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Show that voltage gain in a transistor amplifier in CE mode is negative and hence obtain the expression for the voltage gain. |
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Answer» Solution :Circuit diagram. w.k.t DC biasing condition to be SATISFIED is `V_(BB) = V_(BE) + I_(B) R_(B)` when a signal voltage `V_(1)` is supplied, `V_(i) = Delta I_(B) R_(B) + Delta I_(B) r_(i)` i.e., `"" V_(i) = Delta I_(B) (R_(B) + r)` `V_(i) = r Delta I_(B) R_(B) "" `where, r = `R_(B) + r_(i)` In the AMPLIFIER, circuit , for a change in base CURRENT, there will be a large variation in the collector current. i.e., ac current gain `A_(i) = beta_("ac") = (Delta I_(C))/(Delta I_(B)) ~~ beta_(dc) = (i_(C))/(i_(B))` By applying Kirchhoffs law to the output loop, `V_("CC") = V_(CE) + I_(C) R_(L)` Since change in base current causes change in collector current, `Delta V_("CC") = Delta V_(CE) + R_(L) Delta I_(C).` But `V_("CC")` is fixed. Hence `Delta V_("CC")` = 0. i.e., `'"" Delta_(CE) = -R_(L) Delta I_(C) = V_(0)` the voltage gain of the amplifier`A_(v) =(V_(0))/(V_(i)) ` i.e.,`""A_(v) =(V_(0))/(V_(i)) = (-beta_(ac) R_(L) Delta I_(B))/(r Delta I_(B))` or `"" A_(v) = (-beta_(ac) R_(L))/(r)` where, `R_(L) = R_(c) + r_(0)` 
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