Input voltage is not a truth-table caption; it creates a region and then an output. Exact arithmetic here means exact results for the stated model inputs; measured inputs still carry uncertainty and significant-figure limits.

highlighted = computed this step

Input one volt leaves the output high

Input 1 V gives base current 0 A and collector current 0 A. The checked region is cutoff, so the output is 12 V and the state is HIGH.

Vin=1 VIC=0 A,Vout=12 VV_\text{in}=1\ \text{V}\Rightarrow I_C=0\ \text{A},\quad V_\text{out}=12\ \text{V}
Inverter output scanThe output label is produced from the checked region chain.base 0 Adrive 0 Acollector 0 Ainput 1 Vdrop 1 Vload 3 Aoutput 12 VcutoffHIGH

Input three volts creates an analog middle

Input 3 V gives base current 1/2 A and collector current 3/2 A. The checked region is active, so the output is 6 V and the state is MIDDLE.

Vin=3 VIC=32 A,Vout=6 VV_\text{in}=3\ \text{V}\Rightarrow I_C={3\over 2}\ \text{A},\quad V_\text{out}=6\ \text{V}
Inverter output scanThe output label is produced from the checked region chain.input 3 Vdrop 1 Vbase 1/2 Adrive 3/2 Aload 3 Acollector 3/2 Aoutput 6 VactiveMIDDLE

Input five volts saturates the pull-down path

Input 5 V gives base current 1 A and collector current 3 A. The checked region is saturation, so the output is 0 V and the state is LOW.

Vin=5 VIC=3 A,Vout=0 VV_\text{in}=5\ \text{V}\Rightarrow I_C=3\ \text{A},\quad V_\text{out}=0\ \text{V}
Inverter output scanThe output label is produced from the checked region chain.output 0 Vinput 5 Vdrop 1 Vbase 1 Adrive 3 Aload 3 Acollector 3 AsaturationLOW