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 = μAN2/length

(meters)

Capacitance = Aμ/d

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

## Some physical laws

Coulomb's Law (Electric force)
Force=q2/(4πεr2)

and Ampere's Law (Magnetic force)

Force/L =4πμ i2/r

Note: i=q/t and
this leads to c2 = 1/(με)

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

f = 1.59*105/(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???

$\alpha =n$e2 f2

$CdV$outdt = Ib &tanh;(κ(Vin-Vout)2)