You're trying to compare a BJT and MOSFET in switching application.
First you need to define your application (power, frequency, current, voltage...etc), then you compare the two transistors according to the datasheet of each one.
Bipolar transistors are more adapted to a general purpose switching applications, for high power MOSFET are more suitable because they are faster than BJT. for the MOSFET see the switching parameters in the datasheet. BJT loss current when they are used in high power applications.
While there are relays that will do what you want, these are not the ones.
Better would be either a suitable low Rdson MOSFET per battery, or a properly rated relay.
Relays of this type can generally be expected to be junk for practical purposes at LV DC and dangerous junk at mains voltages. See the pinout footprint in the datasheet below - isolation is much reduced by placing the common contact between the coil contacts. OKish on LV, terrible for mains.
You need to say what Voltage and MAX current you are switching.
The relay is dated at 28V DC at:
10A in SPST mode with resistive load.
7A in SPDT mode with resistive load.
5A in SPST mode with inductive load.
3A in SPDT mode with inductive load.
What current MAXIMUM does your copter draw?
What is your operating voltage?
What voltage drop do you measure across the relay contacts during operation?
[If you have not measured it, why are you asking questions about it?]
At 10A the contacts MAY drop 1V.
At 50A = 5V !!! :-( (and the relay would die)
While the 100 milliOhm contact resistance specification (see below) is a
.
I found this data sheet - the same in details as Edsign links to but not as nicely presented.
The contact resistance is shown as "not more than 100 milliOhm" - Agh !!!!!!!!!!!!!!
At 10A that would drop 1V at maximum allowed resistance.
At 50 A = 5V drop.
Added:
OP advises.
Ioperating ~= 50A
Battery = 3S LiIon.
Vbattery = 3.6V x 3 = 10.8V nominal. ( 9V min, 12.6V max)
The chosen relay is totally unsuited to the task. You either need
A MOSFET at 50A will dissipate
I^2 x R = 2.5 Watt per milliOhm of Rdson.
So even a 5 mOhm FET dissipates 12.5+ Watts - not undoable but almost makes a properly rated relay attractive. (But, see below, a 0.4 milliOhm Rdson MOSFET - 1 Watt at 50 A.
You can buy 30V, 300A, 0.4 milliOhm Rdson MOSFEts for $US3.43 each i nquantity 10 (!!!). Infineon IPT004N03L. Utterly awesome. Interesting package . 8 power SFN - looks unusual but has drain tab on 1 side, 8 pins on other. Pin 1 = gate, 2-8 = source.
Good starting search here
Best Answer
Those MOSFET formulas don't take into account any switching; only the time in between - I'm pretty sure you knew that, but anyway.
When the transistor is on, it has 2v across it and 25A going through it.
But duty cycle is 50%, so during the on period it dissipates 50W, but overall, 25W. Actually, it's a but less than that, since there is the turn off / on time of 100ns.
The total cycle time = 1 / 100kHz = 10us, so with switch time being 100ns, that's 1% each way. So, your transistor is only fully on for 49% of the time. Substitute as above, and you get 24.5W for the fully on portion.
When off, no current flows, therefore no power is dissipated. So that's zero watts.
Finally, the rise and fall time. This could get very tricky depending how far you want to go; your exercise gives little hint as far as I gather what's expected; only the "inductive" load for which AFAIK detail are missing; that, and that the model of the transistor is highly simplified, so they're trying to get you to focus on something other than that. So, guessing, but maybe they want the following.
So, rise and fall: it will dissipate a lot if power during this transition period; consider when half the voltage is across the transistor, and the other half across a resistive load. The current through both would be 12.5A. At 300v, each will dissipate 12.5 x 300 = 3750W (at that instant). That would be the peak power dissipating in the transistor (impedance matching) so that gives you an upper bound for switching losses. 2% of the time, no more than (2% x 3750W) = 75W would be lost. So the total losses would no more than 99.5W. But as I said, that assumes a resistive load.