Hacked my Laptop battery and charger to make a Portable power supply. My old Laptop was irreparably damaged leaving only a good battery. It is a high performance 12 volt Lithium Ion rechargeable battery that can backup more than 2 hours a laptop and many hours for circuit boards and LED lights. So I made a Portable power supply using the battery. So don’t discard the battery and charger if your Laptop fails permanently. Both battery and the charger costs around Rs. 2000.It can be utilized for making a Power supply or LED lamp. So it is the  good power supply I ever made because it is running on battery and output is fully regulated.

What I did is shown in the images. The power supply gives 4 output voltages with more than 1 Ampere current.

1. 18 Volt – The charger is rated 18.5 volts so one power out was directly from the charger positive going to the battery. A 10 Amps diode is provided here to separate the battery and the 18 volt power supply. So when the charger is on, 18 volt will be available in the socket.

2. 12 Volt – It is tapped directly from the battery.

3. 5 Volt – For this, 7805 Regulator IC is used. The 12 volt from the battery enters into the regulator IC and it gives 5 Volt regulated power. Capacitors are provided on either side of the regulator IC to suppress noise and transients.

4. 3.6 Volt– Two IN 4007 diodes are used to drop the 5 volt DC from the regulator IC to get 3.6 Volts DC. Each diode has a forward voltage drop of 0.7 Volts. So both will drop 1.4 Volts and reduces 5 Volts to 3.6 volts. This low volt is similar to two pen cells so it can be used for Microcontroller or Arduino projects.

Low battery indicator.

The Laptop battery is not necessary to charge frequently because it has great ability to hold charge. So to indicate the low battery voltage, a circuit was provided with a Red LED. The LED lights when the battery voltage drops below 9 volts. The circuit works like this.

When the battery voltage is above 9 volts, Zener conducts and gives base current to T1 and it conducts. When T1 conducts, it pulls the base of T2 to ground. So T2 will be off keeping the LED off. When the battery voltage drops below 9 volts, Zener turns off followed by T1. So T2 gets base current through the 4.7 K resistor and it conducts. This turns on LED. At this time, charger can be connected. As the battery voltage rises above 9 volts, LED turns off. Preset VR can be used to set the Zener break down point.

LED light.

The battery has high power so an LED light can be connected to work in night or as an emergency light. I used a ready made SMD LED light but 6 White LEDs with 100 Ohm resistors will be sufficient. A switch is provided for the LED light for on/off.
You can also make this Portable power supply with a 12 volt 4.5 Ah UPS  battery and suitable charger.

Advantage I found
The Laptop battery pack has an internal short circuit protection circuit so that ,when a short occurs momentarily in the power supply outlets or during charger plug in, the battery shut off the power supply for 10 seconds. If the short circuit is permanent, the battery will be in shut off mode. So this protects the Bread board assembly or during trouble shooting the circuits.

Many people leave the outdoor lights switched on as a safety measure while going for a long journey. The light will be on during day time also wasting lot of power and money. So here is a simple solution to avoid this. This circuit lights the premises of your home even if you are not there. It is fully automatic and turns on  Bulbs or CFLs at Sunset and turns off at Sunrise. So no need for daily switching. The circuit can run 250 watts load so many lamps can be connected. It is too compact and costs only Rs. 50 without power supply and the total cost will be around Rs.150. So once it is connected, we need not remember to switch on / off the outdoor lights. It will do its job at the right time.

Power supply for the circuit should be 12 volt Transformer power supply or Capacitor power supply. I used a 12 volt Strip LED driver commonly available at the cost of Rs. 100.

How the circuit works?

It is the simple trick of an LDR. Light Dependent Resistor (LDR) is a kind of variable resistor and its resistance varies according to the intensity of light. In dark, LDR offers very high resistance around 10 Meg while in bright light; its resistance reduces to 100 Ohms or less. So the low cost (Rs.4) 5 mm LDR is an ideal choice as light sensor.

For automatic switching, a Bistable using IC 555 is used. Its Threshold pin 6 and Trigger pin 2 are tied together and connected at the junction of LDR and 10 K resistor ( R1). So during day time, LDR conducts and takes the pins 6 and 2 of IC1 to a high level. This keeps the output pin 3 of IC1 low. As a result, T1 and Relay will be off. The lamp’s Phase line is connected through the Common and NO (Normally Open) contacts of the relay. So when the Relay is off during day time, the NO contact will be open and breaks the power to the lamp.

When the sunlight reduces in the evening, resistance of LDR increases. As a result, pin 6 and 2 of IC1 turns low. Suddenly, the output of IC1 turns high and triggers the Transistor (T1) and the Relay. Relay contacts join and lamp gets power and it turns on. This condition remains as such till morning. When sunlight again falls on LDR, lamp turns off. Red LED indicates that the Relay is on. Capacitor C2 eliminates the Relay chattering during the transition of light to dark and vice versa. Capacitor C2 maintains a steady voltage at the base of T1 so it will switch only when sufficient voltage is available .Diode. D2 removes back e.m.f during relay switch off and protects T1 from damage.

