Saved Bookmarks
| 1. |
Determine V_(CE) in the following silicon based transistor circuit |
|
Answer» 6.8V  USING KVL to CE circuit `V_(C C) = i_(C) R_(C ) + V_(CE) + i_(E) R_(E)` `=i_(C) R_(C) + V_(CE) + i_(C) R_(E) " " ( :. i_(E) ~~ i_(C))` `rArr V_(CE) = V_(C C) - i_(C) (R_(C)+ R_(E))`…(i) By Ohm.s law, current `i_(1)` is: `i_(1) = (V_(C C))/(R_(1) + R_(2)) = (10V)/(15 k Omega) = (2)/(3) mA` Now, `V_(2) = i_(1) R_(2)` `= (2)/(3) mA xx 5 k Omega = (10)/(3)` `V_(2) = 3.3V` KVL to base circuit gives, `V_(2) = V_(BE) + i_(E) R_(E)` For silicon transistor, `V_(BE) = 0.7 V rArr i_(E) = (V_(2) - V_(BE))/(R_(E))` `=(3.3 - 0.7)/(526) ~~ 5mA` So, `i_(C) ~~ i_(E) = 5mA` HENCE, from Eq (i) `V_(CE)` is `V_(CE) = V_(C C) - i_(C) (R_(C) + R_(B))` `=10 - 5mA (1 k Omega + 526 Omega)` `~~2.4V` |
|