answers in both threads were extremely helpful and were along the lines I was thinking, namely using a transistor or FET to switch the resistor on when power is off. This would drain the smoothing capacitor quickly and allow maximum power while running, both of which I need.

It was mentioned that if I provided a schematic, then my question can be better addressed. Here is the schematic.

It is a simple capacitive power supply for 120 V AC. I've racked my brain trying to find a way to connect the transistor or FET in this particular circuit. Is it even possible?

I'm hoping for a simple solution; something relatively small (< 1 cm^3) and cheap (< $1).

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  • \$\begingroup\$ It seems to me that your question was answered in your previous thread. \$\endgroup\$
    – user207421
    Oct 10 '12 at 6:55
  • \$\begingroup\$ possible duplicate of Bleeder resistor: switch resistance on when capacitor is powered off? \$\endgroup\$
    – Dave Tweed
    Oct 10 '12 at 12:44
  • \$\begingroup\$ The resistor 5 seems to be wrong placed in this schematic. \$\endgroup\$
    – johnfound
    Nov 5 '13 at 5:37
  • \$\begingroup\$ Why you are trying to discharge the capacitor? Because of the safety? Or other reason. How about the capacitor 2? \$\endgroup\$
    – johnfound
    Nov 5 '13 at 5:38
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Use a relay driven by the rectified/filtered power. Connect the NC terminal to the bleeder resistor.

schematic

simulate this circuit – Schematic created using CircuitLab

Find a relay with suitable coil voltage to not need separate regulation/resistance for your output voltage.

Btw: Your diagram is DANGEROUSLY WRONG. With 120V AC input, the load/device you're showing marked as "64V" will see up to 180V, as the 120V is RMS-to-zero, not peak-to-peak, and if there is no load, the capacitor will keep the voltage at the peak value rather than the RMS value.

If you really need 64V unregulated, I'd recommend a 2.5:1 transformer added to your circuit. Also, the resistor in the diagram (please label your components!) seems like a waste. It will just burn off power if you draw power, and do nothing if you don't. If it's there to lower the voltage to 64V, then that will only happen if the load is very even and well-characterized. And the resistor will still get very warm -- power equals voltage squared divided by resistance. If it has to drop 100 Volts at 60 mA, it's about 1.7 kOhm, and 60 mA through 1.7 kOhm is over six watts!

I'm tempted to ask what probably you're actually trying to solve here. Working with mains power is pretty dangerous, and extra so if you're going to hook something into power permanently, or if anyone other than you will ever go near the device.

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Use an NC/NO switch, like this:

enter image description here

and connect the resistor to the NO part while having the NC on the supply bus.

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  • \$\begingroup\$ In what way does this answer have anything to do with the question, which is about automatically discharging the capacitors when power is removed. \$\endgroup\$
    – Dave Tweed
    Oct 10 '12 at 11:23
  • \$\begingroup\$ "Power removed" means unplugged, to me. Since we're talking about power and its interruption, we may as well talk about a switch, to turn it on/off. This simple idea means that when the power is on, the bleeder is floating, and when it's off, the bleeder is connected, discharging the capacitor, which is what OP wants, only in a cheaper version. It can also be done with a jack-like connector, for the "automatical" part. The drawing is just to suggest what I'm talking about, flashy drawings tend to distract sometimes. Simple and to the point is my aim. Everything has to do with OP's question. \$\endgroup\$
    – Vlad
    Oct 10 '12 at 14:47

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