# Calculate Collector Voltage In Collector feedback bias Configuration

I've tried using KVL and KCL but I always end up with two or more variables, and I've got another question, how can I know if this transistor is in the saturation region?

• May I assume just DC analysis here? I see a capacitor, which is why I'm asking. – jonk Oct 28 '17 at 22:07
• Yes, just DC analysis. Specially when the capacitor is fully charged . – Xggggg Oct 28 '17 at 22:21
• Looks like two equations and two unknowns: $\frac{V_C}{R_6}+\frac{V_C}{R_7}+315\cdot I_B=\frac{9\:\textrm{V}}{R_6}+\frac{700\:\textrm{mV}}{R_7}$ and $\frac{700 \: \textrm{mV}}{R_7}+\frac{700 \: \textrm{mV}}{R_8}+I_B=\frac{V_C}{R_7}$. Those solve out pretty easy. It isn't saturated. Does that make sense to you? Or do you need a write-up? – jonk Oct 28 '17 at 22:33
• I think you used KCL and substitute with Ohm, am I right? Would you write down the currents you considered? – Xggggg Oct 29 '17 at 0:32
• I used KCL for both equations. You can solve the 2nd one for $V_C$ and then substitute that into the first equation for $V_C$, then solve that resulting equation for $I_B$. I get a base current of $I_B\approx 10.916\:\mu\textrm{A}$. What do you get? – jonk Oct 29 '17 at 0:45

Call the collector node $C$ and call the voltage there $V_C$. Call the base node $B$ and call the voltage there $V_B$. We know that $V_B=700\:\textrm{mV}$, by definition. We also know that $\beta=315$ and therefore that $I_C=\beta \:I_B$, by definition.

Direct Route:

You can immediately compute $I_8=\frac{V_B}{R_8}$. That current, plus the base current must flow through $R_7$. So $I_7=I_B+\frac{V_B}{R_8}$. That current, plus the collector current must flow through $R_6$. As $I_C=\beta \:I_B$, so $I_6=\frac{V_B}{R_8}+\left(\beta+1\right)\:I_B$. The sum of the voltage drops across the three resistors must be your voltage source, $V_{CC}=9\:\textrm{V}$. So it must be the case that $I_6\:R_6+I_7\:R_7+I_8\: R_8=9\:\textrm{V}=V_{CC}$. From this information we have:

\begin{align*} V_{CC}&=\left(\frac{V_B}{R_8}+\left[\beta+1\right]\:I_B\right)\:R_6+\left(I_B+\frac{V_B}{R_8}\right)\:R_7+\frac{V_B}{R_8}\: R_8\\\\ \end{align*}

Solving for $I_B$ you should get:

$$I_B =\frac{V_{CC}-V_B\left(1+\frac{R_6+R_7}{R_8}\right)}{R_6\left(\beta+1\right)+R_7}$$

That method is pretty straight-forward.

Using KCL:

This is using nodal analysis. It will get you to the same place, but through a slightly more complex route.

Then, by KCL at each node I get:

\begin{align*} \frac{V_C}{R_6}+\frac{V_C}{R_7}+I_C&=\frac{V_{CC}}{R_6}+\frac{V_B}{R_7}\\\\\frac{V_B}{R_7}+\frac{V_B}{R_8}+I_B&=\frac{V_C}{R_7} \end{align*}

The first equation is just putting all the currents "spilling away" from node $C$ on the left and all the currents "spilling into" node $C$ on the right. The two must equal each other, of course.

The second equation is just putting all the currents "spilling away" from node $B$ on the left and all the currents "spilling into" node $B$ on the right. The two must equal each other, of course, again.

The above solves out easily as:

\begin{align*} V_C&= \frac{V_{CC}\:R_7+V_B\:R_6\left(1+\beta\left[\frac{R_7}{R_8}+1\right]\right)}{R_6\left(\beta+1\right)+R_7}\\\\ I_B &=\frac{V_{CC}-V_B\left(1+\frac{R_6+R_7}{R_8}\right)}{R_6\left(\beta+1\right)+R_7} \end{align*}

As you can see, the equation for $I_B$ works out the same way. It's just that this also gets you $V_C$ along the way.

Since you already know that $V_B=700\:\textrm{mV}$ and all the resistor values are known as is $V_{CC}=9\:\textrm{V}$, I think you should be able to make the calculations here. You should also be able to come up with those equations.

In the end, I think you will find that $V_C$ is large enough that the transistor cannot be saturated.