BJT in saturation mode

Question

I have a hard time understanding BJTs in saturation. I was answering this problem but I can’t get the answer. I did was like this

The givens were only that so I’m really don’t get the answer

Answer 1

I very much dislike the question. Given the answer you show as correct, there is an implication that both the active mode and saturation mode \$\beta=100\$. Even as \$V_{_{\text{CE}\left(\text{SAT}\right)}}=200\:\text{mV}\$. This is never the case. Not ever.

The only logical remedy is to believe that there is an active mode \$\beta\$ that is higher than 100 and that the stated \$\beta\$ is for the case where \$V_{_{\text{CE}\left(\text{SAT}\right)}}=200\:\text{mV}\$. But I don't believe it. Just doesn't happen that way.

So I don't like it.

But granting it, then:

• \$V_{_{\text{CE}\left(\text{SAT}\right)}}=200\:\text{mV}\$

• \$V_{_\text{BE}}=700\:\text{mV}\$

• Saturation condition is \$\left(\frac{\beta+1}{\beta}\cdot 680\:\Omega+2.7\:\text{k}\Omega\right)\cdot I_{_\text{C}}=12\:\text{V}-V_{_{\text{CE}\left(\text{SAT}\right)}}\$

  Solve to find \$I_{_\text{C}}\approx 3.484\:\text{mA}\$

• \$V_{_\text{E}}=\frac{\beta+1}{\beta}\cdot I_{_\text{C}}\cdot 680\:\Omega\approx 2.393\:\text{V}\$

• \$V_{_\text{B}}=V_{_\text{BE}}+V_{_\text{E}}\approx 3.093\:\text{V}\$

• \$R_{_\text{B}}\le\frac{12\:\text{V}-V_{_\text{B}}}{\frac{I_{_\text{C}}}{\beta}}\$ or \$R_{_\text{B}}\le\left(\frac{8.907\:\text{V}}{34.84\:\mu\text{A}}\approx 255.6544\:\text{k}\Omega \right)\$

Rounded to 5 places this matches the given answer.

Answer 2

You failed to account for the voltage across the 680Ω resistor, making your assertion \$I_C=\frac{12-V_{CE}}{2700}\$ incorrect. By KVL, the corrected voltage \$V_{RC}\$ across the resistor \$R_C=2700\Omega\$ will be:

$$ \begin{aligned} V_{RC} &= 12-V_{CE(SAT)} - V_{RE} \\ \\ I_CR_C &= 12 - 0.2 - I_ER_E \\ \\ 2700I_C &= 11.8 - 680I_E \\ \\ \end{aligned} $$

KCL:

$$ \begin{aligned} I_E &= I_C + I_B \\ \\ &= I_C + \frac{I_C}{\beta} \\ \\ &= I_C(1+\frac{1}{\beta}) \\ \\ &= 1.01I_C \end{aligned} $$

Plug \$I_E\$ into the KVL equation:

$$ \begin{aligned} 2700I_C &= 11.8 - 680\times1.01I_C \\ \\ 2700I_C &= 11.8 - 686.8I_C \\ \\ 3386.8I_C &= 11.8 \\ \\ I_C &= 3.4841mA \end{aligned} $$

I cringe to quote so many significant figures, but since your "official answer" has that many, I'll just grin and bear it.

Now you have \$I_C\$, you can work out \$V_{RE}\$ and \$I_B\$, and derive a corresponding value for \$R_B\$. Again you neglected the voltage across the emitter resistor in your calculation for the voltage across the base resistor. The correct KVL equation is:

$$ \begin{aligned} V_{RB} &= 12-V_{BE} - V_{RE} \\ \\ I_BR_B &= 12 - 0.7 - 680I_E \\ \\ \frac{I_C}{\beta}R_B &= 11.3 - 680(1+\frac{1}{\beta})I_C \\ \\ \end{aligned} $$

I'll leave finding \$R_B\$ up to you.

Answer 3

It appears (after simulation), that the assumptions surrounding this question are: -

• Constant β (irrespective of reality)
• Base-emitter forward voltage exactly and always 0.7 volts

So, if this is modelled in a simulation tool (micro-cap) you find this: -

• And, by manipulation of R3, I get an answer of exactly 200 mV between collector and emitter.

• R3 is 255.64915 kΩ.

• The correct answer is apparently 255.65 kΩ.

• Nowhere in your question did you state what method should be used.

However, should you decide retrospectively that this problem has to be solved using math then, I hope you can see that the simulation above indicates where your problem currently is. Here's an example that assumes R2 is zero ohms: -

Now, the new value for R3 is 258.55933 kΩ but, because your handwriting is so bad I can't confirm that this is what you did although I suspect it is.