Solve the following problem below with pen $($handwrite it $)$ and please show the formulas used step by step. Let the NPN BJT Q$\_($N$)$ have the following properties: $\{($V$\_($BEN$)=0\mathrm{.}7($V$))$:$\}$ when BE junction is ON$,\backslash$beta $=$

$100$ and $\{$:V$\_($CENsat $)=0\mathrm{.}1($V$)\}.$ Let R$\_($B$1$N$)=$R$\_($B$2$N$)$ and each has resistance value in the range of tens of

K$\backslash$Omega $.$ Let R$\_($CN$)$ and R$\_($EN$)$ have resistance values in the range of a few K$\backslash$Omega $.$

$1.$a Choose all the resistors so that I$\_($CN$)=1$mA and V$\_($CEN$)=3$V$.$ For your convenience, assume that

R$\_($B$1$N$)$ and R$\_($B$2$N$)$ create an "almost$-$voltage$-$divider". Then you can easily determine V$\_($B$)$ and V$\_($E$)$ and

choose R$\_($EN$)$ to set the desired I$\_($CN$)$ spec and then set R$\_($CN$)$ to meet the V$\_($CEN$)$ spec. Of course, don't

forget to justify the Active Mode assumption and the "almost$-$voltage$-$divider" assumption.

$1.$b Now, change one of the R$\_($BN$)$ resistors $($to another finite and nonzero value$)$ to drive the BJT

into Saturation Mode. Once you violate the Active Mode assumption, while still keeping the BE

junction forward biased, solve the circuit that now operates in Saturation Mode.

$1.$c In your design of $1.$a offer a change of one of the R$\_($BN$)$ resistors $($to another finite and nonzero

value$)$ in order to drive the BJT into Cutoff Mode. Solve the circuit in the Cutoff Mode.