====== Formulas ======
===== Electronics =====

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

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

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

$$X_L = \omega L$$

$$Z_R = R$$

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

$$Z_L = j\omega L$$

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

<figure center|passive_mixer>
{{ ::passive-avg.png?400 |}}
<caption>Passive mixer</caption>
</figure>

$V_*$ in {{ref>passive_mixer}} 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. 

==== Ebers-Moll BJT model ====
When the following conditions are met (reverse polarities for PNP):

  - $V_C > V_E$
  - the B-E junction is forward biased, and the C-B junction is reverse-biased
  - 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. 

{{tag>electronics math}}