![]() ![]() 477mA, giving about 4.78 V to the gate once charged, activating the mosfet. If the transistor is off, the circuit is a voltage divider, 470 + 10000.I have thought about the common emitter config, and so far have come up with this schematic (attached). ![]() I put the R15 100 ohm resistor to limit the gate charge current, as to limit the ‘surge’ in the optocoupler transistor. The 470ohm resistor is to limit the current through the transistor to about 10mA or arounds there. When the optocoupler transistor is on, the mosfet gate charges through R15 and will quickly gain full charge and have no current, meaning we will have (5 - Vforward transistor), ( 5 - 0.7V) = 4.3V or so on the gate, Correct? When the optocoupler transistor is off, the gate of the mosfet goes to ground, through the 2 resistors, shutting it off, so power loss on the 5V or uC failure means mosfet = off. The reason I did the schematic that way (correct me if I am wrong I am still new at this) I think it would be equivalent to the one I put in my schematic, except for the base. The optocoupler I am going to use is a PS2501-4A, 16 pin, 4 channels.The mosfets I am going to use are similar but not the one on the schematic: Either FQP30N06L or AOI518 ( both logic level N channel, 2.6V max Vgs(th) ). ![]()
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