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In figure , the   \(V_{\mathit{BB}}\text{ supply can be varied from }0\) V to \(5.0V\) . The Si transistor has \(\beta _{\mathit{dc}}=250\)  and \(R_B=100k\Omega ,R_C=1KV_{\mathit{CC}}=5.0V\)  Assume that when the transistor is saturated, \(V_{\mathit{CE}}=0V\)  and \(V_{\mathit{BE}}=0.8\mathit{V.}\)  Find the ranges of \(V{_1}\)  for which the transistor is 'switched off and 'switched on'.
Question: In figure , the   V_{mathit{BB}}text{ supply can be varied from }0 V to  5.0V . The Si transistor has  beta _{mathit{dc}}=250  and  R_B=100kOmega ,R_C=1KV_{mathit{CC}}=5.0V  Assume that when the transistor is saturated,  V_{mathit{CE}}=0V  and  V_{mathit{BE}}=0.8mathit{V.}  Find the ranges of  V{_1}  for which the transistor is 'switched off and 'switched on'.

Answer

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, \(I_C{\leq}\beta I_B\)
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