Timeline for When two transistors connected which one is on and which one is in cutoff from given circuit condition?
Current License: CC BY-SA 3.0
6 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Apr 24, 2017 at 7:28 | vote | accept | user146551 | ||
| Jun 3, 2017 at 15:39 | |||||
| Apr 24, 2017 at 6:47 | comment | added | jonk | @user146551 That works in your first schematic's case. The way I wrote about the issue is more general and will work in schematics that look more complex. Also, if the base voltages were a lot closer to each other in your first schematic, then the way you write probably wouldn't be useful anymore. But the way I write it still would be. For example, suppose the -2 V base were changed so that it was instead +2.94 V. Then both BJTs would be on, but the left one may have almost 10 times the collector current. Which would be important to know, sometimes. But whatever works. | |
| Apr 24, 2017 at 6:21 | comment | added | user146551 | I am trying to say that transistor with most positive Vbe will on first, considering it on we will check for other transistors Vbe if it will >0.7 volt then it will on too. | |
| Apr 24, 2017 at 6:15 | comment | added | jonk | @user146551 I can't parse your writing well. I apologize for that. What I am saying is that the magnitude, \$\vert V_{BE}\vert\$, can't get to much more than a volt because of the extremely large currents that implies. Every \$60\:\textrm{mV}\$ increase means 10 times the collector current. As the base-emitter voltage adds a little at a time, the collector current multiplies by large factors. This behavior causes one BJT to simply "take over" in the cases you show. | |
| Apr 24, 2017 at 5:44 | comment | added | user146551 | Are you saying that one with most positive Vb - Ve (for NPN) and most positive Ve - Vb (for PNP) will be on first and considering one of them on check for others Vbe (is it more than 0.7 V or not) | |
| Apr 22, 2017 at 7:53 | history | answered | jonk | CC BY-SA 3.0 |