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I have attempted the following question on this circuit with a bjt. As I have learned in my electrical engineering class, I can replace the transistor with the following circuit and assume it is in linear mode. Once I solve it using mna, I can determine whether it is in cutoff mode, or saturated. Accorded to my calculation it is in neither? Could someone please assist me as I am new to this topic and I may lack some understanding somewhere.

enter image description here enter image description here

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    \$\begingroup\$ Do you know how to calculate Rb? \$\endgroup\$ – Trevor_G Nov 12 '17 at 1:41
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There are probably a few ways to do this, but I would do it this way.

Calculate \$R_B = R_E * (H_{FE} + 1) = 44 * 101 = 4444\Omega\$

Your circuit then becomes

schematic

simulate this circuit – Schematic created using CircuitLab

You should then be able to calculate \$V_B\$.

From superstition theory...

\$V_B = V_{CC} * \frac{R2||R_B}{R1 + R2||R_B} + V_{BE} * \frac{R1||R2}{R_B + R1||R2}\$

where || means in parallel with.

Now you know that, you can calculate \$I_B\$

\$I_B = \frac{V_B - V_{BE}}{R_B}\$

\$V_E\$ is simply \$V_B - V_{BE}\$

To check your work calculate \$V_E\$ the other way

\$V_E = I_B * (H_{FE} + 1) * R_E\$

and verify \$I_B + I_{R2} = I_{R1}\$

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Just another approach.

First transformation is obvious:

schematic

simulate this circuit – Schematic created using CircuitLab

For the right hand side schematic, you know that:

$$\begin{align*} V_{TH}&=V_{CC}\frac{R_2}{R_1+R_2}\\\\ R_{TH}&=\frac{R_1\:R_2}{R_1+R_2} \end{align*}$$

A very simple KVL yields:

$$V_{TH}-I_B\cdot R_{TH}-V_{BE}-I_E\cdot R_E=0\label{kvl}\tag{$KVL$}$$

Assuming the BJT is not saturated and instead in its active range, \$I_E=\left(\beta+1\right)\cdot I_B\$, and then it follows from above that:

$$\begin{align*} V_{TH}-V_{BE}&=I_B\cdot R_{TH}+\left(\beta+1\right)\cdot I_B\cdot R_E\\\\ V_{TH}-V_{BE}&=I_B\cdot \left[R_{TH}+\left(\beta+1\right)\cdot R_E\right]\\\\ &\therefore\\\\ I_B &= \frac{V_{TH}-V_{BE}}{R_{TH}+\left(\beta+1\right)\cdot R_E} \end{align*}$$

We can reference the \$\ref{kvl}\$ equation above to compute the emitter voltage as:

$$V_E=V_{TH}-I_B\cdot R_{TH}-V_{BE}$$


Normally, after going through a process similar to the above, you'd also then go through some basic "sanity checks." If the sanity checks fail, then the BJT would instead almost certainly be saturated and therefore the assumption above (that \$I_E=\left(\beta+1\right)\cdot I_B\$) has instead failed. This would then direct your analysis in a different direction.

But in your case, it is not possible for the transistor to be saturated. So you don't need to perform those sanity checks.

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