HERE is a new and improved version of the linear load circuit.
First, some answers to your questions.
The previous circuit could draw ~75 Amps at 24 volts with the knob in the full load position. I believe the circuit would only go up to about 37 Amps on 12v though. I have and will continue work on this. Note that the simulator stops at 75 Amps on 24v, and I am getting 59.1 Amps from 12v. But the real circuit would go much much farther than this.
The current sense resistors are not a mistake, they don't need to be very high Watts because they are very low resistance. Our Watts are turned in to heat by what is called Joule heating. These Watts are dissipated in the total resistance of the circuit. before we were using resistors that took up a larger fraction of the total resistance of the circuit, so they were where all the Watts went. The MOSFET's for that circuit were either fully OFF, so no current was going through them, thus no power. Or they were fully ON, so all the current was going through them, but they had almost no resistance for heat to be generated in.
With this new circuit, we are making the MOSFET's have a significant resistance compared to the whole resistance, so almost all the heat is going to be made inside them. The sense resistors will continually have much lower resistance compared to the MOSFET's, and thus generate much less heat. In fact, the resistors and MOSFET's have sort of switched places as far as resistances in concerned.
The current sense resistors purpose is a bit tricky to explain in lay persons terms. But it sorta acts like a "speed bump" or "backs up" the flow of electric current just a bit. We are not doing this to slow it down or anything like that. This "speed bump" is to create a voltage that gets bigger with more current (Amps) going through it, because more current "spills over" so to speak. In the end it is nothing more than a way to measure current. It's output goes to the (-) side of the OP-Amp and is only millivolts.
Here is a schematic of the new and improved circuit. And the following is a bit of explanation on it.
View attachment 65904
How It Works
Basically, all the things in line across the top of the simulation are your MOSFET's. We are using them as if they were our resistors now. As stated above, heat is lost in a circuits resistance. We can change the resistance of these MOSFET's by changing the voltage applied to their gate, more voltage means less resistance and vice-versa. The problem we have is that MOSFET's are WILDLY sensitive and do not change flatly by them selves (non linear). So we need something to stabilize them out a bit if we are to use them reliably. Enter the Op-Amp.
The triangle thing in the center is an Operational Amplifier (Op-Amp for short). Op-Amps are very flat (linear) amplifying devices. They have one output, and two inputs. The best part about them is that their inputs are "differential". Without going into too many cloudy hazy details, we can take advantage of this differential nature to make the circuit keep the two inputs at exactly the same levels. You will see this level out behavior if you move the slider in the simulator, then hover the mouse pointer over the (+) and (-) wires on the Op-Amp. We achieve this effect with what is known as "feedback".
As stated before, the current sense resistors give us a small voltage, based on how much current is going through the circuit. This voltage gets sent into the inverting (-) input of the Op-Amp. The inverting input will cause the output to do the opposite of whatever it is seeing. So, if the current rises, the voltage of the inverting input will also rise. This makes the output drop. The output is connected to the MOSFET's gates, and as stated, they will increase in resistance when the voltage on the gate goes down.
Since the current going through the whole circuit is controlled by the MOSFET's, and the MOSFET's have just increased in resistance because of the drop in gate voltage, the current would of course drop too, in accordance with Ohms law (Volts/Ohms=Amps). If the total current drops, so too must the current on the sense resistors. Finally, if the sense resistors are what are feeding the inverting input it's voltage, and that voltage just dropped, the output would increase again. The opposite of what had started this.
So to summarize, if current goes up, sense output goes up, then inverting input goes up, output thus goes down, so gates voltage is lowered, resistance is raised, current goes back down because of Ohms law. This is simple negative feedback. With out this feed back, a slight breeze could change the circuit enough to cause massive damage to the power parts. And we would like to avoid this.
This is all well and good, but we are missing one critical detail. "At what current does this feedback prefer to hover around?" Well, that is what the other input is for. It happens that by virtue of how Op-Amps work when in a feedback configuration, it will tend to try to make the the two inputs the exact same voltage. This means that what ever voltage we apply on the (+) side, the Op-Amp will try to create on it's (-).
The voltage that the current sense resistors output is very small, but that doesn't matter to the OP-Amp, as they are very sensitive devices. However, we need to create an equally small voltage for the control signal, or the Op-Amp will try to match 24V and destroy things. We make this small voltage with a voltage divider. On the right side of the Op-Amp is the voltage divider. The top resistor is 47k Ohms, and the bottom is 1K. This alone makes ~500mV at 24v (At 12v it's about half this). That ~500mV is farther divided by the fact that the second resistor is actually a variable potentiometer. By adjusting this pot, you can make any voltage between ~500mV and 0V. And the higher this voltage, the more current the system must pass to match it with the current sense voltage. Thus giving you control over current passed through the system.
My explanation wasn't as small as I had hoped it would be, but that is close to how the circuit works without going into more details.
Must be a problem with IE.
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I hear you on that. I have to give you credit for a good schematic nonetheless.