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I need some help. This can be a dumb question as I am not an electrical engineer nor an electronics engineer, but need to solve a problem as given below. I tried to solve it but not sure if it is true because spice simulations give different result.

Where is my mistake?

Problem

Choose Emitter Resistance \$R_E\$ for the given problem for the max Emitter Current \$I_E\$ that is possible. Take \$\beta=100\$.

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My solution approach

\$I_C=\frac{V_+-V_{out}}{R_c}\Rightarrow V_c=R_cI_c+V_0\\ I_E=I_B+I_C=(\beta+1)I_B\\ \\ \frac{V_{out}-V_{CE}+V_+}{R_E}=I_E\\ \\ \frac{R_CI_C+V_+-V_{CE}+V_+}{R_E}=I_E\\ \\ I_C=\left( \frac{\beta}{1+\beta}\right)I_E\\ \\ \frac{R_C\left( \frac{\beta}{1+\beta}\right)I_E+V_+-V_{CE}+V_+}{R_E}=I_E\\ \$

which gives

\$ I_E=\cfrac{2V_+ -V_{CE}}{R_E-R_C \cfrac{\beta}{\beta+1}} \$

for the maximum current \${R_E-R_C \cfrac{\beta}{\beta+1}}\$ should be zero, thus,

\${R_E=R_C \cfrac{\beta}{\beta+1}}\$

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    \$\begingroup\$ Seems like a trick question or a mistype of the requirement. Does it require equations to justify the result? \$\endgroup\$ – alexan_e Jan 9 '14 at 11:23
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    \$\begingroup\$ What the devil is \$V_0\$? I thought you were using \$V_0\$ for \$V_{out}\$ but then you used them both in the same equation. Other than that, I agree with the others that this doesn't make sense. \$\endgroup\$ – Joe Hass Jan 9 '14 at 12:16
  • \$\begingroup\$ @JoeHass You are right \$V_0\$ and\$V_+\$ is same. Corrected equations. \$\endgroup\$ – Burak ER Jan 11 '14 at 5:05
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Is this a trick question? To me, \$I_e\$ is maximum when \$R_e = 0\$. All the other variables besides \$R_e\$ are fixed, so there is no way to increase \$I_e\$ by increasing \$R_e\$. What I am missing?

In that case, \$I_e\$ is determined by:

$$ \min(100 I_b, \frac{2V_+ - V_{ce}}{R_c}) $$

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  • \$\begingroup\$ I also think \$R_E\$ should be zero for the max current. But, what about calculations? \$\endgroup\$ – Burak ER Jan 9 '14 at 10:59
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    \$\begingroup\$ I agree, the question doesn't make sense. \$\endgroup\$ – Andy aka Jan 9 '14 at 11:00
  • \$\begingroup\$ @BurakER No calculations are needed - maximum collector (and by virtue of the circuit emitter) current is obtained when the transistor is turned-on as much as possible and this is when Re is zero thus forcing maximum base current. \$\endgroup\$ – Andy aka Jan 9 '14 at 11:02
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There is an error in your formula. In the first step, you say: $$ I_C = \dfrac{V_+ - V_{out}}{R_c} $$

and this is correct, but this implies

$$ V_+ = I_c \cdot R_c + V_{out} $$ instead of $$V_C = I_c \cdot R_c + V_{out}$$ and the rest of the mathematical development is not right.

In my opinion, the Kirchhoff laws for the ouput circuit are wrong, and this is transferred to the rest of the developing.

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    \$\begingroup\$ Voltage drop on \$R_c\$ should be \$V_C=I_C\cdot R_C\$. Therefore, I think my calculations are correct I think. \$\endgroup\$ – Burak ER Jan 11 '14 at 5:09
  • \$\begingroup\$ @BurakER That equation is very confusing, I would have expected \$ V_C \$ to represent the collector voltage and \$ V_{Rc} \$ the voltage drop on the collector resistor, therefor \$ V_{Rc}=I_{Rc} \cdot R_c \$ \$\endgroup\$ – alexan_e Jan 11 '14 at 8:13

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