Whether heatsinks must be isolated depends on the situation..... IE What else is connected to the heatsinks, are they connected to/integral to the case?? It is usually against all kinds of UL and CE safety ratings to have a large heatsink electrically live like that LOL.... Of course it all depends on the voltages involved, and what the particulars of the situation are... Use common sense to determine.
You asked if electrically isolating the heatsinks would decrease the thermal performance... YES IT WILL... It has a large impact on thermal performance and care should be given in calculating heat rise of the transistor to NOT FORGET the increased thermal resistance added by the isolator..... You can find pretty decent estimates of the thermal resistance introduced by the various kinds of isolators by searching Google....
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As to your other questions:
1.Surface mount MOSFETs are certainly NOT the best for high power applications..... Specifically, the more effectively you can cool a transistor, the more current it can handle.... If you had some magic way to infinitely cool it's semiconductor junctions, then you could push an infinite current through it. This is the idea with superconductors.
Through hole components are almost always MUCH more effectively cooled via the proper use of heatsinks than any surface mount component can ever be through it's circuit board.....
Calculate how much current needs to be handled under the worst case scenario.. IE Highest temperature, motor fully loaded or even seized, THEN determine whether it makes sense from an economic and design perspective to use surface mount components or not. In certain cases where the final size of the device is critical, metal core circuit boards are made to help cool the SMT parts..... Most of the time for a home project, you can give yourself a lot more "elbow room" and ease in design concerning the thermal parameters of the board, in addition to a wider selection of high powered parts by using traditional components.
2. Short answer: This is totally unreasonable. Long answer: While a PCB trace can theoretically carry any amount of current when properly sized, a 1 OZ PCB trace would have to be at LEAST a couple of inches wide in order to carry 100 amps of current.... Also, how would you attempt to make a connection to this trace???? This is problematic. You can easily calculate the {DC} current carrying capacity of a PCB trace of a given width and thickness by considering the conductivity of the copper of which it is made. 1 OZ PCB uses a copper conductor which is about 35 micrometers thick.
Keep your high current paths as short as possible to try to eliminate as much heat production as possible, and consider that if this board is inside a case, that it is not a simple matter of the temperature rise in the copper being defined as the temperature rise of an uninsulated conductor in free space. All the heat builds up in the case, which causes higher resistance, which causes MORE heat to build up..... This same idea applies to ALL of your other components such as capacitors and transistors as well.
Also, at the point that if you have soldered a big wire all along the trace, why not just use an insulated wire to begin with?????
3. Switch current is the total current flowing through the MOSFET..... This is of course dominated by the current from drain to source, usually abbreviated I{subscriptDsubscriptS} Most of the time, when considering the switch current, the gate to source current is proportionally so small that it is disregarded for "quick and dirty general use" calculations...
In most cases, you figure the switch current as the duty cycle ratio multiplied by the current from drain to source while fully on.
......... I think that what you are trying to get at here is the gate current, as that is where your gate charge becomes important, and is the figure you need to know when deciding on a proper driving method to use a MOSFET. The gate current is usually largely affected by the use of a resistor in series in between your driver {In this case, the pin from your microcontroller} and the gate of the transistor. This series gate resistor will prevent your microcontroller pin from attempting to source too much current at once and burning out. The higher the current it can safely source, the lower a gate resistor you can use, and the faster the MOSFET will switch, leading to less "switching losses" during the turn on of the MOSFET.
Remember, the magic of MOSFETs is that they have very small resistance from drain to source when they are "fully on" and saturated. While they are in the linear region of their operation --IE where the voltage drop from drain to source is linearly inversely proportional to the gate to source voltage-- they have high losses. You want to turn these things "on and off" as quickly as possible to avoid dissipating large amounts of power and heat inside the transistor.
This is largely why special mosfet driver circuits and ICs are used in these applications rather than just driving the gate directly from a microcontroller. Higher current being used to drive the MOSFET = shorter time to overcome gate capacitance = faster MOSFET turn on = higher efficiency overall.
Also, do not forget the use and purpose of a pull-down connection at the gate of the MOSFET as well in order to help it turn off quickly and "completely" once the drive voltage is removed.... Forgetting this can cause the MOSFET to slowly turn off, or otherwise to oscillate and cause all kinds of problems.
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Go ahead and use logic level components if it is going to make more sense in this particular design. Although they have some disadvantages, they have upsides too, and otherwise they wouldn't be manufactured at all. Some of them are designed to switch quickly, having a low gate charge, and with tolerable "on state resistance" as well..... Look at the Fairchild FDB7030BL for example.... Logic level MOSFET comes as either a surface mount part or through hole, with a 60 amp continuous current rating....
Here's the datasheet:
https://www.fairchildsemi.com/ds/FD/FDB7030BL.pdf
ALSO please keep in mind that when a datasheet says something like 60 amp continuous current rating, that this is ENTIRELY dependent on ideal conditions as far as thermal dissipation and switching profile is concerned, and that actual power dissipation and continuous current will be lower depending on your heatsink and particulars like the applied gate voltage and switching frequency.....
There is a lot to know when you get into working with high power semiconductors, especially when it's at higher frequencies and not purely being used like a light switch or something.....
Good luck my friend, I hope this has been helpful!