BJT transistor sum

Question

Here I considered it is in active region and considered Vbe=0.6 V.

I did this:

I considered it is in active region but it Vce is less than 0.2 V. What should I do next?

Answer 1

I've checked your work and provided my own calc results next to yours:

So I agree with you.

But this only applies when the BJT is in active mode where the assumption of the active-mode \$\beta\$ applies.

Just as you also calculated, the voltage drop across the collector resistor is very large. I agree here, too. The fact that you got a negative value for the collector voltage, when the entire circuit sits between two rails of \$0\:\text{V}\$ and \$10\:\text{V}\$, is all you need to know that the result is nonsense. It's wrong. So the assumption of active mode and \$\beta=50\$ fails.

You know, a priori, that the collector voltage must be between \$0\:\text{V}\$ and \$10\:\text{V}\$. So you need to change your active mode assumptions into saturated mode. Here, \$0\:\text{V} \le V_{_\text{CE}}\le 400\:\text{mV}\$ and the collector now acts like a voltage source, not a current source. So, here you apply two KVL equations:

$$\begin{align*} 10\:\text{V}-2.2\:\text{k}\Omega\cdot I_{_\text{C}}-V_{_\text{CE}}-100\:\Omega\cdot I_{_\text{E}}&=0\:\text{V} \\\\ 10\:\text{V}-10\:\text{k}\Omega\cdot I_{_\text{B}}-V_{_\text{BE}}-100\:\Omega\cdot I_{_\text{E}}&=0\:\text{V} \end{align*}$$

But you also know that \$I_{_\text{E}}=I_{_\text{C}}+I_{_\text{B}}\$, as that is still true. So:

$$\begin{align*} 10\:\text{V}-2.2\:\text{k}\Omega\cdot I_{_\text{C}}-V_{_\text{CE}}-100\:\Omega\cdot \left(I_{_\text{C}}+I_{_\text{B}}\right)&=0\:\text{V} \\\\ 10\:\text{V}-10\:\text{k}\Omega\cdot I_{_\text{B}}-V_{_\text{BE}}-100\:\Omega\cdot \left(I_{_\text{C}}+I_{_\text{B}}\right)&=0\:\text{V} \end{align*}$$

That solves out to:

$$\begin{align*} I_{_\text{C}}&=\frac{10\:\text{k}\Omega\cdot\left(V_{_\text{CC}}-V_{_\text{CE}}\right)+100\:\Omega\cdot\left(V_{_\text{BE}}-V_{_\text{CE}}\right)}{100\:\Omega\cdot 2.2\:\text{k}\Omega+100\:\Omega\cdot 10\:\text{k}\Omega+2.2\:\text{k}\Omega\cdot 10\:\text{k}\Omega} \\\\ I_{_\text{B}}&=\frac{2.2\:\text{k}\Omega\cdot\left(V_{_\text{CC}}-V_{_\text{BE}}\right)-100\:\Omega\cdot\left(V_{_\text{BE}}-V_{_\text{CE}}\right)}{100\:\Omega\cdot 2.2\:\text{k}\Omega+100\:\Omega\cdot 10\:\text{k}\Omega+2.2\:\text{k}\Omega\cdot 10\:\text{k}\Omega} \end{align*}$$

Let's assume your figure for \$V_{_\text{BE}}= 600\:\text{mV}\$ and use my own guesstimate of \$V_{_\text{CE}}= 200\:\text{mV}\$. Then we'd find \$I_{_\text{B}}=888.8\overline{8}\:\mu\text{A}\$ and \$I_{_\text{C}}=4.2\overline{2}\:\text{mA}\$. From that, we'd find the saturated mode \$\beta= 4.750\$. Nothing like the \$50\$ that would be the case for active mode.

From the above you can work out that:

$$\begin{align*} V_{_\text{C}}&=10\:\text{V}-2.2\:\text{k}\Omega\cdot I_{_\text{C}}=711.1\overline{1}\:\text{mV} \\\\ V_{_\text{B}}&=10\:\text{V}-10\:\text{k}\Omega\cdot I_{_\text{B}}=1.1\overline{1}\:\text{V} \\\\ V_{_\text{E}}&=100\:\Omega\cdot \left(I_{_\text{C}}+I_{_\text{B}}\right)=511.1\overline{1}\:\text{mV} \end{align*}$$

And note that \$V_{_\text{C}}\lt V_{_\text{B}}\$. So the base-collector junction is forward-biased, now.

That's the saturated BJT solution, required as the active mode solution was falsified.

Here's an LTspice run:

The BJT here was tweaked to get close to the earlier assumptions, but not perfectly. So here in practice they work out to \$V_{_\text{BE}}=602.611\:\text{mV}\$ and \$V_{_{\text{CE}_\text{SAT}}}=202.033\:\text{mV}\$. Plugging these actual results into the formulas above, we'd then have found \$I_{_\text{C}}=4.22135\:\text{mA}\$, \$I_{_\text{B}}=888.639\:\mu\text{A}\$, \$V_{_\text{C}}=713.032\:\text{mV}\$, \$V_{_\text{B}}=1.11361\:\text{V}\$, and \$V_{_\text{E}}=510.999\:\text{mV}\$. Which is a perfect match against the results of simulation. So the concepts applied earlier for analyzing the saturated case appear validated by simulation.