Regarding the linearity issue:

Usually when you have a potentiometer to set the output voltage you want the voltage to change in a linear fashion as you turn the pot clockwise.
Vout = k x rotation angle, where k is some constant and rotation angle is the fraction of the pot's total rotation angle

With the LM317, unless you add an extra negative supply for the adjustment voltage, you can't set the output to less than the difference between the adjustment pin and the output pin, which is fixed at nominally 1.25 V. Our equation now becomes
Vout = 1.25 + (k x rotation angle)
This still is not too bad, because the increase in voltage is linear as you turn the pot - or at least that is what you would like to have happen.

Try calculating the output voltage of the LM317 (just ignore the current boost parts) with the circuit as you have shown it with the potentiometer at 1/4, 1/2, 3/4 and full rotation. Then try the same calculations with the fixed resistor removed and a 1.5k pot, which is what the parallel combination of the 4.7k pot and 2.2k fixed resistor yields.



Protecting against overcurrent with a fuse:

Fuses take time to blow and it is often nearly impossible to select a fuse that will blow fast enough to protect semiconductor devices from damaging overcurrent. When you need to do something like allowing a large capacitor to charge it becomes much more difficult. Electronic current limiting is far superior, but more complex. The fuse then becomes something that will protect against fire or such severe damage to the circuit board (burned copper tracks, for example) that you can't repair it. Most semiconductor devices will fail short-circuit if subjected to damaging overcurrent or overvoltage, at least initially. If enough power is available, the shorted device may then fail open circuit (in an IC, the fine "bonding wires" that connect the actual semiconductor die to the package leads may act like fuses; high power devices sometimes will practically explode). A shorted voltage regulator is usually a very bad thing because it will lead to other things in the circuit being damaged due to excessive voltage.

Things like high power rectifier diodes and thyristors may be protectable with fuses, but the fuses required can be really expensive.

 

I saw a really fast blow fuse used in an electric vehicle development place. It was a piece of normal fuse wire immersed in a pot of water.
While the current is relatively low the water carries heat away from the wire. When the current is large enough the water immediately around the wire boils and the locally generated steam insulates the wire from the water. As the current is now way above the 'normal' fusing current for the wire it blows very fast - fast enough to protect the semiconductors down the line. This is so cheap, simple and really cute.

 

"This is so cheap, simple and really cute."

Very clever idea, though a bit problematic for portable equipment.

 

"This is so cheap, simple and really cute."

Very clever idea, though a bit problematic for portable equipment.

Also only for high currents. They were using a 1000A power supply powering a prototype motor.
I was there to mend that supply which had gone to max output and destroyed their very expensive motor - oops!

 

Wouldn’t a 5 amp fuse stop the cuircuit from being destroyed if too much current was being drawn?

Not necessarily.
The old rule of thumb is that the transistor is always faster than the fuse.

 

Q2 is not a current limit - it is how the external transistor shares the current with the LM317.

Oops! You are correct. Thank you.

 

Not necessarily.
The old rule of thumb is that the transistor is always faster than the fuse.

So i’ve been looking for a current limiting circuit I could add that’s not too complex (seeing as I already soldered up the thing). I found one that uses a mosfet. I don’t have the schematic but the operation is fairly simple to explain. The gate is connected to the wiper of a pot that is connected across the 12v battery. Varying this pot allows me to control the current flowing from drain to source. The drain is connected directly to the +12v and the source is connected to a 10 ohm resistor ( I know 1.2 amps would blow an led so it’s perfect to test if I could limit the current). The 10 ohm resistor is connected to the +ve side of the led and the -ve side of the led goes to ground.

When I connect this circuit it works fine as an led dimmer and I’m able to adjust the current within the range of a about 300 ma without any problem. But once I remove the led and connect an ammeter to test the full range of current, my mosfet starts over heating at around 450 ma. It’s a power mosfet btw rated to handle 30 amps. I thought this 1/2 watt resistor would be the one to get hot but it’s the mosfet.

I really don’t understand transistors, can you tell me why this is happening?

 

If the MOSFET is the pass element in a current limiting circuit, you probably need to heatsink it.

 

If the MOSFET is the pass element in a current limiting circuit, you probably need to heatsink it.

It got so hot it started smoking, at 450 ma too, I want to limit the circuit at 4 Amps so if definitely won’t survive... I found a different circuit diagram I plan to try