In figure , the \(V_{\mathit{BB}}\) supply can be varied from \(0 \;V\) to \(5.0\;V\) . The Si transistor has \(\beta _{\mathit{dc}}=250\) and \(R_B=100\;k\Omega ,R_C=1\;KV_{\mathit{CC}}=5.0\;V\) Assume that when the transistor is saturated, \(V_{\mathit{CE}}=0\;V\) and \(V_{\mathit{BE}}=0.8\;\mathit{V.}\) Calculate the minimum base current, for which the transistor will reach saturation. Hence, determine \(V{_1}\) , when the transistor is switched on.
Answer
Given at saturation \(V_{\mathit{CE}}=0\;V,V_{\mathit{BE}}=0.8\;V\)
\(V_{\mathit{CE}}=V_{\mathit{CC}}-I_CR_C\)
\(I_C=\frac{V_{\mathit{CC}}}{R_C}=\frac{5.0\;V}{1.0\;k\Omega} =5.0\;\mathit{mA}\)
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 \(V_{\mathit{IH}}=V_{\mathit{BB}}=I_BR_B+V_{\mathit{BE}}\)
\(=20\mu A\times 100k\Omega +0.8V=2.8V\)
\(V_{\mathit{CE}}=V_{\mathit{CC}}-I_CR_C\)
\(I_C=\frac{V_{\mathit{CC}}}{R_C}=\frac{5.0\;V}{1.0\;k\Omega} =5.0\;\mathit{mA}\)
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 \(V_{\mathit{IH}}=V_{\mathit{BB}}=I_BR_B+V_{\mathit{BE}}\)
\(=20\mu A\times 100k\Omega +0.8V=2.8V\)
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