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}\$?

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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.


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?

  • \$\begingroup\$ I think the error may be -ve sign in (6) you may also compare with youtube.com/watch?v=lBubD87dhwo \$\endgroup\$ Commented Jul 9, 2017 at 0:17
  • \$\begingroup\$ Also youtube.com/watch?v=mC5zhDpQtzI , but in 40 years of using transistors I have never used this formula, yet I know how to optimize stability from experience \$\endgroup\$ Commented Jul 9, 2017 at 0:24
  • \$\begingroup\$ speaking of ∇ @jonk have you watched the TV series Genius where Einstein's wife to be, beat him in a math entrance test and she later quoted the curl equation in the 1st episode. \$\endgroup\$ Commented Jul 9, 2017 at 0:29
  • \$\begingroup\$ @TonyStewart.EEsince'75 Nope. I haven't watched TV of any kind since... 1992 or so? PBS shows were becoming junk as well as starting to have ads (Clinton was the reason for the changes.) I finally just wrote the whole enterprise off. They don't get access to my brain anymore. Cut off. But familiar with divergence and curl -- basic stuff for vector fields and calculus. I just didn't know what symbol the OP wanted, while the OP was fighting with you about what they didn't want. Figured I'd let the dust settle between you before reading the Q again. ;) \$\endgroup\$
    – jonk
    Commented Jul 9, 2017 at 0:53
  • 1
    \$\begingroup\$ partial derivatives, ∂ lead to dx/dy or \$\Delta X/\Delta Y\$ anyways. I prefer Sensitivity derivatives rather than Stability as it shows direct effects on output for input variance. \$\endgroup\$ Commented Jul 9, 2017 at 4:49

1 Answer 1


Derivation for stability factor of Voltage divider

Other factors for voltage divider and other biasing circuits can be obtained following a similar procedure. Hope it helps.

  • \$\begingroup\$ Good academic answer. Considering β has a dynamic range of 3:1 (max:min) (i.e. 300% tolerance on high side) let β+1=β and simply let Rt <=10% of βRe for good stability, not Re>Rth \$\endgroup\$ Commented Jul 23, 2019 at 20:57
  • \$\begingroup\$ When self bias with negative feedback(NFB) the Av gain << β then H bias design becomes obsolete and NFG Av depends more on R tolerances and Iq. with > 10 x better THD. When self bias gain << β Av depends more on R tolerances and Iq. THD is the asymmetry of large signal output and reduces by THD(nfb)~THD(Hbias)*Av/β This done by moving your R2 to Rcb and choosing Rin such that Av is slightly less than Rcb/Rt. \$\endgroup\$ Commented Jul 23, 2019 at 21:11

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