ELECTRONIC REMINDERS

ELECTRONIC REMINDERS
====================

# Ohm's Law

* V = I x R
* R = V / I
* I = V / R

>Voltage = Current(I) x Resistance (Ω=Ohm)
>Units: Volts = Amps x Ohms

## Power's Law

* P = V * I
>Units: Watts = Volts x Amps

* P = I² * R
* P = V² / R

* R = P  / I²
* I = √P / R
* I = P  / V

## Samples

* Computing the resistance:
    9V  Battery + LED 3.4V using 20mA ==> (9V-3.4) / 20mA ==> 5.6 / (20/1000) => 5.6 / 0.02 => 280 Ω
    Power: 9V * 0.02 = 0.18 Watts

    9V  Battery + LED 3V using 20mA ==> (9V-3) / 20mA ==> 5 / (20/1000) => 5 / 0.02 => 250 Ω
    
    18V Battery + LED 3.4V using 20mA ==> (18-3.4)/(20/1000) => 730 Ω
    Power: 18V * 0.02 = 0.36 Watts
    

## Serie vs Parallel
    - In Serie, Voltage drop (consumption increase), but current is constant across the circuit.
    - In Parallel, Voltage is constant, but current consumption increase.

## Resistance in serie

 -------------
 |   |
5V Battery      ~ R1 100 Ω
 |   |
 |   |
 |   ~ R2 200 Ω
 |   |
 -------------

    R = R1 + R2    
    I = 5V / (100+200)Ω => 5 / 300 => 17mA
    Current is constant across the circuit
    Voltage Drop 
        R1: V1 = 17mA * 100 = 1.7V
        R2: V2 = 17mA * 200 = 3.4V
            1.7V + 3.4V ~= 5V

## Resistance in parallel

 -------------
 |   |
5V Battery  I1 /\ I2
    |         R1 R2      R1 100Ω,  R2 200Ω
 |     \/ 
 |   |
 -------------

    R = 1 / ( (1/R1) + (1/R2) ) => 1/((1/100)+(1/200)) => 66.7 Ω

    I = 5V / (66.7)Ω => 75mA
    Current is constant across the circuit
    Current across resistor
        I1 = 5V / 100Ω = 50mA
        I2 = 5V / 200Ω = 25mA
            50mA + 25mA = 75mA

    2 100Ω Resistors in parallel => 100Ω/2 = 50Ω

## RC Circuit - Resistor Capacitor

* Capacitor resist change in voltage
* R x C (Capacitance) = Time

                       (-t/RC)
   * Vout = Vin x (1 - e       )
   * e = 2.7182
   * t = time
   * if t==0 then Vout == 0

* What is t when Vout is half Vin?
    * t = 0.693 x RC
* At 1 x RC the capacitor is charged at 63.2%
* At 2 x RC the capacitor is charged at 86.59%
* At 3 x RC the capacitor is charged at 95.9%

### Sample
    9V Battery + 1kΩ + 470uF + LED (3V, 20mA)

    RC = 1000Ω * 470uF == 1000Ω * (470/1000/100) = 4.7 Seconds
    RC = 1000Ω *  47uF == 1000Ω * ( 47/1000/100) = 0.44.67 Seconds

* http://en.wikipedia.org/wiki/RC_circuit


## Capacitor in parallel
        
## Capactor in serie


## Diode


## Examples

* 1V = 1A x 1Ohm

## References

* [wikipedia - Ohm's_law](http://en.wikipedia.org/wiki/Ohm's_law)
* [Voltage current resistance and ohms law](https://learn.sparkfun.com/tutorials/voltage-current-resistance-and-ohms-law#voltage)
* http://vapor.rigsmedia.com/wp-content/uploads/2012/09/ohms_lawVIPR_relationship.jpg





# Others

## Characters

* √ ¼ ½ ¾ 2º 2¹ 2² 2³ Ω

# Markdown

* http://daringfireball.net/projects/markdown/syntax

 Arduino Tutorial
  https://www.youtube.com/watch?v=js4TK0U848I&list=PLA567CE235D39FA84


# Voltage  Divider
 
 -------------
 |   |
5V Battery      ~ R1 Omh
 |   |
 |   ----> Voltage out
 |   |
 |   ~ R2 Omh
 |   |
 -------------

 Vout =  (R2 / (R1+R2)) * Vin

 R1 100Omh R2 100Omh  =>   100/(100+100) *  5v = 2.5v

 Light

 5v
 |
 10k Resistor
 |
 -----  Vout
 |
 * Ligth Censor (Resistor based on light)

Voltage Regulator:
 9V  -- VR -- 5v
 |        |   |
 |        |  |
    *Cap1mF-Gnd--*Cap22mF

Decoupling Capacitor needed

PWM:                               _   _   _
 Pulse With Modulate Signal : _| |_| |_| |_

Transistor:
 NPN:
  ---------
  |   |   |
  C   B   E (Collector, Base, Emettor)
   Current flow from C to E based on the state of B is high

 PNP: Opposite when B is low current flow



Pull Up/Down Tutorial
http://www.ladyada.net/learn/arduino/lesson5.html