i want to replace a relay to switch Ac load(heater ) that will draw about 20 Amps . i used IRFP250 and bridge rectifier as in the picture . i still didn't do it . the voltage gate come from a plc (programmable logic controller) 24 Volt positive or negative .
will this circuit need any thing else ? will the mosfet need heat sink to act as a switch ?
Damn, this gona blow ))) Man, don't even think about turn all of this on.
Use **broken link removed** if you want so, but please, don't show this schematic anybody else.
You trying to supply AC current motor with DC current.
Your diode bridge (right one) will have to dissipate 20A*2*(~1.5V)=60W power in heat.
And somehow you should find isolated DC voltage source to open the mosfet.
I have no idea what will blow first, but it could be very impressive.
The schematic makes no sense. You say "heater load", but the schematic shows a motor.
A heater could be possibly supplied with DC voltage, but why you should? AC switching is much easier with triac and triac optocoupler, you also get rid of the auxilary supply.
The schematic makes no sense. You say "heater load", but the schematic shows a motor.
A heater could be possibly supplied with DC voltage, but why you should? AC switching is much easier with triac and triac optocoupler, you also get rid of the auxilary supply.
The schematic has many vague points.
why you have add a full-bridge rectifier? (Right-side of the picture)?
in this configuration voltage over the heater is DC.
I recommend using solid-state relay. i have seen some in TE site which works with 24V command voltage.
you could also build one, yourself.
all in all, this circuit strongly is not recommended.