Assembling and setting

You can assemble the circuit on a small piece of common PCB. Use an adapter box with pins for housing the circuit so that it is easy to plug in. Drill two holes for LDR and LED in the case. LDR should get day light (not direct sunlight), so the unit should be fixed outdoor. Porch is an ideal place. Light from the lamp should not fall on LDR at night so orient the LDR’s position accordingly.

Advantage over Triac
In the place of the Relay a Triac can be used. But it has the following draw backs.
1. Triac conducts from Main terminals 1 and 2  , so the lamp should be connected in series – Phase – Lamp – Triac – Neutral. So only Bulb can be connected. CFL works in series connection, but its internal circuit will damage easily.
2. Triac needs correct Firing angle to give full brightness for the bulb. During the transition of light on LDR, Triac suffers Black lash effect so bulb will dim during light transition period and becomes bright only in full darkness. Bulb will also flicker during the light transition. To eliminate the back lash effect, we have to add a Snubber network along with Triac. The dimming, flickering etc will permanently damage the CFL if connected.
3. The relay eliminates all these problems because it works like a switch in the phase line only. So any type of lights can be connected.

Here is a simple circuit to monitor the voltage level in Solar panels. Usually a volt meter test is done during the installation of solar panel and then after, the panel remains unattended. If the solar panel is not providing sufficient voltage for charging the battery, that will permanently destroy it. A 12 volt Solar panel gives maximum of 22 volts DC in bright sunlight, which drops to around 16 volts when the load is connected. So a voltage level around 16 volts is necessary for charging the 12 volt Lead Acid battery. But the voltage output from the solar panel may reduce due to the accumulation of dust on the panel and corrosion of terminal contacts. These remain unnoticed ,so only when the battery fails ,we note this. This simple circuit solves the problem. If the solar panel  gives above 15 volts with the battery connected, the LED remains off. When the voltage from solar panel drops below 15 volts, LED turns on to indicate low voltage. This usually happens when the sunlight is low, but if the LED remains on in bright sunlight also, it indicates low output from the solar panel. The circuit can also be used as Battery level monitor.

Working of the circuit is simple. IC1 ( uA 741 ) is wired as a voltage comparator. Its inverting input (Pin 2) gets a Reference voltage of 15 V provided by the Zener diode ZD. At the same time, the Non Inverting input (Pin 3) gets a variable voltage through VR. So if the voltage from the solar panel is above 16 volts, Zener conducts and provides 15 volts to the Inverting input. The voltage at Pin3 is set slightly higher than 15 volts by adjusting VR. In this condition, output of IC1 will be high because pin 3 of the comparator gets higher voltage than pin 2. LED is connected from the positive rail to the output of IC1. So both the sides of LED will be high and it remains off.

When the voltage from the solar panel reduces below the level set by VR, Voltage at pin 3 drops below pin 2 and the output of IC1 turns low and LED lights due to the current sinking in to the output of IC. Diode D1 separates the circuit from battery so that, it indicates only the solar panel voltage.Connect this circuit to the output of solar panel observing correct polarity.

The same circuit can also be used as Voltage level monitor for different types of batteries. Select appropriate value of Zener as shown in the table below.

This is the simplest type of a Night lamp that gives ample light in a confined area. It is a White LED lamp directly powered by 230 Volt AC. It uses only four components and can be enclosed in a three pin plug that can be plugged into the socket. It also functions as Power on Indicator, Cupboard light, Baby’s bed lamp etc. Its cost is just Rs.20 and power consumption is very low.

You need the following components 
1. 33 Kilo Ohms 1 watt resistor – 1
2. IN 4007 Diodes – 2
3. 8 mm ½ Watt White LED – 1
4. 3 Pin plug

How it works?

As you know, the domestic power supply is 230 volt AC oscillating at 50 Hz. This high voltage is dangerous for an LED because it works only between 2 – 3 volts DC. Therefore it cannot be connected directly to AC and will burn instantly if so. So the 230 volt AC must be reduced to low volt AC first. The 33K resistor will do it and it drops 230 V AC to around 5 volt AC. Since the LED works only in DC, this low volt AC must be Rectified. Diodes D1 and D2 will rectify the low volt AC in both the alternating cycles. So the LED gets low volt DC and it lights.

You can select the resistor value between 22K to 47 K. If the resistor value increases, brightness of LED decreases. If the resistor value decreases brightness of LED increases but its life will be reduced. So 33K is an ideal choice. Just connect the components as shown in the diagram and image. Normally the 33 K 1 W resistors will heat slightly and if more heat appears, increase its value to 2 Watts. Check the heat of resistor only after removing the unit from AC mains.

