In the figure,given that V_{BB} supply can va | vedclass

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(A) For a transistor to reach saturation,the collector-emitter voltage $V_{CE}$ must be $0\,V$.Applying Kirchhoff's Voltage Law $(KVL)$ to the output

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$25\,\mu A$ and $3.5\,V$

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(A) For a transistor to reach saturation,the collector-emitter voltage $V_{CE}$ must be $0\,V$.Applying Kirchhoff's Voltage Law $(KVL)$ to the output loop:$V_{CC} - I_C R_C - V_{CE} = 0$Since $V_{CE} = 0$ at saturation:$I_C = \frac{V_{CC}}{R_C} = \frac{5\,V}{1\,k\Omega} = 5\,mA = 5 	imes 10^{-3}\,A$.The relationship between collector current and base current is $I_C = \beta_{dc} I_B$.Therefore,the minimum base current required for saturation is:$I_B = \frac{I_C}{\beta_{dc}} = \frac{5 	imes 10^{-3}\,A}{200} = 25 	imes 10^{-6}\,A = 25\,\mu A$.Now,applying $KVL$ to the input loop:$V_{BB} - I_B R_B - V_{BE} = 0$$V_{BB} = I_B R_B + V_{BE}$$V_{BB} = (25 	imes 10^{-6}\,A)(100 	imes 10^3\,\Omega) + 1.0\,V$$V_{BB} = 2.5\,V + 1.0\,V = 3.5\,V$.Thus,the minimum base current is $25\,\mu A$ and the input voltage is $3.5\,V$.

In the figure,given that $V_{BB}$ supply can vary from $0$ to $5.0 \,V$,$V_{CC} = 5\,V$,$\beta_{dc} = 200$,$R_B = 100\,k\Omega$,$R_C = 1\,k\Omega$ and $V_{BE} = 1.0\,V$. The minimum base current and the input voltage at which the transistor will go to saturation,will be respectively:

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