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?
\$\begingroup\$
\$\endgroup\$
4
-
\$\begingroup\$ You can also put a schematic of the emitter follower using the inbuilt schematic builder/as an image. \$\endgroup\$– Mitu RajNov 30, 2020 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.dNov 30, 2020 at 19:04
-
\$\begingroup\$ To clarify what are the three terms in your equation. What are the voltages you speak about. \$\endgroup\$– Mitu RajNov 30, 2020 at 19:06
-
\$\begingroup\$ Those are quite typical for any transistors. However, if you wish I'll add it. \$\endgroup\$– gavrilikhin.dNov 30, 2020 at 19:08
Add a comment
|
1 Answer
\$\begingroup\$
\$\endgroup\$
2
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.
-
\$\begingroup\$ I suggest that most of the loss is at collector-base junction (when \$Vcc > Vb + Vbe\$), am I right? \$\endgroup\$ Nov 30, 2020 at 19:14
-
\$\begingroup\$ Yes, if the total voltage drop is significantly more than a diode drop, as will typically be the case. \$\endgroup\$– FrogDec 1, 2020 at 9:03