Caution: The circuit is directly connected to high volt AC so it can give a fatal shock, if handled carelessly. Do not test or troubleshoot when the unit is connected to mains. When it is enclosed in the 3 pin case, it is safe. LED leads must be sleeved to prevent shock on accidental touching on it.
Always keep these kinds of units away from the direct access of children

Ohms law is a must in circuit design to calculate, Current, Resistance and Voltage in the different sections of a circuit. Ohms law simply represents the relationship between voltage, current and resistance. Let us see how Ohms law is applied to select the current limiting resistor for LED.

Ohms law can be expressed like this
V (Voltage) = I X R
I (Current) = V / R – in Amps
R (Resistance) = V / I – in Ohms
Suppose the input voltage is 8 volts, current is 2 Amps and resistor is 4 Ohms. Then
V = I x R = 2 x 4 = 8 Volts
I = V / R = 8 / 4 = 2 Amps
R = V / I = 8 / 2 = 4 Ohms
This is the basic method to calculate voltage, current and resistance in circuit design.

LED Series Resistor
A current limiting resistor is must for LED ,otherwise it will burn instantly. The value of the resistor must be proper to get sufficient brightness and to increase the life of LED. Different LEDs have different forward voltage drop across them. These are the voltage drops of general type LEDs

Red – 1.7 V
Green – 1.8 V
Yellow – 1.7 V
Blue – 3.1 – 3.6 V
White 3- 3.6 V
But this may slightly vary depending on the type and make of LEDs

Current through the LED must be between 10 to 25 Milli Ampere. Optimum current for good brightness is 20 mA. If the value of the resistor increases, current reduces followed by a reduction in the brightness. But this will increases the life of LED. If the value of resistor decreases, current increases followed by an increase in brightness.
Suppose the input voltage is 12 volts and the LED is standard diffuse type 5mm Red LED. It has a voltage drop of 1.7 volts. Fix the current through the LED as 20 mA (0.02 Amps)
Formula for selecting the resistor is
R = Vs – Vf / If
R is resistance in Ohms, Vs is input voltage , Vf is the Forward voltage drop of LED and If is the current through LED.
Out of the four values, we know three values
Vs = 12 V
Vf = 1.7 V
If = 0.02 A
So the value of resistor is
R = Vs – Vf / If = 12 V – 1.7 V / 0.02 A = 10.3 V / 0.02 A = 515 Ohms
So when we select 515 Ohms resistor, exactly 0.02 Amps or 20 Milli Ampere current flows through the LED. But 515 Ohms resistor is hard to get, so we can select either 470 Ohms or 560 Ohms. Then the current will be
I = V / R = 10.3 / 470 = 0.021 Amps or 21 mA.
I = V / R = 10.3 / 560 = 0.018 Amps or 18 mA .
So if 470 ohms resistor is used, LED gets 21 mA current and brightness slightly increases. If it is 560 ohms, LED gets only 18 mA so brightness reduces. So the exact value is 470 ohms. You can apply this formula for all LEDs.
White , Blue, and Pink LEDs
White, Blue and Pink LEDs can handle up to 40 mA current and the brightness will be sufficient only if 25 to 30 mA current is provided. More over the forward voltage drop of these LEDs is around 3 volts. So they need lower value resistor.
Vs- Vf / If = 12 V – 3 V = 9 V = 9 V / 0.03 A = 300 Ohms. Use 270 Ohms resistor
1 Watt White LED

The new type 0.5 and 1 watt White LED requires 100-350 mA current. Forward voltage drop is 3.6 V. So the value of the resistor must be very low. Since high current flows through the resistor, the Wattage of resistor must be ½ or 1 Watt otherwise, the resistor will heat up and burn. Suppose the current required is 250 mA, then the resistor is

R = Vs- Vf / If = 12 – 3.6 / 0.25 A = 8.4 / 0.25 A = 33.6 Ohms. Use 33 Ohms 1 watt resistor

If only 100 mA is required, then 8.4 V/ 0.1 A = 84 Ohms. Use 82 Ohms ½ W resistor
If 350 mA is required, then 8.4 / 0.35 = 24 Ohms. Use 27 Ohms 1 W resistor.

Note: All the calculations given above are based on 12 volt supply. If the supply voltage is 9, 6, 4.5 etc, use the formula Vs – Vf / If to get correct value of the resistor.

In a circuit diagram, we can see a number of Resistors connected in series or parallel between the positive and negative rails. These resistors are controlling the voltage and current to the particular sections of the circuit. When we design a circuit, we should have an idea about the voltage and current requirements for each section of the circuit. Then only we can select the proper values of resistors. If you know this, you can easily answer the questions in Project Viva regarding the selection of resistors.

First understand the Series and Parallel resistors. Series resistors means two or more resistors connected in series. This mainly appears as potential dividers in comparators, transistor amplifiers, oscillators etc. Parallel resistors mean two or more resistors connected parallel between the positive and negative rails.

