Show that voltage gain in a transistor amplifier in CE mode is negative and hence obtain the expression for the voltage gain.

Explanation

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)`