constant-currentcurrent measurementcurrent-limiting
All current limiting circuits I've found until now use some sort of analog Logic with Op-Amps to limit/control the current.
(Why) is no digital current limiting used?
I would imagine a shunt resistor for measurement and a MOSFET for controlling being a flexible solution, when you already have a MCU in your build.
The circuit shown will work, but the transistor and Rsense create a voltage drop that must be taken into account.
What you are seeing is the effect of this:
At 480mA the voltage drop across the 10Ω resistor would be 4.8V, which leaves no "room" for the transistor saturation voltage or the Rsense voltage drops.
So the current will be (Vsupply - Qsat - Vrsense) / Rload. To fix this, raise the supply by a couple of volts and try the 0Ω and 10Ω tests again. Also, lower Rdefend considerably (<10Ω)
You should hopefully see (almost) no difference.
For better results, the more gain you have the better. Another thing to note is (as Dave mentions in his answer) that Rbias needs to have a higher limiting point than the Rsense setting, otherwise it will dominate. If the transistor has a gain of 65, and you want Rsense set for 500mA, then Rbias must be set to allow more than 500mA. At 500Ω, it will set the absolute limit at 65 * ((5V - 1.4V) / 500Ω) = 468mA, so even if Rsense were set for 500mA you won't get it. To avoid this set Rbias for e.g. 250Ω, or as mentioned below use a MOSFET for Q1 and then the value isn't as important (10kΩ will do)
Another option is to use a common opamp constant current circuit:
Simulation with a supply of 4.8V, current limited to 500mA, Rload swept from 1mΩ to 50Ω and current through it plotted relative to this (note current stays flat at 500mA whilst limited):
This meets your requirements of a solid 500mA limiting at 4.8V supply, and is easily adjustable by varying the opamp non-inverting via input voltage R2/R3 divider. The formula is V(opamp+) / Rsense = I(Rload) For example, the 1V reference is divided by 20 to provide 50mV at the opamp+ input, so 50mV / 100mΩ = 500mA.
A MOSFET is used to avoid base current errors complicating matters (a MOSFET with low Vth can also be used in the original transistor circuit to improve things)
So I'm not too sure I follow this first question here, but I'll take a stab at it: if Vin = 3.3V (i.e. your P-channel mosfet does not connect source and drain) then "TRIGGER" will essentially be disconnected, yes. This assumes that "TRIGGER" is connected to a high-impedance source, like an input pin on a standard microcontroller. As a quick note "essentially disconnected" means that there is a high impedance threshold to get across the source-drain junction, but it is not infinite. If Vin = 0V (i.e. source and drain are connected), then trigger will be close to (but slightly less than) 3.3V
So an input pin on a PIC24F is high impedance. This means that it will behave like a resistor with around 1MΩ of resistance to ground. If you check the "Electrical Characteristics" section in the "DC Characteristics: I/O Pin Input Specifications" subsection in the datasheet for the chip you're using, you should find a table which gives you the "Input Leakage Current." Using the dsPIC33FJ256GP710A as an example (it was a datasheet I happened to have on my machine), it lists 0 to 3.5 μA as the leakage current into or out of a pin within the limits of those pins.
So as a note, you'll also see that in the "Electrical Characteristics" section under the "Absolute Maximum Ratings" subsection, there will be a list which says something to the effect of "Voltage on any pin that is not 5V tolerant with respect to VSS(4) .... -0.3V to (VDD + 0.3V)". This is telling you that you can't put more than ~3.6V on any pin without damaging the pin. This doesn't mean that you should put a current-limiting resistor in front of the input pin, this means you should put a voltage divider in front of the input pin to limit the maximum voltage to less than VDD + 0.3V.
To tie this into question #1: this means that when a digital input pin is connected to a high-impedance output source like Fig. 1, you will see oscillating input values (1 and 0 randomly) based on when that very small drain removes or adds enough electrons to jump from one digital state to another. This is called a floating input
So I just answered this question in the previous one, but I should tell you: the "Maximum output current" is not internally limited. So although the PIC24F can "only output 18mA" this means that if you connect an output pin which is high directly to ground, it will output lots more than 18mA before it burns out! The 18mA is the maximum it can safely output without overheating/damaging the chip. This is true for both sourcing and sinking, there is no fuse which will stop you at 18mA, only the magic blue smoke.
Just to reiterate: you can safely connect any voltage to a digital input pin within its range (-0.3V to VDD+0.3V for the dsPIC33F). It would not blow the pin to connect it directly to +3.3V, nor would it blow the pin to connect directly to ground.
Just answered this in #3, but again, it's not the output impedance of the pin which limits this... Or, more accurately, it is the output impedance, but it will limit the currently by melting.
And yes, you can safely remove the 1kΩ resistor from Fig. 2 without damaging your PIC.
Best Answer
If that is all the MCU is doing, then maybe. Otherwise it is not fast enough. Even an inexpensive op amp may be found that will be faster. Besides, no buggy software to worry about (if it works once, it will work every time).
The shunt resistor and series element (bjt or mosfet) has to be there, whether you have op amp or MCU. Hooking up an op amp to that is simpler than hooking up a MCU. [And you might end up with one or two op amps anyway if the MCU has no DAC.]
The analog and digital system does not differ in whether the series element is controlled with an op amp. Both use an op amp to control the BJT/mosfet. The difference is in the way the reference input to the op amp is provided. In analog, the reference is a manual potentiometer. In digital control, it is usually a DAC. A digital pot can be used, but is hardly ever done (they are expensive, and less flexible). This DAC can be substituted with PWM and a low pass filter. Depending on the accuracy vs speed required, the low pass filter may be a simple RC filter, or an active 2nd or 4th order butterworth filter (which requires one op amp for every 2 stages).