Help - Problem

179

02:05:21

The capacitor network below charges because the reverse biased diode (diode in parallel to a FET) is shorted by a FET. What I want is to "unshort" this diode after the capacitor charge reaches a certain value, say 10V. The Idea is that the network will be attached to power line of household, and to avoid additional power consumption, it has to be disconnected once the charge is complete.

Any ideas on how I might attain this?

published 4 years ago

BillyT

4 years ago

There are a few problems with this circuit, the Max Voltage between gate & source should only be about 15V & the zeners require a working current, (13.6mA in EC). The diode is internal to the FET & thus could not be un shorted.

PrathikP

4 years ago

@erics... That makes no sense. Why would you want to disconnect the capacitor from the grid? You will have to reconnect it within seconds because the capacitor will loose charge through the load and on its own. You will not save any power by disconnecting it from the grid. And @billyT, you are addressing completely the wrong thing.

PrathikP

4 years ago

To answer your question @erics..., You need to use two back to back MOSFETs to "nullify" the effect of the body diodes. The FETs would require an isolated gate drive. That would make it a solid state relay, which consumes some power of its own, so whatever little power you save by disconnecting the capacitor is lost in the MOSFET switching circuitry.

PrathikP

4 years ago

But anyways, if you want to learn how a solid state relay works, check this http://everycircuit.com/circuit/5169011642597376 or paste the number at the end in the search box if the link does not work

BillyT

4 years ago

@PrathikP, my statement only addressed the obvious faults that had to be overcome before his query could be started to be addressed.

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