Transistor Relation Cross Scans

Base Input Current Three-Row Scan

Holding the base resistor fixed turns input-minus-drop voltage into a visible current scan. Exact arithmetic here means exact results for the stated model inputs; measured inputs still carry uncertainty and significant-figure limits.

highlighted = computed this step

At the drop, the base has zero drive

Input 1 V minus the stated 1 V base-emitter drop leaves 0 V on the base resistor. With 4 ohm fixed, the base current is 0 A.

1 V−1 V=0 V,IB=0 A1\ \text{V} - 1\ \text{V}=0\ \text{V},\quad I_B=0\ \text{A}
Base input scanInput, drop, excess voltage, and base current are checked.base 0 Ainput 1 Vdrop 1 V

Two volts of excess gives one half ampere

Input 3 V minus the stated 1 V base-emitter drop leaves 2 V on the base resistor. With 4 ohm fixed, the base current is 1/2 A.

3 V−1 V=2 V,IB=12 A3\ \text{V} - 1\ \text{V}=2\ \text{V},\quad I_B={1\over 2}\ \text{A}
Base input scanInput, drop, excess voltage, and base current are checked.input 3 Vdrop 1 Vbase 1/2 A

Four volts of excess gives one ampere

Input 5 V minus the stated 1 V base-emitter drop leaves 4 V on the base resistor. With 4 ohm fixed, the base current is 1 A.

5 V−1 V=4 V,IB=1 A5\ \text{V} - 1\ \text{V}=4\ \text{V},\quad I_B=1\ \text{A}
Base input scanInput, drop, excess voltage, and base current are checked.input 5 Vdrop 1 Vbase 1 A