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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:

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

Soumee
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  • surely you also need to tell us what $I_C$ is: this does not appear in your first two equations – A. Goodier Mar 02 '18 at 12:39
  • @rbird https://math.stackexchange.com/questions/2673441/mathematics-partial-differentiation-arising-in-an-electronics-engineering-ques/2673458#2673458 – Soumee Mar 02 '18 at 12:40
  • is $S''$ a product of $\frac{1}{\beta(1+\beta)}$ and $I_C\frac{(R_T+R_E)(1+\beta)+\beta SR_E}{R_T+R_E}+SI_{CO}$, or the brackets imply a different operation? – John Glenn Mar 02 '18 at 14:28
  • @JohnGlenn It denotes the product. – Soumee Mar 02 '18 at 17:43
  • Is $S''$ correct? Because based on the values of $S$, $I_C$ and $I_B$, what I get is: $$S''=\frac1{\beta(1+\beta)}S(I_C-I_{CO})$$ – John Glenn Mar 02 '18 at 17:45
  • I'll post my computation, tell me if there are any mistakes in the definitions and in the arithmetic – John Glenn Mar 02 '18 at 17:47
  • @JohnGlenn There might be an error in the sign before $\beta S R_E$ in the final expression. I actually posted this as an electronics engg. question. I am attaching the link for your reference: https://electronics.stackexchange.com/q/316246/115973 – Soumee Mar 02 '18 at 17:57
  • Okay, I am posting my answer soon, please evaluate my computation afterwards – John Glenn Mar 02 '18 at 17:58
  • @Soumee do the calculations hold? – John Glenn Mar 02 '18 at 19:37

1 Answers1

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$\require{cancel}$

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: $$S''=\frac{S(\frac{\cancel{1+\beta}}{\cancel{1+\beta a}})(I_C-I_{CO})\cancel{(1+\beta a)}}{\beta(1+\beta)^{\cancel{2}}}$$ $$S''=\frac{S(I_C-I_{CO})}{\beta(1+\beta)}$$ $$\therefore S''=\frac1{\beta(1+\beta)}S(I_C-I_{CO})$$

John Glenn
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  • @Soumee please check for arithmetic and definition errors – John Glenn Mar 02 '18 at 18:30
  • I don't think there's an error in your solution. It seems correct. – Soumee Mar 03 '18 at 07:56
  • But I don't see it being the same with your value for $S''$, there could be something – John Glenn Mar 03 '18 at 07:58
  • Might I ask on what topic/problem in physics this is? – John Glenn Mar 03 '18 at 07:59
  • This is about Bipolar Junction Transistors. The topic is 'stability factors of a voltage divider bias circuit.' I have found out $S$ and $S'$. This is the calculation for $S''=\frac{\partial I_C}{\partial\beta}$ , provided $V_{BE}$ and $I_{CO}$ are constant. – Soumee Mar 03 '18 at 08:06