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I have a 12V DC circuit that I want to switch on/off, based on 8V AC signal.

Can I do it with a diode connected to the gate of a transitor?

EDIT: The AC circuit is a door intercom. I'm trying to couple it with a wireless circuit that works with a 12V battery. The DC circuit should stay on only when the the bell (AC) of the intercom is on.

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  • \$\begingroup\$ do you want it turned on & stay on when the AC is applied? do you want it to go on-off-on-off when the AC is applied. How quickly do you want it to turn-off when the AC is removed \$\endgroup\$ – JonRB Dec 16 '17 at 21:52
  • \$\begingroup\$ Please check my edit. The DC circuit should stay on only when the AC circuit is on. How fast it should make the transition is not important for this application. \$\endgroup\$ – cinico Dec 16 '17 at 21:59
  • \$\begingroup\$ A simple peak detect circuit to drive the gate of a FET to then switch on/off this 12V circuit should do then: D-RC-Fet \$\endgroup\$ – JonRB Dec 16 '17 at 22:02
  • \$\begingroup\$ @JonRB Why should I use the RC in this case? \$\endgroup\$ – cinico Dec 16 '17 at 22:39
  • \$\begingroup\$ to set a timeconstant to track teh peak. OTHERWISE it will turn off everyttime the AC signal drops below the gate threshold \$\endgroup\$ – JonRB Dec 16 '17 at 22:42
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schematic

simulate this circuit – Schematic created using CircuitLab

Here is a sample of the circuit that should do this: The diode does the rectification, R1*C1 provide current limiting/filtering and R2 discharges the cap when AC is off. Things to keep in mind: 1) Time constant of the circuit is large enough to not alter the gate voltage substantially during the cycle of the AC 2) The rectified voltage is less than the max Vgs of the transistor but still allows M1 to fully turn on 3) Depending on the load, you may want to implement the circuit as a high or low side switch (see dashed boxes). 4) You on time after AC is off will be determined by C1R2 time constant and the turn off voltage (Vgs) of M1

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