| ??? 07/09/09 13:11 Read: times |
#166927 - Of course, it does! But... Responding to: ???'s previous message |
Richard said:
The source would have to be connected to supply adequate to drive the relay. If his relay loses 5 volts at rated current, then he has to have at least enough supply to support that voltage. If his pMOSFET has an on-resistance of 100 mOHMS, and his coil current has to be, say, 100 mA, then he won't lose enough from a 5-volt supply to be missed. After all, it's the current that does the work. So, the source is connected to 5V and not 3.3V. Richard said:
As for driving the gate ... if he provides a resistor divider at the gate that limits the 5 volts to 3V3 at the port pin,... How shall this divider look like? A resistor from source to gate and a resistor from gate to 0V, with the gate connected to the port pin? This would divide the 5V at the port pin to 3.3V. But how to turn on and off the pMOSFET then? Emitting low state (0V) at the port pin would cause a gate source voltage of -5V and turn on the pMOSFET. Ok. But emitting high state (3.3V) would cause a gate source voltage of 3.3V - 5V = -1.7V which is also suited to turn on a low threshold pMOSFET, like the BS108, e.g. Richard said:
Since the coil current is likely to be small, he can choose a transistor that has relatively low gate capacitance, and will suffer little in performance. The BS108 shows a gate capacitance of less than 400pF. In combination with an impedance of 4k7 in the divider this results in a time constant of about 2µsec. What is really affecting the turn on time of the relay is not the gate capacitance but the turn on time of the relay itself, which is in the msec range, usually. Kai |
