Mathematical calculations arising in Physics

Question

Can anyone please tell how can I use the following two equations:

$$S=\frac{1+\beta}{1+\frac{\beta R_E}{R_T+R_E}} \tag5$$

and

$$S''=\frac{\frac{V_T-V_{BE}}{R_T+R_E}+I_{CO}(1-\frac{R_E}{R_T+R_E})}{(1+\frac{R_E\beta}{R_T+R_E})^2}$$

to arrive at:

$$S''=\frac{1}{\beta(1+\beta)}[I_C\frac{(R_T+R_E)(1+\beta)+\beta SR_E}{R_T+R_E}+SI_{CO} \tag7 ]$$ ?

Answer 1

$\require{cancel}$

Let: $$a=\frac{R_E}{R_T+R_E}$$ Given: $$S=\frac{1+\beta}{1+\frac{\beta R_E}{R_T+R_E}}$$ $$S=\frac{1+\beta}{1+\beta a}; 1+\beta a=\frac{1+\beta}S $$ From (2) and (3): $$I_B=\frac{V_T-V_{BE}-I_CR_E}{R_T+R_E} \tag2$$ $$\frac{V_T-V_{BE}}{R_T+R_E}=I_B+I_Ca$$ $$I_C=\beta I_B+(1+\beta)I_{CO} \tag3$$ $$ I_B=\frac{I_C-(1+\beta)I_{CO}}\beta$$

Then: $$S''=\frac{\frac{V_T-V_{BE}}{R_T+R_E}+I_{CO}(1-\frac{R_E}{R_T+R_E})}{(1+\frac{R_E\beta}{R_T+R_E})^2}$$ $$S''=\frac{I_B+I_C+I_{CO}(1-a)}{(1+\beta a)^2}$$ $$S''=\Biggl(\frac{S}{1+\beta}\Biggr)^2(I_B+I_Ca+I_{CO}(1-a))$$ $$S''=\Biggl(\frac{S}{1+\beta}\Biggr)^2\Biggl(\frac{I_C-(1+\beta)I_{CO}}{\beta}+I_Ca+I_{CO}-I_{CO}a\Biggr)$$

Kindly check this because the equation above neatly equals to: $$S''=\Biggl(\frac{S}{1+\beta}\Biggr)^2\Biggl(\frac{I_C-(1+\beta)I_{CO}+(I_C -I_{CO})a\beta+I_{CO}\beta}\beta\Biggr)$$ $$S''=\Biggl(\frac{S}{1+\beta}\Biggr)^2\Biggl(\frac{I_C-I_{CO}-I_{CO}\beta+I_Ca\beta -I_{CO}a\beta+I_{CO}\beta}\beta\Biggr)$$ $$S''=\Biggl(\frac{S}{1+\beta}\Biggr)^2\Biggl(\frac{I_C(1+a\beta)-I_{CO}(1+\cancel{\beta}+a\beta-\cancel{\beta)}}\beta\Biggr)$$

$$S''=\Biggl(\frac{S}{1+\beta}\Biggr)^2\Biggl(\frac{(I_C-I_{CO})(1+\beta a)}{\beta}\Biggr)$$

Resulting to: $$S''=\frac{S^2(I_C-I_{CO})(1+\beta a)}{\beta(1+\beta)^2} $$ Then using the value of $S$ can be written as: $$S''=\frac{S(\frac{1+\beta}{1+\beta a})(I_C-I_{CO})(1+\beta a)}{\beta(1+\beta)^2}$$ Notice how the terms cancel out: $$S''=\frac{S(\frac{\cancel{1+\beta}}{\cancel{1+\beta a}})(I_C-I_{CO})\cancel{(1+\beta a)}}{\beta(1+\beta)^{\cancel{2}}}$$ $$S''=\frac{S(I_C-I_{CO})}{\beta(1+\beta)}$$ $$\therefore S''=\frac1{\beta(1+\beta)}S(I_C-I_{CO})$$