Voltage across the Series resistor chain

See image. Two resistors R1 and R2 are connected in series across the 12 volt supply. So the total resistance is
R1 + R2 = 10,000 Ohms ( 10K) + 1000 Ohms ( 1K) = 11,000 Ohms
The relationship between Current, Resistance and Voltage as per Ohms law is Current I = V / R , IR = V, V = IR
Where I is the current in Amperes, V is the input voltage and R is the resistance in Ohms.
Now use basic algebra to change Ohms law to solve for voltage instead of Current.
I = V/R , IR = VR / R , IR = V, V= IR
So current I through R1 + R2 is = V/R = 12 V / 11, 000 Ohms
= 0.001,090,91 Amps
Voltage Drop across R1 and R2
We know Current and Resistance.
Current is 0.001,090,91 Amps
Resistance is 11, 000 Ohms
So Voltage V = I x R
Voltage drop across R1= V1 = 0.001,090,91 Amps x 1,000 Ohms ( 10K) = 10. 9091 V
Voltage drop across R2 = V2 = = 0.001,090,91 Amps x 1000 Ohms ( 1K) = 1.09091
So total voltage drop V1+V2 across R1 and R2 is
10.9091 + 1.09091 = 12 Volts. That is equal to the input voltage

That means, 10 K resistor allows 1,09091 volts and 1K resistor allows 10.9091 volts to pass.

See below images. The theory above explained exactly matches with the real situation.



Voltage across Parallel Resistors

See image. Here two resistors ( 10K and 1 K ) are connected in parallel across a 12 volt power supply.
So total voltage is V1 + V2
The total voltage through the resistor remains same as the entire circuit irrespective of the value of the resistors in parallel.
Total current through the resistors
Total current will be the sum total of the current running through each parallel path.
I total = I1 + I2
Here the resistors are 10,000 Ohms (10K) and 1000 Ohms (1K) and input voltage is 12 volts.
So 1 / R total
That is 1/10,000 + 1/ 1000 = 0.0001 + 0.001 = 0.0011
So R total is 1 / 0.0011 = 909.090,909 Ohms
Now Voltage across a Resistor
V = I x R
Input voltage (V) – 12 Volts
Current (I) = V / R = 12 V / 909.090,909 Ohms = 0.0132 Amps
So Voltage (V) across the resistors is I x R
That is 0.0132 A x 909.090,909 Ohms = 12 V

As you know, when a resistor is connected across the positive and negative rails of a power supply, some voltage will be dropped through the resistor. This is due to the resistance offered by the resistor. Let us see one example.

Voltage drop means the less voltage from the resistor than the supply voltage. It is the supply voltage minus the voltage across the Resistor.
For example, two resistors R1 and R2 are connected in series across a 12 volt power supply. R1 is 10K ( 10,000 Ohms) and R2 is 1K (1000 Ohms).
Now the Current through the series resistors
I = V / R1+R2 = 12 / 11,000 = 0.00109091 Amps
Now VR1 = IR1
VR2 = IR2
Voltage drop across R1
VR1 = IR1 = 0.00109091 Amps x 10,000 Ohms =10.9091 volts
Input voltage – 12 V
Voltage drop across R1 – 10.9091V
Voltage output from R1– 1 V

Voltage drop across R2
VR2 = IR2
That is 0.00109091 Amps x 1000 Ohms = 1.09091 Volts
Voltage output from R2 – 10.9091V

Note that slight difference of 0.6 volts. It is because the input supply is 11.7 volts

Note – Here R1 and R2 are reversed for measurement

Software result 

So total voltage drop across R1 and R2 = 10.90909 +1.09091 = 12 Volts
That means the 12 volt supply voltage splits in to two through R1 and R2 and finally the total values reach same as the supply voltage. That is, the voltage drop in one resistor is compensated by the voltage drop in the other in series connection.

Zener is a kind of Diode widely used in Voltage regulation and Switching applications. Zener is similar to other signal diodes but it conducts only if a specific voltage is provided to it based on its voltage rate. For example, a 5 volt Zener conducts only if it gets more than 5 volts. When the voltage to the zener rises above its value, its p-n junction undergoes breakdown and conducts. This breakdown point is known as Avalanche point. Unlike other diodes, current through the zener must be controlled, otherwise it will fail. So a series resistor must be connected to control the current through it. Here comes the importance of Ohms law .

Like other diodes, Zener diode also has Anode and Cathode ends. The Rectifier and Signal diodes are connected with Anode to the positive line and Cathode to the negative line. That means, the current flow is in the Anode – Cathode direction. But Zener is usually connected in the “Reverse bias” mode. That is, Cathode to Positive and Anode to Negative. It is easy to identify the Anode and Cathode of Zener. The method is same as that of other diodes. A white or black ring will be present at one end of the diode. It is the Cathode side.

How to select the correct Zener resistor?

Generally, the Zener requires minimum 5 Milli Ampere current for its breakdown. The resistor in series with the zener will protect it from excessive current.

