Voltage rise during circuit closure

[shaiko]

A flyback diode between the Source and Drain of a transistor is used to manage the voltage rise between the terminals of the switch during circuit opening.
But how do we manage voltage rise during circuit closure ?
V = L * (dI/dt)
When "I" rises suddenly the Vds of the transistor increases accordingly and can damage the switch - right?
How do we manage this problem?

 

[jiripolivka]

A flyback diode between the Source and Drain of a transistor is used to manage the voltage rise between the terminals of the switch during circuit opening.
But how do we manage voltage rise during circuit closure ?
V = L * (dI/dt)
When "I" rises suddenly the Vds of the transistor increases accordingly and can damage the switch - right?
How do we manage this problem?

Any inductive load creates an overvoltage as you described. To protect the switching element, a diode is connected in parallel with the switch to swallow the pulse. Some FET switches have such diode built in, use a 1N4007 diode . If your switch has a positive Vcc connect diode positive to collector of NPN transistor or to FET drain, and negative to ground.

 

[shaiko]

This isn't what I asked.
My question was regarding circuit connection (not disconnection)

 

[crutschow]

A diode across the transistor cannot protect against turn-off transients for an inductive load in the drain since the transient generates a positive spike at the drain (for an N-MOSFET). For that protection you need a reverse-biased diode directly across the inductor (or from the drain [anode] to the plus supply [cathode]).

You don't need any protection during turn-on of an inductive load, since that produces no voltage transients. The transistor applies V to the inductor and the current rises according to di = V *dt / L.

 

[dick_freebird]

Connecting, in a discontinuous conduction mode, you
should not need a diode - current is zero, dI/dt voltage
is nicely shunted by the FET on resistance. Maybe in a
continuous conduction mode there could be a sliver of
the cycle where circulating current wants to take the
drain negative and would wake up the body diode with
undesirable results; a low-Vf antiparallel diode is good
for that, but it's not the flyback diode (which shunts
the load, not the switch).

Disconnecting is where you have the current all wound
up and then take away its path, leaving it to find the
weakest link unless you provide one explicitly.