I am trying to derive the stability factors of a voltage divider bias circuit. I could derive S,S', however, I'm facing some problem with the derivation of S''. Please help me in this regard.
How to find \$\frac{\partial I_C}{\partial \beta}\$?
The equations given in my book are these:
For a voltage divider circuit:
Kirchhoff's voltage law at the base circuit gives:
$$V_T=I_BR_T+V_{BE}+(I_B+I_C) R_E \tag1$$ $$I_B=\frac{V_T-V_{BE}-I_CR_E}{R_T+R_E} \tag2$$ The collector current \$ I_C \$ is given by: $$I_C=\beta I_B+(1+\beta)I_{CO} \tag3$$
Substituting \$ I_B \$ from 2nd eq to 3rd eq yields:
$$I_C=\beta\frac{V_T-V_{BE}-I_CR_E}{R_T+R_E}+(1+ \beta)I_{CO} \tag4$$
$$ I_C(1+\frac{\beta R_E}{R_T+R_E})=\frac{\beta(V_T-V_{BE})}{R_T+R_E}+(1+\beta)I_{CO} \tag {4a}$$
I understand and verified that the above equations came as a result of manipulating equations derived from Kirchhoff's voltage law and other current equations.
In my book, they haven't derived the following equations(5,6,7), but have jotted down the equations as given below:
Then the stability factor can thus be expressed as: $$S=\frac{\partial I_C}{\partial I_{CO}}=\frac{1+\beta}{1+\frac{\beta R_E}{R_T+R_E}} \tag5$$
$$S'=\frac{\partial I_C}{\partial V_{BE}}=-\frac{\beta}{(1+\beta)R_E+R_T} \tag6$$
$$S''=\frac{\partial I_C}{\partial\beta}=\frac{1}{\beta(1+\beta)}[I_C\frac{(R_T+R_E)(1+\beta)-\beta SR_E}{R_T+R_E}+SI_{CO} \tag7 ]$$
I could derive equation eq 5 and 6 however I am facing the following issues with equation 7:
1) I am getting $$ \frac{\partial I_C}{\partial\beta}=\frac{1}{\beta(1+\beta)}[I_C\frac{(R_T+R_E)(1+\beta)-\beta SR_E}{R_T+R_E}-SI_{CO} $$ instead of $$\frac{\partial I_C}{\partial\beta}=\frac{1}{\beta(1+\beta)}[I_C\frac{(R_T+R_E)(1+\beta)+\beta SR_E}{R_T+R_E}+SI_{CO} $$
2) Why do we need to devive equation (7) ,in which stability factor \$S''\$ is in terms of other stability factors. Why wouldn't it be sufficient to say that \$ S'' \$ is given by eq 9a? or $$\frac{\partial I_C}{\partial \beta}=\frac{V_T-V_{BE}-I_CR_E}{(R_T+R_E)(1+\frac{\beta R_E}{R_T+R_E})}+\frac{I_{CO}}{(1+\frac{\beta R_E}{R_T+R_E})}$$
3) How does eq 7 or 9a account for the fact that the Voltage divider circuit provides better stability against \$ \beta \$.?
4)If anyone has come across these equations, eq 5,6 and 7 in any standard book, then please mention the name of the book.
DERIVATION(MY ATTEMPT):
Partial differentiating eq 4a with respect to \$ \beta \$:
$$\frac{\partial I_C}{\partial \beta}(1+\frac{\beta R_E}{R_T+R_E})+I_C\frac{R_E}{R_T+R_E}=\frac{V_T-V_{BE}}{R_T+R_E}+I_{CO} \tag 8$$ $$\frac{\partial I_C}{\partial \beta}(1+\frac{\beta R_E}{R_T+R_E})=\frac{V_T-V_{BE}-I_CR_E}{R_T+R_E}+I_{CO} \tag 9$$
$$\frac{\partial I_C}{\partial \beta}=\frac{V_T-V_{BE}-I_CR_E}{(R_T+R_E)(1+\frac{\beta R_E}{R_T+R_E})}+\frac{I_{CO}}{(1+\frac{\beta R_E}{R_T+R_E})}\frac{1+\beta}{1+\beta} \tag {9a}$$
From eq 4a:
$$ I_C=\frac{\beta(V_T-V_{BE}}{(R_T+R_E)(1+\frac{\beta R_E}{R_T+R_E})}\frac{1+\beta}{1+\beta}+\frac{(1+\beta)I_{CO}}{(1+\frac{\beta R_E}{R_T+R_E})} \tag {10}$$
$$I_C=\frac{\beta(V_T-V_{BE})S}{(1+\beta)(R_T+R_E}+SI_{CO} \tag{11}$$ $$=> (V_T-V_{BE})=\frac{(1+\beta)(R_T+R_E)}{\beta S}(I_C-SI_{CO}) \tag{12}$$
Substituting eq 12 in eq 10
$$\frac{\partial I_C}{\partial\beta}=\frac{\frac{(1+\beta)(R_T+R_E)}{\beta S}(I_C-SI_{CO})-I_CR_E}{(R_T+R_E)(1+\frac{\beta R_E}{R_T+R_E})}+\frac{S}{1+\beta} I_{CO} \tag{13}$$
$$ \frac{\partial I_C}{\partial \beta}=\frac{\frac{(1+\beta)(R_T+R_E)}{\beta S}I_C-I_CR_E}{(R_T+R_E)(1+\frac{\beta R_E}{R_T+R_E})}+\frac{S}{1+\beta}I_{CO}-\frac{\frac{(1+\beta)(R_T+R_E)}{\beta S}SI_{CO}}{(R_T+R_E)(1+\frac{\beta R_E}{R_T+R_E})} \tag{14}$$
$$=\frac{1+\beta}{1+\beta}\frac{(1+\beta)(R_T+R_E)I_C-\beta SR_EI_C}{\beta S(R_T+R_E)(1+\frac{\beta R_E}{R_T+R_E})}+\frac{S}{1+\beta}I_{CO}-\frac{S}{\beta}I_{CO} \tag{15}$$
$$=\frac{SI_C[(1+\beta)(R_T+R_E)-\beta SR_E}{\beta(1+\beta)S(R_T+R_E)}+SI_{CO}[\frac{\beta-1-\beta}{\beta(1+\beta)}] \tag{16}$$
$$\frac{\partial I_C}{\partial \beta}=\frac{I_C[(1+\beta)(R_T+R_E)-\beta SR_E}{\beta(1+\beta)(R_T+R_E)}-\frac{SI_{CO}}{\beta(1+\beta)} \tag{17}$$
$$\frac{\partial I_C}{\partial \beta}=\frac{1}{\beta(1+\beta)}[I_C\frac{(R_T+R_E)(1+\beta)-\beta SR_E}{R_T+R_E}-SI_{CO} ]\tag{18}$$
The sign of the last term should be +ve. Where might have I gone wrong?