The diagram below shows the typical zener application to produce a regulated voltage. The supply voltage used is 9 volts 500 mA. From this 9 volt, we have to produce a fixed 5 volts. So we need a 5 volt (or 5.1V) zener and a correct series resistor. The Cathode of zener is connected to the resistor and it goes to the positive rail. The Anode of zener is connected to the negative rail. 5V regulated DC is tapped from the junction of the resistor and the cathode of zener.

Suppose we need 5 volt DC and 25 mA current to the load. So let us check.
Input voltage – 9V (V in)
Voltage for the load – 5.1 V (V out)
Load current – 25 mA (Iin)
Zener current – 5 mA

The formula for calculating the resistor value is
V in – V out / I in = 9 – 5.1 / 30 mA x 1000 = 120 Ohms
So the current flowing through the zener is
9 – 5.1 V / 120 Ohms = 32.5 mA.
So 32.5 mA passes through the zener. Out of it, minimum 7.5 mA passes through the zener – resistor chain ( 32.5 – 25 = 7.5 ). So the output current to the load will be 25 mA at 5.1 volt.
Power dissipation in Zener
Power rating of zener is very important. When excess current flows through the zener, it heats up and burn. Zener is available in different power ratings like 400 Milli Watt, 1 Watt, 2 Watts , 5 Watts etc. So it is important to calculate the power dissipation of the zener before selecting it.
Power dissipation in Zener
Zener voltage x Zener current

In the above example, 7 .5 mA current passes through the zener (total current – load current). Let us round up it as 10 mA. So,
5.1 V x 10 mA = 51 Milli watts.
So up to 350 Milli Watts, you can use 400 mW zener. Above that, use 1 W or 2 W based on the power dissipation.
Special note – For the proper breakdown of the zener, the input voltage to it should be 1-2 volts excess than its voltage rating. For example, a 5 volt zener requires more than 6 volts.

Special note – For the proper breakdown of the zener, the input voltage to it should be 1-2 volts excess than its voltage rating. For example,  a 5 volt zener requires more than 6 volts.

This is one simple but highly useful circuit for you. Most of us use different types of Rechargeable batteries in Portable devices, Inverter, Emergency lamp etc. If the battery is deeply discharged, sometimes it will not get fully charged later. This is the most common reason for the failure of Emergency lamp. So we need an indicator to show the safe level of battery voltage. It is too simple to make it and needs only three components. Just hook up it in the battery and it will remind you when it needs charge.

Each type of rechargeable battery has a safe voltage limit. The 12 V Lead Acid battery attains full charge at 13.8 volts and Tubular battery at 14.8 V. The 9 V battery shows full charge as 9.6 V and 6 V battery as 6.8V. So how much it can discharge. Some battery can discharge to 50% but is ideal to keep 80% charge in it. That means, charging requires when 20% discharge occurs. But the condition of NimH and Li-Ion batteries used in Mobile phone, Digital Camera etc is different. These batteries attain full charge even if they are discharged completely.

How the circuit works?

It is a simple Voltage controlled indicator. A Zener diode (ZD) is used as a voltage controlled switch. In the circuit, the Cathode of Zener is connected to the Positive terminal of battery. To its Anode, a Red LED with a 100 Ohms series resistor is connected. The other end of resistor is connected to the Negative terminal of the battery.

Now let us see what happens to this circuit. As you know, the Zener conducts only when it gets a higher voltage than its value. For example, a 5 Volt zener conducts when it gets 6 volts and below 5 volts, it remains off. So here a Red LED is selected as the indicator. It has a forward voltage drop of 1.7 volts. This makes the trick to fix the “charging reminding point”. That means

Zener value + 1.7 V is the reminding point. When the voltage in the battery is above this value, Zener conducts and LED remains lit. When the battery voltage reduces below this, LED turns off to indicate the need of charging. For example, take the example of a 12 volt Lead acid battery. It attains full charge at 13.8 volts. So take a 10 V Zener.

10 V (Zener voltage) + 1.7 (LED forward voltage) = 11.7 V

So till battery retains voltage above 11.7 volts, LED remains on. When it turns off, it indicates that, the voltage reduced below 11.7 volts and needs charging.


You can select the Zener value in another way. Minus 1.7 V from the full charge voltage. It will give the Zener value. But this has one problem. The battery voltage may vary from 12 V to 13.8 V. If we take the Zener value like this, it may change the charge remind point. So fixing the “reminding point” is best as per the method explained. Note that, 1.7 V is for Red LED only. If you use Green LED, it will be 2 volts.

This simple technique is used in Inverters for Auto charging and , as Auto cut off. When the battery voltage reduces below 11.5 or 11 V, a relay turns on to charge the battery.

The only drawback of the circuit is that, the LED consumes some current even when the load is off. No problem. Battery needs both charging and discharging to maintain its health.

So try this. The value of Zener for different batteries is given below.

You can make a simple LED light to read and type in Computer or Laptop during night without disturbing other’s sleep. It is also useful as a mini Emergency light in the event of a power failure. It is too simple and powered from the USB port. USB lights are available in the market, but costs around Rs.150. You can make this USB light at the cost of just Rs.5.

