Sunday, August 5, 2012

LED Flasher using NE555

LED Flasher circuit using NE555 timer IC

LED FLasher

  • 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.

USB Lamp

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

Thursday, August 2, 2012

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  Series Resistors


    • 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)) PotentialDivider





    • 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.

Tuesday, July 31, 2012

10 Stage LED Sequencer

Circuit diagram

Simple LED Chaser


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.

Saturday, July 21, 2012

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
  •  

Sunday, July 15, 2012

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
20 Watt Inverter

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

Saturday, July 7, 2012

Lamp Dimmer

12v LAMP Dimmer


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