## BJT Transistor Nodal Analysis

### Basic BJT Equations:

$i_C = \alpha i_E \hspace{20 mm} i_B = (1- \alpha ) i_E = \frac{i_E}{\beta + 1}$ $i_C = \beta i_B \hspace{20 mm} i_E = ( \beta + 1) i_B$ $\beta = \frac{ \alpha }{ 1 - \alpha } \hspace{21mm} \alpha = \frac{ \beta }{ \beta + 1 }$

It is also important to know that $|V_{EB}| = |V_{BE}|$ can be modeled as $.7V$.

These equations are not very informative by themselves so a few examples are demonstrated below. In both examples we will assume $\beta$ is very large. What this means for our calculations is $i_B \approx 0$. Since $i_B \approx 0$ we also assume that $i_C \approx i_E$.

### Finding missing voltages in a BJT circuit

Example 1. Solve for V3:

### There are several ways to find $V_3$. The more “difficult” way is to first find the emitter current, $i_E$, and then use Ohm’s Law. Since we know $i_C \approx i_E$, we can find the collector current, $i_C$, and then solve for $V_3$.

$\frac{-4-(-10)}{2.4k } = i_C = 2.5mA$ $12 - (i_E)(5.6k) = 12 - (i_C)(5.6k) = V_3$ $V_3 = 12 - (2.5mA)(5.6k) = -2 V$

The easier way to find $V_3$ is to recall that $|V_{EB}|$ behaves like a diode. For this pnp BJT: $V_{EB} = V_E - V_B = .7$.

We know that $V_B = -2.7V$ so $V_E = V_3 = -2 V$

$\beta$ may not always be a very large number. Had that been the case here, we would have started by finding the collector current (since it’s voltage drop and resistance are given) and since $i_B \neq 0$ anymore, we would use the formulas above to the find the base and collector current.

### Finding BJT Bias Voltages and Currents

Example 2 Solve for V2 and I1:

Here we will want to start by finding $I_1$. $I_1$ also equals $I_E$ which approximately equals $I_C$ and this collector current will allow us to find $V_2$ .

$I_1 = \frac{10.7-.7}{10k} = 1mA$ $V_2 = (1mA)(10k) - 10.7 = -.7 V$

Notice that $V_E$ was given as $.7 V$.  If this had not been given, we would have been able to find it because $V_{EB} = V_E - V_B = .7 V$ and $V_B = 0 V$.

Similar to the previous example, if $\beta$ was not a very large number. We would first find the emitter current and then use the equations in the table to find the other branch currents.

Note that both of these examples used pnp BJTs. The difference in an npn BJT is the base-emitter voltage is reveresed. You would use $V_{BE} = V_B - V_E = .7 V$.

### General Rule of Thumb

Most of these problems are very simple to solve. Typically $\beta$ is given and you will need to use Ohm’s Law to identify one of the currents. After one of the currents is found you will be able to solve for the other currents using the basic equations listed above. If one of the currents is not immediately obvious, the base-emitter voltage is likely needed. Most problems have you deduce the emitter voltage from the base, but it is easily possible to find the base voltage from the emitter voltage and then use that to find the base current.