\(\text { and } I _{ E }=\frac{ I _{ C }}{\alpha}\)
\(I _{ E }=\frac{24 mA }{0.8}=30\,mA\)
\(\therefore I _{ B }= I _{ E }- I _C\)
\(=6\,mA \text { (into the base) }\)
$A$ | $B$ | $Y$ |
$0$ | $0$ | $1$ |
$0$ | $1$ | $0$ |
$1$ | $0$ | $1$ |
$1$ | $1$ | $0$ |
આઉટપુટ $Y$ માટેનો સંબંધ. . . . . . . . થશે.