The transistor in the circuit below has a very high value of $$beta$$.

 The value (V) of the collector voltage $$V_o$$ is most nearly:

A.  6.0
B.  4.0
C.  2.0
D.  1.3

 

To solve this equation, we first use KVL to set up an equation down the right side of our circuit. Since our beta value is very high, and since we know $I_B+I_C=I_E, I_C=\beta I_B$, we can conclude that $I_C\approx I_E$:

$$\begin{gathered} V_{cc}-\left(4k\Omega \right)\left(I_C\right)-V_{CE}-(1.3k\Omega )\left(I_E\right)=0 \\ 10-\left(4k\right)\left(I\right)-V_{CE}-(1.3k)\left(I\right)=0 \\ (-5.3k)\left(I\right)-V_{CE}=-10 \end{gathered}$$

In order to continue solving this equation, we must first calculate the value of $V_{CE}$. We can do this according to these equations:

$$\begin{gathered} V_{CE}=V_{CC}-I_{EQ}(R_C+R_E) \\ {I}_{EQ}=\frac{V_{BB}-V_{BE}}{{\displaystyle \frac{R_B}{\beta +1}}+R_E} \end{gathered}$$

$V_{BB}$can be found by using the equation for a voltage divider. Since we know our beta value is very high, we do not need to bother finding $R_B$ since the expression will be approximately 0:

$$\begin{gathered} V_{BB}=V_{CC}\left(\frac{R_2}{R_2+R_1}\right) \\ {V}_{BB}=\left(10\right)\left(\frac{20k}{80k+20k}\right) \\ {V}_{BB}=2V \end{gathered}$$ 

Plugging these values back into our original equations to find $V_{CE}$:

$$\begin{gathered} I_{EQ}=\frac{V_{BB}-V_{BE}}{{\displaystyle \frac{R_B}{\beta +1}+R_E}} \\ {I}_{EQ}=\frac{2-0.7}{{\displaystyle \frac{R_B}{\beta +1}+1.3k}} (\beta \approx \infty ) \\ {I}_{EQ}=\frac{2-0.7}{{\displaystyle 1.3k}} =\frac{1.3}{1.3k}\approx 1mA \end{gathered}$$

$$\begin{gathered} V_{CE}=V_{CC}-I_{EQ}(R_C+R_E) \\ {V}_{CE}=10-(.001)(4k+1.3k) \\ {V}_{CE}=10-(5.3) \\ {V}_{CE}=4.7V \end{gathered}$$

Now we can plug in our $V_{CE}$ value into our original KVL equation to get our current value.

$$\begin{gathered} (-5.3k)\left(I\right)-V_{CE}=-10 \\ (-5.3k)\left(I\right)-(4.7)=-10 \\ (-5.3k)\left(I\right)=-5.3 \\ I = 1mA \end{gathered}$$

Finally, knowing the current travelling through $R_C=4k\Omega$, we can find the collector voltage, $V_o$.

$$\begin{gathered} V_{cc}-\left(I\right)\left(4k\Omega \right)=V_o \\ 10-(.001)\left(4k\right)=V_o \\ 6V= V_o \end{gathered}$$

Therefore, our answer is A, 6.0.