Simple doorbell Using UM66

Below is a simple circuit of doorbell using UM66. The circuit can be assembled in a general purpose PCB and enclosed in any toys. 

Parts

C1 - 220 µF 10v
VR1 - 100 k pot
VR2 - 10 k pot
T1 and T2 - BC548
IC1 - UM66

Speaker - 8 ohms
Power - 3v

LED Torch with adjustable brightness

LM3914 is used as the basis of a 10 step variable brightness current-regulated white LED torch. This circuit has only four components in the control and regulation circuit: R1,R2, VR1 and LM3914.

Part List

1. IC - LM3914

2. LED1-10 - 10 white led

3. R1 - 620 ohm

4. R2 - 100 ohm

5. VR1 - 100 k pot

6. C 1 - 10 uf 16 v

7. Power - 3-4.5 volt

8. Suitable cabinet and a general purpose PCB.

Mains Powered LED Lights

Mains powered single LED



Figure1

A single LED can be illuminated by using a 100n or 220n capacitor with a rating of 400v

• The power diode in the first diagram is designed to discharge the 220n during one half of the cycle so that the capacitor will charge during the other half-cycle and deliver energy to the LED. The 1k resistor limits the peak in-rush current when the circuit is first turned on and the main happens to be at a peak.    
• Two LEDs can be driven from the same circuit as one LED will be illuminated during the first half cycle and the other LED will be driven during the second half of the cycle.

Mains Powered Multiple LED's

When 50 to 80 white LEDs are connected in series, a resistor can be used. For 50 white LEDs, use a 4k7 2watt resistor to provide 10mA average current.For 100 white LEDs, use a 2k2 1watt resistor to provide 10mA average current.

Figure 2

The above circuit can be modified with a bridge rectifier to provide DC input to LED. This will prevent from flickering or strobing. 

Mains light with 10 LED

The following circuit illuminates a column of 10 white LEDs. The 10u and 100 R prevents flicker.



Audio Amplifier using LM386

Audio amplifier - LM386

LM386 is a power amplifier designed to be used in low voltage consumer applications. Typical applications are like portable AM/FM amplifier, TV sound systems. It requires minimum external parts and operate from wide range of voltage starting from as low as 4 volt.

Below is the version of the circuit designed by me in one of my projects. It has a Gain of 50 with Bass boost.

Part list

1. C1 - 100 µf 

2. C2 - 10 µf

3. C3 - 0.01 µf

4. C4 - 0.1 µf

5. C5 - 0.01 µf

6. C6 - 220 µf

7. R1 - 10 K

8. R2 - 1.2 K

9. R3 - 10 

10. VR1 - 10 K

11. 9 volt Battery

12. IC - LM386 

C3 and C4 can be replaced by 0.05 or 0.047 µf capacitor. 

This amplifier is ideal for Ipod and cell phones and can be assembled in small box. 

White LED on 1.5 v Supply

2 White LED on 1.5 v Supply

Parts

T1 - BC338

R1 - 2.7 K

C1 - 10 n

D1,D2 - Super bright LED

Transformer - 60 T, 40 T

S1 - On/off Switch

Battery 1.5 V

White LED Flasher

Parts

T1 - BC338

T2 - 2N2222

R1 - 2.7 K

R2 - 47 R

R3 - 330 R - 1K

C1 - 10 n

C2 - 100 

D2 - 1N4148

Battery 1.5 v

1 Watt Audio Amplifier

Audio Amplifier

Parts

Q1 Transistor - MPSW45A
R1- 50 ohms
R2- 10 K
R3,R4 - 100 K

9 volt battery and Speaker

Flashing LED

Alternate red and green flashing LED using 9 volt battery



Part list 

1. R1,R2 - 100 Ohm
2. R3,R4 - 105 K
3. C1,C2 - 10 μF 25 Volt
4. T1-T2 - BC 548
5. D1-D5 - Red LED
6. D6-D10 - Green LED
7. 9 Volt Battery
 

Dark Sensor

Dark sensor using 2 transistor BC547

When light falls on LDR, its low resistance drives the transistor T1 into conduction. This keeps transistor T2 cut-off due to low base bias. The LED does not get power as long  as ambient light falls on it.

Parts list

1. T1 and T2 - BC 547
2. LDR  -1 nos
3. Resistance  R1 - 1k
4. R2 & R3 - 330 Ohms
5. 6 volt battery

Modifying the above circuit to use with Relay and switch on / off

You can also modify the above circuit in controlling AC appliances. Remove R2 and LED from the circuit and also remove the positive connection to Pole of the relay. Connect the AC bulb or any other appliance as shown



.

Electrical Appliance Protection

This is a simple circuit 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

1. Diode 1N4007 - 3


2. Capacitor C1 - 1000 25v


3. Relay - 6 V 100


4. Step Down Transformer = 6-0-6 300mA


5. Push to On Switch - 1

Very-short-duration interruptions or fluctuations will  not affect the circuit because of presence of largevalue 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

USB Reading Lamp

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

Basic Electronics - Resistor Combinations

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.

V_out= V_in(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

Circuit diagram

Components

1. IC1- CD4017


2. IC2- NE555


3. C1 - 1μ


4. C2- 0.01 μ


5. R1 - 470 Ω


6. R2 - 100 KΩ


7. R3- 100 Ω


8. LED1-10 - RED LED


9. 9volt DC power supply.

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

Variable DC Power Supply using LM317

Variable 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

1. Transistor – BC338 and TIP 3055


2. Resistance – 47 K, 47 R, 180 R, 2R2


3. Variable Resistance – 100k


4. Capacitors – 100u 16v, 100n


5. On/Off Switch


6.  1 ferrite rod 10mm in 8mm long


7. 30 m winding wire .28mm dia


8. 4 m winding wire .61mm dia


9. 2* 20 watt tube or 1* 40 watt tube


10. 12 v DC power supply

Lamp Dimmer

12v LAMP Dimmer


[caption id="attachment_82" align="aligncenter" width="308"] 12v Lamp Dimmer[/caption]

Parts

1. IC 1 – NE555


2. Transistor – 2N2955 -1


3. Resistance – 1k(2 no.), 100 Ω


4. Variable Resistance – 50k


5. Capacitor – 0.1 µF


6. Diode – 1N4001 – 3


7.  12v 2 amp Bulb

Input Voltage is 12v. To create your own bench top power supply use the circuit shown in http://electronichobbyist.wordpress.com/2012/06/17/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.

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.

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



3. USB Reading Lamp

http://electronichobbyist.wordpress.com/2012/08/04/usb-reading-lamp/

Regulated DC Power Supply Circuit

Below is the regulated DC power supply( 12, 8, 5 v) circuit. This circuit can be customized by adding 9v and 6v voltage regulator.

Also, the 4 diodes can be replaced by bridge rectifier.



If you are looking for a variable power supply then click here

Basic Power Supply Circuits - Part 2

Simplest DC power supply circuit using 78XX is shown below.



Parts

D1, D2, D3, D4 - 1N4007

IC1 - 7812

IC2 - 7805

C1 – 1000 mf

C2 – 330 mf

C3 – 10 mf

T1 - 12 v Step down transformer, 1 A.

Laptop Audio Amplifier

Laptop Audio Amplifier.