I guess it makes sense that 54.4C is above ambient since (16W * 3.6C/W) + 84.4C would be 142C, close to the 150C target junction temperature. Is that right?
Yes, give or take rounding I guess.
When I set my PSU to 10V and current limit to .5A, the voltage goes down to 0v and the current limit led turns on. I guess is trying to supply more current than it can, therefore drops the voltage to produce more current? Why doesn't it just leave the voltage at 10V and only delivers the 500mA? That would be 5W. Isn't that what current limiting should do? It doesn't matter how much current the circuit is asking for, if it's set to 500mA it should only deliver that?
It can't deliver 10V AND 0.5A because the power supply would have to somehow reach out and change the load resistance to 10V / 0.5A = 20 ohms. It's not smart enough to do that... and frankly, neither am I!
The current drawn from a voltage source is determined by the load resistance. Or alternately, the voltage developed by a current source is. If it's a current- and voltage-regulating type supply, it will choose whichever option yields less power (so it doesn't get overloaded).
In other words, if you plot the points of (V, I) that the power supply can deliver (its load line), you will find it traces half of a square on the graph. For load resistances greater than 20 ohms, it will be a straight vertical line segment (constant V from 0-0.5A), and for lesser, it will be a straight horizontal line (constant I from 0-10V).
Note: such a bench supply is NOT an ideal Thevenin voltage source (or Norton current source). Though it has a well-defined open circuit voltage (10V) and short-circuit current (0.5A), the equivalent output resistance is NOT 10V / 0.5A = 20 ohms. It is not a linear function of V(I), but a piecewise linear function.
(Which is, in turn, a crude approximation of the real case. If you measure closely, you will see the voltage isn't constant along its segment, but it varies by some millivolts; likewise the current slope will not be perfectly orthogonal. The crossover region between the two segments will also not be an ideal sharp corner, but probably rolls off, following a curve, going gently from one to the other. The origin of this curve is probably related to some exponential function, such as a diode-OR function. And if you look *really* closely, you may see drift, noise, hysteresis and all other sorts of nasty effects.)
What the supply is doing is, for currents less than 0.5A, it is intended to approximate a voltage source, which really means it has an equivalent output resistance significantly less than the operating range (i.e., << 20 ohms). Real regulators are in the milliohm range, so that, over a span of amperes, the voltage only changes by some millivolts -- a pretty stable voltage for most purposes. But if you try to draw more current, the voltage ceases to be ideal, and the approximation breaks down rather dramatically.
Anyway, back to dissipation;
It would be impossible which I guess I makes sense since the datasheet specifies the max power dissipation is 41W at 25C ambient and 175C junction.
(175C - 25C) / 41W = 3.6C/W which is the junction to case thermal resistance, in which case you would need an ideal heatsink with a coefficient of 0C/W.
So the specs gotta be lower than that.
Correct. So if you want to dissipate more power, not only do you need a larger heatsink, but you need a better transistor, or more of them.
I still don't understand how to tie power, voltage, current and Ron resistance. The calculations I've done above don't take the 36mOhm resistance into consideration.
A transistor is a poor approximation of a resistor to begin with (Rds(on) is only nearly true when the MOSFET is fully saturated), and you are purposely using it in the non-resistive region for this project -- instead, there are many ways you could equivalently describe a transistor in this state. A voltage-controlled resistor would be one poor description, but suffice it to say: the effective load your power supply sees will NOT be Rds(on), but something much larger than that.
The only relevance Rds(on) holds in this circuit, is how low the voltage can go while maintaining the requested load current. Since you said the shunt resistance is 1 ohm, it clearly doesn't do you much good if Rds(on) is significantly smaller than 1 ohm -- if it were 0.1 ohm instead of 0.035 ohm, you can only go down to 1.1V instead of 1.035V -- but who cares, they're both pretty damn close to 1V, and pretty low in any case (would you ever really need it to work at less than 2, or even 3V?).
Tim