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.