Given at saturation \(V_{\mathit{CE}}=0V,V_{\mathit{BE}}=0.8V\)
\(V_{\mathit{CE}}=V_{\mathit{CC}}-I_CR_C\)
\(I_C=\frac{V_{\mathit{CC}}}{R_C}=\frac{5.0V}{1.0k\Omega} =5.0\mathit{mA}\)
Therefore \(I_B=\frac{I_C}{\beta} =\frac{5.0\mathit{mA}}{250}=20\mu A\)
The input voltage at which the transistor will go into saturation is given by \(V_{\mathit{IH}}=V_{\mathit{BB}}=I_BR_B+V_{\mathit{BE}}\)
\(=20\mu A\times 100k\Omega +0.8V=2.8V\)
The value of input voltage below which the transistor remains cutoff is given by \(V_{\mathit{IL}}=0.6V,V_{\mathit{IH}}=2.8V\)
Between \(0.0V\) and \(0.6V,\) the transistor will be in the 'switched off state. Between \(2.8V\) and \(5.0V,\) it will be in 'switched on' state.
Note that the transistor is in active state when \(I_B\) varies from \(0.0\mathit{mA}\) to \(20\mathit{mA}\) . In this range, \(I_C=\beta I_B\) is valid. In the saturation range, \(I_C{\leq}\beta I_B\)