Farad = one coulomb per volt.

Henry = one volt per amp

Coulomb = charge carried by one amp in one second

ε = 4π10^{-7} H/meter

μ = 8.85 pF/meter

L = μAN^{2}/length

(meters)

Capacitance = Aμ/d

where A = area of capacitor and d = plate separation.

Coulomb's Law (Electric force)

Force=q^{2}/(4πεr^{2})

and Ampere's Law (Magnetic force)

Force/L =4πμ i^{2}/r

Note: i=q/t and

this leads to c^{2} = 1/(με)

frequency, f = 1/(2πRC), where R and C are resistance and capacitance.

f = 1.59*10^{5}/(RC) where R is in ohms and C is in microfarads

E and M basics

Conventional current flows from + to minus

In conven current, electrons flow other way

Negative voltage pushes electrons away, toward + anode. Creates conventional current from + to -

Diode

P and N type junction. P has holes, N has electrons excess.

Apply battery to junction with plus to the p and minus to the n, pushes electrons away from N toward P and thus conventional current flows from P side to N side.

Transistor

Diode sandwich

PN junction basic part of transistor

Battery anode = negative

cathode = positive

If wire connects, conventional current flows from cathode to anode

BUT electrons flow from A to P

Apply + voltage to P type gives conventional current flow (bias) from A to P???

$\alpha \; =n$

$CdV$_{out}_{b}
&tanh;(_{in}-V_{out})