# BJT in saturation mode

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

• In line 1 you are missing the 680 ohm resistor Aug 6 at 9:17
• @periblepsis The definition reads in the picture, Vcesat = 0.2V. Aug 6 at 9:35
• mama, do you intend on going any further with this question? Aug 20 at 11:56

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.

• Where did the equation come from that is multiplied to Ic in the saturation condition? Aug 6 at 12:33
• Where did also b+1/b came from? Aug 6 at 12:37
• @mamab The emitter current is (b+1)/b larger than the collector current. The collector current times the collector resistor is its voltage drop. The emitter current times the emitter resistor is its voltage drop. Those two drops plus the saturation Vce must equal the supply voltage. Aug 6 at 12:41
• Is it okay to show on how did you derive the (b+1)/b? Aug 6 at 12:53
• @mamab The collector current Ic requires a base current of Ic/b, under the silly assumption your problem requires. I granted that but did not like it and said so. Regardless, the emitter current is the sum of the base and collector currents. So Ie=Ic+Ib=Ic+Ic/b=Ic*(b+1)/b. Aug 6 at 12:59

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.

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.