Presumably you are okay with a small drop, provided it is less than 0.5V.
Here is a circuit that will limit at 100uA and will drop less than 100mV before it limits.
simulate this circuit – Schematic created using CircuitLab
The MCP6001 is an inexpensive rail-to-rail input/output op amp that will operate from a 5V supply. The op-amp will saturate at ground until the load current reaches about 98uA nominally (with the values shown). The supply thus 'looks like' 5V with ~1K in series (the MOSFET contributes less than 10 ohms with Vgs =-5V), so it will drop between 0 and 100mV for load resistances of infinity down to 50K.
For lower load resistances the circuit regulates the output current to ~98uA.
The circuit draws about 200uA from the 5V supply, in addition to the load current of 0~100uA.
You have sortof the right idea:
But the capacitor is in the wrong place. For slew rate control, it should be between the drain and the gate, not the source and the gate as you show it. Putting it between drain and gate causes feedback so that when the drain rises quickly, it turns the FET off more.
Just a cap between drain and source can be good enough. The timing relies on some parameters that are usually poorly known, and the slope limiting doesn't kick in until the gate gets to near its threshold voltage.
Here is a more sophisticated slope-limiting power input circuit I've used a few times.
This device connects to the rest of the system via two CAN bus lines, ground, and 24 V power. It can be hot-plugged at any time. It can't be allowed to suddenly draw a large pulse of current when plugged in.
CANPWR is the direct connection to the 24 V power bus, and 24V is the is the internal 24 V power in this device. The purpose of this circuit is to make 24V rise slowly enough to limit the inrush current to a acceptable level. After that, it should get out of the way as much as possible.
A rising voltage slope on 24V causes current thru C2, which turns on Q3, which turns on Q1, which tries to turn off the gate drive to Q2, the power pass element. Note that this kicks in with less than 1 V on 24V.
Slope limiting feedback occurs when there is enough voltage across R4 to turn on Q3. Figure that's about 1.5 V, considering the drop across R5 required to turn on Q1. The slope limit is therefore what it takes to pass (1.5 V)/(10 kΩ) = 150 µA thru C2. (150 µA)/(1 µF) = 150 V/s. To rise 24 V should therefore take about 150 ms. I remember measuring a few 100 ms of rise time with a scope, so that all checks out.
Once the 24V net has risen, R3 holds Q2 on, and D2 keeps its gate-source voltage within the allowable range.
Best Answer
It should be symmetrical to the original but not like that. The circuit you need is following:
simulate this circuit – Schematic created using CircuitLab
Decreasing voltage of V1 will make BE junction constant at cca 0.6V, what means BJT will be in active region. As \$I_B\$ increases so do the gate voltage and source voltage, therefore \$V_{GS}\$ will be constant as well as your drain current.