## Electrical Appliance Protection

This simple circuit of electrical appliance protection  is to protect your electrical appliance during irregular power supply. Normally when power goes off we forget switching off our appliances. Especially in India where power cuts are very frequent we tend to leave our appliances on. In such case there are very high chance of getting it damages by high voltage surge.

The circuit below helps in this and does not switch on unless we press reset button.

#### Components – Electrical Appliance Protection

• Diode 1N4007 – 3
• Capacitor C1 – 1000 25v
• Relay – 6 V 100
• Step Down Transformer = 6-0-6 300mA
• Push to On Switch – 1

Very short duration interruptions or fluctuations will  not affect the circuit because of presence of large value capacitor. Thus the circuit provides suitable safety against erratic power supply conditions.

## LED Flasher using NE555

LED Flasher circuit using NE555 timer IC

• R1, R2, C1 and the supply voltage determine the flash rate. For a variable flash rate, replace R1 with a 1 MΩ pot in series with a 22k resistor.
• The purpose of R3 and R4 is to limit current through the LEDs to the maximum they can handle (usually 20 milliamps). 470 ohms works well with a supply voltage of 9-12 volts.
• The duty cycle of the circuit (the percentage of the time LED 1 is on to the time it is off during each cycle) is deterimed by the ratio of R1 to R2. If the value of R1 is low in relationship to R2, the duty cycle will be near 50 percent.
• The NE555 timer chip can be damaged by reverse polarity voltage being applied to it. You can make the circuit fool proof by placing a diode in series with the supply leads.

USB reading lamp is powered using USB port. The USB port provide 5 v and 100mA which is sufficient for this circuit. Cut the one end of the USB cable and use RED(pin1) and Black(pin4) for 5v positive and negative respectively.

Parts
1. c1,c2 – 100 mF 25 v
2. Zener diode – 4.7v 400m
3. R1- 220 r
4. R2- 100 r
5. T1 – SL100
6. White LED – 5 nos.
7. USB Cable

## RESISTOR COMBINATIONS

• When we do not get specific resistor values we have to either use variable resistors such as potentiometers or presets to obtain such precise values. Pots are too expensive to use forevery case.
• Another scheme is to combine two or more resistors to obtain the necessary precise values.Such resistor combinations can cost as little as 50p or so only.
• Then the question arises as to how one should combine these resistors, because, they can be combined in two different ways.
• These are called “Series” and “Parallel” combinations.

## Series Combinations

R Total= R1+ R2

• Calculating values for two or more resistors in series is simple, add all the values up.
• The connection ensures that the SAME current flows through all resistors.
• In this type of connection RT will always be GREATER than any of the included resistors.

Even if we have more than two resistors the total resistance is the sum of all the resistors connected in series:

R Total= R1+ R2+ R3 +•••••

• Total Applied voltage is divided by two resistors
• Current in the circuit is  I = V/(R1+R2)
• Voltage across R1 and R2 are from OHMS law.

V1= I*R1

V2=I* R2

Total voltage V=V1+V2

For Example if V=12v and the 2 resistors are 1k each, then the current in the circuit is

12/2k=6mA

The voltage across each resistor is 6v

Thus the series combination is characterized by

• The same current flows through all the resistors connected in series.
• The resultant resistor is SUM of all the resistors in series
• Series resistors divide the total voltage proportional to their magnitude.

## Resistors in Parallel

In Parallel combination, 2 paths are available for current, hence the current divides but the voltage across the resistors is same.

1/R total =1/R1 + 1/R2   or

R total = (R1*R2) / R1 + R2

• If the two resistors are equal, the current will divide equally and total resistance will be exactly half.
• For example if voltage is 12v and there are 2 resistance for 1k each,

The current through each resistance will be 12v/1k= 12 mA. Hence the total current is 12 mA.

Effective resistance is 0.5k

Thus the parallel connection is characterized by

• The same voltage exists across all the resistors connected in parallel, and
• The reciprocal of resultant resistor is the sum of reciprocals of all resistors in parallel, and
• Parallel resistors divide the total current in an inverse proportion to their magnitude.

## Potential Divider

Since series resistors divide voltage, this idea can be used to get smaller voltage from a power supply output. For example, we have a power supply with 10V fixed output. But we want only 5V from it.

Vout= Vin(R2/(R+R2))

• The Current I=Vin/R1+R2
• Since the current I flows through R2, voltage developed across it from Ohm’s law is Vo=I*R2=( Vin/R1+R2) * R2                                                                                                   Vo = (R2/R1+R2) Vi
• If R1=R2, then Vo=Vi/2

R1 and R2 cab be 100k or 100 ohm. Which one to be used?

If we need more current through load then R1 must be small. But too small a value will cause energy drain on the power supply. So the value must be chosen very carefully.

### Note:

• When two resistors are in parallel then their overall power rating is increased.
• If both resistors are the same value and same power rating, then the total power rating is doubled. If parallel resistances are not equal, then the resistors with smaller values will be required to handle more power.
• Four identical 0.25W resistors can be wired in parallel to give a resistor with one fourth the value in ohms, but four times the power rating. (1.0W). This is most useful when we require higher power handling, but don’t want to go out and buy more expensive (and physically larger) resistors.
• We have already seen earlier, that the power (in watts) can be calculated by multiplying voltage by current. P=V * I
• By using ohms law, the parallel or series resistor formulas and the above formula, a minimum power rating for a certain resistor can be calculated. If this is exceeded the resistor is likely to get hot and hopefully quietly breakdown.

