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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?

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  • \$\begingroup\$ May I assume just DC analysis here? I see a capacitor, which is why I'm asking. \$\endgroup\$ – jonk Oct 28 '17 at 22:07
  • \$\begingroup\$ Yes, just DC analysis. Specially when the capacitor is fully charged . \$\endgroup\$ – Xggggg Oct 28 '17 at 22:21
  • \$\begingroup\$ 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? \$\endgroup\$ – jonk Oct 28 '17 at 22:33
  • \$\begingroup\$ I think you used KCL and substitute with Ohm, am I right? Would you write down the currents you considered? \$\endgroup\$ – Xggggg Oct 29 '17 at 0:32
  • \$\begingroup\$ 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? \$\endgroup\$ – jonk Oct 29 '17 at 0:45
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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.

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