Formulas

$$R_{series}=\sum R_i$$

$$R_{parallel}=\frac{1}{\sum \frac{1}{R_i}}$$

$$X_C = \frac{1}{\omega C}$$

$$X_L = \omega L$$

$$Z_R = R$$

$$Z_C = -\frac{j}{\omega C}$$

$$Z_L = j\omega L$$

Given two resistors $R_1$ and $R_2$ with $R_2=nR_1$, the parallel combination has total resistance $\frac{nR1}{n+1}=\frac{R2}{n+1}$.

$V_*$ in Figure 1 is related to $V_A$ and $V_B$ by: $$V_* = \frac{V_A\cdot R_B + V_B\cdot R_A}{R_A+R_B}$$ Note that each input voltage is weighted by the resistor opposite $V_*$. This generalizes to $n$ voltages and $n$ resistors as: $$V_*=\left(\sum_{k=1}^n \frac{V_k}{R_k}\right) \cdot\left(\frac{1}{\sum_{k=1}^n \frac{1}{R_k}}\right),$$ that is, the short-circuit current times the total parallel resistance.

When the following conditions are met (reverse polarities for PNP):

1. $V_C > V_E$
2. the B-E junction is forward biased, and the C-B junction is reverse-biased
3. the device's maximum ratings for $I_C$, $I_B$, and $V_{CE}$ are not exceeded

then $I_C$ is related to $V_{BE}$ by $$I_C = I_S(T)\left(e^{V_{BE}/V_T-1}\right)$$ or equivalently $$V_{BE}=\frac{kT}{q}\ln \left(\frac{I_C}{I_S(T)}+1\right)$$ where

• $V_T=\frac{kT}{q}\approx 23.5\mathrm{mV}$ at room temperature (293 K)
• $q=1.60\times 10^{-19}~\mathrm{coulombs}$ is the electron charge
• $k=1.38\times 10^{-23}~\mathrm{J/K}$ is Boltzmann's constant
• $T$ is absolute temperature in Kelvin
• $I_S(T)$ is the saturation current of the given transistor (roughly $10^{-15}\mathrm{A}$ for a small-signal device like the 2N3904)

Then $I_B\approx\frac{I_C}{\beta}$, though it is still heavily dependent on $V_{BE}$. In the active region, $I_C >> I_S$, so the $-1$ term can be neglected.