Electronic – Current limiting for at the input side of the transimpedance amplifier ( photodiode protection)

Your first circuit appears to be a current limiter. If the resistor values are properly chosen, the circuit will clamp the collector current (the current through the transistor and thus your load) at some value. It will not cut power.

The second circuit image you posted is not a complete circuit. The box labelled "Drive Circuit" represents some logic block that you would have to design. It takes feedback and directly turns power on and off to the load. That could be done any number of ways.

If you want something that acts more like a fuse that can be easily reset, something like a latching relay in series with the load could work. I'd probably use a comparator to trigger the relay when the voltage across a shunt resistor went above a threshold.

^{simulate this circuit – Schematic created using CircuitLab}

The relay is a latching type. Rshunt is an appropriately chosen shunt resistor. R1 and R2 are chosen such that they divide to the voltage that triggers the comparator when the current through the load is 1.5A or higher: Vtrigger = 1.5A * Rshunt = Vcc * R2/(R1+R2)

Here are some considerations:
1. Rshunt should be small so it does not steal too much power from the drive circuit.
2. This circuit assumes Vcc is a stable voltage source. If it's weak and changes voltage under load, the comparator circuit will not trigger correctly.
3. If you're driving a motor, there might be a temporary current surge on start up that exceeds 1.5A for a short period of time. If that's allowable, you may need to build a low-pass filter into the non-inverting input of the comparator.
4. If you choose a double-coil latching relay (not shown), you can rig a button up to the second coil to reset the relay after it triggers.
5. The coil of the relay may not be compatible with Vcc.

The advantage of an incandescent light bulb over a resistor is that its cold resistance is about 10 times less than when lit up to normal brightness. Therefore you can choose a wattage that limits short circuit current to a safe value, without excessive voltage drop at lower currents. Also, since it takes some time for the filament to heat up and increase resistance, short duration current surges cause much less voltage drop.

You should choose a lamp with voltage rating close to the power supply voltage (a little lower won't hurt) and wattage that gives an acceptable short circuit current. Power = Volts x Amps, so if you wanted to limit a 12V supply to 0.5A then you would need a 12V 6W lamp.

The maximum normal operating current will then be determined by how much voltage sag is acceptable. This is tricky to calculate because the lamp's change in resistance with current is very non-linear. A 12V 15W festoon lamp that I tested had a resistance of 0.9 Ohms when cold, and 9.8 Ohms at full brightness. At 0.3A (24% of the bulb's nominal current draw) the output voltage dropped from 12.0V to 11.5V. Here is a graph showing output voltage vs current for my lamp compared to using a 10 Ohm resistor.

The 'soft' current limiting that a lamp provides is good for protecting against continuous short circuits without affecting normal operation, but won't protect sensitive electronic components which cannot withstand short current surges or slightly higher than normal operating current.

If you need a fast acting current limit with sharper cutoff then you could use a bipolar transistor (PNP for positive voltage, NPN for negative) with its Emitter connected to the power supply, Collector to the load and a high value resistor going from Base to Ground. Maximum collector current will then be determined by the Base current multiplied by the transistor's current gain (which varies from unit to unit, so you may have to adjust the resistor value to get a precise current limit).