Will it damage an IRF740 MOSFET to have a positive 108 V potential connected to it's drain before a conductive signal is connected to it's gate?
Electronic – MOSFET question
mosfet
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\$R_{DS(ON)}\$ is an important parameter, and many datasheets start with mentioning values for them.
For the FDC885N two values are mentioned in the Features section at the start of the datasheet:
Max \$R_{DS(ON)}\$ = 27m\$\Omega\$ at \$V_{GS}\$ = 10V, \$I_D\$ = 6.1A
Max \$R_{DS(ON)}\$ = 36m\$\Omega\$ at \$V_{GS}\$ = 4.5V, \$I_D\$ = 5.3A
From the same datasheet:
So, yes, \$R_{DS(ON)}\$ varies with \$V_{GS}\$, and yes, it's higher at higher temperatures.
If your manufacturer can't give you the information and you really need it, move on to another manufacturer.
A MOSFET always consumes power from the circuit. It has no mechanism to convert energy from some other form to electrical energy.
Therefore, the currents through a MOSFET always flow from a higher potential to a lower one.
This means, for an n-channel FET, if the drain is biased higher than the source, current will flow from drain to source (through the channel). If the source is biased higher than the drain, current will flow from source to drain (through the body diode).
Now, Lets say I have connected Drain to higher than Source and current is flowing through the body diode, But at the same time I want to turn on the MOSFET Source to Drain channel to allow current to flow from the channel from source to drain parallel with body diode.
If you turn on the FET, you may get parallel conduction through the channel and body diode, but both currents will flow from drain to source, because the drain is at a higher potential.
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Best Answer
If you mean that the gate is left open, then yes the part could be damaged.
When the gate is left floating it can get a high enough \$V_{\text{gs}}\$ to turn on. If the source of \$V_{\text{ds}}\$ has sufficient current the FET could be damaged. Some FETs have a ratio of \$C_{\text{gs}}\$ to \$C_{\text{dg}}\$ that applying a voltage to the drain with open gate will turn on the FET.