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Every BJT needs a current limit on the base. It may be a resistor or any other circuitry.
In your case R2 and R3 and the BJT´s gain will limit the base current to a save value.
hello
i am wonder what is this circuit . and why you feed the LM317 unregulated input pin which is 1.5 Amp max. current output from two BJT 2N3055 connected in parallel those can supply more than 3Amp .
and i have calculated the the output regulated voltage of LM317
VO = 1.25 (1 + R4/R10) =1.25 (1+0.1/100)=~1.25 . why this ratio R4/R10
anyhow any BJT transistor needs base current only on "class C amplifier" (which is for R.F amplifier ) class C amp. works on cutoff region .
Hello @00kam ,
This circuit is a linear power supply. It uses a preregulator for LM317, which has the role to keep the LM317 input at about 4-5V above LM317 output. The preregulator is made using 2 2N3055. I used 2 x 2N3055 because the short circuit current through the LM317 is abot 2.7A (max) and I wanted to keep the 2N3055 in the SOA somehow. Regarding the VO, in the schematic is just a simulation, in reality there will be a pot connected (2k pot).
LM317 has an internal short-circuit current limiting, so theoretically, will never be in situation to have 2.7A at its input.
Is no reason to use two 2N3055 in parallel BEFORE a LM317 regulator.
Anyway, there are many other simple solutions to build this kind of linear regulator.
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