Required Emitter Resistance \$R_E\$ for the Maximum Emitter Current for a BJT

Question

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\$.

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}}\$

Answer 1

7 years late, but this was a fun one to do some forensic math on! And it actually may be a bit of a trick question! The trick comes down to that you're given an input current \$I_B\$, not an input voltage.

First: The error was on line 3 when you calculated \$I_E\$. You actually found \$I_C\$ instead. You have to add \$I_B\$ to that to get \$I_C\$.

Here's the shortcut solution. If the BJT is in the active mode: $$ I_E=(\beta+1)I_B $$

That means that it doesn't matter what \$R_E\$ is, the current will just be \$101\times I_B\$. We can explore this a little bit more thinking about Ebers Moll

^{simulate this circuit – Schematic created using CircuitLab}

Let's say we start off with \$I_B=0\$. At this point CE is positive, reverse biasing D2 - so no current flows anywhere. This is in cutoff. It doesn't matter what \$R_E\$ is in this case either.

Let's turn on \$I_B\$ a little bit. D1 is forward biased and D2 is still reverse biased (so it's in the active mode). The current through D1 is \$I_{D1} = I_B+\alpha_F I_{D1}\$. Solving for the current through D1 is \$I_{D1}=I_E= I_B/ (1-\alpha_F)I_B = (\beta +1)I_B\$. Again, the value of \$R_E\$ does not matter.

Ok, so our BJT is in the active state. That means: $$ V_BE=0.7\\ I_C=\beta I_B\\ V_o = V_+ - \beta I_B R_C=V_C\\ V_E = (\beta+1)I_B R_E-V_+\\ V_B \approx 0.7 - V_+ +(\beta+1)I_B R_E\\ V_{BE} \approx 0.7\\ V_{BC} \approx -2V_+ + 0.7 + (\beta+1)I_B R_E + \beta I_B R_C $$

That means that at some point \$I_B\$ gets large enough that the \$V_{BC}\$ becomes positive, and our diode D1 becomes forward biased. We are now in saturation mode. At that point, no matter how much we increase IB, \$I_C\approx I_E\$. \$\beta\$ is effectively reduced, but it's still basically \$I_E=(\beta_{reduced}+1)I_B\$.

So it's not exactly a trick question, but it's a question about controlling a BJT with current instead of voltage.

Answer 2

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:

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

Answer 3

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.