A NPN bipolar junction transistor (BJT) is connected in common emitter (CE) configuration as shown in the circuit diagram below. The amplifier is operating in the saturation regime. The collector-emitter saturation voltage (\(V_{CE,sat}\)) is 0.2 V. The current gain \(\beta = 100\). The maximum value of base resistance \(R_{BB}\) is \rule{1cm{0.15mm} k\(\Omega\). (in integer)}
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To ensure a transistor is well into saturation, designers often use a base current that is 5 to 10 times larger than the calculated \(I_{B,sat}\). This is called "overdrive." The question asks for the maximum resistance, which corresponds to the boundary of saturation, so we use \(I_B = I_{B,sat}\) exactly.
Step 1: Understanding the Concept:
For a BJT to operate in the saturation regime, the base current \(I_B\) must be large enough to cause the collector current \(I_C\) to reach its maximum possible value, known as the saturation current \(I_{C,sat}\). The maximum value of the base resistance \(R_{BB}\) corresponds to the minimum base current required to just achieve saturation (\(I_{B,sat}\)). Step 2: Key Formula or Approach:
1. Apply Kirchhoff's Voltage Law (KVL) to the collector-emitter loop to find the collector saturation current, \(I_{C,sat}\).
\[ V_{CC} - I_{C,sat} R_C - V_{CE,sat} = 0 \]
2. Use the current gain \(\beta\) to find the minimum base current required for saturation, \(I_{B,sat}\).
\[ I_{B,sat} = \frac{I_{C,sat}}{\beta} \]
3. Apply KVL to the base-emitter loop to find the maximum base resistance \(R_{BB}\) that allows this minimum saturation current to flow.
\[ V_{BB} - I_{B,sat} R_{BB} - V_{BE} = 0 \]
Step 3: Detailed Explanation:
Given values: \(V_{CC}=5\) V, \(R_C=3\) k\(\Omega\), \(V_{CE,sat}=0.2\) V, \(V_{BB}=1\) V, \(V_{BE}=0.68\) V, \(\beta=100\). 1. Calculate \(I_{C,sat}\):
From the collector loop KVL:
\[ I_{C,sat} = \frac{V_{CC} - V_{CE,sat}}{R_C} = \frac{5 \text{ V} - 0.2 \text{ V}}{3 \text{ k}\Omega} = \frac{4.8 \text{ V}}{3000 \, \Omega} = 1.6 \times 10^{-3} \text{ A} = 1.6 \text{ mA} \]
2. Calculate \(I_{B,sat}\):
This is the minimum base current to put the transistor in saturation.
\[ I_{B,sat} = \frac{I_{C,sat}}{\beta} = \frac{1.6 \text{ mA}}{100} = 0.016 \text{ mA} = 1.6 \times 10^{-5} \text{ A} \]
3. Calculate the maximum \(R_{BB}\):
Using this minimum required base current in the base loop KVL:
\[ R_{BB, max} = \frac{V_{BB} - V_{BE}}{I_{B,sat}} = \frac{1 \text{ V} - 0.68 \text{ V}}{1.6 \times 10^{-5} \text{ A}} = \frac{0.32 \text{ V}}{1.6 \times 10^{-5} \text{ A}} \]
\[ R_{BB, max} = 0.2 \times 10^5 \, \Omega = 20 \times 10^3 \, \Omega = 20 \text{ k}\Omega \]
Step 4: Final Answer:
The maximum value of the base resistance \(R_{BB}\) is 20 k\(\Omega\).
