Base Drop Boundary Three-Row Scan - Electronic Circuits XI: Transistor Switching and Device Regions Step by Step

Moving the base-emitter drop changes the same fixed input's positive drive. Exact arithmetic here means exact results for the stated model inputs; measured inputs still carry uncertainty and significant-figure limits.

highlighted = computed this step

Zero-volt drop passes two volts to the base resistor

Input 2 V minus drop 0 V leaves positive drive 2 V, so base current is 1/2 A.

2\ \text{V}\ -\ 0\ \text{V}\rightarrow I_B={1\over 2}\ \text{A}

One-volt drop leaves one quarter ampere

Input 2 V minus drop 1 V leaves positive drive 1 V, so base current is 1/4 A.

2\ \text{V}\ -\ 1\ \text{V}\rightarrow I_B={1\over 4}\ \text{A}

Two-volt drop reaches the cutoff boundary

Input 2 V minus drop 2 V leaves positive drive 0 V, so base current is 0 A.

2\ \text{V}\ -\ 2\ \text{V}\rightarrow I_B=0\ \text{A}