Im having trouble getting my head around using two mosfets to pass AC.
I have reduced my schematic to its bare bones to help keep it un-cluttered. The aim of this circuit is to use PWM to dim a bulb running AC. the PWM circuitry and gate drive is omitted.
I am taking 25V AC rms through a bridge rectifier to allow me to maintain a DC level at the gate through the voltage divider with the aid of a cap.
The above circuit caused my lower fet to fry and also the insulation on the wire from the source to my bridge negative terminal melted.
Any assistance is greatly appreciated!
EDIT: I have amended the circuit to eliminate the short.
But i am unclear as to how the VGS limit of 20V is not exceeded. The output of the Divider is around 7V (more than enough for my fet). So if the top rail initially peaks at 34V and charges the cap, then the cap will hold the gates at 7V relative to the source ( and cathode of the cap). For the opposing cycle, the lower rail peaks at 34V, or relative to the top rail/capacitor -34V (i believe this is where my logic is incorrect). When the bottom fet opens up, this -34v potential is passed to the source before travelling through the body diode of the top fet to the load. if this is true, the VGS potential would be 7V (positive at the gate) + the negative 34V peak at the source passed from the drain = 41V.
I have been told in a forum elsewhere this reasoning is incorrect and a simulation proves the VGS potential to maintain at 7V.
I do not doubt the simulation may be correct, however i am failing to prove to myself either way what is happening.
If you think at how the bridge works, you see that it will always connect + to the AC terminal having the highest potential and - to the AC terminal with lowest potential. And let's call "ground" the lowest AC potential, the one that is connected to - of the bridge.
The source of M2 (top terminal of your M2) is therefore connected to this "ground". In the other hand both positive and negative terminals of AC power will have a potential that will be equal or above this "ground" (alternatively half cycle equal, half cycle above). But one terminal of AC power is connected to M2 drain (bottom terminal of M2).
M2 is always ON because gate to source voltage is positive, you are therefore shorting AC power through M2 during half cycle.
One way of doing it, as I can think of using MOSFET is showed in the following schematics. D1 and D2 avoids reverse conduction. M1 (an NMOS) allows conduction during positive cycle of AC and M2 (a PMOS) allows conduction during negative cycle. Also M1 only conducts if gate is positive in regards to AC bottom rail, and M2 only conducts if gate is negative in regards of AC bottom rail. Then by setting + and - as showed in the schematics the bulb would be 100% ON. By reversing + and - voltages in the gates, you would get a 0% ON.
One other possibility using only NMOS could be this one:
By the way, R1 allows some current to flow through the bridge so that the diodes in the bridge are ON, and D1 (which is indeed necessary) disconnects the capacitor from AC when its voltage is above AC. At this moment D1 is OFF but not the diodes in the bridge so that (-) terminal is always connected to the lowest AC potential. Then the voltage divider gets about 15 V or so, if your AC voltage is 110 V, this is the "ON" signal.
During positive AC cycle, current goes from top to bottom through body diode of M3, and through M1's channel. During negative AC cycle, current goes bottom to top through body diode of M1 and through M3's channel.
This circuit idea could work, but I've maybe overlooked something. It is better to try it with a reduced voltage at first.
It is necessary also to pay attention at the absolute maximum ratings. For example when AC voltage is 100 V during positive cycle, VGS of M3 would be around -85V, its channel will be off as we expect, but the transistor will be likely destroyed! Therefore it is necessary to clip M3 "ON" signal voltage to avoid going too much negative.