## 10 Stage LED Sequencer

### 10 Stage LED Sequencer

Components

• IC1- CD4017
• IC2- NE555
• C1 – 1μ
• C2- 0.01 μ
• R1 – 470 Ω
• R2 – 100 KΩ
• R3- 100 Ω
• LED1-10 – RED LED
• 9volt DC power supply.

For power supply you can use 9 volt battery or can design separate power supply using step down transformer and 1N4007 diodes.

## Variable DC Power Supply using LM317

Below is the circuit for variable dc power supply

R1- 240 Ω

R2- 5K VR

R2 can be replaced by fixed value resistor for fixed power supply. Following formula can be used to calculate output voltage.

Vo=1.25(1+R2/R1)

Output voltage should be2 voltage greater than input.

Parts for current setup

•  D1, D2 – 1N4001
• C1-0.1 μ
• C2 – 10 μ 50v
• voltage regulator – LM317

## 20 Watt Inverter

This circuit will drive a 40 watt fluorescent or two 20-watt tubes in series. The transformer is wound on a ferrite rod 10mm dia and 8cm long. The wire diameters is 0.61mm wire for the primary and 0.28mm wire for the secondary and feedback winding.

The circuit will take approx. 1.5amp on 12v, making it more efficient than running the tubes from the mains. A normal fluorescent takes 20 watts for the tube and about 15 watts for the ballast.

Note: Do not remove the tube when the circuit is operating as the spikes produced by the transformer will damage the transistor.

Parts list

• Transistor – BC338 and TIP 3055
• Resistance – 47 K, 47 R, 180 R, 2R2
• Variable Resistance – 100k
• Capacitors – 100u 16v, 100n
• On/Off Switch
•  1 ferrite rod 10mm in 8mm long
• 30 m winding wire .28mm dia
• 4 m winding wire .61mm dia
• 2* 20 watt tube or 1* 40 watt tube
• 12 v DC power supply

## Lamp Dimmer

### 12v LAMP Dimmer

Parts

• IC 1 – NE555
• Transistor – 2N2955 -1
• Resistance – 1k(2 no.), 100 Ω
• Variable Resistance – 50k
• Capacitor – 0.1 µF
• Diode – 1N4001 – 3
•  12v 2 amp Bulb

Input Voltage is 12v. To create your own bench top power supply use the circuit shown in

http://digitalab.org/2012/06/regulated-dc-power-supply-circuit/

## LED Basics

Today LED has become an integral part of consumer electronics.

LED TV, LED Display, LED Lights and so on. These are becoming very popular because of there low power consumption.

What is LED?
LED stands for Light emitting diode.

A light emitting diode is essentially a PN junction semiconductor diode that emits a monochromatic(single) colour light when operated in a forward biased direction.

For detail in technical evolution refer the following url

http://en.wikipedia.org/wiki/Light-emitting_diode

Early LEDs were only bright enough to be used as indicators, or in the displays of early calculators and digital watches. More recently they have been starting to appear in higher brightness applications.

## LED Basics – Characteristics voltage drop

When a LED is connected around the correct way in a circuit it develops a voltage across

it called the CHARACTERISTIC VOLTAGE DROP. A LED must be supplied with a voltage that is higher than its “CHARACTERISTIC VOLTAGE”  via a resistor – called a VOLTAGE DROPPING RESISTOR or CURRENT LIMITING RESISTOR

How LED works?

LED and resistor are placed in series and connected to a voltage.As the voltage rises from 0v, nothing happens until the voltage reaches about 1.7v. At this voltage a red LED just starts to glow. As the voltage increases, the voltage across the LED remains at 1.7v but the current through the LED increases and it gets brighter. As the current increases to 5mA, 10mA, 15mA, 20mA the brightness will increase and at 25mA, it will be a maximum.

This is just a simple example as each LED has a different CHARACTERISTIC VOLTAGE DROP and a different maximum current.

In the diagram below we see a LED on a 3v supply, 9v supply and 12v supply. The current-limiting resistors are different and the first circuit takes 6mA, the second takes 15mA and the third takes 31mA. But the voltage across the red LED is the same in all cases.

## LED Basics – Head Voltage

As the supply-voltage increases, the voltage across the LED will be constant at 1.7v (for a red LED) and the excess voltage will be dropped across the resistor. The supply can be any voltage from 2v to 12 or more. The resistor will drop 0.3v to 10.3v. This is called HEAD VOLTAGE.

The voltage dropped across this resistor, combined with the current, constitutes wasted energy and should be kept to a minimum.

Most supplies are derived from batteries and the voltage will drop as the cells are used.

Here is an example of a problem:
Supply voltage: 12v
7 red LEDs in series = 11.9v
Dropper resistor = 0.1v
As soon as the supply drops to 11.8v, no LEDs will be illuminated.

Example 2:
Supply voltage 12v
5 green LEDs in series @ 2.1v = 10.5v
Dropper resistor = 1.5v
The battery voltage can drop to 10.5v
Suppose the current @ 12v = 25mA.
As the voltage drops, the current will drop.
At 11.5v, the current will be 17mA
At 11v, the current will be 9mA
At 10.5v, the current will be zero

Many batteries drop 1v and still have over 80% of their energy remaining. That’s why you should design your circuit to have a large HEAD VOLTAGE.

Some Basic circuits using LED

1. Polarity Tester

2. Continuity Tester