Need expert solution

Task $1$ Large$-$Signal Analysis and Design $[25$P$.]$

$V_{Tp1}=V_{Tp2}=-1V$

$V_{Tn3}=+1V$

$K_{p1}={}_p{C}_{ox}(\frac{(W)}{(L)}{)}_1=50\frac{(A)}{V^2}$

$K_{n3}={}_n{C}_{ox}(\frac{(W)}{(L)}{)}_3=200\frac{(A)}{V^2}$

The values might be unrealistic and are chosen for the simplicity of calculation.

Fig. $1.$

In the schematic of Fig. $1,$ a $V_{BE}-$multiplier is implemented using transistor $Q_1,R_1$ and $R_2.$ All transistors, $Q_1$ to $Q_4,$ are working in forward active region and have similar characteristics with $V_{BE}=V_{BE,ON}$ and $.$ The early effect is neglected $(V_A).$

a$)$ Calculate the current through the resistors $R_1$ and $R_2.($Hint: consider the base current of $Q_1$ to be zero.$)[2$ P$.]$

b$)$ Derive a general formula for $V_{B2}$ based on the parameters in $V_{BE}-$multiplier and $V_{EE}.$ Hint: In this part it is sufficient to just analyze the $V_{BE}-$multiplier and $V_{EE}.[4$ P$.]$

c$)$ Derive the equation for the current $I_{\text{R}\text{e}\text{f}}$ based on the equation of $V_{B2}$ in part $($b$).[4$ P$.]$

d$)$ Find the ratio of $R_2$ to $R_1,$ which results in an output current $I_{\text{R}\text{e}\text{f}}=\frac{V_{EE}}{R_{E4}}.[2$ P$.]$

e$)$ Considering the result of part $($d$),$ what is the advantage of this implementation. $[2$ P$.]$

$"$ NEED COMPLETE SOLUTION OF ALL THE QUESTIONS"