Understanding NPN Transistor Base-Emitter Junction Behavior

An NPN transistor has three terminals: Collector (C), Base (B), and Emitter (E). The junction between the Base and the Emitter is essentially a P-N junction (Base is P-type, Emitter is N-type in an NPN transistor, or vice versa depending on the doping layers, but the junction behavior is like a diode). Like any diode, this base-emitter junction requires a certain minimum forward voltage to conduct a significant current.

For silicon transistors, which are very common, this minimum forward voltage, often called the cut-in voltage or turn-on voltage, is typically around \(0.7 \, \text{V}\). When the voltage applied across the base and emitter terminals (\(V_{BE}\)) is less than this value, the junction is either reverse-biased or forward-biased but with very little voltage, so only a negligible leakage current flows.

Applying 0.7V to the Base and Emitter

When a voltage of about \(0.7 \, \text{V}\) is applied across the base (positive with respect to emitter for NPN), this forward biases the base-emitter junction. This voltage is the typical threshold required to overcome the potential barrier of the silicon P-N junction.

• If \(V_{BE} < 0.7 \, \text{V}\), the base-emitter junction conducts very little current. It is essentially 'off'.
• If \(V_{BE} \approx 0.7 \, \text{V}\), the base-emitter junction starts to conduct current. This is the point where the transistor begins to turn 'on'.
• If \(V_{BE} > 0.7 \, \text{V}\) (slightly), the current through the base-emitter junction increases significantly.

Therefore, applying approximately \(0.7 \, \text{V}\) across the base and emitter of an NPN transistor forward biases the junction and causes a small, but significant, current to flow between these two terminals. This base current is crucial because it controls the larger current flow from the collector to the emitter.

Analyzing the Options

• Option 1: A small voltage will flow between the two terminals. Voltage is a potential difference, not something that "flows". Current flows. This option is incorrect.
• Option 2: A small current will flow between the two terminals. Applying \(0.7 \, \text{V}\) (the typical turn-on voltage) forward biases the base-emitter junction, allowing a current to flow from the base to the emitter. This is the correct behavior.
• Option 3: There will not be any current flow between the two terminals. This would be true if the voltage were significantly less than \(0.7 \, \text{V}\) or reverse biased. At \(0.7 \, \text{V}\), current definitely starts flowing. This option is incorrect.
• Option 4: There will not be any voltage flow between the two terminals. Again, voltage doesn't flow. Current flows. This option is incorrect.

Based on the behavior of the base-emitter P-N junction in a silicon NPN transistor, applying approximately \(0.7 \, \text{V}\) causes a current to flow.