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I can't find out how this transistor fuse works. I know that the R1 is there because of voltage drop 0,7 V when the voltage on output is at its highest possible value. But I don't understand why there is the diode and reverse biased. And my next question is how the transistors operate. How is it possible that there is output immediately after it is turned on, shouldn't there be an small current to open the T2? But the base of T2 is grounded.

Thank you

schematic

simulate this circuit – Schematic created using CircuitLab

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First of all, note that T1 and T2 are both PNP transistors. To turn a PNP transistor on, the base must be more negative than the emitter.

When the circuit first powers up, T1 is off, and T2 is turned on by the current through R3 and R4.

As more current is drawn by the load (R5), the voltage drop across R1 increases, making the base of T1 more negative with respect to its emitter. When this voltage drop reaches about 1.3V — the forward voltage drop of T1's B-E junction plus the forward drop of D — current begins to flow through D and T1, turning T1 on. This reduces the base drive to T2, causing it to start to turn off.

As T2 turns off, the voltage across the load drops, causing current to flow through R2. Since this current can't flow through D, it must flow through T1, turning it on harder, and turning T2 off more. This process continues until T1 is saturated, T2 is completely cut off because there's no voltage across R3, and there's only a trickle of current flowing through T1, R2 and the load. The collector current of T1 is flowing through R4 to ground. Now the fuse is "tripped". The load (or the supply) must be removed altogether in order to reset the fuse and restore power to the load.

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R1 is the current sense element. When the voltage across R1 reaches about 1.4V (The Vbe of T1 and the Vf of the diode) T1 starts to turn on. This robs base drive from T2 causing T2 to start to turn off. The falling voltage on the output across R5 is fed back through R2 to provide positive feedback accelerating the turn off and maybe latching the output off if all the component values are right.

At turn on, presumably there's not enough base drive to T1 to keep T2 from conducting and bringing the output voltage up. Seems like things could get tricky over process and temperature so if this were a production circuit some careful calculation and simulation might be in order.

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  • \$\begingroup\$ Thank you for your answer but I still do not understand why the diode is on positive branch connected through the cathode how the current can flow through the diode and how there can be Vf at this moment. \$\endgroup\$ – Matej Dec 29 '14 at 16:10
  • \$\begingroup\$ The cathode is positive relative to ground, but negative relative to the other side of R1, which is where it's measuring across. \$\endgroup\$ – Majenko Dec 29 '14 at 16:25

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