Assuming \(V_{CEsat} = 0.2\) V and \(\beta = 50\), the minimum base current (\(I_{Bmin}\)) required to drive the transistor in the figure to saturation is
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To find the saturation current for a BJT switch, first calculate the maximum possible collector current as if it were a closed switch (\(I_{Csat} = (V_{CC} - V_{CEsat})/R_C\)). Then, divide this by \(\beta\) to find the minimum base current needed to achieve it.
Step 1: Define the condition for saturation.
A BJT is in saturation when it is fully "on". In this state, the collector-emitter voltage drops to its minimum value, \(V_{CEsat}\), and the collector current is limited primarily by the external circuit components.
Step 2: Calculate the collector current at saturation (\(I_{Csat}\)).
Using Kirchhoff's Voltage Law (KVL) on the collector-emitter loop:
\[ V_{CC} = I_C R_C + V_{CE} \]
At saturation, \(V_{CE} = V_{CEsat}\). So,
\[ I_{Csat} = \frac{V_{CC} - V_{CEsat}}{R_C} \]
Substitute the given values:
\[ I_{Csat} = \frac{3 V - 0.2 V}{1 k\Omega} = \frac{2.8 V}{1000 \Omega} = 2.8 \text{ mA} \]
Step 3: Calculate the minimum required base current (\(I_{Bmin}\)).
To ensure saturation, the base current must be large enough to support the saturation collector current. The relationship is given by the transistor's current gain, \(\beta\).
\[ I_C = \beta I_B \implies I_B = \frac{I_C}{\beta} \]
The minimum base current is the one that just puts the transistor at the edge of saturation:
\[ I_{Bmin} = \frac{I_{Csat}}{\beta} = \frac{2.8 \text{ mA}}{50} = 0.056 \text{ mA} \]
Converting to microamperes: \(0.056 \text{ mA} = 56 \text{ } \mu A\).