USB ( Universal Serial Bus ) is the port that gives 5 volts and around 400  milli ampere current. We can use this power to light the 0.5 watt White LED.

Procure the USB male from a discarded Pen drive or USB cable. Remove the wires or the Pendrive circuit connected to it.It has four pins. The pins at the ends are + ( +5 V ) and – ( Ground ) while the middle pins are Data In ( D + ) and Data Out ( D – ). We need only the + and – pins to power the LED.

Connect 100 Ohms ¼ Watt resistor to the + pin and the other end of resistor to the Anode of LED. Connect the Cathode of LED to the – pin. Note that, first identify the + and – pins of USB male by connecting it in the USB port and checking with a multimeter or simply using an LED with 1K resistor.

How much current can be provided to 0.5 W White LED.
The 0.5 W White LED works well between 50 mA to 250 mA current. As the current increases, brightness also increases. But if the LED takes more than 100 mA current from USB port, it will over load the USB hub. In the PC, all the USB ports are sharing the same current of 350-500 mA. Mouse and other devices are usually connected to ports. If LED draws more current, over loading occurs and the devices connected to the ports will not work. So use minimum current of 50- 100 mA from USB port for the LED. You can apply Ohms law to fix the LED current. First fix LED current and select Resistor using Ohms law.
I = 50 mA or 0.05 A
V = 5 Volt
R = V / I = 5 / 0.05 = 100 Ohms
Other resistors and current at 5 V
82 Ohms – 60 mA
68 Ohms – 73 mA
47 Ohms – 106 mA

Note – In the diagram, 47 Ohms resistor is shown and it gives 106 mA current to LED. In the Prototype, 100 Ohms resistor is used which gives only 50 mA current to LED.

If the Pendrive has a transparent case, you can enclose the LED inside, to get a professional touch.

Soldering may sometimes make nuisance because the Iron will be either too hot or cold. If it is switched on continuously, it becomes too hot, wasting power and the bit gradually erodes. If it is switched off during a work, it takes time to become hot. So what is the solution to solve this problem? Do it, your Soldering Iron will always be ready in its two heat conditions. Simple, see how the trick works.

Most of the Soldering iron Tips or Bits now available are Copper core surrounded by Iron. It is coated with Nickel or Chromium to prevent the sticking of solder. If we switch on the soldering iron for long period, the excess heat may damage it permanently. If the soldering work is intermittent, it is ideal to keep the tip half warm. This circuit performs this function. It has a switch and if it is in the off position, the soldering iron tip remains half warm and if the switch is in the on position, within seconds, it will become fully heated
The Soldering Iron uses an AC outlet to power it. So just do this arrangement. A IN 5404 or IN 5204 diode is connected in series with the Phase line. An AC switch bypasses the diode. A Neon bulb with 82 K resistor runs parallel to the diode. So the current going to the AC socket has two paths.

1. Direct Full wave AC when switch is closed.
2. Half wave AC through the diode when switch is opened because, the diode will block the other half of AC.

So when you are doing the soldering work, just switch on the AC switch. Full wave AC passes to the Soldering Iron through the switch and it works in full heat. At this time, Neon turns off and indicates that, the Iron is in normal heat. Suppose you want a break during the work, switch off the AC switch. Now the current flows through the Diode and Neon turn on. The Iron will then get Half AC so that, it remains half warm and consumes, only half power. When the switch is turned on again, within seconds, the Iron will turn to normal heat.

The indications of the Neon are:
1. Neon off with switch on – Load (Soldering Iron) not connected to socket.
2. Neon off with switch on – Full AC to Soldering Iron and normal heat
3. Neon on – with switch off – Half AC to Soldering Iron and half heat

Caution : High volt AC in the circuit can give a fatal shock. Do not touch, or trouble shoot when connected to mains.

Title of the circuit is deliberately given because it is used for testing the Photo diodes, Infrared LEDs, Photo transistors etc working in “Silent Mode”. How can we confirm an Infrared LED is working or not. It gives No visible indication. It is working and emitting Infrared rays but the IR rays are beyond our spectral response and we can’t see it. So this simple tester will help you to check whether such devices are good or bad before using in a circuit or even during trouble shooting. Not only one function the tester doing . It can test, all diodes, Capacitors, LEDs etc. Its cost is just Rs.10 but it is a worth tool.

These are the common types of IR LED, Phototransistor, Photodiodes etc now commonly available. See, all are alike and very difficult to identify visually whether one is Transmitter or Receiver. So this tester will make identification cum testing very easily.

