Here’s how industrial equipment is started and stopped using push buttons(On-Off Switch). The circuit is called a “locked-out relay.” It uses an ordinary relay and the ON switch activates the relay to close the terminals. When the switch is released, the relay remains activated and the load is powered via the contacts of the relay. When the OFF switch is pressed, the relay is de-energised and the contacts open. This removes power to the load.
Here is a very simple example of a low-cost battery charger for 12 v 7Ah lead-acid batteries.
As shown in the figure, diodes 1N4001 (D1) and 1N4001 (D2) form a full-wave rectifier. The rectified voltage appears across SCR1 and the battery to be charged. At low battery voltages, SCR2 is in cut-off state. SCR1 is triggered by the gate current flowing through resistor R1 and diode D3, and the battery charging starts.
At the start of charging, the reference voltage Vr of the battery is determined by a voltage-divider circuit. This voltage is too low to drive zener diode into conduction. Thus, the zener diode is effectively open and SCR2 is in cut-off state as the gate current is zero. Capacitor C1 prevents accidental triggering of SCR2 due to voltage transients in the circuit.
As charging continues, the battery voltage rises to a level high enough to turn on the zener diode and trigger SCR2, which brings down the voltage to a level too low at the junction of R1-R2 and SCR1 cuts off. This occurs when the battery is fully charged and the open-circuit state of SCR1 cuts off the charging current. The regulator starts recharging the battery whenever the reference voltage Vr drops below the zener diode’s breakdown voltage and prevents overcharging when voltage Vr is equal to the breakdown voltage of zener diode ZD1.
Diode D1 and D2 1N4001
SCR1 and SCR2- 2P4M
Zener diode ZD1 12v 1 watt
R1,R3 – 1K
R2 – 6Ω 5 watt
R4 – 680 Ω
VR1 – 1M
C1 – 0.1 µ
Transformer – 230 v Primary to 12-0-12 v secondary 2A
The Led torch with adjustable brightness circuit uses LM3914 as the basis of a 10 step variable brightness current-regulated white LED . This circuit has only four components in the control and regulation circuit: R1,R2, VR1 and LM3914.
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 μf 16 v
7. Power – 3-4.5 volt
8. Suitable cabinet and a general purpose PCB.
How it works?
The LM3914 is set to operate in bar graph mode so that the LEDs light progressively as its input signal increases.This signal comes from VR1, which provides a variable voltage between 0V and 4.5 v(supply voltage) to pin 5 of the LM3914. The internal resistor ladder network of the LM3914 has its low end (pin 4) connected to ground and the high end (pin 6) connected to the supply voltage via R2. Resistor R1 (620Ω) on pin 7 of IC1 sets the current through each LED to about 20mA. Set value of R1 as per input voltage. As VR1 is rotated from the 0V position (all LEDs off) to the supply voltage position (all LEDs on), the LEDs will progressively light. When all LEDs are off, the circuit will draw about 5mA. Whereas when all LEDs are illuminated, it will draw about 205mA and dissipate 307mW with a 4.5V supply.
Reference voltage at pin5 is given by- Vout=1.25(1+R2/R1)
The device dissipation will depend entirely on the input voltage and LED forward voltage. A resistor (R3) is inserted in series with the positive supply, chosen so that the LM3914’s dissipation is limited to about 500mW.
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.
T1 and T2 – BC 547
LDR -1 nos
Resistance R1 – 1k
R2 & R3 – 330 Ohms
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
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.
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.