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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\$.

enter image description here

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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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

schematic

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.

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  • \$\begingroup\$ I think, there is an error in your reasoning: When the B-C junction is forward biased, the base current wiil be drastically increased (this increase is an INDICATION of saturation - not its cause) and Ie will NOT be app equal to Ic. Therefore, Ie=(beta + 1)Ib does NOT hold anymore. \$\endgroup\$ – LvW Feb 26 at 8:47
  • \$\begingroup\$ It still works, it's just that \$\beta\$ is effectively reduced a lot. The equations still hold, and it still doesn't matter what \$R_{E}\$ is. \$\endgroup\$ – KD9PDP Feb 26 at 19:57
  • \$\begingroup\$ I can see that this question drove you to solve it. Excellent ! +1 vote. . And I think your solution is elegant and correct. \$\endgroup\$ – Marla Feb 26 at 21:35
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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
  • \$\begingroup\$ The question as asking about a constant base current, but this answer is answering for a constant base voltage. That's why the question doesn't seem to make sense if you try to answer it this way. It actually does not matter what RE is at all, IE will be same no matter what. If RE is zero, in fact, you get the same current as it was 100k. (this is a case where calculations are needed to see that). \$\endgroup\$ – KD9PDP Feb 25 at 21:49
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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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