To make the tester, you need the following:
1. A small piece of Dot type Common PCB
2. Three resistors – 1K, 10K and 100 Ohms
3. One BC 547 NPN transistor
4. A Bicolour LED – Red and Green – Common Anode type
5. A connector pin or probes as test points
6. A 9 V battery snap

The Bicolour LED is used for the visible indication of the invisible working of IR LED. A common Anode type Bicolour LED is required. It has two chips –Red and Green– inside with separate Cathodes. Anode is common for both the LEDs. You can easily identify the leads. The longest middle lead is Anode. A long lead is the Cathode of Red Half and the short lead is the Cathode of Green half.

Circuit connection
See image. It is too simple. Test point is marked as A and K. Point A is for connecting the Anode of the IR LED and K for its Cathode. Base of the BC 547 transistor is connected to the K point with a 10 K resistor going to ground. Common Anode of the Bi colour LED is connected to the Positive rail through the 100 Ohms resistor. Note the Cathode connections. Cathode of Red half is connected to the Collector of T1 while the Cathode of Green half is connected to ground directly.

Circuit working
When the circuit is connected to the Battery, Bicolour LED lights as Green because, it’s Green half gets direct current path from positive to negative through the 100 ohms resistor. So Green colour indicates that the tester is ok. Now what about the Transistor T1 and Red half of Bicolour LED? The base of T1 is floating because there is no IR LED connected to the test point. So T1 will be off and the Red half of Bicolour LED remains dark because it has no current path.

When an Infrared LED (Tx) is connected with Anode to point A and Cathode to point K, current flows to the base of T1 and it conducts. Bicolour LED turns Red because it gets current path through T1. At the same time, the Green LED turns off because, most of the current passes through the Red LED and T1. So if the bicolour LED turns Green to Red when the IR LED is connected, it indicates that, the IR LED is forward biasing and good. If the Bicolour LED is not changing Green to Red, it indicates that the IR LED is bad and not forward biasing.

This test is applicable to all diodes and LEDs with Anode in point A and Cathode in point K.

But for IR Receiver LED (Rx), Photo diodes, Phototransistor etc, the leads should be reversed because, these devises show “Reverse biasing” like the Zener diodes. So connect Cathode to point A and Anode to point K.

See the test procedure given below.

For Capacitor testing, connect the +ve pin to point A and – ve pin to point K. Capacitor charges quickly and the colour of Bicolour LED gradually turns from Green to Orange and then to Red. This change depends on the value of the capacitor. For example, 1 uF capacitor has short duration for the colour change and a 1000 uF capacitor takes long time for the colour change. If this happens, the capacitor is good.

This simple circuit tells you whether your Interior plant requires water or not. When the soil dry up, the Red LED lights to inform you that, the plant needs water. So this saves the life of your valuable interior plants.

Circuit is too simple. IC1 is the commonly available Op Amp uA 741. It is used as a Comparator. Its Inverting input (Pin2) is connected to a voltage divider R2,R3 that gives half supply voltage (4.5V) to it. The Non Inverting input (Pin3) of IC is connected to one of the probe. The other probe gets 9 volts directly from the battery. The comparator gives high output when its Non inverting input gets higher voltage than the Inverting input. Probes are two nails inserted in the soil with a gap of 2 cm. So when the soil is moist with water, 9 volt passes from probe 1 to probe 2 through the moist soil. So pin 3 of IC gets higher voltage and its output remains high.

IC1 Pins

Pin 7 Vcc
Pin 4 Ground
Pin 6 Output
Pin 2 Inverting Input
Pin 3 Non Inverting Input

T1 is a PNP transistor, so it remains off when its base gets high voltage from the comparator. LED also remains off. When the soil dry up, the electrical path between the probes break and pin3 of IC becomes low. As a result, output of IC1 turns low and T1 conducts to light the LED.

This simple Circuit Breaker can be added to the Work Bench power supply, to break power to the load when the current flow through the load increases more than 470 mA due to short circuit or power surge. It immediately breaks the power to the load and the load again gets power only if the current flow returns to normal.

Circuit working is simple. Power to the load is connected through the Common and NO contacts of the relay with a bypassing push switch for relay activation. Current from the negative of the load passes through the Current Sensor resistor R2 (1.5 Ohms 2 Watt). A Medium power NPN transistor (T1) is used as a switch to break the power. When the Push switch is pressed and released, Relay turns on and makes the contacts closed. Load gets power. When the current passing through R2 increases more than 470 mA, T1 turns on. As a result, current reduces through the relay and the contacts break. If the over load is ceased, T1 turns off. The Relay can be turned by pressing SW1.

Selection of Resistors R1 and R2 needs some calculations
Tripping current value through R2
R2 = 0.7 / It
Where 0.7 is the voltage across R2 due to T1 and It is the tripping current value.
Selection of R1 for different supply voltages
R1 = (V supply – VR) / IR
Where VR is the Relay voltage and IR is the Relay current

For example, in the circuit, a 6 Volt 500 Ohms relay is used , so 240 Ohms resistor is selected as R1 and for 470 mA Tripping current, R2 is selected as 1.5 Ohms.

