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In DC-DC converters, a "peak switch current" is usually set to turn off the switching transistor when the current through it hits a certain peak value. This is done to prevent an overload or an initial overcurrent. Close the switch to see what happens without current limiting.
In this circuit, peak current limiting is achieved by saturating the switch (spoiler: it's not the best method. More about it later). The MOSFET is put into saturation by an "auxillary" BJT, which turns on and saturates the MOSFET when the drop across the current sense resistor reaches the base emitter voltage of the BJT, just like in this Constant-Current supply: http://everycircuit.com/circuit/5204466287247360
The peak current I = 0.7/Rse. In this circuit, Rse=0.33ohms. So I=0.7/0.33=2.12A.
Is this a good method? It's simple and does the job, right? Well, not really. Yes, it's simple and yes, it does the job, but it doesn't do the job in an efficient way. First of all, sending a huge current through a saturated MOSFET is very inefficient. Second, the MOSFET will overheat if not cooled properly.
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