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I'd wanted to know what happens in an NPN emitter follower with \${\delta}V=Vcc-Vee\$. At which junction does voltage drop? May this drop be fatal to a transistor?

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  • \$\begingroup\$ You can also put a schematic of the emitter follower using the inbuilt schematic builder/as an image. \$\endgroup\$ – Mitu Raj Nov 30 '20 at 19:02
  • \$\begingroup\$ Is it necessary? I wanted to draw it at first but it will clarify nothing in the question and only add complexity. \$\endgroup\$ – gavrilikhin.d Nov 30 '20 at 19:04
  • \$\begingroup\$ To clarify what are the three terms in your equation. What are the voltages you speak about. \$\endgroup\$ – Mitu Raj Nov 30 '20 at 19:06
  • \$\begingroup\$ Those are quite typical for any transistors. However, if you wish I'll add it. \$\endgroup\$ – gavrilikhin.d Nov 30 '20 at 19:08
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There will be a loss of one diode drop in the base-emitter junction, and the remainder of the drop is in the collector-base junction. Yes this can cause overheating and destroy the transistor.

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  • \$\begingroup\$ I suggest that most of the loss is at collector-base junction (when \$Vcc > Vb + Vbe\$), am I right? \$\endgroup\$ – gavrilikhin.d Nov 30 '20 at 19:14
  • \$\begingroup\$ Yes, if the total voltage drop is significantly more than a diode drop, as will typically be the case. \$\endgroup\$ – Frog Dec 1 '20 at 9:03

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