Look at Rds(on) - this will give you the Static Drain-Source on resistance, Typically 25milliohm. If we look at the maximums - If we push 6A through a 25millohm resistor ohms law give us a dissipation of 150mW, well within the 2W maximum.
P=I^2R
6 amps through 0.025 ohms is 0.15 volts. 0.15 volts times 6 amps is 0.9 watts.
It is still comfortably within the maximum power ratings.
Sorry, you confused the hell out of me, until I Googled the actual formula:
V = I * R
P = I * V
Therefore P = I * I * R = I^2 * R
So to confirm my understanding of it. If I pick a transistor that has 0.0021ohm Rds(on) and I have a 50A load. The power loss would be:
V = 50A * 0.0021ohm = 0.105V
Technically not power, but that's nitpicking I don't want to go in to.
Because there is a diode in a transistor, I have to add 0.7V which gives me a total of 0.805V. So:
P = 50A * 0.805V = 40.25W of power
The diode is probably reverse biased, so I don't see how adding 0.7V would help.
Moreover, getting a diode drop of 0.7V at 50A is a
very good diode.
You need to multiply 50A x 0.105V = 5.25W loss.
Using a RthJC of 2.0 C/W. The heat generated by the transistor is Max 2 C/W * 40.25 W = 80.5C (or Typical 1.4 C/W *40.25W = 56.35) above Ambient. Given that the current temp in Brisbane is 29C, the actual temp will be 80.5 + 29 = 109.5C Max (85.35 Typical). And if the Max temp for the MOSFET is 150C, I am well within the specification.
RthJC = thermal resistance Junction-Case. That is not ambient.
You need the RthJA figure, which is the total "series" thermal resistance from junction to a typical use-case of free-standing TO-220 for example. That is usually in the order of 60 C/W.
As a general rule of thumb, assume you can dissipate 1-2W in side a single TO-220 with no heatsink, but at 2W it's
really really hot.
If you install a heatsink, you will modify the RthJA figure and can't use it anymore. In that case you need to calculate your own, which is the sum of your heatsink, plus any mounting material (like heating compound or isolation pads), plus the Junction-Case specification.
So a heatsink of 10C/W with perfect contact will have a RthJA of 10+2=12C/W.
In addition when calculating , I would assume that ambient remains constant and work towards the die temperature. For example:
Tj = Ta + (RthJC + RthHeatsink) * P = 29 + (2 + 10) * 5.25 = 83.5C
And to get the thing to turn on (Vgs = 1.2V), I need a min D and S of 0.805V, and 1.2V on G.
Have I understood all of this correctly?
The gate voltage is measured against the source voltage voltage.
For a low-side driver the gate is tied to ground. 1.2V (compared to the gate.. so compared to GND) the MOSFET will open very slightly (note: the Rds on is huge and you will blow the arse out of the MOSFET with any significant current).
For a high-side driver N-channel mosfets are usually a bit of a pain, because you need to drive a voltage above the supply voltage (e.g. VCC is 12V, you would need 13.2V for the threshold voltage). A P-channel MOSFET or MOSFET driver are quite useful for that, but you may need to take into consideration if there are any duty cycle limitations.
VDS is a maximum voltage the MOSFET can withstand when it's OFF. If you exceed it, you will kill the MOSFET. The diode that is drawn is more of an "artifact" during the production processes than actually desired, and you need to remember that it's there (i.e. it is not a bi-directional switch) but I personally would not intend on using it.