Hum is a nuisance in Amplifier circuits which is mainly due to improper filtering of AC fraction from the power supply and improper grounding. Hum will reduce the quality of sound from the speaker. You can easily eliminate the hum from the amplifier to get clear voice.

Usually we use a 12 volt 1-5 Ampere Step down transformer as power supply in High power amplifiers. The output DC must be well filtered to remove the AC fraction from the rectified DC. So do these techniques to remove the hum.

1. Transformer body should be grounded to the Metal chasis of the Amplifier.

2. Power supply section should be away from the Preamplifier section. If needed, isolate it from the amplifier circuit.

3. Amplifier Negative should be grounded to the Metal chasis of the amplifier.

4. Use a high value filtering capacitor like 2200 uF 50 V or 4700 uF 50 volt. As the value of the filtering capacitor increases, filtering efficiency also increases.

5. If hum continues, try one of the circuits given below. Figure 1 uses a diode and a capacitor to give an additional rectification and filtering. It should be between the Power supply and the Pre Amp.

6. Second circuit uses a medium power transistor to give regulated power supply. It should be between the Power supply and the Pre Amp.

Sometimes we need higher sound output from Audio based circuits using either Buzzer or Speaker. This simple Transistor based Audio amplifier will give loud sound from the Speaker or Buzzer.

Two Transistors are used as ” Darlington Pair” Q1 is a general purpose PNP Transistor and Q2 is the Medium power NPN Transistor. So their combination amplify the sound to its maximum.

A simple circuit to test the TV remote whether it is working or not. TV remote uses Infrared LED to emit pulsed IR rays which are invisible to human eye. So it is difficult to confirm the working of IR Led of remote. This Remote Tester blinks an LED when it receives IR rays from the TV remote.

TV uses Photomodule TSOP 1738 as IR sensor. It has a Photodiode , amplifier circuit and an output transistor inside the package. It works off 5 volt DC only and gives 5 V output when it is not receiving the IR rays and sinks current when it receives IR rays. That means it gives 5V in its output pin when no IR ray in its front. It gives 0 V when it receives IR rays.

In the circuit, the output of the IR sensor is directly connected to the base of the PNP transistor. PNP transistor remains off when its base is positive and conducts when its base is negative. So when the IR sensor is in standby, the high output from it, keeps the PNP transistor off . So LED also remains off. When the Remote is focused on to the IR sensor and any button is pressed, PNP transistor conducts and LED blinks indicating that the Remote is working.

Pin connection of TSOP 1738

Pins of BC 557

Pins of LED

Here is a Power Bank for charging your Mobile phone while traveling or in the event of power failure. This power bank can deliver 4.5 Volts with reasonable current to charge the Mobile phone quickly. It can charge the Mobile phone 3-4 times . The Power bank uses a 4.5 V Rechargeable battery that can be charged using the Mobile phone charger or from the USB port.

4.5 V Rechargeable battery is now available with a rating of 1 Ah, 1.5 Ah, 2 Ah etc for less than Rs.100. This battery can be used or use a high efficiency Lithium battery.The Mobile phone battery is 3.6 V so 4.5 V from the Power bank is sufficient for charging. Transistor T1 act as a current regulator during charging the Mobile battery.

4.5 V Lithium Polymer Battery

4.5 V Alkaline Battery

Use suitable Sockets for connecting with the charger and Mobile phone.

This simple add on circuit prevents misuse of Land Phone and also indicates whether the Phone lines are Ok or not .It is an ideal tool to test the Telephone lines. It lights LEDs to confirm the signals coming in the Telephone lines. It can also be used as a simple Telephone lock to block outgoing calls.

Just connect the circuit to the Telephones lines as shown in the diagram. LED indications as follows.

1. Handset lifted – One LED turns on – Phone is OK
2. Dialling – LED blinks
3. When the other phone attends the call, other LED lights.
4. LED blinks without using the phone – Misuse. Somebody is tapping the lines.
5. When switch SW1 is in OFF position ,no outgoing calls made. Phone gives Incoming calls only.
6. If the switch is in Off position, it should be turned to On position before attending the call to hear voice.

This Tiny circuit can protect your LCD TV from Voltage surge at power on. It takes 2 minutes delay before connecting power to the TV. Within that time, mains power will become stable. Heavy inrush current at power on or when power resumes after a power failure can damage the TV permanently. If few minutes delay is provided, the AC will become stable. This is the delay technique used in Stabilizers.

Circuit working is simple. 9 Volt power supply is needed for the circuit so use a 9 volt adapter for it. When power is switched on, Capacitor C1 charges slowly through R1 and VR1. When the voltage in the capacitor C1 rises above 3.1 volts, Zener conducts and triggers the Relay driver transistor T1. When T1 conducts, relay energize and provides power to TV. Charging of capacitor and Zener conduction takes around 2 minutes, so relay turns on only after that.

Connect the Phase line to the Common contact of the relay. Phase to the TV should be from the NO (Normally Open) contact of relay. Neutral should go directly to TV. Use a 2 pin socket for connection.